Form 7 Notes

FORM 7

NOTES , EXAMPLES , PROBLEMS & SOLUTIONS

Table of Content

Chapter Topic Page

1. Mechanics …………………. 2

2. Gravitation …………………. 61

3. Direct Current ………………….. 71

4. Electrostatics …………………. 90

5. Simple Harmonic Motion ……. 116

6. Waves ………………………… 128

7. Magnetism …………………. 156

8. Electromagnetic Induction …… 162

9. Atomic Physics …………. 184

10. Answers …………. 194

MECHANICS

KINEMATICS of LINEAR MOTION Uncertainties of Measurement

Power of Ten | Prefix | Abbreviation | 10-18 | atto | a | 10-15 | femto | f | 10-12 | pico | p | 10-9 | nano | n | 10-6 | micro | μ | 10-3 | milli | m | 10-2 | centi | c | 103 | Kilo | K | 106 | Mega | M | 109 | Giga | G | 1012 | Tera | T | 1015 | Peta | P | 1018 | Exa | E |

* Whenever a quantity is measured, there is an amount of uncertainty in the value obtained. This may be due to the instrument being used, the method used or the person making the measurement. * Uncertainty refers to the fact that a measurement is an estimation of the true value. The two main types of uncertainty in a measurement are: 1. Random Uncertainty Is one for which the measurement is just as likely to be larger or smaller than the true value. This type of uncertainty shows up when the same quantity is measured several times. The measured values are spread randomly around an average value. The slight difference may be due to changes in pressure, temperature, etc. 2. Systematic Uncertainty Is one which consistently makes the measurement larger or smaller than the true value. This effect can result from: i) an incorrectly adjusted measuring instrument. ii) an instrument that has zero error. iii) a measuring procedure used by the observer that constantly gives a bias reading (parallax error).

Calculations in Uncertainties Eg: 2.6 ± 0.1 cm (Measured value) (Absolute uncertainty) (i) absolute uncertainty to % uncertainty abs.uncert.measured val.x1001 = 0.12.6 x 1001= 3.85% (ii) % uncertainty to absolute uncertainty % uncert.100xmeasured val.1 = 3.35100 x 2.61 = 0.1 Two principles to be followed: i) Express the absolute uncertainty to one significant figure. ii) Express the measured value to the same number of decimal places as the absolute uncertainty. 1. Addition and subtraction of quantities with errors. Whenever quantities are added or subtracted, the absolute uncertainties are added. Example: A = (6.4 + 0.5) cm , B = (4.2 + 0.2) cm A + B = (6.4 + 0.5) cm + (4.2 + 0.2) cm = ( 10.6 + 0.7 ) cm. 2. Multiplication and division of quantities with errors. When measured quantities are multiplied or divided, the percentage errors are added. Example: A = (6.4 + 0.5) cm , B = (4.2 + 0.2) A B = (6.4 + 0.5)(4.2 + 0.2) cm2 (convert the absolute error to % error and add) = (6.4 +)(4.2 +) = (6.4 + 7.8 %)(4.2 + 4.8 %) cm2 = ( 26.9 + 12.6% ) = ( 26.9 + 3.4 ) cm2 = ( 27 + 3 ) cm2 The absolute error should always be one significant figure. Note should be taken of number of significant figures in calculation.

3. Power calculations For measurements raised to a power, the percentage uncertainty is multiplied by the power. Example: (i) Find V; V = y3 y = (1.56 + 0.02 ) cm V =y3 = (1.56 +0.02 )3 = (1.56 + 1.3 % )3 = ( 3.80 + ( 3 1.3 % )) = ( 3.80 + 3.9 % ) = ( 3.8 + 0.2 ) cm3 (ii) Find = = = = = 1.25 0.01 4.Oscillation of uncertainties When several people take a series of measurement of an item, there is bound to be a set of results. To determine the average result with an associated error; Find the average of the result and the difference of the data and averages. The biggest difference is the uncertainty. Example: The length of a slide is measured and a series of readings are taken in cm; 10.1, 10.2, 10.1, 10.0, 10.3, 9.9, 10.4 ( let the values be x ) Find the best estimate of the length of the slide with its absolute error. The average = = 10.1 cm The biggest difference is = (10.4 – 10.1) = 0.3 Length = ( 10.1 + 0.3 ) cm

5. Uncertainties in graphical analysis y m1 m line of best fit m2 lines of worst fit

x Experimental data when plotted graphically usually does not fall in a perfectly straight line. So it becomes necessary to draw the lines of best and worst fits to determine the slope. To obtain a value of slope with an associated uncertainty, calculate the numerical values of m, m1and m2. Slope = Example; If m = 10, m1= 11 and m2 = 8 in the graph shown before, the slope with error will be Slope = ( 10+ 2 )

Power Relationships If a set of data analysed graphically gives a curve, then this is a power relation. Power relations can be analysed better using logarithms to obtain a linear graph. Example: To obtain a linear graph for the relation a =Kbn Taking logarithms on both sides gives log a = log ( K.bn ) ( log x.y = log x + log y ) log a = log K + n.log b ( log bn = n.log b ) log a = n.log b + log K y = mx + c

So if a graph of log a versus log b is plotted, it will give a linear graph, From this graph, the slope will be n, and the antilog of the y-intercept will give K.

a log a

slope = n

y-intercept = log K

b log b Newton’s Laws of Motion First Law: Law of Inertia “An object will remain in a state of rest or uniform motion unless an external force acts upon it.” Second Law: “When an unbalanced force acts upon an object, it will be accelerated in the direction of the force with acceleration proportional to the force.” F = ma Third Law: “To every action there is an equal and opposite reaction.” Examples in force/acceleration problems. Example 1. A system of masses in a plane. A system of two masses is connected using a light inextensible string on a frictionless surface. A force of 12 N acts on the 2 kg mass. 2 kg
1kg
m1 m2 12 N

Calculate: (i) the acceleration of the system (ii) the tension in the string Solution: Since the system is accelerating, use Newton’s second law equation; (i) Fnet = ma , a = = = 4 m/s2 (ii) To calculate the tension in the string, isolate the system into its individual masses. Draw a free-body diagram and then use F = ma equation to solve. Using the 1 kg mass: Fnet = m1a, T = m1 a = (1 kg) (4 m/s2) T = 4 N Fnet

Using the 2 kg mass: Fnet = m2a T m2 12 N 12 N – T = (2 kg)(4 m/s2) T = 12 N – 8 N T = 4 N Fnet Example 2. A system of masses in two planes. A system of two masses is connected using a light inextensible string on a frictionless surface over a smooth pulley as shown. m2 string 4 kg pulley Calculate: m1 (i) the acceleration of the 4 kg mass. 10 kg (ii) the tension in the string.

Solution: (i) The system accelerates so use, F = ma; Fnet = m1g = (m1+m2) a, a = = = 7.14 m/s2 (ii) Using m2: Fnet = m2a, T = m2 a = (4 kg) (7.14 m/s2) T = 28.6 N Fnet

Using the mass m1 : Fnet = m1a T (100 N) – T = (10 kg)(7.1 m/s2) T = (100 – 71.4) N m1 T = 28.6N Fnet 100 N

Forces on an inclined plane. Case 1: A system of two masses on a frictionless inclined plane.

m2gcos

m2gsin

normal m2

m1 m2gsin m2g m1g

If m1g m2gsin : The acceleration is given by: Fnet = m1g – m2 gsin = (m1 + m2) a a =

The tension in the string can be calculated by: Using m1: Fnet = m1 a T m1g – T = m1a m1 Fnet T = m1 ( g – a ) m1 g

Or T =m1 Using m2: Fnet Fnet = m1a T T - m2gsin = m2a T = m2 ( a + gsin) m2gsin Case 2: The mass accelerating downhill. If m2gsin m1g : The motion will now be down the inclined plane and the sine component of the weight of m1 would be responsible for the acceleration of the system. The acceleration can be calculated by: Fnet = (m1 + m2) a m2gsin - m1g = (m1 + m2) a a =

The tension in the cable can be calculated using any of the masses individually: Using m1: Fnet = m1a T Fnet T - m1g = m1a T = m1 ( a + g ) m1g

Using m2: Fnet Fnet = m2a T m2gsin - T = m2a T = m2 ( a + gsin) m2gsin

Example Consider the same inclined plane as discussed earlier but with friction. The expressions for acceleration in both cases are as given below. Case 1: If m1g m2gsin : The acceleration is given by: Fnet = (m1 + m2) a m1g – ( m2gsin + f ) = (m1 + m2) a ( f = frictional force (N) ) a =

Case 2: If m2gsin m1g : The acceleration can be calculated by: Fnet = (m1 + m2) a m2gsin - ( m1g + f ) = (m1 + m2) a ( f = frictional force (N) ) a =

m2g

m2

m1 m2gsin m1g

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PROJECTILE MOTION Case 1
For a complete – parabolic path.( A full projectile motion ) An object with initial velocity v, is projected at an angle , to the x-axis. Calculate: (i) the maximum vertical displacement, y. (ii) the total time of flight of the projectile. (iii) The horizontal range of the projectile (R ). v v sin y v cos

v

0 R x (i) Using the third equation of motion the vertical displacement can be obtained; vf2 = vi2 + 2aS ( acceleration is ‘g’ ) 0 = (vsin)2 – 2gS ( vertical displacement S = y ) v2sin2 = 2gy y =

(ii) Half the time of flight can be calculated by the first equation of motion, using the vertical motion. vf = vi + at , 0 = vsin - gt t = So the total time of flight will be tT = (iii) Since the acceleration in the horizontal direction is zero, the second equation of motion can be used. S = vt R = v cos R = Case 2

For a half – parabolic path. ( A half – projectile motion ) If a mass is projected horizontally at vm/s from a vertical height y, derive expressions to calculate: (i) the time of flight of the mass. (ii) its horizontal range. (iii) its impact velocity.

Solution: v m/s (i) Using the second equation of motion, the time can be obtained.

S = ut + ½ at2 ( uy = 0 ) y = ½ gt2 y y = gt2 y = 5t2 t2 = x t = R

(ii) The horizontal range (R) can be obtained (iii) The impact velocity can be from the second equation of motion. found using vectors. S = ut + ½ at2 ( a = 0 ) vy = gt = (10)= R = vt v = R = v vx v = R = v vy v = tan-1 Case 3

For a three – quarter parabolic path. ( A partial– projectile motion )

If a mass is projected at vm/s from a vertical height y, derive an expression to calculate the time of flight of the mass.

v y1

y

* To derive an expression for y1: vf2 = vi2 + 2as,
0 = (vsin)2 + 2y1 y1 = * An expression to find the time taken to reach maximum height (t1): t1: vf = vi + at1, 0 = vsin + 10t1 t1 =

* An expression to find the time taken from the maximum height to impact with the ground (t2): t2: s = vit + ½ gt2, s = ½ gt2, Total time = t1 + t2 y + y1 = 5t2 t = + y + = 5t2 = = t2 = t2 ttotal = t2 =
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PROJECTILE MOTION PROBLEMS

1. A soccer ball is kicked at 17.5 m/s at 60o above horizontal on a level playing field. It lands on the same elevation as when initially kicked. Calculate: (a) The total time of flight of the ball. (b) The horizontal range of the ball. (c) The velocity of the ball 1.0 s after it is kicked.

2. A cannonball is fired from a cliff top horizontally out to sea with a speed of 74 m/s. The cliff is 320 m above sea – level.

(d) Calculate the time taken by the cannon ball to strike the water. (e) What is the vertical distance dropped by the cannon in 2.0 s ? (f) Determine the horizontal distance travelled by the cannon ball. (g) Calculate the speed of impact of the cannonball with the water.

3. Mere shoots an arrow at 16 m/s from a bow while standing on the roof of the National Stadium, 24 m high. The arrow lands on the ground as shown.

60o

24 m

(a) What is the maximum height reached by the arrow from the ground? (b) Calculate the time the arrow was in the air after release. (c) How far horizontally did the arrow travel before impact with the ground?

4. Jone tries to kick a rugby ball over a pine tree. If he kicks the ball at 25 m/s at 50o above ground level, will he be able to clear the tree that is 30 m away and 18 m in height 5. A B2 dive-bomber is heading downwards at an angle of depression of 30o. It releases a still bomb ( does not have its own propulsion system ) at an altitude of 800 m at a velocity of 120 m/s.
30o
120 m/s

800 m

(h) Calculate the time it takes the bomb to detonate. (i) At what range does the bomb detonate? (j) Calculate the velocity of the bomb upon impact with the ground. 6. A golf ball is chipped from the woods to clear it from a tree.

16.2 m

60 m

(k) Calculate the time the golf ball will be in flight. (l) With what velocity and direction should the player strike the ball to clear the tree?

7. During Diwali, a firecracker is launched at 53o above horizontal with a speed of 100 m/s. After moving for 3.0 s, along its line of motion, accelerating at 20 m/s2, it runs out of fuel and proceeds to fall as a free body. Calculate: (m) The maximum height reached by the firecracker. (n) Its total time of flight. (o) The horizontal range. (p) Impact velocity of the firecracker.

8. A stone is thrown off a building at 20 m/s at an elevation of 30o to the horizontal. If it takes 4.0 s to strike the ground, calculate the height of the building.

9. A helicopter is ascending from a level ground at a constant speed of 20 m/s. At an altitude of 130 m a package is thrown from it horizontally at a speed of 16 m/s.
Calculate: (a) the maximum height above ground reached by the object. (b) the total time of flight of the package after release from the chopper. (c) the velocity of impact of the parcel with the ground.

10. In a ski flying competition, a skier starts from point A and takes off at an angle of 40o at point B. Neglect air resistance in the problem. (q) Calculate the maximum height above the ground reached by the skier. (r) The length of time the skier is airborne.

A

25 m B

5 m 40o

[ Hint: Take energy considerations ]

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ROTATIONAL KINEMATICS Circular Motion An object travelling at a constant speed in a circular path has acceleration towards the centre, called the centripetal acceleration. (Centre – seeking acceleration) The expressions for velocity, acceleration and force are as follows: v = a = F =

(i) To derive an expression for acceleration in terms of r and T; a = since v = , a = = = a =

(ii) To derive an expression for force in terms of r and T; F = ma , a = F = m F =

Example A small mass of 0.2 kg attached to a light cord 0.20 m long, revolves in a horizontal circle. If the mass revolves at 2 revs / sec, find the tension in the cord. Solution: The tension in the cord is the centripetal force. ( Period, T = 0.5 s ) F == = 6.32 N , tension, T = 6.32 N -------------------------------------------------
VERTICAL CIRCLES and ELEVATORS

Vertical circles When a mass is whirled in a vertical circle, the force it experiences in its motion varies. To understand the forces, imagine an elevator ride. When descending, you feel lighter, and while ascending, you feel heavier. The best way to describe these forces is to use a Ferris wheel. Imagine yourself sitting in the Ferris wheel and the forces you experience at the highest and lowest points in the ride.

v P

Q

At the top of the ride (P), you will feel lighter because of the centripetal force. The normal force is the force you will experience from the seat that will make you seem lighter. The two forces acting on you are your weight and the centripetal force as shown. Fnet = Normal force
At P ; F = mg - ( Fnet = mg – FC ) mg FC

At the bottom of the ride you will seem to be heavier because the normal force (the force of the seat on you) and the expression of force acting on you is given by: FC At Q ; F = mg + Fnet = Normal force

mg

Example

A mass of 2 kg is tied to a string of length 5 m and whirled in a vertical circle at a constant speed of 6 m/s. Calculate: (a) the centripetal acceleration of the mass. (b) The tension in the string at the highest and lowest points.

Solution: (a) a = = = 7.2 m/s2 (b) At the highest point, the tension will be less; F = mg - = (2kg)(10 N/kg) - = ( 20 – 14.4 ) N T = 5.6 N.

At the lowest point, the tension will be higher than the weight; F = mg + = ( 20 + 14.4 ) N = 34.4 N

Elevators.

The same principle as applied above can be used to analyze elevators. The normal force can be used to explain the lightness in weight while descending and the increase in weight while ascending.

Example

An elevator that is operated by cables is carrying some passengers and is ascending at 3 m/s2. The combined mass of the elevator and passengers is 750 kg. Calculate: (a) the tension in the cable while ascending. (b) the apparent weight of a 60 kg passenger.

Solution: (a) Fnet = ma
Fnet T T – mg = ma , T = m(g+a) = (750 kg)( 10 + 3 ) N/kg = 9750 N

mg (b) W = mg + ma = ( 60 kg )( 10 + 3 ) N/kg = 780 N

Note: The apparent weight of a mass when the elevator is accelerating upwards is given by:

W = mg + ma

The apparent weight of a mass when the elevator is accelerating downwards is given by: W = mg – ma

Motion of Satellites around Planets For a satellite moving around a planet, the gravitational force of the planet provides the centripetal force to keep it in orbit.

Re

R

Earth v

Fc Satellite

There are two forces acting on the satellite, the gravitational force and the centripetal force and using these two forces, an expression for the orbital velocity of the satellite can be derived. FG = and FC = FG = FC = v2 = so v =

The period of the satellite is given by: For circular motion the velocity is v = Rearranging, we get an expression for the period T, T =

Example Suppose we want to place a satellite in a circular orbit, 300 km above the surface of the earth. Calculate: (i) the orbital speed of the satellite (ii) the period of the satellite (iii) the radial acceleration of the satellite. ( Radius of earth, Re = 6.4 x 106 m, Mass of earth, Me = 5.98 x 1024 kg ) Solution: Using the equations derived we can substitute and calculate; (i) v = = = 7.72 x 103 m/s (ii) T = = = 5.4 x 103 s

(iii) a = = = 8.89 m/s2 Kinematics of Rotation When dealing with translational motion, we deal with objects moving in a straight line and the associated variables in analysing this motion are: displacement (s), velocity (v) and acceleration (a). But objects undergo rotational motion also. For analysing rotational motion the variables used are: angular displacement (), angular velocity () and angular acceleration (). R

The angle in rotational motion is measured in radians. One radian is the angle subtended by the arc of a circle, which has a length equal S to its radius. Angular displacement, is S = R 1 rev = 360o = 2 radians Where: S = distance in metres. radians = 180o R = radius in metres 1 radian = 57.3o = angular displacement in radians ( rad ).

Angular velocity ( ): The average angular velocity is the angular displacement per time. av = units: rad/s For instantaneous angular velocity inst = , inst = Note: 1 rev/s = 2 rad/s Angular acceleration ( ): The rate of change of the angular velocity of the rotating object. = units: rad/s2

Relation between linear and angular systems. 1. S = r Differentiate both sides with respect to time t; = = r v = r Where: v = linear velocity (m/s) = angular velocity (rad/s) 2. v = r Differentiate both sides with respect to time t; = = r a = r Where: a = linear acceleration (m/s2) = angular acceleration (rad/s2)
Example 1 A bicycle wheel of radius 30 cm is turning at a constant speed of 3 rev/s. Calculate: (i) its angular speed in rad/s (ii) the linear speed of a point on the rim. (iii) the angular acceleration if the angular speed is increased to 10 rev/s in 3.5 seconds. Solution: (i) If 1 rev/s = 2 rad/s, then using ratios, 3 rev/s = 6 rad/s (ii) v = r = ( 0.30 m)(6 rad/s) = m/s (iii) = = = 4 rad/s2
Example 2 Determine the angular velocity in rad/s of a crankshaft, which is rotating at 4200 revs/min. Solution: = rad/s So if 1 rev/min = rad/s Then 4200 rev/min = 140 rad/s So = 140 rad/s Graphs of Motions The motion of a rotating object can be described by graphing the motion against time. The way the information is represented is very similar to graphs for translational motion. Graph of Angular displacement against time * Slope(gradient) gives angular velocity(w) θ(rad) θ(rad)

Constantangular velocity angular acceleration

t(s) t(s) Graph of Angular velocity against time * Slope gives angular acceleration(α) * Area under the graph = Angular displacement(θ) W(rad/s)

Slope= α

Area= θ t(s) Example: The graph shows Angular velocity against time for a bicycle wheel. w(rad/s)

50

0 5 10 15 20 25 t(s) Find: i) The angular acceleration during the first 10s of the motion? ii) The angular displacement for the 20 second duration? Solution: i) α = slope= 5 rad/s2 ii) θ = Area under graph = 750 rad

Equations with Linear and angular relations Quantity | Relation | Displacement | s = r | Velocity | v = r | Acceleration | a = r |

The Analogy between Linear and Angular systems. Symbols Variable | Linear | Angular | Speed | u , v | o, | Velocity | u, v | o, | Displacement | s | | Acceleration | a | | Time | t | t |

Relationships Linear | Angular | v = u + at | = o + t | s = ut + ½ at2 | = ot + ½ t2 | v2= u2 + 2as | 2 = o2 + 2 |

Exercise A turntable accelerates uniformly to an angular speed of 10 rev/s from rest in 3 seconds. Calculate its angular speed at the end of 3 seconds in rad/s. Solution: o = 0, = 10 rev/s = 20 rad/s, t = 3s, = ? = o + t , = = = 20.9 rad/s2

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ROTATIONAL KINAMATICS PROBLEMS

1. Convert to radians. (a) 12 rev (b) 210 rev (c) 500 rev

2. Convert to revolutions. (a) 5 rad (b) 0.6 rad (c) 300 rad

3. Convert to rad/s. (a) 60 rev/s (b) 30 rev/min (c) 500 rev/min

4. A wheel turns 50 revolutions in 10 seconds. Find: (a) its angular velocity in rad/s. (b) the linear speed of a point on the rim, if the radius of the wheel is
60cm.

5. A merry-go-round turns 15 revolutions in 20 seconds. It’s radius is 2.5 m. (a) Find its angular velocity in rad/s. (b) Find the linear speed of a point on the outer edge of the merry-go -round.

6. If a point on the rim of a bicycle wheel has a linear speed of 10 m/s, find the angular speed of the wheel. The radius is 50 cm.

7. A cyclist pushes down on the pedals and the angular speed of the wheels increases from 4.5 rad/s to 6.0 rad/s in 2s. Find: (a) the angular acceleration. (b) the linear acceleration of a point 40cm from the centre of one wheel.

8. The speed of a point on the rim of a fly wheel of radius 1.2m, decreases from 5.0 rad/s to 1.0 rad/s in 3.2s. Calculate: (a) the angular deceleration. (b) the angle turned in 3.2s interval.

9. A flywheel initially at rest, accelerates uniformly at 3.0 rad/s2 for 20s. Find: (a) the angle it turns through. (b) the linear acceleration of a point 12cm from the centre of the wheel. (c) the angular velocity after 20s. 10. A wheel turns 6 revolutions in 10 seconds. Find: (a) its angular velocity in rad/s. (b) the linear speed of a point on the rim if radius of wheel is 80 cm.

11. Jave is riding a bicycle and the wheels are turning at 5 rad/s. He starts to pedal faster and attains a speed of 11 rad/s in 3 seconds.
Calculate: (a) the angular acceleration. (b) the linear acceleration of a point 0.50 m from the wheel’s centre.

12. A flywheel at rest is accelerated uniformly at 4.0 rad/s2 for 20 seconds. Calculate: (a) the angle turned through for the duration of 20 s. (b) the linear acceleration of a point 15 cm from the centre of the flywheel. (c) The angular velocity after 20 s.

13. The blade in a blender speeds up from 24 rad/s to 74 rad/s uniformly. During this time it turns through 40 revolutions. Calculate: (a) the angle turned through in radians (b) the angular acceleration (c) the time taken to turn 40 revs

14. If the expression for the angle turned through of a crankshaft is given by: = 4t3 – 2t2 + t + 5 ( radians )

Find the angular velocity and acceleration at time t = 3 seconds.

15. A flywheel is speeded up from 5 rev/min to 11 rev/min in 100 seconds. The radius of the flywheel is 0.08 m. Calculate:

(a) The angular acceleration of the flywheel in rad/s2. (b) The linear acceleration of a point on the rim of the flywheel.

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LINEAR DYNAMICS MOMENTUM(p) When a body of mass m, moves with a velocity v, the momentum is the product of its mass and velocity. p = mv units: kg.m/s Momentum is a vector quantity. Using Newton’s second law: F = ma , ( a = ) F = m F = , ( p = mv ) F = , ( p = pf – pi ) F = F.t = p units: N.s = kg.m/s

The value of the product of force and time over which this force acts is known as the impulse ( J ). From the above equation it can be seen that the impulse is also equal to the change in momentum. N.B: Impulse = Area under a force Vs time graph F(N)

Area=Impulse

t(s)

Example: The force against time for a golf club striking a stationary ball of mass 100g is shown in the diagram below: F(N) 2000

0.01 0.02 0.03 t(s) (a) What is the impulse imparted to the ball? (b) Calculate the average force on the ball? (c) Find the speed of the ball after the strike? Solutions: (a) Impulse = Area = 12 (0.02) (2000) = 20 Ns (b) F. ∆t = 20 F = 200.02 = 1000 N (c) F. ∆t = m. ∆v (0.02)(1000) = (0.1)(v) 200 m/s = v Elastic collision: When in a collision or explosion, the momentum and the total kinetic energy is conserved. J = p = F.t = 0 Inelastic collision: When in a collision or explosion, the total momentum of a system is conserved, but the kinetic energy is not conserved. . J =p = F.t 0 Example 1
Jone of mass 80 kg is traveling in a car at 90 km/hr along the Queens highway near Nadroga and collides with a cow of mass 400 kg sitting in the middle of the road. It takes the car 0.5 seconds to come to rest. Jone flies off through the windshield of the car after impact since he was not wearing any seatbelt. car Jone cow

(a) What is the initial momentum of the car? ( mass of car = 920 kg) (b) Calculate the net force in the collision. (c) With what velocity does Jone leave his seat after the collision? Solution: (a) First convert the initial velocity of the car to m/s. v = = 25 m/s p = mv = ( 920 + 80 )kg ( 25 m/s ) = 25 000 kg.m/s (b) F = = = 50 000 N = 50 kN (c) Using law of conservation of momentum; pi = pf 25 000 kg.m/s = mJone.v v = = 312.5 m/s
Exercise
A 2.5 kg object (A) travelling at 25 m/s collides head-on with a 5 kg mass (B), calculate the velocity of the 5 kg mass if (a) The mass B is initially at rest. (b) Travelling opposite to mass A at 20 m/s. (c) Travelling in the same direction as A at 12 m/s. Example 2 Two cars A and B collide as they approach a junction as shown.

vc

20 m/s A

600 kg

B 8 m/s 1000 kg If the two cars combine together after collision, calculate their combined velocity. Pi = (mv)A + (mv)B = (600 kg)(20 m/s) + (1000 kg)(8 m/s) = 12 000 kg.m/s + 8 000 kg.m/s

= 8 000 kgm/s =

12 000 kgm/s = 14 422 kgm/s = 33.7o Using the law of conservation of momentum: Pf = Pi = 14 422 kgm/s = ( mA + mB )v v = = 9.01 m/s at = 33.7o above horizontal. Example 3 A 6 kg bomb is launched at 30 m/s and it explodes into two fragments as shown below. vA A 30 m/s 30o B

6 kg 30o Mass of A = 2 kg Mass of B = 4 kg vB Calculate the velocities of masses A and B after the explosion. Solution: Since the motion is two-dimensional the directions of the motions are important, as momentum is a vector quantity. pi = pf (6 kg)(30 m/s) = pA + pB

180 kg.m/s = + pA + pB

Sketch a vector diagram. Sine rule:

pA 120o pB Cosine rule: a2 = b2+c2-2bcCosA 30o 30o 180 kg.m/s Using the sine rule, the values for pAandpB can be determined; = , pA = pB = 103.9 kg.m/s (pA = pB , because, the diagram is an Isosceles triangle) vA = = = 52 m/s , vB = = = 26 m/s in the directions stated above.
Center of Mass
This is the point on an object where the total mass acts as if it is concentrated at that point. The center of mass is sometimes called the center of gravity since it is a point of balance and the total weight can be thought to act through this point. When an object is moving, the center of mass travels in a straight line. In the co-ordinate system ( the x-y plane ) the coordinates for the centre of mass is given by: xo = yo =

The above coordinates are for a two-point particle (mass) system. y y2 m2 d2

yo * c.m. d1 y1 m1

0 x1 xo x2 x Example Calculate the centre of mass of a two-particle system. Particle A has a mass of 5 kg and is positioned at (2,3) and mass B is 3 kg placed at (5,4). xo =, yo = = = = 3 = 3 c.m. = (xo , yo) = (3 , 3 )

Velocity of Center of mass In a collision, it is seen that the velocity of the centre of mass of a system of two objects remains constant. The velocity is given by: vc.m =

m1 v1

vcm

m2 v2

Note: 1. The centre of mass is a ‘mass weighted’ average position of the particles of a system. 2. The velocity, v, of the centre of mass of a system of the two objects is unchanged throughout a collision. -------------------------------------------------
MOMENTUM PROBLEMS

1. A locomotive of mass 15 000 kg travelling at 2.5 m/s collides and couples with a set of rail carts of mass 7 000 kg. Calculate: (a) The combined velocity of the train and carts after impact. (b) The loss in the kinetic energy in the collision.

2. A 1 500 kg car (A) travelling to the east collides head-on with a 2 500 kg car (B) travelling due north. If they couple together after collision, and move off with a velocity of 16 m/s at 53o north of east, calculate the individual initial velocities of the two cars ( A and B ).

3. A 200 g hockey puck (A) moving at 2 m/s collides with a stationary, 300 g puck (B) as shown. 1 m/s
A

A 2 m/s 53o
B

vB Find the velocity of puck B after collision.

4. A 80 g bullet is fired into a 2.92 kg ballistic pendulum with a speed of 200 m/s. If the bullet embeds itself in the pendulum, calculate the height to which the ballistic pendulum rises after the collision.

5. A hockey puck B rests on a smooth ice surface and is struck by a second puck A, which was originally travelling at 40 m/s and which is deflected 45o from its original direction as shown in the diagram below. Puck B acquires a velocity at 30o to the original. The pucks are identical with mass of 600 g.

A A 40 m/s 45o

B 30o

(i) Compute the speed of each puck af6ter collision. (ii) Is the collision perfectly elastic? If not, what fraction of the original kinetic energy of puck A is "lost"?

6. When a firework of mass 0.1 kg reaches its highest point, it has a horizontal velocity of, ms-1. At this point it explodes into two parts (A and B) as shown.

40 m/s

A 0.04 kg 0.1 kg m/s 37o 53o

0.06 kg B

vm/s
Calculate the speeds v and .

7. An astronaut of mass 80 kg (including his back-pack) is on a trajectory far above the moon’s surface, moving at 5 m/s upward at an angle of 60 with the horizontal. A short burst of compressed nitrogen is ejected horizontally at 100 m/s. after the ejection of nitrogen thee astronaut is moving vertically upwards. Use the diagrams given below as a guide to answer the questions, which follow.

Before nitrogen ejection After nitrogen ejection astronaut 5 m/s

60 nitrogen 100 m/s (a) Calculate the mass of the nitrogen ejected (b) What is the new speed of the astronaut?

8. A 2 kg mass travelling at 8 m/s collides with a stationery mass, B. After the collision, A moves 40 above horizontal while B moves 30below horizontal. vA

A A 8 m/s 40o 2 kg B 30o vB

Calculate the speed of A after collision

9. A mass A of 0.2 kg move son a frictionless table with a velocity of 1.6 m/s at an angle of 45 to the x-axis. Mass A makes a glancing collision with a stationery mass B of 0.1 kg and then moves at 1 m/s at 20 above the x-axis after the collision.

A y

45o B x

Calculate the magnitude and direction of B’s final velocity.

10. Two cars collide at an intersection and lock together skidding across the road as shown below. The combined velocity of the vehicles after the impact was 15 m/s at 30 east of north. N

E 30o 15 m/s

Car B vB

900 kg

vA Car A

1 100 kg

(a) What was the magnitude of the momentum of the two vehicles after the collision? (b) What were the velocities of the two vehicles before the collisions?

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ROTATIONAL DYNAMICS TORQUE The turning effect of a force is called torque. The torque or ‘moment of force’ about a point is the product of a force magnitude and moment arm ( radius ). The force and rotating distance are always perpendicular. = Fr Units: Nm

Where: F = Force (N) r = radius of turning object (m)
Example

Calculate the torque on the spanner.

F = 900 N 19o

0.80 m

= Fr = (900 cos 19o N)(0.80 m) = 681 N Moment of Inertia ( I )
Inertia is the tendency of a body to resist any changes in its existing state of being stationary or in uniform motion. The rotational inertia of an object depends on the mass and the radius from the center of rotation. Even if the mass remains constant, changing the distance from center of rotation can change the moment of inertia. The general formula of inertia from a point of rotation is given by: I =

Units: kg.m2 Where: m = mass (kg) r = radius (m)
Example

Calculate the inertia (I) of the two mass system about P. P 2 kg 1 kg 0.5 m 0.5 m

I = = m1r12 + m2r22 = (2 kg)(0.5 m)2 + (1 kg)(0.5 m)2 = 0.75 kg.m2 For an object whose mass is concentrated further away from the point of rotation, inertia is greater than for an object whose mass is closer to the point of rotation. Mass 1 Mass 2

Inertia of mass 1 is greater than for mass 2. Moment of Inertia for various rigid bodies: 1. Cylindrical shell I = MR2

2. Solid disks or cylinder R I = MR2

3. Solid sphere ( radius, R ) I = MR2 R

4. Long thin rod ( length, L )

I = ML2

L

Finding a relation between torque ( ) and angular acceleration (): Newton’s second law of motion tells us that an unbalanced force acting on an object causes acceleration. In a similar way, an unbalanced torque on a rotating object causes angular acceleration. This is Newton’s second law of rotational motion. Taking a mass attached to the end of a light rod of length r, undergoing horizontal circular motion with a constant force provided on it as shown: r

F

m

The mass undergoes circular motion following Newton’s second law; F = ma multiplying both sides by r gives; Fr = rma ( Fr = ) = rma ( a = r ; = ) = mr2 ( mr2 = I ) = I Where: I = Inertia ( kg.m2 ) * = angular acceleration ( rad/s2 )

The Falling Bucket

A solid cylindrical pulley of mass M, and radius R, is used to draw water from a well. A bucket of mass m, is attached to a cord that is wrapped around the cylinder. Derive expressions for linear acceleration (a)of the bucket and the tension in the string (T).

r

Cylinder of mass, M

T a

m Using Newton’s second equation, we get the equation: mg - T = ma (1) and torque is given by: = Tr (2) Since = I , and I for a solid cylinder is ( I = ½ MR2 ) we can derive an expression for T; I = Tr ( r = R ), ( = ) ( ½ MR2 ) = TR ½ Ma = T (3)

Substituting (3) in (1) we get; mg - ½ Ma = ma ma + ½ Ma = mg a = mg a =

Rotational Kinetic Energy A rigid particle in a fixed axis will have kinetic energy equal to EK = ½ mv2 ( v = r) = ½ m (r)2 = ½ ( mr22 ) = ½ (mr2) 2 ( I = ) = ½ I 2 EK = ½ I 2 EK = ½ mv2

Rotational kinetic energy Translational kinetic energy Angular Momentum ( L ) The figure below represents a small body of mass m, moving in a plane at constant speed v, the momentum p, is given by p = mv ( kg.m/s )

m v

r

We define the angular momentum of the particle about an axis through the perpendicular to the plane of motion, as the product of linear momentum and perpendicular distance from the line of motion. An object moving in circular motion: m v r

Angular momentum L: 1. L = p.r (1) ( p = mv) L = mvr units: kgm2s-1 Where: m = mass ( kg ) v = velocity ( m/s ) r = radius ( m ) 2. L = mvr ( v = r ) L = m r2 L =mr2 ( I = ) L = I Conservation of angular momentum: “ The total angular momentum of the system remains the same provided no external forces act.”
Example

A turntable of inertia I1 = 0.09 kgm2 spins freely with a constant angular speed of = 4 rad/s. (a) Calculate the angular momentum of the turntable. L = I1 = ( 0.09 kgm2 )(4 rad/s ) = 0.36 kgm2s-1 (b) If a record of inertia ( I2 = 0.03 kgm2 ) is dropped onto the turntable, calculate the new angular speed of the turntable. Li = Lf 0.36 kgm2s-1 = ( I1 + I2 ) c c = = 3 rad/s

Translational and Rotational analogy | TRANSLATIONAL | ROTATIONAL | Displacement | S (m) | ( rad ) | Velocity | v ( m/s ) | ( rad/s ) | Acceleration | a (m/s2) | ( rad/s2 ) | Mass, Inertia | m, M (kg) | I ( kgm2) | Force | F (N) | ( Nm) | Force ( formula ) | F = ma | = I | Momentum | p = mv | L = I |

Kinetic Energy of a Rotating Body When a rigid body undergoes both linear and rotational motion (example- a rolling ball) it possesses both translational and rotational kinetic energy. The total kinetic energy ( EK ) of such a body is the sum of the translational ( EKT ) and rotational ( EKR ) kinetic energies. ET = EKT + EKR Or: ET = ½ mv2 + ½ I2

An experimental set-up to show kinetic energy considerations.

The following experiment was set-up in the lab to study translational and rotational motions. M v

String Pulley

m

h

The data gathered from the above experiment can be used to calculate the inertia of the solid cylinder with mass M.

Mass (m) = 0.100 kg Mass of cylinder (M) = 0.700 kg Radius of cylinder(r) = 5.0 cm Height (h) = 0.600 m Velocity of m at impact = 1.00 m/s The above set of data was gathered from a practical carried out by a group of form 7 Grammar students. To calculate the inertia of the cylinder, the following steps are to be followed. Step 1. Calculate the change in potential energy of mass m. EP = mgh = (0.100 kg)(10 N/kg)(0.600 m) = 0.600 J Step 2. Calculate the gain in translational kinetic energy of the mass system. ( m+M ) EKT = ½ ( m+M) v2 = ½ ( 0.100 + 0.700 )kg(1.00 m/s)2 = 0.400 J Step 3. Determine the rotational kinetic energy of the cylinder. EP = EKT + EKR 0.600 J = 0.400 J + EKR EKR = 0.200 J Step 4. Using the expression EKR = ½ I 2 , find the inertia of the cylinder. EKR = ½ I 2 ( = = = 20 rad/s ) I = = = 1.0 10-3 kgm2 The value for inertia calculated above can be checked using the standard formula for the inertia of a cylindrical disk. Theoretical Inertia of the cylinder: I = ½ MR2 = ½ (0.700 kg)(0.050 m)2 = 0.9 10-3 kgm2 It can be seen that the theoretical value is notexactly the same as the experimental value but they are comparable within the uncertainties of the experiment. Exercise 1 A hollow cylinder with mass, M and radius R, rolls without slipping with speed v, on a surface. Determine an expression for its total energy in terms of its mass and velocity. Solution: ETOTAL = EKT + EKR = ½ Mv2 + ½ I2 ( I = MR2 ) = ½ Mv2 + ½ (MR)22 ( v = R ) = ½ Mv2 + ½ ( MR2) = ½ Mv2 + ½ Mv2 = Mv2 Exercise 2 A spherical ball of mass m, and radius r, starts from rest at a height of 2m and rolls down a slope without slipping. Find the linear velocity of the ball as it leaves the incline. The inertia of the ball:

h = 2 m

Solution: Write an energy equation and derive an expression for velocity. EP = EKT + EKR mgh = ½ mv2 + ½ I2 mgh = ½ mv2 + (mr)22 ( v = r ) mgh = ½ mv2 + ( mr2) mgh = mv2 + mv2 ( m can be eliminated ) gh = v2 v2 = gh v = Substitute the values and determine the velocity. v = = = = 5.35 m/s Exercise 3 A solid cylinder of the same mass m, and radius r, as the ball in exercise 2, is rolled from the same height. In the race between the two objects, which one will win? Solution: Write an energy equation and derive an expression for velocity. Then compare. EP = EKT + EKR mgh = ½ mv2 + ½ I2 ( I = ½ mr2 ) mgh = ½ mv2 + ( ½ )( ½ )(mr)22 ( v = r ) mgh = ½ mv2 + ( mr2) mgh = ½ mv2 + mv2 ( m can be eliminated ) gh = v2 v2 = gh v = = = 5.16 m/s
Since the velocity of the cylinder is less than that of the ball, the ball will reach the bottom of the slope first and so win the race.

h

Acceleration of a yo-yo. A uniform disc of mass m, and radius r, has a moment of inertia ( I = ½ mr2 ) about its axis and is securely tied to the ceiling with a string. Show that the disc has an acceleration of a = g.

Solution: Write a net force equation. ( Using Newton’s second law ) mg - T = ma T T = mg – ma (1) r Write a torque expression. = I ( I = ½ mr2 ) = ( ½mr2 )() ( r = a ) mg = ½ mra (2) = T r (3) Substituting (1) and (2) in equation (3): = T r ½ mra = ( mg – ma ) r ( m and r can be eliminated ) ½ a = g – a ½ a + a = g a = g a = g

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ROTATIONAL DYNAMICS PROBLEMS

1. A cylinder with a diameter of 16 cm and mass of 7.2 kg starts from rest and is accelerated uniformly by a force of 240 N applied to it tangentially. (a) Calculate the rotational inertia of the hollow cylinder. ( I = MR2 ) (b) What is the angular acceleration of the cylinder?

2. A solid cylinder of mass 2.0 kg and radius 6 cm is accelerated uniformly from rest.
After 4 seconds it is rotating at 9.6 rad/s. (a) Calculate the angular acceleration. (b) What is the applied torque? (c) Through what angle has the cylinder turned in the four seconds from rest?

3. The wheel of a car is spinning freely. Torque is applied and the wheel is set turning at a frequency of 10 Hz. When the applied torque is removed the wheel slows down to a stop in 240 s. If the inertia of the wheel is 0.48 kg.m2, calculate the frictional torque, which brought it to rest.

4. A winch of mass 5 kg is uniformly accelerated from rest by a 10 N force applied to its handle at point R. The diameter of the drum is 80 cm. Calculate: 10 N (a) the magnitude of the torque exerted R by the 10 N force. (b) the angular acceleration of the winch 0.2 m if the inertia I = 0.4 kg.m2 .

5. An axe is pressed with a constant force against the edge of a freely spinning grindstone. The radius of the grindstone is 40 cm and moment of inertia is 50 kg.m2. The grindstone initially at 10 rev/s and after 10 seconds it drops to 5 rev/s. (a) Calculate the angular acceleration of the grindstone in rad/s2. (b) What is the torque applied to the grindstone? (c) Determine the applied force. (d) Calculate the time taken for the grindstone to stop rotating.

6. A solid cylinder of mass M = 20 kg radius R = 10 cm and inertia given by I = ½ MR2, rolls down a slope without slipping starting at rest. At B, its translational velocity is 5 m/s. (a) What is the angular velocity of the cylinder in rad/s at B? (b) Calculate the rotational kinetic energy at B. (c) Assuming no energy lost, determine the vertical height, h between A and B.

A

h

B
7. A 5 kg mass rotates a cylindrical drum when it unwinds accelerating downwards at 6 m/s2. Calculate: (a) the tension in the string. (b) The rotational kinetic energy of the drum. (c) the moment of inertia if the radius of the drum is 0.42 m.

Mass = 5 kg P

0.80 m

Q

8. The uniform turntable of a record player is a solid disc of mass 0.8 kg radius 24 cm. The turntable is driven from rest by a motor for 1.5 s, when its angular speed reaches 3 rad/s the motor is turned off. [ Inertia, I = ½ mr2 ] .

(a) Calculate the torque supplied by the motor. (b) What is the angular momentum supplied by the motor.

While rotating at 3 rad/s a record of moment of inertia of 3 x 10-3 kg.m2 is placed on the turntable. (c) What is the new angular speed of the turntable and record? (d) Calculate the decrease in rotational energy of the system.

9. A phonograph turntable driven by an electric motor accelerates at a constant rate from rest to 33 rev/min in a time of 25 seconds. The turntable is a uniform disc of metal of mass 1.5 kg and radius of 13 cm. ( Inertia, I = ½ MR2 for the disk). (a) Calculate the angular acceleration in rad/s2. (b) What torque about the axis is required to drive this turntable?

10. A light inextensible string is wrapped around a wheel of radius 10 cm, which can rotate freely without friction. The free end of the string is tied to a 20 kg mass, which accelerates at 2 m/s2 when released from rest at A. The string comes off the wheel just before the load hits the ground at B. Calculate: (a) The tension in the string when the mass is released. (b) The speed with which the mass strikes the floor. (c) The rotational kinetic energy of the wheel just after the string comes off. (d) The moment of inertia, I of the wheel.

20 kg A

2m B

11. A 12 kg mass is attached to a string wrapped around a wheel of radius, r = 10 cm. The mass accelerates at 2 m/s2 down the incline. Calculate: (a) the tension in the string (b) the moment of inertia of the wheel (I) (c) the angular speed of the wheel 2 seconds after it begins rotating starting from rest.

r

2 m/s2

40o

Equilibrium When a body is in equilibrium, the vector sum of forces acting on it must be zero. ; and

This means that the sum of the translational forces and the rotational forces ( torque ) must be zero. Example 1 A car engine of mass 500 kg is suspended from a chain that is linked at P to two other chains. Find the tension in the three chains. Solution: 60o The weight of the car engine will T3 be equal to T1. T2 T1 P T1 m mg mg T1 = mg = ( 500kg )(10 N/kg ) = 5 000 N To solve for T2 and T3, Write sum of forces equations and Solve simultaneously. T3 ; T2 = T3cos 60o (1) T2 60o ; T1 = T3 sin 60o (2) Taking equation (2) and substituting: T1 T1 = T3 sin 60o T3 = = 5774 N Substituting in equation (1): T2 = T3cos 60o = (5774 N) cos 60o = 2887 N Example 2

A weight of 400 N is suspended from a beam as shown. Calculate the tension in the two cables.

30o 60o Method I T1 T2
Vector components T1 T2 30o 60o 400 N

400 N ; T1cos 30o = T2cos 60o, T1 = (1) ; T1 sin 30o + T2 sin 60o = 400 N (2) Substituting (1) in (2): ( ½ ) + T2 = 400 N , T2 = 346 N Solving for T1 : T1 = = = 200 N

Method 2. Trigonometry Sketch a vector diagram and solve. T1 = 400 cos 60o 30o = 200 N T2 T2 = 400 sin 60o = 346 N 90o 400 N 60o T1

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EQUILIBRIUM PROBLEMS

1. A traffic light of weight 140 N is suspended from a vertical cable tied to two other cables that are fastened to two supports. Find the tensions in the three cables.

35o 45o

T2 T1

2. Find the tension in each cable supporting the 700 N mass.

37o T2

T1

700 N

3. A mass is suspended by two cords. The tension in one cord is 120 N as shown.

120 N 50o 40o T

Weight, W

(a) What is the weight, W of the mass? (b) What is the tension, T in the other cord?

Contact Forces ( FRICTION ) Whenever two bodies interact by direct contact of their surfaces, it is known as contact force. For every surface there is a co-efficient of friction ( ). There are two types of friction; 1. Static coefficient of friction ( s ) 2. Kinetic coefficient of friction ( k ) Static force of friction is the product of the coefficient of friction and the normal force. This force needs to be overcome by the applied force in order to bring about motion or movement. If the force is not overcome, there will be no motion. Fr =s N

Units: N Where: N = the normal force (N) Kinetic force of friction is the force exerted on an object when the object is moving with a force F. Fr =k N

Units: N

The static force is greater than the kinetic force of friction.

Force of Friction

Static Kinetic

Example 1 A man is pulling a crate of mass 50 kg ( weight = 500 N ) with a force of 200 N. The coefficient of kinetic friction between the floor and crate is 0.2. Calculate: (i) the frictional force exerted by the floor. (ii) the acceleration of the box. Solution: (i) Fr = k N (ii) Fnet = ma = ( 0.2 )(500 N) (200 – 100) N = (50 kg) a = 100 N a = = 2 m/s2

Forces acting on the crate: motion Normal (N) = 500 N Applied Force Frictional Force = 100 N 200 N

Weight = mg = 500 N Example 2 A 10 kg block rests on a plane inclined at 30o to the horizontal. The coefficient of friction between the plane and object is 0.4. Calculate: (i) The frictional force ( Fr ) (ii) The acceleration of the block down the incline. ( a ) Normal Force (N) = mg cos

mg

Frictional F

m mgsin

Solution: (i) Fr = N = mg cos = ( 0.4 )(10 kg)(10 N/kg) Cos 30o = 34.6 N

(ii) Fnet = ma, mg sin - Fr = ma (10 kg)(10 N/kg)Sin 30o - 34.6 N = (10 kg) a a = a = 1.54 m/s2

Exercise A two mass system is connected over a light frictionless pulley. It is released from rest and mass A falls a distance of 2 m in one second.

m2g

m2 = 4 kg A

B Fr m1 = 9 kg m2gsin m1g

= 30o Calculate: (i) the acceleration of mass A (a) (ii) the tension in the string. (T) (iii) the frictional force on mass B ( Fr ) (iv) the coefficient of kinetic friction between mass B and the inclined plane. (k) Solution: (i) Can use equation of motion to find the acceleration. s = ut + ½ at2 ( ut = 0 ) 2.00 m = ½ a (1.00 s)2 a = 4 m/s2

(ii) Using mass A: T m1g - T = m1a Fnet(36 N) 90 – T = 36 N m1g = 90 N T = ( 90 – 36 ) N = 54 N

(iii) Using mass B: T T – ( m2g sin + Fr ) = m1a m1g sin Fnet= m1a (54 - 20) N – Fr = 16 N Fr Fr = ( 54 – 20 – 16 ) N = 18 N (iv) Using the friction formula: Fr = k N Fr = k ( mg cos ) 18 N = k ( 34.6 N ) k = 0.52

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Banking of Curves Cars usually slow down while banking a curve. When a car is banking a curve, the frictional force equals the centripetal force. The coefficient of friction between the road and tyre is a factor that decides the maximum speed with which you can bank a curve. Example
Case 1 - Bend on a level road A car travels at a constant speed of 15 m/s on a level circular road (bend) of radius 70 m. Calculate the minimum coefficient of static friction between the tyres and roadway in order for the car to turn without slipping. Fr = N Fr = mg (1) 70m Fc = (2) Fr = Fc mg = , = (3)

Using equation (3) ; = = = 0.32 Case 2 - A Banked road

A car is banking a curve inclined at 30o. If the friction between the tyre and the road is = 0.32 and the radius of the curve is 70 m , what is the maximum speed the car can turn without slipping?

30o mg 30o

The centripetal force is equal to the frictional force;

Fc= , Fr = N , Therefore = N ( Where N = mg cos )

So: v2 = rgcos - General formula.

v = = = 13.93 m/s

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FRICTION PROBLEMS 1. Two masses are connected by a light inextensible string over a frictionless pulley. The coefficient of friction between the 7 kg mass and the plane is 0.25. Calculate the acceleration of the system.

7 kg 12 kg

30o

2. Two masses are connected as shown. The mass m1 = 10 kg and m2 = 5 kg.
The system starts from rest and m2 falls a distance of one metre in 1.2 seconds.
Calculate the coefficient of kinetic friction between m1 and the table. m1 m2

3. Two masses m1 = 4 kg and m2 = 9 kg are connected by a light cable over a frictionless pulley. m1 rests on the floor and when released from rest, it moves up 1 m in 4 seconds. Calculate: (a) The acceleration of m1. (b) The tension in the string. (c) The coefficient of kinetic friction between m2 and the incline.

40o 37o

m2

m1

4. Lying on a table is a 10 kg block for which the coefficient of friction is s = 0.4. Another block of mass m, hangs from point K as shown. T
10 kg 30o s = 0.4 K

m

(a) If the mass m is just sufficient to make the 10 kg block slide, calculate the tension T. (b) Determine the value of m.
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GRAVITATION Newton’s Law of Universal Gravitation “Every particle in this universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.” F =

Units : N

G = Universal gravitational constant = 6.67 10-11 Nm2kg-2

M

m

F F

r

To derive an expression for the acceleration due to gravity for earth: Fg = , Fa = mg Equating the forces we get: Fg = Fa = mg ( m can be eliminated ) g = Where: M = mass of earth r = radius of earth To obtain a value for gravitational acceleration: Mass of earth = 5.98 1024 kg , Radius of earth = 6.4 106 m g = = = 9.74 m/s2 Assuming that earth is perfectly spherical. Graph of weight (F) against distance from earth.

F (N) Area under graph = work done Wd = F.r

r1 r2 Distance (r)

Wd =F.dr

Note: 1. For a uniform field, gravity is constant. 2. For a non-uniform field, gravity varies both in magnitude and direction. 3. Earth’s gravitational field is to be uniform up to a certain altitude – ( g is approximately taken to be constant )

Gravitational Potential Energy ( U ) The potential energy of an object can be found by using the equation EP =mgh

Where – h is the height above or below a reference level. This equation is only valid when the object is near the earth’s surface. For objects high above the earth’s surface, such as satellites, an alternative equation is used. The reference level is redefined. The reference or zero level is placed at an infinite distance from the centre of the earth. When a particle of mass m, is at distance r, from the centre of mass, the potential energy of the system is same as the work done in the direction of the field to move the particle from Infinity ( ) to r.

Taking Infinity () as our frame of reference (0 J), we can derive an expression for potential energy.

r m ( EP )

WD = EP = U WD = = U = GMm = GMm = GMm Taking , we get; = GMm U = Where: r = distance from centre of earth

Graph of U against r r

As U , As , U

Escape Velocity This is the velocity required by a projectile to escape the earth’s gravitational field. To derive an expression for the escape velocity, we equate the kinetic energy with its gravitational potential energy. EP = U = EK = ½mv2 v = Where: r = radius of earth (m). The Kinetic Energy of a projectile at a distance rfrom earth; The gravitational potential energy of a projectile is the energy gained by bringing it from infinity to a distance r from the earth. The EK acquired is the same as work done on the projectile from the centre of the earth to infinity. EP = 0 , EK = max g = 0 r

WD = EK = = = GMm = GMm = GMm = GMm EK = Where: r = distance from centre of earth If a projectile has energy equal to E = , the force of gravity at no time will be able to reduce the EK to zero. At the surface of the earth: U = EP = , EK = ETotal = EP(U) + EK = + = 0

If the total energy ETotal> 0, the projectile is able to escape. ( EK> EP ) If the total energy ETotal< 0, the projectile is not able to escape. ( EP> EK )

Example Find the escape velocity of a rocket of mass 5 tons from the surface of the earth. ( Mass of earth = 5.98 1024 kg, Radius of earth = 6.4 106 m ) v = = = 11 090 m/s Note: The escape velocity is independent of mass of rocket.

Satellite in circular orbit around earth

v Me FC satellite

The velocity of the satellite as derived earlier is: v = Or: v2 =

The kinetic energy of the satellite: EK = ½ mv2 EK= ½m EK =

The gravitational potential energy ( EP ) of the satellite is: EP = U = At the surface of the earth, the EP is large and negative. ( much less than points further away) The total mechanical energy of a satellite in a circular orbit will be the sum of its gravitational potential energy and its kinetic energy. ETotal = U + EK = + = = ETotal = So the total energy is the negative of the kinetic energy. - = - EK ET = - EK

Graph of Energy of satellite versus distance

Energy

EK ETotal EP ** The negative energy means that the projectile or satellite cannot escape the planet. ( It will remain in a circular orbit. )

Binding Energy Binding energy is the energy needed for a projectile or a satellite to escape the gravitational field. ETotal 0 for a projectile. In order for the projectile to escape, its total energy should be positive. If the satellite is to be in a circular orbit, the total energy is ETotal = Since in order for the satellite to escape, the total energy has to be 0 means that ET Therefore the binding energy for a projectile in circular orbit =

Example (a) What is the gravitational potential energy of a 50 kg satellite in a circular orbit 500 km above the surface of the earth? (b) Determine the kinetic energy of the satellite. (c) Calculate the total energy of the satellite. [ Mass of earth = 5.98 1024 kg , Radius of earth = 6.4 106 m ] Solution: (a) U = = = - 2.85 109 J (b) EK = = = 1.43 109 J (c) ET = = - EK = - 1.43 109 J

Kepler’s Laws of Planetary Motion Law I “All the planets move in elliptical orbits having the sun as one focus.”

F F’

Law II “A line joining any planet to the sun sweeps out equal areas in equal times.”

O B C

D A

( The time to travel AB is the same time it takes to travel CD, Areas AOB = COD ) A planet moves along its orbit at a rate such that a line from the sun to the planet sweeps over areas, which are proportional to the time intervals. Note: 1. The value for average radius for an ellipse is half the minor axis. 2. The value for average radius of a circular orbit is the radius of the circle. Law III “In a circular orbit, the square of the orbital period ( T2 ) is directly proportional to the cube of the radius ( r3 ) of the orbit.”

Proof: Fcentripetal = Fgravitational = v2 = = = 4 2 r3 = GM T2 T2 =r3

Since = a constant;
T2 r3

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GRAVITATION PROBLEMS

1. A satellite of mass 500 kg is launched from a site on the earth’s equator into an orbit of 200 km above the surface of the earth. Assuming the satellite orbit is circular. Calculate: (a) the speed of the satellite (b) the orbital period of the satellite (c) the minimum energy needed to place the satellite in orbit. 2. Astronomers believe that the Milky Way Galaxy is rotating and the sun takes 200 million years to make one complete revolution (orbit). (Take one year = 365 days)
The sun is in one of the spiral arms of the galaxy at a distance of 5 x 1021 m from the galactic centre. (a) Calculate the speed at which the sun orbits the galactic centre. (b) Assuming that most of the galactic mass is at its centre, calculate the mass of the galaxy.

3. Use Newton’s Laws to derive an expression for the period, T of a satellite in a circular orbit of radius, r about the earth in terms of the radius, r and mass of earth, M.

4. The mean radius of the earth’s orbit around the sun is 1.5 x 108 km and it takes 365.25 days to complete the one revolution. Calculate the mass of the sun.

5. At what distance from the surface of the earth does the moon orbit the earth if the orbital period of the moon is 27.3 days?

6. How far above the earth’s surface will the acceleration due to gravity be exactly half of that at the surface? Give your answer in terms of the radius of the earth. ( Re ).

7. A satellite of mass ms is moving in a circular orbit of radius r and period T around the earth. The centripetal force on the satellite is equal to the gravitational force of attraction between the satellite and the earth.

Show that T2 is directly proportional to r3.

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DIRECT CURRENT

Drift Velocity

Whenever an electric field is established in a conductor, charges start to flow and this is said to be the current. The current is defined as the rate at which charges flow through a conductor. If Q is the amount of charge passing through an area in a time interval, t, the current is given by I=
The charges flowing through in a circuit can be negative, positive or both.

A By convention we take the direction of current in the direction of the positive flow of charges. x

vd

e-

vd t

Consider the current flowing in a uniform metal conductor of cross sectional area A.
(The charges flowing are electrons, q = e-)
The charges (e-) move with a speed called drift velocity (vd). The distance travelled by the charges in a given time, t, is x = vd t.
The number of mobile charges in the section of length x, is given by nAvdt.
Where n is the number of mobile charges per unit volume. The value of charge, Q, is given as Q = ( nAvdt )e ( q = e ) If we divide the above equation by time (t), we get an expression for current I = nevA

Where: I = current in the conductor (A). n = number of charges ( electrons ) flowing in the interval given. v = vd = drift velocity ( m/s ). A = cross sectional area of conductor ( m2 ).

The average rate of movement of the charges is called the drift velocity. This value is not the same as the speeds of the individual charges because the charges undergo random motion with a very high velocity in the presence of an electric field. The charges do not travel along the electric field lines in the conductor. The charges undergo collisions with the atoms of the conductor and this results in a very complex motion. ( Just like the Brownian motion of a gas ) This continuous collision of the charges with the atoms increases the vibrational energy of the atoms and therefore causes an increase in the temperature of the conductor whenever current is flowing through it.

Example.

A 0.1 m long silver wire has a diameter 1 mm and transfers 90 C of charge in 75 minutes.
Silver contains 5.8 x 1028 free electrons per cubic metre. (a) Calculate the current in the wire. (b) What is the drift velocity of the electrons in the wire?

Solution:

(a) Since by definition, current is charges per second, I = Q/t;

I = = = 0.02 A

(b) We can use the expression I = nevA , to calculate the current.

A = r2 = (0.5 x 10-3 m)2 = 7.85 x 10-7 m2

I = nevA , v = [n = 5.8 x 1028x πr2h = 4.56x1021e-s] = = 34.91 m/s

Current Density

If a current is uniformly distributed over a cross-sectional area A, the current density ( J ) is given by J = , J =

Therefore J = nev

Where: J = current density ( A.m-2 ) n = number of mobile charges e = electronic charge ( C ).

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Galvanometer conversion

A galvanometer is a device that operates when a torque acts on a current loop in the presence of a magnetic field. A galvanometer is used in the construction of both ammeters and voltmeters.

The Ammeter
A galvanometer has an internal resistance of approximately 100 and is not suitable to be used as an ammeter because a resistance this large can considerably alter the amount of current in the circuit. Another limitation in a galvanometer is that it has a very small full-scale deflection. (The limit of measurement is very small.)

Example.
We want to convert a galvanometer of internal resistance 60 and full scale deflection ( fsd) of 1 mA to an ammeter that has a full scale deflection of 2A. Calculate the resistance needed for the conversion.

A shunt resistance ( Rs ) has to be connected to the galvanometer in parallel, to carry the bulk of the current since the galvanometer can only take 1 mA of current at fsd.

G

Galvanometer, 60

a b

Rs

As the resistance of the galvanometer is 60 , the potential difference across ab is given by V = IR, V = ( 1 x 10-3 A ) ( 60 ) = 0.06 V The resistance of the shunt resistor (Rs) can now be calculated as we know that the voltage across it is 0.06 V and the current at fsd in Rs will be (2 - 0.001) A = 1.999 A

Rs = = = 0.030

The value of the Rs is very low because it provides a path of very low resistance for carrying the bulk of the current.

The Voltmeter

The voltmeter is used to measure the potential drop in a circuit and it is connected externally, across two points in the circuit. This means that the internal resistance of the voltmeter should be very high so that it does not allow any significant current to pass through and alter the circuit.

Example.
We want to convert a 60 galvanometer with fsd of 1 mA into a voltmeter that can read up to 12V. Calculate the resistance needed for the conversion.

A resistance has to be connected to the galvanometer in series to increase its total effective resistance.

G

R a b c

V

The current flowing through the galvanometer and the resistance will be same ( 1 mA ).
The voltage across the galvanometer can be calculated using Ohm’s Law

Vab = IR = (0.001 A) (60 ) = 0.06 V

Since the fsd for the voltmeter is 12 V ( Vac ) , the potential difference across the resistance ( Vbc ) is Vac - Vab , Vbc = (12 – 0.06) V = 11.94 V

The value of the resistance, R, is therefore

R = = = 11 940

The value of the resistance, R is very high because a voltmeter should not provide a path for the current to flow.

Kirchoff’s Laws Simple circuits are analysed using Ohm’s law. More complex circuits containing several sources of emf and resistances are analysed using Kirchoff’s laws.

Kirchoff’s Rule 1. ( The junction rule )

“The sum of currents entering any junction must be equal to the sum of the currents leaving that junction.”

This junction rule is based on the conservation of charge.

The schematic diagram below illustrates Kirchoff’s junction rule. This rule requires that whatever current enters a junction must leave that junction. This is analogous to liquid flowing through a pipe and splitting at a junction. The total amount of liquid entering the junction in the pipe should equal the liquid carried by the branching pipes.

I2 I1 Flow in Flow out

I3

(a) Schematic diagram (b) A mechanical analogue

If we apply the junction rule to the above schematic diagram, we get I1 = I2 + I3 .

Example.

Calculate the unknown current in the circuit given.

I1 = 3.1 A Using the junction rule, we get: I2 + I3 = I1 + I4 I2 = 2 A I3 = x x = I1 + I4 - I2 I4 = 2.4 A x = ( 3.1 + 2.4 – 2 ) A x = 3.5 A

Kirchoff’s rule 2. ( The loop rule ) “The sum of the potential differences across all elements around any closed-circuit loop must be zero.”

The loop rule is based on the principle of conservation of energy.
Any charge moving around any closed loop in a circuit must gain as much energy as it loses. Problem solving strategy.

1. Assign symbols and directions to the currents in all the branches of the circuit. If the direction of the current is labelled wrongly then after analysing, the value will be negative but the magnitude of the current will be accurate. 2. Choose a junction and use the first rule to relate all the currents. 3. Choose any closed loop in the network and designate a direction (clockwise or counter clockwise). It doesn’t have to be in the direction of the assumed current. 4. Start from a point and traverse the loop and write the elements as emf’s using the following the rules: a. An emf is positive if you traverse it from – to + and negative if traversed from + to –. b. An IR product is negative if your path is in same direction as the assumed current and positive if direction of current is opposite to chosen path. 5. Solve the equations simultaneously for the unknown quantities. Be careful in your algebraic steps, and check your numerical answers for consistency.

\

Travel Travel

- + + -

+ -

Travel Travel

R
R
+ - - +
- IR + IR I I

Example

Find the currents I1, I2 and I3 in the multi-loop circuit given.

14 V e f + -

4 I2

- + I1 b c 10 V 6 I3

2 a d

Solution.

To find the three unknown currents, we apply the junction rule at one junction and the loop rule twice to obtain three equations and so be able to solve for three variables.

Step i. Choosing junction c and applying the Kirchoff’s first rule we get

I1 + I2 = I3 (1)
Step ii. The circuit has three loops but only two is needed, so lets take loops abcda and befcb and traverse in a clockwise direction

Loopabcda: 10 V – 6 I1 – 2 I3 = 0 (2)

Loopbefcb: -14 V + 6 I1 – 10 V – 4 I2 = 0

Loopbefcb simplifies to - 24 V + 6 I1 – 4 I2 = 0 (3)

Step iii. We now have to use the equations 1, 2 and 3 and solve them simultaneously. Take equation 1 and substitute in equation 2. ( The units can be ignored in the calculations) 10 – 6 I1 – 2 ( I1 + I2 ) = 0

10 – 8 I1 – 2 I2 = 0 (4)

Step iv. We have to use the equations (3) and (4) and eliminate one of the variables. Take equation (3) and divide throughout by 2.

12 – 3 I1 + 2 I2 = 0 (5)
Step v. Add equation (5) to (4) to eliminate I2, gives

12 – 3 I1 + 2 I2 = 0

(+) 10 – 8 I1 – 2 I2 = 0 22 – (11) I1 = 0 , (11) I1 = 22 , I1 = 2 A Step vi. Substituting I1 in (5) results in a value for I2

12 – 3 ( 2 ) + 2 I2 = 0 , I2 = - 3 A

Step vii. Finally use equation (1) to calculate I3

I3 = I1 + I2 , I3 = - 1 A

The values for the currents are:

I1 = 2 A I2 = - 3 A I3 = - 1 A

The negative values of I2 and I3 indicate that the directions of the currents are opposite to that designated initially.

Resistance, Resistivity and Ohm’s Law
For a uniform conductor, the voltage applied across a certain length is directly proportional to the measured current provided the physical conditions are not altered. The constant of proportionality is said to be the resistance.
The SI unit of resistance is the ohm ( ). R =
In a conductor, the electrons flow under the influence of an electric field and collide with its atoms and progress slowly as current. This random collision of the charges creates an internal friction in a direction opposite to the current flow and can be said to be the resistance.
The resistance in a conductor is directly proportional to its length ( l ) and inversely proportional to its cross-sectional area ( A ). R =

Where: R = Resistance ( ) l = length of conductor ( m ) A = cross-sectional area ( m2 ) = Resistivity ( .m )

Resistivity of a substance is a constant value and is characteristic of that material.
Examples of resistivity of some materials in .m :

Material | Resistivity (.m) | Copper | 1.7 x 10-8 | Aluminium | 2.8 x 10-8 | Carbon | 3.5 x 105 | Glass | 1010 - 1014 | Ohm’s Law

The voltage is directly proportional to the current in a conductor provided the physical conditions like temperature is constant. V = IR

Current Density

Current density as a function of resistivity:

Wd = F.l F = , Wd = Vq F = , V = E l ( l = d ) E = , E = [ R = ] E = E = nev , [ J = nev ]

E = J
Where:
E = Electric field ( N/C ) = Resistivity ( .m ) J = current density ( A.m-2 )

The Wheat stone Bridge

This is a special circuit that can be used to determine the resistance of an unknown resistor by comparison with three other resistances.

b

P I1 NB: The resistance with the same current are divided.
Q

G a I1 d N I2
M

S2 I I c I2

+ - S1

When both the switches ( S1 and S2 ) are closed and the galvanometer reads zero, the circuit is said to be balanced. Let P be the unknown resistance, and the other resistances can be varied to obtain the balance point. At balance point the precise values of the resistances Q, M and N can be taken. Then the unknown resistance can be calculated as outlined below.

At balance point, the galvanometer reading, G = 0.

Taking loop abca : - I1P + I2 M = 0

I1P = I2 M (1)

Taking loop bdcb : - I2 Q + I2 N = 0

I1 Q = I2 N (2)

Dividing equation 1 by 2, we get:

=

= (3)

Example.

A bridge circuit is setup to determine the resistance of an unknown resistor. Three variable resistances are arranged and setup as shown in the previous diagram. These resistances are varied till the bridge is balanced. At balance point the resistances of Q, M and N are read to be ( Q = 3 , M = 2.5 and N = 4 ). Calculate the value of P.

Using equation (3), we can substitute directly;

=

P = Q , P = ( 3 ) P = 1.875

The Slide wire Potentiometer

The potentiometer is an instrument that can be used to measure the emf of a source (Example; a dry cell) without drawing any current from the source. Basically this device balances an unknown potential difference against an adjustable, measurable potential difference.

+ - 1 R

l1 l2

Slide wire jockey G

x + - r

When the galvanometer reads zero, balance point is obtained. At this point there is no current flowing in the unknown emf source (x ) .

We can then write two equations involving the individual emf’s.

1 = I R(l1) (1) , x = I R(l2) (2)
Dividing equation (1) by equation (2) we get:

= (3)

since R = So the ratio becomes =

Equation 3 then becomes: =

Making x the subject leaves x = 1

EMF and internal resistance of a cell

EMF is the voltage measurement of power sources. An example of a source of emf is a dry cell or battery. The circuit diagram below shows a cell of internal resistance r, connected to a load resistance R.

Terminal Voltage, VT

a - + r b

I I

d load resistance c
R

From the above circuit diagram, the terminal voltage across the cell is given by the potential difference between a and b. Using Kirchoff’sLoop rule we get:

VT = Vb – Va , VT = - IR (1)

In the expression (1) the emf is equal to the terminal voltage when the current is zero. This is called the open-circuit voltage. The voltage across the load resistance follows Ohm’s law and is, Vcd = IR . (2)
Combining equations (1) and (2) we get:

* = IR + Ir (3)

Solving for current gives:

I = (4)

The above expression (3) shows that the current in a simple circuit is dependant on both the external resistance and the internal resistance to the battery. But if R is much greater than r, then r is neglected in analysis of a circuit.

The internal resistance of a dry cell can be calculated practically in a lab. The expression (3) can be used to graphically analyse the emf, and internal resistancer.

The expression VT = - Ir can be rewritten as

VT = - Ir + (5)

The corresponding values of current and voltage can be measured across a cell and a graph plotted.

Graph of Voltage versus Current

Voltage (V)

Current (A)
From the above graph the y-intercept will represent the emf , and the slope the negative of the internal resistance, r.

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INTERNAL RESISTANCE AND EMF’S PROBLEMS

1. A cell of emf 1.5 V is connected to a 4 resistor. The voltage across the cell is then observed to be 1.2 V. (a) How many coulombs of charge pass through a cross-section of the circuit in 5 seconds? (b) How many joules of energy is supplied by the cell to the external part of the circuit in one second? (c) Calculate the internal resistance of the cell.

2. A storage battery of emf 6.4 V and internal resistance of 0.08 is being charged by a current of 15 A. Calculate: (a) the power loss in the internal heating of the battery. (b) the rate at which energy is stored in the battery. (c) its terminal voltage.

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CONDUCTION THEORY and BAND STRUCTURES

Electrical conductivity is understood on the basis of the mobility or lack of mobility of electrons in a material.

Metallic crystals

The valence electrons are not bound to individual lattice sides but are free to move through the crystal. Thus metals are good conductors.
Example: Mg – 1s2 2s2 2p6 3s2 electrons in 3s2 are free to move

Covalent crystals

The valence electrons are involved in bonds responsible for the crystal structure and are therefore not free to move. Thus there are no mobile charges for conduction and such materials are insulators. Example: CO2 , O C O

Ionic crystals

The charges are held in fixed positions when in the crystalline solid state and so is an insulator. However in the molten state the ions ( charges ) are mobile and so is a conductor.
Example: NaCl ( common salt )

Temperature and Conductors

| Temperature | Resistivity | Conductivity | Insulator | Increase | Decrease | Increase | Conductor | Increase | Increase | Decrease | Semi-conductor | Increase | Decrease | Increase |

The resistivity ( ) of a good insulator is much greater than that of a good conductor by an enormous factor. Example: ( copper ) = 1.7 x 10-8 m ( Pyrex ) = 1012m

Metals
Lattice vibration increases with raise in temperature and provides a larger surface area for the electrons to collide. The increased rate of collision of electrons increases the resistivity and therefore decreasing the conductivity ( conduction ).

Insulators
In insulators, what little conduction does take place is due to electrons that have gained enough energy from thermal motion of the lattice to break away from their “home” atoms and wander through the lattice.

Semi – conductors
This is a class of materials intermediate between conductors and insulators in their ability to conduct electricity. ( Example: Silicon (Si) and Germanium (Ge) ) – group 4 elements in the periodic table. All the electrons are involved in bonding but a relatively small amount of energy is needed to break electrons from the lattice and to become mobile. So as the temperature increases, the conductivity of a conductor increases.

Band Structures
1. Insulators
For insulators, band 1 is completely filled but band 2 is too far above band 1 energetically to permit any appreciable number of band 1 electrons to jump the energy gap. ( E – gap ).

Energy Band 2

3s

E – gap 3 – 6 eV

2p

Band 1 2s

1s

2. Semi – conductors In semi – conductors such as silicon, band 1 is completely filled but band 2 is so close energetically that electrons can jump easily after absorbing energy into the unfilled levels of that band. Energy

Band 2 3s 0.7 – 1 eV

2p

Band 1 2s

1s
3. Conductors

Conduction in conductors such as copper have their band 2 only partially filled so that the electron can move easily to the higher energy levels and thus travel through the solid. Energy

Band 2 3s

2p

Band 1 2s 1s

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DIRECT CURRENT PROBLEMS

1. A copper conductor has a uniform cross-sectional area of 3 x 10-6 m2, and carries a current of 21 A. (a) Determine the current density in the wire. (b) Find the drift velocity of the electrons, assuming that the free electron per unit volume is 1029 electrons.

2. An aluminium wire has a length of 10 m and a cross-sectional area of 5 mm2. The resistivity of the given wire is 1.8 x 10-8m. (a) Calculate the resistance of the aluminium wire. (b) What current flows through the wire if a potential difference of 1 V is maintained between the ends of the wire? (c) How does the resistance vary when the current through it is reduced?

3. The density of free electrons per cubic metre. When an electric field of 0.35 V/m is applied across a 50 m long piece of copper wire, the electrons have a drift velocity of 1.6 x 10-3 m/s. (a) Calculate the current density in the wire. (b) Find the resistivity of the wire. (c) Calculate the resistance of the 50 m wire if its cross sectional area is 1 x 10-6 m2.

4. An electric motor operates on 250 V and a current of 1.8 A raises a load of 234 N at a steady speed of 0.6 m/s. Calculate the efficiency of the motor.

5. We require an ammeter for a current range of 0 to 1.0 A. A galvanometer is available that has a resistance of 100 and gives a full deflection at current of 1.0 mA. (a) Draw a circuit diagram to show how the shunt resistance is connected. (b) Calculate the value of the shunt resistance required.

6. An electric toaster has a resistance of 8 and is connected to a a power supply and carries a current of 15 A. Calculate the amount of heat energy dissipated by the toaster in a time of 360s.

7. A galvanometer of internal resistance of 120 , which has a full-scale deflection of 2 mA, is to be converted to a voltmeter to measure up to 12 V. (a) Draw a circuit diagram to show how the resistance is connected. (b) Calculate the value of the resistance required for the conversion.

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KIRCHOFF’S LAWS PROBLEMS

1. Study the circuit diagram given below and answer the questions that follow:

3 Ω 12 V B A

I3

4 Ω 2 Ω I2 C F

I1

2 Ω 8 V D E

(i) Using Kirchoff’s first law, write an equation for junction, C.
(ii) Using Kirchoff’s second law, write an equation for the loops.

1. ABCFA 2. EDCFE

(iii)Use the equations obtained in (i )and (ii) to solve for I1, I2 and I3.

2. Use the circuit below to answer the questions that follow.
I A 12 I1 (i) Use the Kirchoff’s Laws to calculate the values of I, I1 and I2. 2 (ii) Find the potential difference between
Points A and B. 4 V 6 V 2 V

I2 2 B
3. A multiloop circuit is given below.

7B 5
(i) Using junction B, write down the equation relating the currents. I3

(iii) Calculate values of I1 , I2 and I3 . I1 4 V I2
Comment on significance of any negative values obtained. 6V 8V

(i) Find the current through the 6 resistor. (ii) Calculate the values of E1 and E2 . (iii) Hence determine the potential difference between points A and B.

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ELECTROSTATICS

Electrostatics is stationary accumulation of excess charge.

Charge density ( )

The amount of charge per unit area is defined as the charge density.

= Units: C.m-2

Example.
Find the charge density, if the field lines from a positive charge of 0.1 nC emanate from a sphere of radius 20 cm. ( n = nano = 10-9 )

Area of a sphere is A = 4 r2 = 4 ( 0.2 m )2 = 0.503 m2

= = = 1.99 x 10-10 C.m-2 , can also be written as = 1.99 C.m-2
Where: [ 1 Angstrom () = 10-10 ]

Coulomb’s Law

Charles Coulomb established the fundamental law of electric force between two stationary charged particles. The electric force has the following properties:

1. It is directly proportional to the product of the magnitudes of the charges. 2. It is inversely proportional to the square of the separation (r) between the two particles. 3. It is attractive if the charges are of opposite sign and repulsive if the charges have the same sign. Coulomb’s Law formula: F = k

Where: k = = 9 x 109 Nm2.C-2 o = is the permittivity of free space
= 8.8 x 10-12 C2.N-1m-2

Charge and mass of the electron, proton and neutron.

Particle | Charge ( C ) | Mass ( kg ) | Electron | - 1.6 x 10-19 | 9.11 x 10-31 | Proton | + 1.6 x 10-19 | 1.67 x 10-27 | Neutron | 0 | 1.67 x 10-27 |

Principle of Superimposition

“When two or more charges are present, the resultant force on any one charge equals the vector sum of the forces exerted by the individual charges that are present.”

Examples. Linear Superimposition 1. Three charges are positioned as shown below.

Q1 = +2 nC Q2 = +5 nC Q3 = +3 nC

2 m 2 m

Calculate the total force exerted on Q2 due to the charges Q1 and Q3.

Solution. Since force is a vector quantity, the total force on Q2 will be the vector sum of the individual forces exerted by Q1 and Q3 .

Q1 = +2 nC Q2 = +5 nC Q3 = +3 nC

F1 F2

F1 = k F2 = k = (9x109 Nm2C-2)(2x10-9C)(5x10-9C) = (9x109 Nm2C-2)(5x10-9C)(3x10-9C) (2 m)2 (2 m)2

= 2.25 x 10-8 N = 3.38 x 10-8 N

Fnet = F1 + F2 = 2.25 x 10-8 N + 3.38 x 10-8 N

= 1.13 x 10-8 N

2. Two Dimensional Superimposition

Consider three point charges at the corners of a triangle.
The charges Q1 = 6 nC, Q2 = - 2 nC and Q3 = 5 nC.

Q2 37o

3.0 m 5.0 m

F1 F2 Q1 4.0 m Q3

Find the resultant force on Q3 by the other two charges.

Solution. It is important to find the directions of the forces being exerted on Q3. Force F1 is attractive since the charges are opposite and force F2 is repulsive since the charges are identical in nature. To find the net force, find the individual forces using Coulomb’s law and add them vectorially.

F1 = k = (9x109 Nm2C-2) = 3.60 x 10-9 N 37o

F2 = k = (9x109 Nm2C-2) = 1.69 x 10-8 N

Finding the components of F1 and add them to F2:

F1 = 3.60 x 10-9 N

F1y = (3.60 x 10-9 N) sin37o
F1y = 2.17 x 10-9 N 37o F1x = (3.60 x 10-9 N) cos37o = 2.88 x 10-9 N F1x Fx = F1x + F2 = 2.88 x 10-9 N + 1.69 x 10-8 N = 1.40 x 10-8 N

Fy = F1y = 2.17 x 10-9 N

Using Pythagoras theorem Fnet = Fx + Fy Fnet Fy = 1.42 x 10-8 N Fx

= tan-1 = 8.8o

Electric field ( E )

Electric field is defined as a region in space where a test charge will experience a force.

Expression for electric field: E = Units: =

The electric field is a vector quantity. The direction of the E field at a point is given as the direction of the electric force exerted on a positive test charge placed at that point.

Expression of electric field at a distance r from a charge:

The force between two charges Q and q is: F = k (1) and the electric field at q is E = and so F = Eq (2)

Equating (1) and (2): Eq = k gives E =

Example 1

Calculate the electric field generated at P by the charges S and T.

+2 C P +2 C S T 2 cm 3 cm ES ET

ES = ET =

= =

= 4.5 x 107 NC-1 = 2.0 x 107 NC-1

Therefore net electric field at P is E = ES + ET

= 4.5 x107 NC-1 + 2.0 x 107 NC-1

= 2.5 x 107 NC-1

Example 2
Three identical positive charges are positioned at the ends of an equilateral triangle as shown.

Q1 x = 3 cm

Q = + 2 nC

x x

Q2 Q3 x (a) Calculate the net force exerted on the charge Q1 by the other two charges. (b) If Q1 is removed, find the electric field at the position of Q1.

Solution.

(a) The net force on Q1 will be the vector sum of the forces exerted by Q2and Q3.

Fnet = F1 + F2

F1 = F2 =

= = = 4 x 10-5 N = 4 x 10-5 N Since the charges are positioned as an equilateral triangle, all the interior angles are 60o. The component method can be used to find the net force. Trigonometry can also be used.

Vector component method.

Fnet = F1 + F2

60o 60o

F2 F2x = F2cos 60o = 2 x10-5 N

60o F2y = F2 sin 60o

= 3.5 x 10-5 N

F1 F1x = F1cos 60o

60o = 2 x 10-5 N

F1y = F1 sin 60o

= 3.5 x 10-5 N
Fx = 0

Fy = 7 x 10-5 N

Therefore Fnet = 7 x 10-5 N

Trigonometric method.

Take the vectors F1 and F2 and sketch a vector diagram. The result will be a triangle. Find and label as many sides and angles as possible.

F2 = 4x10-5 N 30o

Fnet 120o

30o F1= 4x10-5 N

Since we need to calculate a side, the sine rule can be used. = , Fnet= 7x 10-5 N

E1 E2
(b)
P

+
+

Q2 Q3

The net electric field at P will be the vector sum of the E fields generated by Q1 and Q2.

E1 = k E2 = k

= ( 9109 Nm2C-2 ) = ( 9109 Nm2C-2 ) = 2 104 NC-1 = 2 104 NC-1

Enet = E1 + E2

E2 = 2104 N/C 30o

120o Using sine rule: 30oE1= 2104 N/C = E = 3.46 104 NC-1

Electric Flux ( E) Electric flux is defined as the product of the electric field and the perpendicular area over which it acts. Area = A

E= EA

Units: E

Electric Flux through a sphere:

E=EA , E = , Area of a sphere is: A = 4r2 E= ( 4r2 ) , k =

= ( 4r2 )

= , E=

The spherical area enclosed at a distance r, is known as a Gaussian surface.

This means that the net electric flux through any closed Gaussian surface is equal to the net charge inside the surface divided by o. This is known as Gauss’s Law

r

q Gaussian surface

A spherical surface of radius r, surrounding a point charge q. When the charge is at the center of the sphere, the electric field is normal to the surface and constant in magnitude everywhere on the surface.
Electric Potential ( V )

Electric potential is the work done per unit charge. Positive charges move from a point of higher potential to a point of lower potential. The negative charge moves from a point of lower potential to higher potential. Electric potential V, is a scalar quantity since potential energy is scalar.

E

Q+

A B

+q r VA VB

If the positive test charge is moved from B to A, the potential difference between points A and B is VAB =Vb - VA = = == Er ( V = Ed )

Electric potential is given as V = Er ( r = d ) And E = V =r

V =

Where: k = = 9 x 109 Nm2.C-2

r = separation distance between charges (m)

Example 1 Two charges X and Y of magnitudes 4 nC and –3 nC are placed 20 cm apart. What is the electric potential at point P, which is midway between charges X and Y?

-
+
X 10 cm P 10 cm Y

4 nC - 3 nC

Vx = Vy =

= =

= +360 V = - 270 V

Therefore total voltage at P: V = Vx + Vy

= ( +360 + ( - 270 ) ) V

= 90 V

** Electric potential (Voltage) is a scalar quantity.
Uniform Electric Field

When a potential difference is applied across a pair of parallel plates, a uniform electric field is established which is dependent on the separation of the plates.

V = Ed

Where: V = potential difference across the plates ( V ) E = electric field strength ( N/C = V/m ) d = separation distance of the plates ( m )

Uniformly accelerated motion between parallel plates.

+ + + + + + +

E q

V d

_ _ _ _ _ _ _

The charge experiences gravitational and electrical forces as shown;

FE

Fg

If the charged particle is accelerated between the plates, and the acceleration is much greater than the gravitational acceleration, the acceleration will be:

Fa = FE

ma = Eq

a =

Example 1

An electron of charge ( e- = 1.602 10-19 C ) and mass ( me = 9.11 10-31 kg ) is accelerated between a pair of parallel plates 5 cm apart as shown.

5 cm + _ + _ A B

250 V

(a) Calculate the electric field strength between the plates. (b) What is the energy of the electron if moved from B to A? (c) Determine the acceleration of the electron. (d) Calculate the speed of the electron upon reaching A.

Solution:

(a) V = Ed

E = = = 5 000 Vm-1 (b) W = F.d = Eqd = ( 5 000 V/m )( 1.602 x 10-19 C )( 0.05 m ) = 4.0 x10-17 J

(c) a = = = 8.8 x 1014 m/s2

(d) The work done on the electron from B to A will be the gain in kinetic energy,

Ek = ½ mv2

4 x 10-17 J = ½ ( 9.11 x 10-31 kg ) v2

v2 = 8.8 x 1013 J/kg

v = 9.4 x 106 m/s

Example 2 A proton of mass 1.67 x 10-27 kg, moving horizontally with a velocity of 2.5 x105 m/s, enters a uniform electric field at right angles as shown.

+ + + + + + 30 V

Proton 2 cm * 2.5 x105 m/s

_ _ _ _ _ 0 V 10 cm
Calculate:
(i) the time taken for the proton to move through the field. (ii) the electrostatic force on the proton. (iii) the acceleration of the proton. (iv) the vertical displacement of the proton as it leaves the plates.

Solution:

The path taken by the proton will be parabolic and is analogous to the motion of a projectile in the earth’s gravitational field. All the formulae used for analysing projectile motion can be applied in this situation.

(i) t = = = 4 x 10-7 s

(ii) F = Eq , E = = = 1 500 V/m F = ( 1 500 V/m )( 1.602 x 10-19 C ) = 2.4 x 10-16 N

(iii) a = = = 1.44 x 1011m/s2

(iv) s = ut + ½ at2 ( since u = 0, vertical component of velocity)

s = ½ at2 = ½ ( 1.44 x 1011 m/s2 )( 4 x 10-7s )2 = 0.012 m

Equipotential Surfaces

Equipotential surfaces are defined as a surface on which all points are at the same electric potential(V). The potential difference between any two points on an equipotential surface is zero. Since potential difference (pd) is the work done per unit charge, the work done in moving a test charge from one point to another on the same surface should be equal to zero.

Equipotential surface for any isolated point charge q :

q

The dashed lines show equipotential surfaces. The electric field at any point on an equipotential surface is perpendicular to the surface.

Example.

Find the work done in moving a charge q = 5C from point A (16 V) to a point B (5 V).

V = , W = Vq = ( 5 – 16 )V (5 x 10-6)C = - 5.5 x 10-5 J -------------------------------------------------
CAPACITORS

A capacitor consists of insulated conductors, which stores electric charge and releases it sometime later. C Symbol

Units : Farads ( F ) pF = picofarad 10-12 F nF = nanofarad 10-9 F F = microfarad 10-6 F

Capacitors are devices, which store charge. They give a steady discharge of electrons. The charge build-up occurs at the dielectric material or vacuum between the plates after being subjected to an electric field. Capacitors are used in electrical circuits as * Tuning the frequency of radio receivers * Automobile ignition systems * Camera flashes etc.

A parallel plate capacitor consists of two parallel plates, each of area A, separated by a distance d. The plates carry equal and opposite charges.

+Q - Q

Area = A d

The capacitance C, of a capacitor is defined as the ratio of charge on either plate to the potential difference across the plates. C =

This capacitance formula can be rearranged to give:

Q = CV Where: Q = charge on each plate (C) C = capacitance (F) V = potential difference across the plates (V)

The parallel plate Capacitor.

V

_ + Area = A

d
The capacitance of the capacitor is given by:

C =o

Where: o = the permittivity of free space
= 8.8 x 10-12 C2.N-1m-2 A = area of the plate ( m2 ) d = plate separation (m)

Derivation: We know that, E=EA , and E= as shown before; That is E=EA = EA =

Substituting and in the above equation gives:

A = = , rearranging, C =o

Example 1
The parallel plates of a capacitor are 5 mm apart and 2 m2 in area. The plates are in vacuum and a potential difference of 15 kV is applied across the plates. Calculate: (i) the capacitance of the capacitor (C) (ii) charge on each plate (Q) (iii) the electric field between the plates (E).

Solution (i) C =o = (8.8 x 10-12 C2.N-1m-2) = 3.6 x 10-9 F = 3.6 nF

(ii) Q = CV = (3.6 x 10-9 F)( 15 000 V ) = 5.4 x 10-5 C (iii) E = = = 3x 106 V/m

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Combinations of Capacitors

Capacitors in parallel.

In the parallel circuit below, the total charge supplied by the battery is shared by the individual capacitances. VT QT

Q1 C1

Q2 C2

Q = CV , QT = Q1 + Q2 ( CV )T = ( C1V1 ) + ( C2 V2 ) , ( V1 = V2 )

CV = C1V + C2V CT = C1 + C2

Capacitors in series.

In the series circuit below, the sum of the voltage across the individual capacitances is equal to the voltage across the power supply.

VS

C1 C2

V2
V1

VS = V1 + V2 = + , ( QT = Q1 = Q2 ) Q = Q

= +

Example 2

A combination of two capacitors of capacitances 3 pF and 6 pF are arranged in series across a potential difference of 1000 V.
Calculate:
(i) total capacitance of the circuit. (ii) the total charge in the circuit. (iii) the potential difference across the 3 pF capacitor.

VS 1000 V

C1 = 6 pF C2 = 3 pF

Solution.

(i) = + = + =

CT = 2 pF

(ii) QT = CT . VS = (2 x 10-12 F)( 1000 V ) = 2 x 10-9 C = 2 nC

(iii) V2 = = = 667 V

Example 3.

Three capacitors of capacitances 2 F, 3 F and 4 F are arranged in parallel and connected to a 250 V power supply. Calculate: (i) the total capacitance of the circuit. (ii) the charge in the 2 F capacitor.

250 V 2 F 3 F 4 F

Solution:

(i) CT = C1 + C2 + C3 = ( 2 + 3 + 4 ) F = 9 F

(ii) Q = C V = ( 2 x 10-6 F )( 250 ) = 5 x 10-4 C

Example 4

Design networks of three 1 F capacitors that have total capacitances of: (i) F (ii) 1.5 F
Solution:

(i) (ii)

Example 5

Calculate the total effective capacitance between a and b of the combination shown below.

2 F 5 F 10 F a b

1 F

parallel series

C = C1 + C2 = + = + = = ( 2 + 1 ) F CT = F = 3 F

The circuit can now be reduced to:

a b 3 F F The reduced circuit can be treated like a simple series combination and added according to the formula:

= + = = CT =F

Energy stored in a capacitor

A capacitor stores energy in the form of an electric field. A plot of voltage versus charge for a capacitor is a straight line with slope = .

V

Q q

The total work required to move a charge of q through a potential difference of V across the capacitor plates is given by the area under the curve ( area of the triangle formed ):

E = ½ QV

Mathematical derivation:

As derived earlier, the energy stored by the capacitor is the area under the curve so;

= = = =

E =

Note: 1. The energy stored in the capacitor ( ½ QV ) is only half the energy provided by the voltage source ( QV ). This is because during the charging process, there is a current in the circuit and half the energy provided by the power supply is dissipated as heat while the other half is stored in the capacitor as an electric field.

2. The energy stored in the capacitor can be expressed in different forms:

E = ½ QV , E = ½ CV2 , E =

Capacitors with Dielectrics

A dielectric is an insulating material such as waxed paper, rubber or glass. When a dielectric is inserted between the plates of a capacitor, the capacitance increases. This increase is by a factor (), which is called the dielectric constant.

For capacitors that have vacuum or air between the parallel plates, the capacitance is given by:

C =o

But for capacitors that have a dielectric material with a dielectric constant (), the capacitance is given by:

C =o

Where = dielectric constant

Some common dielectrics:

Material Dielectric constant ()

Vacuum, Air 1.00 Pyrex glass 5.6 Paper 3.7 Water 80

-------------------------------------------------
Charge and Discharge of a Capacitor

Charging a capacitor through a resistor

A completely uncharged capacitor, resistor, switch and battery (power supply) are connected in series as shown.

S C R

When the switch is closed, a current starts to flow in the circuit in charging the capacitor till a maximum charge of Qo = CV ( V = ) . Once the capacitor becomes fully charged, the current in the circuit becomes zero.

The charge in the capacitor varies with time according to the relation:

q = Qo ( e = base of natural logarithms )
The term RC in the equation above is called the time constant () .

= RC unit: seconds (s)

The time constant represents the time needed for the charge to increase from zero to 63% of maximum charge. q Qo

0.63 Qo

t

Discharging a capacitor through a resistor.

A fully charged capacitor is hooked up to a resistor and switch.

C _ +

S R

When the switch is closed, the capacitor will start to discharge through the resistor. Since the capacitor has an initial charge Qo , current will start to flow through the resistor that will decrease to a minimum ( very close to zero ) after a while.
The charge q, will vary with time according to the equation: q = Qo q Qo

0.37 Qo

t Note:

1. During charging, as the voltage across the capacitor increases, the current decreases.
( The curves are growth and decay – exponential curves )

Voltage vs. time Current vs. time

V I Io Vo

t t

2. During discharging, the capacitor is the only source of emf in the circuit, and so as the capacitor discharges ( charges decrease ), the voltage across it decreases and so the current will decrease as well. V I Vo Io

t t

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ELECTROSTATICS PROBLEMS

1. Two point charges of 20 x 10-8 C and –5 x 10-8 C are separated by a distance of 10 cm. P is the mid point between the two charges. (i) Calculate the electric field intensity at P. (ii) Calculate the force on a 4 x 10-8 C point charge placed at P.

2. Three charges are located at the corners of a rectangle as shown.

6 cm

3 cm 8 C R

3 cm

2 C P - 4 C

(i) Find the net force on the 2 C charge due to the other two charges. (ii) Find the electric field strength at point P due to the 2 C and the -4 C charge. (iii) Find the electric potential at point P.

3. An electric charge q of mass 2 x 10-25 kg travels horizontally through an electric field between the charged plates A and B as shown.

A B - 5 cm +

- + Electric field strength - q + E = 1000 Vm-1

- +

If the charge has velocities of 5 x 106 m/s and 6 x 106 m/s at plates A and B respectively, (i) find the work done by the electric force on q by moving it from A to B, (ii) calculate the electrical force experienced by the charge, (iii) hence determine the polarity and the magnitude of the charge.

4. An electron of mass m kg and charge e coulomb is projected with initial velocity of v m/s along the x – axis and midway between two parallel horizontal plates each of length l metres. The electric field intensity between the two plates is E N/C directed downwards.

Derive a formula for the vertical displacement yof an electron just as it leaves the deflecting plates.

5. A capacitor consists of two parallel plates separated by a layer of air 0.4 cm thick, the area of each plate being 202 cm2. (iii) Calculate the capacitance C of the capacitor. (iv) If the capacitor is connected across a 500 V source, find the charge q on each plate and the energy stored in the capacitor.

6. The diagram given below shows a combination of capacitors connected to a 12 V power supply. 3F

3 F

12 V 6 F 6 F

(i) Determine the equivalent capacitance for the combination. (ii) What is the total charge flowing in the circuit? (iii) Determine the voltage across the 3 F capacitor. 7. An electron has a velocity of 107 m/s at right angles to the electric field between a pair of parallel plates in a CRO. The plates are 8 mm apart and 20 mm long and they are maintained at a potential difference of 50 V, the upper plate being positive. (v) Describe the path of the electron between the plates. (vi) How long does the electron take to travel the length of the plates? (vii) Calculate the deflecting force on the electron. (viii) What distance is the electron deflected vertically during its passage through the field?

8. An 8.0 F parallel plate capacitor is connected to a 100 V battery. (ix) Calculate the energy stored by the capacitor. (x) The battery is now disconnected and Luke pulls the plates of the capacitor apart, doubling the separation. Show that the final value of the stored energy is now twice its initial value.
(Do not calculate the values) (xi) Where does this excess energy come from?

9. A parallel plate capacitor is 100 cm by 50 cm and the plates are separated by 10 cm. They are connected to a 12 V battery. Calculate the:
(a) capacitance of the capacitor.
(b) electric field strength.
(c) energy stored in the capacitor.
(d) charge stored in the capacitor.

10. Two charged metal plates in a vacuum are placed 15 cm apart as shown.
The electric field is E = 3.0 x 103 N/C. An electron is released from P at rest. A B
Calculate the: (a) force on the electron. + _ (b) acceleration of the electron. P (c) time taken by the electron to reach plate A. (d) work done by the electron to reach A. 15 cm

11. A 2 pF and 6 pF capacitors are connected to a 120 V power supply in parallel.
Calculate:
(a) the total equivalent capacitance of the circuit. (b) the total charge coming out of the power supply. (c) the charge on each capacitor. 12. If the set-up for question 11 was in series, calculate the variables as above. 13. The diagram below shows a combination of capacitors.

C1 = 1 F C3 = 1 F

C2 = 3 F C4 = 3 F

24 V

(a) Calculate the effective capacitance of the circuit. (b) Calculate the charge supplied by the power supply. (c) Find the energy stored in the capacitor C1.

SIMPLE HARMONIC MOTION

Oscillatory Motion

Any motion, which repeats over in regular cycles, is called a periodic motion. One common type of periodic motion is called simple harmonic motion (SHM).

Example

Pendulum Mass On a spring

SHM: The motion where the acceleration of an object is opposite to its displacement from a central position as an object oscillates along the same path between two extreme points.

Using the spring:

F= ma F= -kx

So : ma = -kx

a = | Fmax

m vmax

amax

The Characteristics of SHM

1. The mass undergoing SHM oscillates between 2 extreme positions on either side of the central point. 2. The displacement of the mass will be taken from the central point, the position the mass will take if allowed to come to rest. 3. The oscillating mass takes exactly the same time to complete one cycle, this time being the period (T) of SHM. 4. The oscillating mass is fastest when passing through the central point and momentarily at rest at the extreme points of the path. 5. The acceleration is always centrally directed and increases to a maximum towards the extremes and is zero at the centre. Any motion caused by a force that is proportional to the negative of the displacement must be simple harmonic.

Spring with same force constant mass

Frictionless-x O +x surface A A

Equilibrium position

1. Amplitude ( A ) – maximum displacement from the central position. 2. Frequency ( f ) – the number of oscillations per second. 3. Period ( T ) – time taken for one complete revolution (s).

The displacement in a simple harmonic motion is given by:

y = A sint |

Where: y = displacement (m) A = amplitude (m) = angular frequency (rad /s) t = time ( s )

To derive the expressions for velocity and acceleration:
Velocity;
y = A sin t ( Differentiating sin t givescost ) = A cost ()

v = A cost |

Acceleration;

= v = A cost = A()(-sint) a = -2 A sint , ( sincey = A sin t )

a = -2y | NOTE:

1. If the motion is sometimes after the mass has swept through an angle (the phase angle). Then the displacement angle is given in radians. y = A sin (t + ) |

Phase angle =

2. The expression derived for acceleration a = -2 y shows that acceleration is opposite displacement in SHM.

The velocity of a simple harmonic oscillator is given as:

v = A Cost = Cos + Sin = 1 = Cost + Sint = 1 = Cost = 1- Sint = ( y = A sin t ) = v =

The above equation enables calculation of velocities of SHM knowing (angular speed), amplitude and displacement from central position. The velocity is maximum at the central point where y = 0.

Maximum velocity is given by : v = ( taking y = 0, gives)

v = |

Expression for PERIOD (T) of springs - To derive an expression for period (T) in SHM using acceleration

x = y k = m | ω = = ; using k and expanding gives , T = 2 |

= 4
( taking square roots )

Expression for PERIOD (T) of simple pendulum

l θT

x

s

mg

ax = Fxm = mgsinθm = g sinθ = g xl

But we have a problem. a and x are not parallel. Hence if θ is small we can take them as parallel.

a = - ( gl) x (for small θ) * w2 = gl * w = gl * w = 2πT * T = 2πlg mg = T cosθ sin θ= rl

Energy in SHM

The total energy of a SHM is constant and is the sum of its potential and kinetic energy. This may be found by considering the maximum energy of the system.

Q A ET = EP = ½ kA2

m O ET = EK = ½ mv2

P A ET = EP = ½ kA2

Graph of Energy verses Displacement

Energy P Q ET

EK

EP

-A O A y
Example 1
An equation of SHM is given : y = 2.0 sin ( 2.3t )
Calculate:
(a) the amplitude (A) (b) the period (T) (c ) the frequency (f) (d) displacement after (e)
Solution: From the SHM equation it can be seen that, amplitude, A = 2m and = 2.3 rad/s.

(a) Amplitude = 2m

(b) (c) f = (d) y = 2.0 Sin (2.3t) ( = rad) T= = y = 2.0 Sin

= f = 0.36 Hz y = - 0.9m

T= 2.73 s
Example 2. A SHM has the equation y = 6.8 Sin . (a) What is the initial phase angle ? (b) What is the displacement at t = 0 sec? (c) Calculate velocity at time 0. (d) What is the maximum velocity? (e) Sketch a graph of the motion (phasor diagram).

Solution: (a) y = A sin (t + ) (b) y = 6.8 Sin () = rad = 6.8 () y = 4.8m

(c) v = = 6.8 cos (4t + (4) = 27.2 cos (4t + (t = 0) = 27.2 cos () v = 19 m / s

(d) v = 27.2 [cos (14t + )] ( max velocity will be obtained when cos (14t + ) = 1 ) v = 27.2 m/s

(e) /2

0 2

Example 3
A 1.6 kg mass on a spring executes SHM with equation y = 3.0 sin
Calculate:
(a) the velocity and acceleration of mass at 1.0 sec.
(b) the initial displacement. (c) initial speed of the mass. (d) the force experienced by the mass at 1.0 sec.
Solution:
(a) ; v = == 0 m/s

Since the expression for velocity obtained above is = = a = - 7.40 m/s (b) t = 0 sec ; = 0 m

(c ) = 4.71 m/s ( t = 0 )
(d) m = 1.5 kg , a = - 7.40 m/s , F = ma = (1.5) (- 7.40) = - 11.1 N

Example 4
A SHM has equation x = 5.2cos.
Calculate:
(a) Period (b) Amplitude (c) Phase constant (d) Displacement at time t = 0 sec, t =1 sec (e) The velocity and acceleration at t = 3 sec. (f) The maximum speed and acceleration. (g) The maximum force acting on the mass (m = 2.5kg). (h) The force constant k (use = k/m). (i) The total energy of the mass.

Solution:

(a) = ; x = A cos(t + )
T = = = 1 sec

(b) Amplitude =5.2 m ( c ) x = A cos (t + ) = (d ) t = 0 sec : x = 5.2 cos = 5.2 cos () = 2.6 m t = 1sec : x = 5.2 cos = 5.2 cos () = 2.6 m (e) x = 5.2 cos () v = = - 5.2 (2) Sin (2t + /3) = -10.4 Sin (2t + /3) ( t = 3 sec ) = (-10.4 ) (Sin (2t) (3) + /3) = (-10.4 ) () v = -28. 30 m/s x = 5.2 cos () a = = x = (2 )(5.2 cos (2t + ) = 4 [5.2 cos (2t + )] t = 3 sec = 4 (2.6 ) = 102.64 m / s ( f ) = = m/s

amax= 4 [5.2 cos (2t + )] = (4) (5.2) = 205.3 m/s2 (g ) m = 2.5kg amax = 205.3 m/s Fmax = ?

F = ma = (2.5) (205.3) = 513.2 N

(h ) = , m = 2.5 kg , k = ? = k/m k = m = ( 2 ) (2.5) = 98 .7 N/m

(i) at x = 0 m , ET = EK Ek = = (1/2) (2.5) (-32.67) = 1.3 10J OR: At x = A, ET = Ep

Ep = ½ kA2 = ½ (98.7 N/m)(5.2m)2 = 1.3 x 103 J

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FORCED OSCILLATIONS

A B ( Driven pendulum )

( Driver pendulum )

If an object (B) is free to move, it can be caused to oscillate by a driving force ( A ). This is called forced SHM. The frequency of the motion always matches the frequency of the driving force, but the amplitude varies greatly.
When the driving frequency equals the natural frequency the amplitude is very large. This condition is called resonance. Graph of amplitude versus driving frequency

Amplitude

( A )

Frequency fn ( natural frequency )
[Q] Why aren’t soldiers allowed to march on bridges?

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DAMPED SHM

When friction is introduced in SHM, the amplitude gradually decreases and the oscillation comes to a stop. This effect is called damping.

1. Mass in air. A

O t

m O

-A

2. Mass in water A

O t

m O

-A

3. Mass in oil

Mass comes to a stop after a few oscillations.

* This effect is used in shock absorbers in cars to quickly dispense SHM effect caused by pot – holes, bumpy roads etc…

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SHM PROBLEMS

1. In a SHM an object completes 10 vibrations in 2 seconds. Find the angular frequency of the SHM.

2. A 0.4 kg mass hangs from a spring with spring constant of 80 N/m. It is set in SHM with amplitude of 0.1m. Calculate the acceleration at its maximum displacement.

3. An object performs SHM according to the equation y = - 15 sin 0.5t (cm),
Find: (a) the amplitude of the motion
(b) the angular frequency

4. A 4 kg mass is suspended by a spring, which is stretched 0.16 m when the mass is attached. The mass is then pulled down an additional 0.1 m and released. It starts performing SHM. Calculate:

(a) the spring constant, k (b) the period T. (c) the maximum speed of the mass

5. The piston in a car engine moves in approximate SHM with amplitude 40 mm and frequency of 120 Hz. Calculate:

(a) the maximum acceleration. (b) the pistons speed at a displacement of 20 mm. 6. A 0.5 kg mass is connected to a spring with a constant of 20 N/m and undergoes SHM with amplitude of 3.0 cm. Calculate:

(a) the maximum velocity of the mass (b) the velocity when the mass is at a displacement of 2.0 cm.

7. A 1.5 kg mass is oscillating at the end of a steel spring with an amplitude of 15 cm. The spring constant, k, is 600 N/m. Calculate:

(a) the angular frequency (b) the maximum velocity of the oscillating mass (c) the total energy of the oscillating mass.

8. A pendulum of length 1.2 m oscillates with amplitude of 0.2 m. (a) What is the period of the pendulum? (b) Find the velocity at the midpoint of the swing. (c) If the mass of the pendulum bob is doubled, calculate the new period.

9. The equation of a SHM is given as x = 6.5 cos. Calculate:

(a) the period, amplitude and phase constant of the motion. (b) the displacement at t = 0s and t = 1s. (c) the velocity and acceleration at t = 3 s. (d) the maximum speed and acceleration.

10. An object undergoing simple harmonic motion has its displacement, x, at time t seconds Given by the equation x = 0.4 cos 4πtmetres Find its: (a) period. (b) maximum speed.

11. A spiral spring is suspended vertically. When a 100g mass is attached to its lower end, the spring extends by 4 cm. The mass is then pulled a further 5 cm and then released so that it oscillates in simple harmonic motion.

(i) find the spring constant. (ii) calculate the frequency of the oscillation. (iii) find its speed when the mass has a displacement of 3 cm. (iv) find the maximum kinetic energy of the mass. (v) Explain the meaning of the term ‘damped oscillation’.
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WAVES

Periodic waves

A periodic wave is one in which each particle in a medium executes SHM about a mean rest position. Example: Water waves, sound waves, periodic pulses along a string, radio waves.

Mathematically a periodic SHM wave can be represented into waves.
(a) ( displacement y as function of time t )
(b) ( displacement y as function of distance x )

Where the first equation gives the displacement of each particle in a medium with respect to time and the second gives the snapshot view of the displacement of all particles at a given instance.

Wave Velocity = distance travelled per second.

v = f | Unit : m/s or cm/s

Where : f = frequency ( Hz = ) = wavelength ( m )

The travelling wave equation: |

Where : A = amplitude (m) = angular frequency (rads)

k = or k = = = , k = The wave equation given above is for a wave travelling to the right. For a wave travelling to the left, the equation becomes; y = A sin (t + kx )

General Equation

If the position kxterm has a negative sign, then the wave is travelling to right and if it has a positive sign, the wave is travelling to the left.

Example 1
A wave travelling through a string along the x-axis has the equation: y = 0.05sin (0.2x + 4t). (a) State the direction of wave. (b) Calculate the value of:
(i) Amplitude (A) (ii) Frequency (f) (iii) Wavelength () (iv) Velocity of wave (v)

Solution:
(a) Direction is to the left because the kx term is positive.

(b) (i) y = A sin (t + kx ) A = 0.05m

(ii) = 4 rad /s ; = 2f ; f = =

(iii) k = 0.2 = ; = = 10 m

(iv) v = f = () (10 ) = 20 m/s

Example 2

A wave travelling to the right has equation y = 2.0 Sin (5.0t – 2.0 x) (a) Write down the equation for an identical wave travelling to the left. (b) Calculate the speed of the wave.

(a) Left y = A Sin (t + kx )

y = 2.0 Sin (5.0t + 2.0 x)

(b) = 2f ; 5.0 = 2f ; f = = Hz

k = ; 2.0 = ; = = m

f = Hz = m v= ?

v = f = = 2.5 m/s

Example 3
The equation for a travelling waveis y = 3.6 Sin ( . Find: (a) Its frequency (b) Its wave length (c) Speed of the wave.

(a) = 2f ; f = = = Hz

(b) k = ; = = = m

(c) f = Hz , = m , v = ?

v = f = () () = 0.375 m/s

Example 4
A wave travelling to the left has amplitude 0.6 m , wave length 0.5 m and frequency 5 Hz .
Write the wave equation.
Solution: The general wave equation is y =A Sin (t + kx)

We have to calculate the values for A, , and k. A = 0.6 m = = 4 m

= 2f = 2 ( 5 Hz ) = 10 rad/s

Substituting in the general equation, we get,

* y = 0.6 Sin (10t + 4x )
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Exercise 1

The displacement from equilibrium of a wave travelling parallel to x- axis is given by

y = 0.6 Cos 2t + x )

(a) What is the wavelength of the wave?
(b) What is the velocity of the wave?
(c) Sketch the wave at time t = 0.005 sec.

Solution: (a) = 2 m (b) v = 100 m/s.
Exercise 2
For the travelling wave given, the particles vibrate at a frequency of 10 Hz.
Determine: (i) the velocity of the wave. (ii) the wave number (k) (iii) the wave equation.

0.6 y (m)

0 1 2 3 4 x (m)

-0.6

Solution: (i) v= 20m/s (ii) k = m (iii) m

Exercise 3
The equation of a travelling wave is given as , Calculate: (a) the period, T of the wave. (b) the wavelength. (c) the speed of the wave.

Since:

Solution:

(a) T = 0.02 sec (b) = 0.4 m (c) v = 20 m/s.
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Energy Transfer in Waves

Waves can carry energy without the transfer of matter. Example; Sound waves, radio waves, and water wave. A medium which transmits waves must be elastic so it can be stretched / compressed.

In the case of electromagnetic waves (E.M), energy transfer occurs by collimating electrical and magnetic field vibration. Energy transfer by waves involves changes in elastic potential energy.

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WAVES PROBLEMS

1. The displacement y of a particle x metres from an arbitrary origin is given by y = 0.8 Sin 4 ( t – x )
Where tis the time is in seconds after the travelling wave starts to pass through the medium.

(i) In which direction is the wave travelling? (ii) What is the displacement of a particle 0.2 m from the origin after a time of 0.1 s ? (iii) Calculate the amplitude, frequency, wavelength, and the velocity of this travelling wave.

2. A progressive wave has the equation y= 5 Sin ( 0.01 x – 4.00 t )

Where x and yare expressed in cm and t in seconds.

(i) Determine the amplitude, angular frequency and speed of the wave. (ii) Another wave has half the amplitude, and twice the period than the above wave and travels in a direction opposite to it. Write down the equation of this wave.

3. The equation of a travelling wave is given by y = 0.01 Sin 2 ( 0.75x + 0.8t )

Where x and y are in metres and t is in seconds.

(i) Determine the wavelength of the wave. (ii) Calculate the wave velocity and indicate the direction of the wave. (iii) Calculate the displacement of the wave at x = 0.1 m and t = 1 s.

4. A travelling wave has the equation y = 0.04 Sin 2π(50t -2x). (i) Determine the wavelength. (ii) Find the frequency of the travelling wave. (iii) Hence, calculate the speed of the travelling wave. (iv) Using the equation of the travelling wave given above, write the equation of the simple harmonic motion of a particle in the medium, 0.5 m from the origin.

5. The equation of a certain travelling wave is given as:

y = 0.1 sin π (10x-80t) (i) Calculate the wavelength, frequency and speed of the wave. (ii) Write the equation of another wave with the same amplitude, frequency and wavelength but moving in the opposite direction. (iii) If the twowaves mentioned above form standing waves, find the distance between the two adjacent nodes. (iv) Sketch onecycle of the first wave at time t = 0, showing its amplitude and waventh clearly.

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THE STANDING WAVE

Standing wave is formed by the interference of two waves of same frequency and amplitude moving in opposite directions.
Example: superposition of an incident travelling wave and its reflection from a solid barrier.

When the forward wave is reflected, it undergoes a phase change of half cycle so that it is always out of phase with the incident wave. The resulting interference pattern is a non – traveling standing wave where some points are always at rest. These points are called nodes. Between the nodes, the particles of the medium undergo SHM. The point of maximum displacement is called an antinode. The frequency of the SHM is high enough to give the impression of a stationary point.

The Standing Wave Equation

y = A Sin (t – kx ) y = A Sin (t + kx )

If the two waves were superimposed their displacement is given by : y = = A Sin (t – kx ) + A Sin (t + kx ) According to the trigonometric identity: ( Sin (x + y) = Sin x Cos y + Sin y Cos x )

then : y = (A Sin t Cos kx) – (A Cos t Sin kx) + (A Sin t Cos kx) +(A Sin kx Cos t) y = 2A Sin t Cos kx.

General equation

Amplitude(Always with x)

Example 1
An equation of a standing wave is given as: y = 0.5 Sin 4x Cos 46t

(a) Say if the above equation is true for a standing wave. (b) Calculate the amplitude at x = 0 (c) What is the frequency?

Solution:

(a) Since the kx term and t term are separate the above equation is true for a standing wave. (b) The amplitude term is 2A Cos kx therefore A = 0.5 Sin 4x
At (x = 0 ) therefore A = 0 m. (c) = 2f f = = = Hz. Example 2
Equation of a standing wave is given as : t
Calculate:
(a) Frequency of the wave. (b) Wavelength, (c) The equation for the vertical displacement of a point of position from start. (d) Using your answer to part (c ) , sketch the shape of the standing wave of m for t = sec
Solution:
(a) = 2f ; ; f = =

(a) k =, = = = 1 m

(c ) =

(d) t | 0 | 1 | 2 | 3 | 4 | y | 0 | -0.1 | -0.26 | -0.1 | 0 |

0 2 4 t (s)

y

-0.26

Example 3 A wire is stretched between two supports on a sounding board and when it is plucked a standing wave results. The equation is given by :

(a) For which values of x is the displacement zero? (b) For which values of t is the displacement zero?

Solution: (a) 0 = ; x = 0, x = , x = x = ( 0 , ½ , 1 , 1 ½ , 2 ,……..)

(b)

Standing Waves in String and Air Columns

When a string or air column is vibrated, it will resonate and form a standing wave. The type of standing wave formed is different, depending on whether a node or antinode exists at the ends.

String instruments, such as the guitar or violin have a node at each end of the string because the strings are fixed at the ends and cannot move.

Example: A

N N

A

Wind instruments such as a pipe contain a column of air with one open end. The open end forms an antinode and the closed end forms a node.

A

N

A Other instruments such as a saxophone contain pipes open at both ends. In these, antinodes form at each end.

A A

N

A A

For each situation, a standing wave can form several different frequencies or harmonics. The lowest frequency standing wave is called the fundamental or the first harmonic; the higher frequencies are called overtones.

Modes of Vibration of String Instruments

MODE | STRING | FREQUENCY | Fundamental or first harmonic | L | | 2nd harmonic or 1st overtone. | | | 3rdharmonic or 2nd overtone | | | 4th harmonic or3rd overtone | | | 5thharmonic or 4th overtone | | |

** Note that L is the length of all the strings.

Mode Of Vibration Of Wind Instruments

MODE | WIND | FREQUENCY | Fundamental orFirst harmonic | L | | 3rdharmonic or 2nd overtone | | | 5thharmonic or 4th overtone | | | 7thharmonic or 6th overtone | | |

* Only odd harmonic are present in a closed pipe.

** Note that L is the length of all the closed pipes.

Mode of Vibration of Columns

MODE | For COLUMNS | FREQUENCY | Fundamental or first harmonic | L | | 2nd harmonic or 1st overtone | | | 3rdharmonic or 2nd overtone | | | 4thharmonic or 3rd overtone | | | 5thharmonic or 4th overtone | | |

** Note that L is the length of all the open pipes.

For strings with both ends fixed, and air columns (pipes) with both ends open, the general formulae for frequency is given by: |

Where n = 1,2,3,…………

For a pipe closed at one end and open at the other. There are only odd harmonics. The general formula for frequency is given by :

|

Where n = 1,3,5,…………

Speed of a Wave in a String

The wave speed is dependent on the tension of the string. The acceleration and wave speed increase with increase in tension of the string. Also the wave speed is inversely dependent on the mass per unit length of the string. The speed v is given by:

| Where: v = wave speed (m/s) T = tension in the string (N) = mass per unit length (kg/m)

Beats

Beating is a regular pulsing of loudness of a sound that is heard when two sources produce sound of slightly different frequency. Beat frequency (f) is given by:

|

Example on beats

If two tuning forks of frequency, f = 256 Hz and 330Hz are sounded simultaneously, what beat frequency will be heard ?

= Hz = 74 Hz

Example 1
A string vibrates in its fundamental mode of 50Hz. The string is 50 cm long and both ends fixed. (a) Calculate the wavelength. (b) Determine the speed of the wave in string (c) Sketch the first and second harmonics. (d) Find the frequency for 1st and 2nd harmonics.

Solution:
(a) n = 1 f = 50 Hz L = 0.50 m = ? = 2L = (2) (0.50) = 1.0 m

(b) , , v = 50 m/s

(c)

1st harmonic 2nd harmonic

(d) 1st harmonics = 50 Hz 2ndharmonics = = 100 Hz

Example 2
A uniform string has a mass of 0.3 kg and a length of 6m.
Tension is maintained in the string as shown.

T = 20 N

(a) Calculate the speed of a wave in the string. (b) Sketch the fifth harmonic on the string. (c) Find the frequency of the fifth harmonic.

Solution:

(a) m = 0.3 kg , T = 20 N , l = 6m , v = ? ; v = = = 20 m/s. (b)

(c) L = 6m v = 20m/s n = 5 f = ?

= = 8Hz

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STANDING WAVES PROBLEMS

1. The largest pipes in great organs usually have a length of about 5 m. These pipes are open at both ends. Calculate the frequency of the second overtone in the pipe given that the speed of sound in air is 340 m/s.

2. A silencer pipe of a car is open at both ends and has a length of 0.25 m. (a) Draw the resonance wave pattern produced by the fundamental frequency in the pipe. (b) Given that the minimum resonant frequency is 240 Hz, find the speed of sound in air on that particular day.

3. The length of a test tube is 15.0 cm. The speed of sound in air is 330 m/s.

15.0 cm

(a) Draw the fundamental note when the open end of the test tube is blown.
(b) Determine the wavelength of the fundamental note.
(c) Calculate the frequency of the fundamental note.

4. The diagram represents a tube partially filled with water.

2 cm

(i) What is the wavelength of the fundamental standing wave produced in this tube?

The diagram shows a string under tension.

4 cm

(ii) Sketch a standing wave in the loaded string, which has the same wavelength as the fundamental wave in the tube. (ii) The sound produced by the third harmonic in the tube above travels at 330 m/s. Determine the frequency of the sound. (iii) What is the major difference in the way the particles in the air column vibrate as compared with the particles in the string?

4. A tuning fork of frequency 256 Hz is used to tune a sonometer wire of length 0.850 m. The vibrating length of the wire is then shortened to 0.800 m.

(j) What would be the new frequency of the string when plucked? (k) What would be the beat frequency heard when the tuning fork and the shortened wire are sounded together?

5. A silencer pipe of a certain car is open at both ends and has a length of 0.80 m.

(e) Draw the resonance produced by the fundamental frequency in the pipe. (f) Given that the minimum resonant frequency is 210 Hz, find the speed of sound in air.

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SOUND INTERFERENCE

Sound from a speaker enters the tube and splits into two parts at P. Half of the sound travels to the left, and the other half travels to the right. If the two paths (R1) and (R2) are of the same length, a crest of the wave combines again at the ear. This reuniting of the two waves produces constructive interference, and thus the listener hears a loud sound.

Suppose, however, that one of the paths (R1) is adjusted by sliding the right path (R2). In this case, an entering sound wave splits and travels the two paths as before, but now the wave along the left path must travel a distance equivalent to half a wavelength farther than the wave travelling along the right path. As a result, the crest of one wave meets the trough of the other when they merge at the receiver. Because this is the condition for destructive interference, no sound is detected at the receiver. Speaker

P

R1 R2

Q

Receiver

Diffraction Grating

A diffraction grating is a series of many very fine parallel slits, closely spaced on a sheet of glass or plastic. The large numbers of slits produce an interference pattern of light in a similar way to the double slit pattern, but there are some important differences. * The slit are very thin so the light is diffracted through a wide angle
( Almost 180 ) * The slits are very closely spaced so the bright fringes are widely spread. * There are a large number of slits so the fringes are bright.

The formula for the interference pattern from 2 slits also applies to the diffraction grating

| n = 0, 1, 2, 3 … Its calibration is given by N, Example: N = 300 lines/mm
To find the distance between two slits: | d = ( 1 mm = 1 10-3 m ) d =

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Young’s Experiment

In a Young’s experimental set-up, a double slit is used where the thickness between the slits is d, is placed in front of a source of light. A screen is placed a distance L metres away from the double slits. The resulting interference pattern is created on the screen. The fringes are a distance x from the central maximum O.

x S1 d O central maximum

S2

L

For Bright Fringes
: For (Bright Fringe)

| | Sin =

|
For Dark Fringes:

| |

Monochromatic Light is light of a single colour. Coherent means that the light is in phase.

Example 1
In a demonstration of young’s experiment Laser light of 680nm is shown through a pair of slits 0.15mm apart, screen is 2.5 m away. Calculate the distance from the central maximum to: (a) The first maximum. (b) The second dark fringe
Solution:
(a) n = 1 L = 2.5m = 680 10m d = 0.15 10 ; , x = 0.0113 m

(b) n= 2 L = 2.5m = 680 10m d = 0.15 10 ; , x = 0.0170m

Example 2 Light of wavelength of 590nm is shown on a grating with 700 lines/cm. At what angle will the fourth maximum occur? = 1.43 10-3 cm d = 1.43 10-3 cm

d = 1.43 10-5 m

4(590 10m) = (1.43 10m) Sin , = 9.5

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Thin Film Interference

A spectrum of colours is seen when light is reflected by thin layers of oil and water through air. It is caused by the interference between reflection from the top and bottom of the thin film. Interference occurs when there is a phase difference between the light waves. Reflection/Transmission of waves when changing medium is given by:

Incident wave

Less dense medium more dense medium

Reflected transmitted Wave wave

The Soap Bubble film Observer R1 O R2

P S air

d Q soap bubble air Thickness of the bubble = d

When the light is incident on the soap film, both reflection and refraction take place. Phase change due to reflection takes place at P.
1. Phase change due to reflection by the reflected rays are given as: R2 = = 0
2. Phase change due to path difference: R1 = 0, R2 = 2d,

So the total phase change due to the path difference is 2d.

P.d. = +2d |
3. Total Phase Change:

A constructive interference will occur (visible colours) if:

| n = 1,2,3,… For values of thickness of film:

A destructive interference (dark pattern) will occur when:

For values of thickness of film: | | | | | n = 2,3,4,……..

Example 1
Red light of frequency, f = 4.6 10Hz strikes a soap film.
The speed of light is 2.998 10 m/s. Find: (a) the wavelength, of light in air. (b) the minimum thickness of the film to cause constructive interference. (c) the minimum thickness for a dark fringe.

Solution:
(a) f= 4.6 10Hz v = 2.998 m/s = ? ( v = f ) , = = 6.52 x 10-7 m , = 652 nm
(b) Minimum thickness: d = = = 1.63 x 10-7 m
(c) Minimum thickness: d = = = 3.26 x 10-7 m

Example ( Air wedges )
A strand of Willies hair is laid between two glass slides at one end. The two slides touch at the other end. Red light of wavelength 650 nm is shown on the slides and the students see 67 thin parallel red lights (interference fringes) between the ends. Calculate the thickness of the hair.

Bright fringes: R1 R2

So the thickness of hair is 0.0215 mm.

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LIGHT INTERFERENCE PROBLEMS

1. A diffraction grating is calibrated by using the 546.1 nm wavelength line of mercury vapour. It is found that the first order line is at angle of 210. Calculate the number of lines per mm on this grating?

2. How many bight lines can be observed using a diffraction grating of length 2 cm and having 104lines when illuminated normally with light of wavelength 495 nm?

3. A diffraction grating having 5 000 lines per centimetre was used to find the wavelength of a monochromatic light source. The second order fringe was formed at an angle of 30o from the central bright image.

Calculate the wavelength of the light source.

4. A laser produces red light of wavelength 600 nm. A grating with 500 lines/cm diffracts the light and the interference fringes are observed on a screen 3 m away. (a) Calculate the fringe spacing on the screen. (b) Determine the angular deviation of the second interference fringe on the screen.

5. In a Young’s double slit interference experiment, a set of parallel slits 0.10 mm apart is illuminated by light of wavelength 589 nm. The interference pattern so formed is observed on a screen 4.00 m from the slits.
Calculate the path length difference to the screen of: (c) the third order bright fringe (d) the third dark fringe away from the central maximum.

6. Monochromatic light of wavelength 632.8 nm is incident normally on a diffraction grating containing 6 000 lines/cm. What is the diffraction angle at which the second order maximum is observed?

7. In a double slit interference experiment, the wavelength of light used is 600 nm. It is found that the second dark band is located 6 mm from the central maxima on a screen 1 m from the slits. (e) Find the separation, d, of the slits. (f) What changes would be observed in the pattern on the screen if light source of a shorter wavelength were used?

8. A thin film of soap has an index of refraction of 1.33 and is surrounded by air on both sides. It is illuminated with light of frequency 6 x 1014 Hz.

Calculate the minimum thickness that will produce constructive interference in the Reflected light.

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POLARIZATION of LIGHT

The phenomenon of polarization of light maybe defined as that in which a transverse wave associated with many plains of vibration maybe caused to exhibit oscillatory motion in one plane called the "plane of polarization". Polarization is a property associated with transverse waves only.

Mechanical Analogy

Waves in a light rope can be made in several planes at the same time. This wave is said to be unpolarised. From this unpolarised wave, transmission of wave (vibration) in one plane only can be brought about by introducing a ‘polariser’ and analyser through which the rope can be made to pass. (Both the polariser and analyser are Polaroid’s).

y Polariser Analyser

x

Vertically plane Polarised waves.

Polarization of light

Passing the light through Polaroid sheets can show polarisation of light. Polaroid sheets are material consisting of two plastic sheets within which a layer of needle like quinine iodosulaphate crystals which have been aligned by a very strong electric field.

Polaroid sheet

Vertically polarised Waves

Light Observer
Source
Unpolarised Polariser Light Analyser

Light passing through the first polariser sheet becomes plane polarised (vertically). This plane-polarised wave can be detected by using the second Polaroid called the analyser. When the analyser is rotated through 360o, two positions of minimum intensity (no light transmitted) will be observed.
This method of using Polaroids is called absorption polarization. Other methods are reflection from shiny surfaces for plane polarizing light. Using polarization of light, it can also be confirmed that light waves are transverse.

Uses: One of the uses of Polaroid’s is in Polaroid sunglasses that reduce glare by vertically polarised lenses. The horizontal glare from the road is minimised.

Energy in polarized light

The amplitude of a light wave is given by; Amplitude = A cos.
The energy of the wave is equal to the square of the amplitude of the wave thus the transmitted component of light energy is given by E A2 , therefore E = kA2 - incident energy of wave

The transmitted energy is about 50% of the incident energy on any Polaroid. General formula for transmitted energy is: E = cos2

Where: = angle between planes of polarisation of the analyser and polariser.

Polaroid sheet

Vertically polarised Waves

Light Observer
Source

Energy reduced By 50% Analyser E = cos2

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DOPPLER EFFECT

When there is relative motion between a sound source and an observer, the frequency of the sound as heard by the observer changes. General equation for Doppler effect:

f = |

Where: f = observed frequency (Hz) f = frequency of sound (Hz) v = speed of sound in air on a particular day (m/s)
(Varies between 300 and 350m/s) v = velocity of observer (m/s) v = velocity of source (m/s)

Case 1

When the source is stationary and the observer is moving. vo

Observer

f = | (a) If the observer is approaching source,
The apparent frequency will be:

The apparent wavelength will be: =

f = | (b) If the observer is receding from the source,
The apparent frequency will be:

The apparent wavelength will be: =

Case 2

When the observer is stationary and the source is moving.

Observer vs Observer ( Source Receding ) ( Source approaching )

f |
The observed frequency of source moving: (a) Source approaching observer:

f | (b) Source receding from observer:

The apparent wavelengths ( ) of the sounds heard by the observer will be:

(a) Source receding (b) Source approaching

= = Note: * If both approaching towards each other then, f |

* And if both is moving away from each other then,

f |

Example 1

Ruby is travelling at 3m/s towards a stationary siren, which is emitting sound of wavelength 20cm. The speed of sound in air that day is 300m/s. Calculate the frequency of sound heard by Ruby.

v = 300m/s , = 0.20m , f = ? v = f

Solution: The frequency heard on the day is given by v = f, so f = = = 1500Hz

The frequency heard by Ruby (who is an observer approaching a source) is

f= 1500 = 1515 Hz

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DOPPLER EFFECT PROBLEMS

1. A bullet train approaches a stationary observer on a platform at a constant speed of 40 m/s. The train sounds its horn as it nears the platform, producing sound of 500 Hz frequency. (a) What is the wavelength of the waves that arrive at the observer? (b) What is the frequency of the sound from the horn as heard by the observer? 2. A man plays a didgeridoo (an air musical instrument modelled as a pipe open at one end) while on a moving truck. The fundamental frequency of this instrument is 71.0 Hz and the truck’s speed is 6.0 m/s as it approaches Mere standing by the road.

Calculate the frequency heard by Mere as the truck approaches her.

3. A moving train blows its whistle of frequency 1000 Hz. A listener on the platform hears a frequency of 940 Hz.

The speed of sound in air is 320 m/s.

(a) Is the train moving away or towards the listener? (b) Find the wavelength of the sound waves reaching the listener? (c) Calculate the speed of the train.

4. A car is travelling at 12 m/s. It sounds its horn of frequency 960 Hz. The speed of sound in air is 320 m/s. A stationary listener in front of the car hears the horn of the approaching car.

(a) What is the wavelength of the sound reaching the listener? (b) What is the apparent frequency of the sound heard by the listener?

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ELECTROMAGNETISM

( MAGNETISM ) Magnetic Materials

All materials show magnetic effects. In many substances the effects are so weak that the materials are often considered to be non-magnetic however a vacuum is the only truly non-magnetic medium. In general materials can be classified according to magnetic behaviour into * Diamagnetic, paramagnetic, ferromagnetic, anti-ferromagnetic,
Ferrimagnetic and super ferromagnetic.

In diamagnetic materials, magnetic effects are weak. If a diamagnetic specimen is brought near either pole of a strong bar magnet, it will be repelled, an effect discovered by Michael Faraday in 1846. Examples of diamagnetic materials are: silver, lead, copper and water.

In paramagnetic materials magnetic effects are significant. When a specimen is brought near the pole of a strong bar magnet, it will be attracted to it. Examples are air and aluminium.

In a few materials like Iron, cobalt, nickel a special phenomenon occurs which greatly facilitates the alignment process. In these substances called ferromagnetic there is a quantum effect known as “exchange coupling” between the adjacent atoms in the metal crystal lattice of the material, which locks their magnetic moment into a rigid parallel configuration over a region, called domain, which contain many atoms.

Magnetic fields.

Into the page. Out of the page.

x x x . . .

x x x . . .

x x x . . .

Force on charges in a magnetic field.

x x x x xB v x x x x x

+q x x x x x x x x x x The charge entering the B field will experience a magnetic force whose direction will be given by the right hand slap rule.
F = Bvq

The magnitude of the force is given by:

If the B field is extensive, the charged particle (q) will undergo circular motion as shown. x x x x x B q F v x v x x x x F . + q

x x x x x F v x x +q x x x F = Bvq = ; ( where: Fc = ) Fc is centripetal force.

So to find the radius of the circular path travelled by q:
R=
F = Bvq =

Where: m = mass of charged particle. Example 1

A proton with mass 1.6 x 10-27 kg and charge of 1.6 x 10-19 C travels at 3.0 x 106 m/s in a magnetic field of strength B = 0.03 T. Find the radius of its circular motion in the B field. R = = = 1.07 m ------------------------------------------------- PROBLEMS

1. An electron enters a magnetic field of a mass spectrometer with a speed of 1.0 x 106 m/s.

(a) State the direction of the electron in the field. x x x (b) If the magnetic field strength B = 1 x 10-4 T, - Q find the radius of the path. x x x 1. A positively charged ion of mass 6 x 10-24 kg and charge 4.8 x 10-19 C enters a magnetic field of strength 0.5 Wb/m2 with a speed of 2x 106 m/s. The resulting path is circular.
(a) Find the magnitude of the magnetic force acting on the ion.
(b) Find the radius of the circular path. 2. A cathode ray beam is bent in a circle of radius 2 cm by a field of induction 4.5 x 10-3 T. Calculate the velocity of the electrons given, e- = 1.60 x 10-19 C, me = 9.11 x 10-31 kg.
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The Mass Spectrometer

Ions from a source pass through the slits (S1 and S2) forming a narrow beam. Then the ions pass through a velocity selector with crossed E and B fields. Finally, the ions pass through a region of magnetic field, B1 , move in circular arcs with radius R given by the equation R = .
Particles with different masses strike the photographic plate at different points and the value of R can be measured. (The velocity of the ions passing through the velocity selector is v = )

This technique is used to find masses of different isotopes of an element. Ions with larger mass travel in paths with a larger radius. Photographic plate x x x x

x x x x S1 S2 velocity selector B R E x x x x x x x x

Ion Source x x x x x x x x

x x x x x x x x

x x x x x x x xB

S3

Force on a current carrying conductor.

The magnetic force experienced on a conductor carrying a current (I) is given by: F = BIlsin

B I

l

For the conductor, the force will be directed out of the page.
For conductors moving perpendicular to the B field, the equation reduces to F = BIl. (Sin 90o = 1)

= BANI cos
The torque on a current loop is given by:

Where is the angle between the coil and the magnetic field.

F

. x B l F B w

Example

A rectangular loop of dimensions 6 cm x 12 cm is oriented in a B field of 0.50 T at an angle of 30o. The current in the loop is 2.0 A. Find the magnitude of the torque at that instant.

= BANI cos = (0.50 T)(2.0 A)(0.06 x .012 m2) cos30o = 0.0062 Nm.

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MAGNETIC FIELDS

Straight Conductors If the wire is gripped in the right hand with the thumb in the direction of the current, the fingers will curl in the direction of the B field. This is known as the right hand screw rule.

The magnetic field strength at a distance r from a wire carrying a current I is

B=

Where: o = permeability of free space = 4x 10-7 Tm.A-1

SinceB= k = = 2 x 10-7 Tm.A-1 The simplified equation.
Force between parallel conductors

When two parallel conductors carrying currents I1 and I2 respectively are separated by a distance, r, they experience a force since they lie in the magnetic fields created by each other.

The force on the conductor C, is C F = BI1l = r F I1

and the force per unit length of the F conductor is l I2 =

[Sincek = ]

Note: Parallel currents in same direction attract and when in opposite directions repel.

AMPERE’S LAW Ampere’s Law states

“The line integral of B around any closed path equals o times the net current through the area enclosed by the path.”

= o I

1. Field of a long straight wire.

To apply Ampere’s Law to derive B field for a long straight conductor, we take as our integration path a circle with radius, r, cantered on the conductor used in a plane perpendicular to it. At each point, B is a tangent to this circle.

= 2rB =o I
Therefore
B =

2. Field of a solenoid.

A solenoid consists of a helical winding of wire on a cylinder.

Using ampere’s law to find the magnetic field near the Centre of a solenoid.

= BL BL = onI
Therefore
B= onI
B=

Or:

Where: N = number of turns in coil. L = length of solenoid n = ( m-1)
B outside the solenoid is zero.

3. Field of a Toroid.

= 2rB = o n I

B= o n I
2r

Where: r = distance from Centre.
B outside the toroid is zero.

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MAGNETISM PROBLEMS

1. A wooden ring of mean diameter 0.1 m is wound with a closely spaced toroidal winding of 500 turns. Compute the field at a point on the mean circumference of the ring when the current in the windings is 0.3 A. 2. A very long straight conductor carries a current of 2 A current. a. Calculate the magnetic field strength at a distance of 10 cm. b. Another conductor also carrying a current in the same direction is placed parallel to it. Find the force per unit length experienced by the second conductor. c. Sketch a simple diagram showing the cross – section of the conductors in (ii) and the pattern of the magnetic field around them.

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ELECTROMAGNETIC INDUCTION

Magnetic Flux and EMF’s due to motion.

1. Induced EMF ( ) Generator effect.

x x x x x B x x x x x l v R x x x x x

x x x x x

If the length of conductor moves at a uniform speed, v , in the B field; it means that the magnetic force cancels out the electrical force in the conductor.

FElectrical = FMagnetic ( E = ) , d = l E q = B v q E = B v = B v V = B v l
V = B v l

Therefore:

Where: V = = induced emf ( V ) B = magnetic field ( T ) v = speed of conductor ( m/s ) l = length of conductor ( m )

The current induced will be given by the Ohm’s Law equation: V = I R

2. Magnetic Flux ( B )

B= B A

Magnetic Flux ( B ) is the product of the magnetic field ( B ) and the area through which it radiates. Units: T.m2

The relation between Magnetic flux ( B ) and Induced emf ( )

“An induced emf exists in a loop whenever there is a change in flux through the area of the loop. The induced emf exists only during the time that the flux through the area is changing.”

Faraday’s Law:

“The emf induced in a coil with N loops in a changing magnetic flux ( B ) in time ( t ) is given by:
= - N

The negative sign indicates that the emf opposes the change, which produces it.

Lenz’s Law:

“An induced emf is always in such a direction as to oppose the change in magnetic flux that are inducing it.”
= - L

Where: L = self-inductance.

Induced emf by loops with N turns.

= - N , B= BA ( For perpendicular fields ) B = BA sin

For an alternating supply with angular speed, .

B = A B sin t = - N

= - N

= - N A B

= - N A B Cos t
= - NAB Cos t

** The angle (B ) to be taken with respect to the magnetic field lines.

EMF due to changing current in a coil.

Example: A solenoid; BSolenoid = o N I

Therefore B = B A = o N I A

= = = o NA ( i denotes changing current )

For a wire with cross-sectional area, A, A = r2 ; = o Nr2

Note: o = 4x 10-7 Tm.A-1

When Two coils of wire carrying currents I1 and I2 are placed together, with coil 1 carrying a current, i, the emf in the coil 2; EMF ( coil 2 ) No. of coils in coil 2 ( n )

i.e. EMF; = [ onr2N ]

M M is called mutual inductance. Where: M = onr2N
=M

Therefore

EMF induced in the same coil is called self-inductance. ( Back emf ) Self inductance, L = noNA B = Li = - L

This above equation can be modified when using quantifiable values of current over a span of time to: = - L

Energy in an inductor.

The energy in an inductor is given as
EL =LI2

Where: L = inductance ( H ) I = current ( A )
Units: Joules

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Transformers (Revision F6)

A transformeris a device that changes electricity at one electric potential (voltage) into electricity at a different potential. To achieve this, the primary coil of the transformer must have a different number of windings than the secondary coil.

Step – up Transformers Step – down Transformers

* Ns>Np * NP> Ns

Formula

VPVs = ISIP = NPNS or VpIp = VsIs

Example:

A spark generator uses a DC input voltage and a transformer to increase the voltage to the required level.

S

. 15 V . Spark generator

Primary coil Secondary coil

(i) What usual name is given to this type of transformer? (ii) The primary voltage is 15.0 V and the primary coil has 100 turns. Calculate the secondary voltage given that the secondary coil has 100 000 turns. (iii) Briefly explain how a secondary voltage is produced although the secondary coil is not connected to any power supply.

Further investigation into the primary coil shows that when the switch is closed a primary coil current of 12.5 A flows. This current drops to zero in 0.625 seconds, generating an average emf of
-125 V in the primary coil.

(iv) Calculate the self-inductance of the primary coil.

(v) Explain why the generated emf has a negative value.

Solution:

(i) Step-up transformer

(ii) VsVp = NSNP Vs = (100 000) (15)/100 = 15 000 V

(iii) Large emf is induced from circuit break using switch in primary coil, Change in B, ∅.

(iv) V = - L didt L = 125 x 0.62512.5 = 6.25 H

(v) Generated emf opposes current change. ( Lenz’s law)

-------------------------------------------------
ELECTROMAGNETIC INDUCTION PROBLEMS

1. A rectangular loop of area 0.01 m2 is placed on the table where the magnetic field of magnitude 0.4 T is perpendicular to the plane of the loop. (c) What is the maximum flux through the loop? (d) If the magnetic field through the loop is reduced from 0.4 T to zero in 0.2 seconds, what is the magnitude of the voltage induced in the coil?

2. A coil of wire, initially carrying no current, has a current of 40 A established in 5 seconds. The current increases at a steady (constant) rate during this time, and the changing current induces an emf of 16 V in the wire. Determine the self – inductance of the coil.

3. A 0.1 mH inductor is carrying a DC current of 5A. What is the energy stored in the inductor’s magnetic field ?
4. An inductor of inductance 3 H and negligible resistance is connected in series with a 1.5 resistor across an AC generator. At the instant the current in the inductor is increasing at the rate of 2 As-1, the voltage across the inductor is
5. A square solenoid of 50 turns of wires and sides 5 cm long is placed perpendicular to a magnetic field of strength 0.5 T.
(a) How much flux passes through the coil?
(b) The magnetic field is reduced to zero during a time of 0.1 second.
What is the induced voltage in the solenoid? (e) If the current changes from zero to 2 A in 0.1 second, calculate the inductance of the coil. 6. A straight conductor of length l is moved with velocity v at right angles through a magnetic field B. Electrons experience a force in the conductor and start to move until electrostatic forces balances with the magnetic forces. Use this information to show that the induced voltage across the ends of the conductor is V = Blv .

7. An inductor of inductance of 0.18 mH is connected in a circuit in which the current is decreasing at the rate of 50 A/s. Calculate: (a) the self-induced emf in the inductor. (b) The energy stored in the magnetic field of the inductor when the current is 40 A. 8. In a model AC generator, a 500 turn rectangular coil, 8.0 cm x 20.0 cm, rotates at 130 revs per minute in a uniform magnetic field of 0.50 T.

Determine;

(i) the maximum emf induced in the coil. (ii) the instantaneous value of the emf in the coil at t = π30 s, assuming the emf is zero At t = 0 s. (iii) the smallest value of t for which the emf is a maximum.
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ALTERNATING CURRENT (ac)

The symbol for alternating current is

It is important to understand the concept of how ac operates because nearly all appliances that connect directly to the three-pin plug power supply in a household use ac to operate. We will take a look at how ac circuits operate with resistors, capacitors and inductors individually and in combination circuits.

The output of an ac power supply is sinusoidal and varies with time: V = Vmaxsint V = Vmaxsin 2ft A plot of voltage and current versus time across a resistor.

Imax

Vmax

R

V t

From the graph of voltage and current against time it can be observed that voltage and current across a resistor are in-phase. The instantaneous current is given by

I = Iosin 2ft ( Io = peak current )

Usually an average value of current and voltage should be taken in analysis of ac circuit since an ac output is sinusoidal but the average value ( from the graph ) is zero. So another quantity known as the root mean square ( rms) is taken.

rms current Irms = ( Io = Imax )

rms voltage Vrms = ( Vo = Vmax )

Capacitors in an ac circuit

Shown below is a series circuit where a capacitor is connected to an ac power supply.

V V

c
A

C I

Slope = reactance of capacitor ( c )

From experimental results, the voltage across the capacitor is directly proportional to the current and this constant of proportionality is called the reactance of the capacitor (c) .

Therefore the voltage and current can be related by V = Ic
Where c is in ohms ().

When a graph of current and voltage against time is plotted, the voltage reaches its peak value a quarter of a cycle after the current reaches its peak value as shown by the graph below. Io Vo

t

From the graph it can be said that the voltage always lags behind the current by 90o.

The impeding effect of a capacitor on the current in an ac circuit is called the capacitive reactance, c. The capacitive reactance is defined as

c =

Example A 100 F capacitor is connected to a 150 V ( rms ) ac power supply of frequency 50 Hz.
Calculate: (i) the reactance of the capacitor (ii) the rms current in the circuit.

Solution: (i) The reactance of the capacitor is given by

c = = = 31.8

(ii) V = Ic , I = = = 4.72 A

In a capacitor, the current leads the voltage by 90o ( quarter cycle ).

V

I

t

V

In a DC circuit, the voltage across components connected in series with a supply adds up to the supply voltage. The same does not apply to an AC series circuit containing a capacitor and a resistor.

------------------------------------------------- INDUCTION PROBLEMS

1. A square coil of side 4.0 cm and 500 turns is rotating at a constant speed in a uniform magnetic field of strength 0.4 T. If the maximum induced voltage is 120 V, determine: (a) the angular velocity () of the coil. (b) the frequency of the coil.

2. A generator is made from a 200-turn coil, 4 cm 8 cm, and placed in a magnetic field of strength 0.4 T as shown. The coil rotates at 25 revolutions per second.

N S

(a) What is the peak voltage generated by the coil as it rotates? (b) Sketch the graph of the output voltage of the coil if timing begins at the position shown in the diagram. Indicate the appropriate values on the voltage and time scale. 3. A simple AC generator is made from a 60 turn coil, of total resistance of 12 , placed in a magnetic field of 0.8 T. The coil has a radius of 2 cm and rotates at 100 Hz. (a) Calculate the maximum induced voltage in the coil.
(b) Calculate the maximum induced current in the coil.
(c) Hence, write an expression that gives the variation of induced current with time.
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RC Series circuit

VS

R C

VC
VR

V

VR

VC

VC

t

VR

VR leads VC by 90o.

VS = VR + VC

= VR + VC

VR =

VS VC

Using the Pythagoras theorem we get: VS 2 = VR2 + VC 2

Example

A 5 resistor and 500 F capacitor are connected in series with a 50 Hz ac power supply. The voltage across the resistor is measured to be 2 V. Calculate: (i) the rms current in the circuit. (ii) the voltage across the capacitor. (iii) the supply voltage.

Solution.

(i) Using the voltage across the resistor;
V = IR , I = = = 0.4 A (ii) V = Ic , both V and c are unknown, so calculate c first; c = = = 6.37
Then using V = Ic ,

V = Ic = (0.4 A)(6.37 ) = 2.55 V

(iii) VS 2 = VR2 + VC 2 = (2.55 V)2 + (2 V)2 = 3.24 V Impedance( Z )

The quantity which relates the supply voltage (VS) and the current (I) is called the impedance (Z). Impedance is analogous to resistance in a direct current (dc) circuit.

For an RC circuit the current is related to the supply voltage by VS = I Z

VS

R C

VC
VR

VS 2 = VR2 + VC 2 (IZ)2 = (IR)2 + (Ic)2 ( current gets eliminated )

Z2 = R2 + c2

Z =

Example

A 10 V, 50 Hz ac power supply is connected to a 20 resistor and 80 F capacitor in series. Calculate:
(i) the capacitive reactance (ii) the total impedance (Z) of the circuit (iii) the rms current in the circuit (iv) the voltage across the capacitor.

Solution:
(i) c = = = 39.8

(ii) Z = = = 44.5

(iii) Irms= = = 0.22 A

(iv) V = Ic = (0.22 A)(39.8 ) = 8.8 V

Inductors in an ac circuit

It has been shown in dc circuits that inductors always oppose a change in current, so in an ac circuit, where the actual current is continually changing, the inductor will act to limit the amount of current.

V V

A

L I

Slope = reactance of the inductor ( L )

The effective resistance of the coil in an ac circuit is termed the inductive reactance, L .

L = 2fL ( L = L )

And voltage and current are related by: V = IL

In an inductor, the voltage leads the current by 90o ( a quarter cycle ). V

V
I

t

The LR Series circuit

VS

R L

VL
VR

V VL VR

VL

t

VR

VL leads VR by 90o.

VS = VR + VL

=

VS VL

VR
Using the Pythagoras theorem we get: VS 2 = VR2 + VL 2

The reactance of the inductor ( L ) depends on the value of the inductor L, and the supply frequency.
The impedance of an RL circuit can be derived to be Z =
Example 1

A 0.5 H inductor is connected to a 10 V, 50 Hz ac power supply. Assuming the inductor has negligible resistance, calculate: (i) the reactance of the inductor (ii) the current through the inductor.

Solution: (i) L = 2fL= 2 (50 Hz)(0.5 H) = 50 = 157

(ii) V = IL, I = = = 0.064 A

Example 2

An inductor of 2 H inductance and a 940 resistor are connected in series to a 100 Hz power supply. The rms current flowing in the circuit is 3.2 mA. Find: (i) the voltage across the resistor. (ii) the voltage across the inductor. (iii) the supply voltage. (iv) the impedance of the circuit.

Solution: (i) VR = IR = (3.2 x 10-3 A)(940 ) = 3.0 V

(ii) L = 2fL V = IL = 2 (100 Hz)(2 H) = (3.2 x 10-3 A)(1257 ) = 400 = 4.0 V = 1257

(iii) VS 2 = VR2 + VL 2

VS = = = 5 V

(iv) Z = = = 1570

The LRC circuit

The diagram below shows a combination of a resistor, a capacitor and an inductor connected in series. VS

R C L

VR VC VL

The phase relations in the above series circuit are as following: * The instantaneous voltage across the resistor is in phase with the current. * The instantaneous voltage across the inductorleads the current by 90o. * The instantaneous voltage across the capacitorlags the current by 90o.

Since all the voltages are not in-phase, the relation between them is easily interpreted by the use of a phasor diagram.

y

VL

VR VS x VL – VC VC VR

A phasor diagram for an LRC circuit.

The addition of the vectors from the phasor diagram to get a relation with the supply voltage can be deduced from the vector diagram above.

VS =

Eliminating the current gives the total impedance ( Z ) of the circuit.

Z =
The phase angle of an LRC circuit can be derived from the vector diagram.

tan = , or tan =

If in an LRC circuit has VL VC , or ( LC ) , the circuit is said to be inductive.

The phasor diagram will be as shown below and the phase angle will be positive.

VL VS VL - VC

VC VR

If VC VL , or ( CL ) , than the circuit is capacitive. The phase angle will be negative.

VL VR VL - VC VS VC

Example

Analyse an LRC series circuit that has a 250 resistance, 0.60 H inductor and a 3.5 F capacitor connected to a 150 V, 60 Hz power supply. Calculate the reactance of the inductor and capacitor and overall impedance of the circuit, phase angle and voltages across the individual components of the circuit.

Solution: Reactance/impedance: L = 2fL = 2 (60 Hz)(0.6 H) = 72 = 226 c = = = 758

Z = = = 588

The rms current is: I = = = 0.255 A

The phase angle between the current and voltage is: tan = = tan-1= tan-1 = -64.8o
The negative angle shows that the series circuit is dominantly capacitive.

The rms voltages are: VR = IR = (0.255 A)(250 ) = 63.8 V VL = IL = (0.255 A)(226 ) = 57.6 V VC = IC = (0.255 A)(758 ) = 193 V The phasor diagram:

VL = 57.6 V

VR = 63.8 V

= 64.8o

VC = 193 V

Power in an ac Circuit

There is no power lost from the capacitor and the inductor in an ac circuit. The only power lost is from the resistance, which is given by:

PL = I2R

The power dissipated from an ac circuit can also be determined from

PL = VI cos Where: = phase angle

The quantity cosis called thepower factor. ( cos = )

Resonance in LCR In a series LCR circuit the value of the current can be expressed as

I = =
From the above equation it can be seen that the current is highest when Z is minimum. This will occur when L = C(purely reactance) . At this point the impedance Z reduces to Z = R as shown;

Z = = = = R [ Z = R ]

The frequency at which this happens is called the resonant frequency, fo.
If we equate L = C : 2foL = gives fo =

This concept of resonance in a series ac circuit is widely used in applications such as radio tuners, metal detectors, TV tuners etc.

Note:

Ampere’s Law: B.dl | Faraday’s Law: B.dA | Gauss’s Law: E.dA | Time Constant(capacitor)τ= RC | Time Constant(Inductor)τ = LR |

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AC CIRCUITS PROBLEMS

1. A simple AC generator is made from 60 turn coil, of total resistance 12 , placed in a magnetic field of 0.8 T. The coil has a radius of 2 cm and rotates at 100 Hz. (i) Calculate the maximum induced voltage in the coil. (ii) Calculate the maximum induced current in the coil. (iii) Hence write an expression that gives the variation of induced current with time.

2. A LRC series circuit with R = 3000 , L = 60 mH, and C = 0.5 F has an AC source of 50V with angular frequency 10 000 rad/s. (i) Calculate the circuit impedance (ii) What is the circuit current? (iii) Calculate the phase angle (iv) Calculate the voltage across each circuit element and explain why their algebraic sum is not equal to the source voltage 50 V.

3. The ends of the circuit below are connected to an AC outlet of 250 V, 50 Hz.

20 0.2 H 80 F

250V, 50 Hz Calculate: (i) the reactance of the capacitor and the inductor (ii) the total impedance of the circuit (iii) the circuit current (iv) the resonance frequency of the circuit. 4. An LCR circuit is tuned to resonance by varying the capacitance of the capacitor. The current measured through the circuit is 0.02 A. The inductor with negligible internal resistance has 20 V across it. VS

900 L C

VR 20 V VC

(i) What is the voltage across the capacitor, Vc? (ii) What is the voltage across the resistor, VR? (iii) Draw a phasor diagram showing the voltages across the L, C and R. (iv) Calculate the source voltage, Vs. (v) If the inductance is 2 H and the capacitance is 4.5 F, calculate the frequency of the supply. [ Hint : L = c ]

5. A 60 Hz source with a rms output of 100 V is connected in series with a 30 resistor, a 3 F capacitor and a 2 H inductor as shown.

100 V 30

3 F

2 H
Calculate:
(i) the impedance of the circuit (ii) the rms current in the circuit (iii) the voltage drop across the resistor (iv) the phase angle (v) the power dissipated in the circuit.

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RL Circuits in DC

DC power supply switch, S

i

R

a b L

When S is open the current in the circuit is zero. When S is closed the current starts to build up from zero to a maximum value. Let i be at some time t and be its rate of charge at that time. The voltage across the inductor is L , VL = L , Vab = iR

Taking voltage drops around the circuit ( Kirchoff’s Rule ):
- iR - L + = 0 , = iR + L - iR = L = ,

= -

As soon as the switch is closed i is zero and potential difference across the resistor is equal to zero. Therefore the initial rate of change of current is = - The greater the value of inductance, the more slowly the current increases.

As the current increases, the term increases and the rate of increase of current becomes smaller and smaller and when the current reaches its final steady state ( valueI ) , its rate of increase is zero. = 0 , = - ,
0 = - , 0 = - iR , = iR imax =

Example 1.

A coil of inductance 0.1 H and resistance 6 is connected to a 12 V battery.

(a) Find the initial rate of growth of current the moment the switch is closed.

= = = 120 A/s

(b) Find the final current in the circuit after a long time when the switch is closed.

imax = = = 2 A

The current at any given time t is given by the formula i =

The rate of change of current at any time t is given by =

Derivation: i = expanding the equation gives; i = - differentiating throughout gives; ( = constant) = 0 - = =

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INDUCTANCE IN DC Problems

1. An inductor of inductance 0.2 H is connected in series with a 10 resistor and a 1.5 V cell and a switch. Calculate: (a) the initial rate of current growth. (b) the voltage across the inductor when the current is 0.1 A.
Hence find the rate of current growth at this point. (c) the final current. (d) the total energy stored in the inductor.

2. The diagram below shows an inductor L = 0.2 H and a resistor R = 4 connected in series with a 10 V dc power supply.

S 10V

L = 0.2 H R = 4

(a) Find the initial rate of change of current when switch S is closed. (b) What is the current in the circuit when the current is rising at a rate of 30 A.s-1? (c) Find the maximum current in the circuit.
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ATOMIC PHYSICS

The Photoelectric effect When Light is incident upon certain metallic surfaces, electrons are emitted. This phenomenon is known as the photoelectric effect. Circuit diagram for observing the photoelectric effect.

Photocell Light ( E = hf ) .
.
e

Emitter plate Collector plate

A

Retarding potential ( Vc )

V

The energy gained by a photon of light is given by: E = hf Where: h = Plank’s constant = 6.625 x 10-34 J.s f = frequency of light used ( Hz ) The frequency and wavelength are related by: c = f Where: c = speed of light = 2.998 x 108 m/s

Einstein’s interpretation: According to Einstein, the total energy ( hf ) is given to a single photon and so the emitted photoelectron will posses kinetic energy given by: EK = hf -

Definitions: Retarding / Stopping potential / Cut-off voltage ( Vc)
The potential applied to a photocell whereby the current in the circuit becomes zero. At this potential the electrons leaving the emitter plate have zero kinetic energy.

Threshold frequency ( fo )
The minimum frequency of light needed for photoelectric effect to occur.

Threshold wavelength ( o )
The maximum wavelength of light needed for photoelectric effect to occur.

Work function ( )
The amount of energy needed for a photoelectron to eject from the metal surface.

Photons
The extremely small particles in light that are like tiny packets of energy (particle model of light).

At threshold frequency or wavelength the following relation can be used: c = foo

A graph of kinetic energy ( EK ) against frequency ( f ) of incident light in a photoelectric set-up.

EK

Slope = h

fo (x-int.) f

y – intercept = –(work function)

EK =hf -

Analysing the Einstein’s equation we get:

Which corresponds to the linear equation y = mx + c

It can be seen that from the graph that the slope represents the Plank’s constant (h), and the y – intercept is the negative of the work function ().
It can also be deduced that the x – intercept is the threshold frequency ( fo ).

Two other relations can be obtained are:

= hfo and EK = eVc

In photoelectric effect the unit of energy used is called the electron-volt ( eV)

1 eV = 1.602 x 10-19 J , 1 J = = 6.24 x 1018eV.

Work functions of some metals: Metal (eV)

Na 2.46 Al 4.08 Fe 4.50 Cu 4.70

In a laboratory, Vc can be measured along with the frequencies. (for the frequencies, colour filters can be used and the theoretical values for corresponding frequencies can be determined. ) The Einstein’s equation can be further modified to give;

EK = hf - ( since,EK = eVc ) eVco = hf - ( dividing by e throughout gives)

Vco =- Graph of Vco against kinetic energy (Ek)

y = mx + c Vco slope =

fo f

y – intercept = – -------------------------------------------------
PHOTOELECTRIC EFFECT PROBLEMS

1. When light of frequency 7.5 x 1014 Hz is incident on a metal surface, the maximum kinetic energy of the emitted photoelectrons is 9.44 x10-20 J. (i) Calculate the work function of the metal. (ii) Hence find the threshold frequency of the metal. 2. In a photoelectric experiment it is found that a reverse potential difference of 1.25 V is required to reduce the photocurrent to zero. Calculate the maximum kinetic energy of the photoelectrons, giving your answer in electron volts.

3. When an orange light is incident on a clean potassium metal, the photoelectrons emitted have a cut – off voltage of 1.4 V. Which of the following changes will increase the cut – off voltage? A. Using a red light. B. Using a green light. C. Increasing the intensity of light. D. Increasing the distance between the light and metal surface.

4. When UV light of wavelength 2.54 x 10-7 m from a mercury arc falls on a clean copper surface, the electric potential necessary to stop emission of photoelectrons is 0.59 V.

(i) Calculate the energy associated with UV light. (ii) Calculate the threshold wavelength for copper metal.

5. In an experiment, blue light of frequency 7 x 1014 Hz shines on a photoelectric cell and produces a cut – off voltage of 1.63 V.

(i) What is the maximum kinetic energy of the ejected electrons?
(ii) Determine the work function of the metal.

6. The work function for copper metal is 4 eV.
(i) Calculate the threshold frequency for copper.
(ii) If a piece of copper is illuminated by light of wavelength 400 nm, determine whether it would be possible to get photoelectrons.

7. Light of wavelength 300 nm is incident on a metallic plate. If the stopping potential for the photoelectric effect is 2.0 V. (a) Calculate the maximum kinetic energy of the emitted electrons. (b) Calculate the work function of the metal. (c) Hence find the cut-off frequency.

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EARLY ATOMIC MODEL The atom in the days of Newton was a tiny, hard, indestructible sphere. This theory could be used to describe the behaviour of ideal gases, but a new model had to be devised to account for the electrical nature of the atom. J.J. Thomson ( 1856 – 1940 ) suggested that an atom was a mixture of positive and negative charges arranged anyhow in a sphere. This was known as the ‘plum pudding’ model. In 1911 Ernest Rutherford with his colleagues performed the famous alpha scattering experiment revealing that the ‘plum pudding’ model was incorrect. Rutherford then suggested that the atom had a massive positive nucleus surrounded by electrons orbiting the nucleus in a planetary motion. (Like the planets orbiting the sun.) There were two observations that the Rutherford’s model could not explain. * An atom emits certain discrete characteristic frequencies of electromagnetic (EM) radiation and not continuously. * According to Rutherford, the planetary motion of the electrons around the nucleus is undergoing a centripetal acceleration. According to Maxwell, the electrons should radiate EM waves of the same frequency. The second observation cannot be applied to the theory of the atomic model because when an electron radiates energy, it will collapse in the nucleus.

ATOMIC SPECTRA When an evacuated glass tube is filled with a specific gas of low pressure (example- hydrogen), and a voltage applied across the metal electrodes of the tube, an electric current is produced in the tube. The tube then emits light of certain colour depending on the type of gas used in the tube. This phenomenon is called atomic spectra. These spectral lines are called emission spectrum. The wavelengths contained in a given line spectrum are characteristic of the element emitting the spectrum. No two elements have identical atomic spectra and so this is a method of identifying elements. Johann Balmer found that the wavelengths of these emitted light could be identified by an empirical equation = R Where: R = Rydberg’s constant = 1.097 x 107 m-1 In addition to emitting light of specific wavelengths, an element can also absorb light of specific wavelength, and this is known as absorption spectrum. An absorption spectrum can be obtained by passing light through a vapour of an element. Both the emission and absorption spectra of an element are identical. Bohr Model of Hydrogen Since Rutherford’s model of the atom failed to explain the phenomenon of atomic spectra, a new theory had to be developed. In 1913 Neils Bohr proposed a theory that includes some features of the currently accepted theory. Using the simplest atom, hydrogen, Bohr developed four postulates to explain his theory of the atom. The postulates are as follows: 1. The electron moves in circular orbits under the Coulomb’s force of attraction. 2. Only certain electron orbits are stable. These are orbits in which the hydrogen atom does not emit EM radiation. Therefore the total energy of the atom remains constant, and classical mechanics can be used to describe the electron’s motion.

3. The hydrogen atom emits radiation when the electron “jumps” from a more energetic initial state to a lower state. The frequency, f, of the radiation emitted in the jump is related to the change in the atoms energy and is independent of the frequency of the electrons orbital motion. The frequency of the emitted radiation is Ei – Ef = hf Where Ei and Efis the energy in the initial and final states respectively. EiEf, and h is Planck’s constant.

4. The size of the allowed electron orbits are those for which the electron’s orbital angular momentum about the nucleus is an integral multiple of h= . mvr = n h n = 1,2,3, …

Using the above four postulates, the allowed energies and emission wavelength of the hydrogen can be calculated.

Using Bohr’s theory, the energies associated with the allowed energy states of an electron for the hydrogen atom can be calculated. The energy equation is derived to be: E = - eV

The lowest energy state ( also known as ground state ), corresponds to n = 1 and has E =-13.6 eV. The next state is n = 2, E = -3.60 eV and so on.

The energy level diagram for the hydrogen atom. n E (eV) 0.00

5 - 0.54 4 - 0.85 3 Paschen series - 1.51 HHH

2 Balmer series - 3.40

Lyman series

1 - 13.6 The top energy level labelled (infinity) represents the total removal of the electron (ionization) from the atom. The minimum energy required to remove an electron completely from an atom is called the ionization energy. The ionization energy of hydrogen is 13.6 eV.

According to the third postulate of Bohr, whenever an electron jumps from a high energy level to another lower one, it emits a photon with a fixed frequency (f). The energy associated with this transition is the difference of the two energy levels. If the electron jumps from n = 2 to n = 1, then the energy of the emitted photon will be,
Ep = En=2 – En=1 , Ep = ( -3.40 - -13.6 ) eV = 10.2 eV.

The frequency of this photon can be determined using E = hf. This analysis of energy conforms to Balmer’s modified equation:

= R

Where: L = series number ( Lyman = 1, Balmer = 2, and so on..) n = the energy transition. ( line number )

Series | Spectrum | Series #(l) | Line #(n) | Lyman | UV | 1 | 2 | Balmer | Visible | 2 | 3 | Paschen | Infra – red | 3 | 4 | Brackett | Infra – red | 4 | 5 | Pfund | Infra – red | 5 | 6 |

Example1

An electron makes a jump from the second energy level ( n = 2 ) to ground state ( n = 1 ). Calculate the wavelength and frequency of the emitted photon and identify which part of the spectrum the photon belongs to.

Solution: The wavelength can be obtained using

= R = 1.097x 107 m-1 * = 121.5 nm.

This wavelength lies in the ultra violet (UV) region of the EM spectrum.

The frequency can be obtained using, c = f, f = so;

f = = = 2.47 1015 Hz

Example2

The Balmer series for the hydrogen atom corresponds to all the electron transitions above the third energy levels to the second energy level.

(a) Find the energy related to the longest photon emitted in the Balmer series.

Solution: (i) Using the Empirical formulae.

The longest wavelength transition will be from n = 3 to n = 2. So the wavelength can be determined by making the minimum so that the is maximum. = R = 1.097 107 m-1 max = 565.3 nm.

The energy associated with this emitted photon is E =hf = E = = 3.03 10-19 J = 1.89 eV

(ii) Using the energy associated with the energy levels in transition.

Since the longest wavelength photon will have the smallest energy change since, E = ;

E = E3 – E2 = (-1.5- - 3.40) eV = 1.9 eV

(b) Determine the shortest wavelength photon emitted in the Balmer series and its energy.

Solution: The shortest wavelength photon will be when the energy difference will be the largest and this corresponds to the electron transition from n = to n = 2.

= R = 1.097 107 m-1 min = 364.6 nm. Energy; E =hf = = = 5.528 10-19 J = 3.40 eV

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ATOMIC SPECTRA PROBLEMS

1. The diagram shows possible jumps of the electron from a higher to a lower state in the hydrogen atoms.

Energy Energy level 5 -0.85eV 4 -1.5 eV 3 B

-3.4 eV 2

A

-13.6 eV 1

(a) What type of electromagnetic energy is emitted by the electron in jump A and name the spectrum it belongs to. (b) Calculate the energy of the photon emitted when the electron makes jump B. (c) Calculate the wavelength of the photon emitted when the electron makes jump A.

1. Calculate the frequency of the H - line of the Balmer series for hydrogen. This line is emitted in the transition from n = 4 to n = 2.

2. A hydrogen atom initially at ground state ( E1 = -2.18 10-18 J ) absorbs a photon which causes it to be excited to the n = 3 state ( E3 = -2.4 10-19 J ).
Calculate the wavelength of the photon absorbed.

ANSWERS

PROJECTILE MOTION

1. (a) 3 s (b) 26.25 m (c) 5.16 m/s

2. (a) 8 s (b) 20 m (c) 592 m (d) 80 m/s

3. (a) 33.6 m (b) 4 s c) 32 m

4. Yes, since Range = 61.53 m > 30 m

5. (a) 12.65 s (b) 1314.63 m (c) 126.5 m/s

6. (a) 3.6 s (b) 24.53 m/s at 47.20 North of east

7. (a) 318.90 m (b) 6 s (c) 361 m (d) 139.86 m/s

8. 80 m

9. (a) 142.8 m (b) 3.2 s (c) 36 m/s

10. (a) 25 m (b) 2.2 s

ROTATIONAL KINEMATICS

1. (a) 24π (b) 420π (c) 1000π

2. (a) 0.79 (b) 0.1 (c) 47.75

3. (a) 120π (b) π (c) 16 23π

4. (a) 10π rad/s (b) 6π m/s

5. (a) 32 π rad/s (b) 154π m/s

6. 20 rad/s

7. (a) 0.75 rad/s2 (b) 0.3 m/s2

8. (a) -1.25 rad/s2 (b) 9.6 rad

9. (a) 600 rad (b) 0.36 m/s2 (c) 60 rad/s

10. (a) 1.2π rad/s (b) 2425π m/s

11. (a) 2 rad/s2 (b) 1 m/s2

12. (a) 800 rad (b) 0.6 m/s2 (c) 80 rad/s

13. (a) 80π rad (b) 9.75 rad/s2 (c) 5.13 s

14. (a) 97 rad/s (b) 68 rad/s2

15. (a) 1500π rad/s2 (b) 5.03 x 10-4 m/s2

MOMENTUM

1. (a) 1.70 m/s (b) 14914.77 J

2. VA = 25.67 m/s VB = 20. 44 m/s

3. 1.07 m/s

4. 53.33 m

5. (i) VA = 20.7 m/s VB = 29.28 m/s (ii) No, 47240

6. μ=20.03 m/s v = 20.09 m/s

7. (a) 2 Kg (b) 4.44 m/s

8. 4.26 m/s

9. 10.95 m/s

10. (a) 30 000 Kgm/s (b) VA = 15.75 m/s VB = 19.25 m/s

ROTATIONAL DYNAMICS

1. (a) 0.05 Kgm2 (b) 384 rad/s2

2. (a) 2.4 rad/s2 (b) 8.64 x 10-3 Nm (c) 19.2 rad

3. 0 Nm

4. (a) 2 Nm (b) 5 rad/s2

5. (a) –π rad/s2 (b) -50 π Nm (c) -125 π N (d) 10 s

6. (a) 50 rad/s (b) 125 J (c) 1.88 m

7. (a) 20 N (b) 40 J (c) 0.88 Kgm2
8. (a) 0.05 Nm (b) 0.07 Kgm2 (c) 2.65 rad/s (d) 0.0086 J

9. (a) 0.14 rad/s2 (b) 1.77 x 10-3 Nm

10. (a) 160 N (b) 2.83 m/s (c) 319.91 J (d) 0.80 Kgm2

11. (a) 53.13 N (b) 0.27 Kgm2 (c) 40 rad/s

EQUILIBRIUM

1. T1 = 116.45 N T2 = 100.52 N

2. T1 = 928.93 N T2 = 1163.15 N

3. (a)186.69 N (b) 143. 01 N

FRICTION

1. 3.68 m/s2

2. 0.29

3. (a) 0.125 m/s2 (b) 40.5 N (c) 0.23

4. (a) 46.2 N (b) 2.3 Kg

GRAVITATION

1. (a) 7786.94 m/s (b) 5325.46 s (c) 1.52 x 1010 J

2. (a) 4.98 x 106 m/s (b) 1.86 x 1045 Kg

3. in your notes

4. 2.01 x 1021 Kg

5. 377.09 x 106 m

6. 2.55 x 106 m + Re

7. in your Notes

INTERNAL RESISTANCE AND EMF’s

1. (a) 1.5 C (b) 0.36 J (c) 1 Ω

2. (a) 18 W (b) 96 W (c) 5.2 V

DIRECT CURRENT

1. (a) 7 x 106 A/m2 (b) 437.5 μ m/s

2. (a) 0.036 Ω (b) 27.8 A (c) increases

3. (a) 2.048 x 107 Am-2 (b) 1.7 x 10-10Ωm (c) 0.0085 Ω

4. 0.312
G

5. (a) (b) 0.10 Ω

Rs

6. 648 KJ
Rs
G

7. (a) (b) 5880 Ω

V

KIRCHOFF’s LAWS

1. (i) I3 = I1 + I2 (b) 12 = 3I3 + 6I2, 8 = -2I1 + 6I2 (c) I1= 0 A, I2 = I3 = 43 A

2. (i) I = 2 A, I1 = 1 A, I2 = -1 A (ii) 4 V

3. (i) I1 = I2 + I3 (ii) I1 = 0.3 A, I2 = 2.4 A, I3 = -2.1 A (opposite direction)
Type equation here.
ELECTROSTATICS

1. (i) 90 000 N/C (ii) 0.036 N

2. (i) 161.2 N, 7.10 east of south (ii) 60 x 106 N/C (iii) -600 x 103 V 3. (i) 1.1 x 10-12 J (ii) 2.2 x 10-11 N (iii) 2.2 x 10-14 C, negative 4. y = Eql22mv2 5. (i) 44.44 pF (ii) 22.22 nC 6. (i) 4 μF (ii) 48 μC (iii) 8 V

7. (i) Parabolic Shape and moving upwards to the positive plate. (ii) 2 ns (iii) 1 x 10-15 N (iv) 2.197 mm 8. (i) 0.04 J (ii) W = F.d=F.(2d) = 2 F.d = 2W (iii) Stored in the capacitor 9. (a) 44.21 pF (b) 120 V/m (c) 3.2 nJ (d) 0.53 nC 10. (a) 4.8 x 10-16 N (b) 5.27 x 1014 m/s2 (c) 2.38 x 10-8 s (d) 7.2 x 10-17 J 11. (a) 8 pF (b) 9.6 x 10-10 C (c) 2.4 x 10-10 C-2pF, 7.2 x 10-10 C-6pF 12. (a) 1.5 pF (b) 1.8 x 10-10 C (c) same 13. (a) 2 μF (b) 4.8 x 10-5 C (c) 2.88 x 10-4 J

SIMPLE HARMONIC MOTION

1. 10π rad/s

2. -2000 m/s2

3. (a) 15 cm (b) 0.5 rad/s

4. (a) 250 N/m (b) 0.795 s (c) 0.79 m/s

5. (a) 2.27 x 104 m/s2 (b) 26.1 m/s

6. (a) 0.19 s (b) 0.14 m/s

7. (a) 20 rad/s (b) 3 m/s (c) 6.75 J

8. (a) 2.18 s (b) 0.58 m/s (c) same(2.18 s)

9. (a) 0.67 s (b) 4.60, -4.60 (c) -43.32, -408.26 (d) 61.26, 577

10. (a) 0.5 s (b) 5.03 m/s

11. (i) 25 N/m (ii) 0.397 s (iii) 0.776 m/s (iv) 0.0313 J

WAVES

1. (i) right (ii) 0.017m (iii) A = 0.8 m, f = 2π Hz, λ = 2π m, v = 39.48 m/s

2. (i) 5cm, 4, 989.46 cm/s (ii) y 2.5 sin (-0.005x – 2.00t)

3. (i) 1.33 m (ii) 1.67m/s, right (iii) 9.58 x 10-4 m
4. (i) 0.5 m (ii) 20 Hz (iii) 25 m/s (iv) y = 0.04 sin 2π (50t -1)

5. (i) 0.2 m, 40 Hz, 8 m/s (ii) y = 0.1 sin (-10πx - 80πt) (iii) 0.1 m

(iv) y 0.1

0.05 0.1 0.15 0.2 x (m) -0.1

STANDING WAVES

1. 102 Hz

2. (a) Notes (b) 120 m/s

3. (a) Notes (b) 0.6 m (c) 550 Hz

4. (i) 0.08 m (ii) Notes (iii) 12375 Hz

5. (a) 272 Hz (b) 16 Hz

6. (a) Notes (b) 336 m/s

LIGHT INTERFERENCE

1. 6.56 x 108 lines/mm

2. 9

3. 500 nm

4. (i) 0.09 m (ii) 3.440

5. 49.40 or 0.86 rad

6. (a) when the light waves interfere from the two sources, and the pd = nλ, there is constructive interference and a bright band is formed, but if pd = (n - 12 ) λ, there is a destructive interference and thus a dark band is formed.

(b) (i) fringe spacing increases (ii) increases

7. (a) 1.5 x 10-4 m (b) fringe spacing decreases

8. 9.4 x 10-8 m

DOPPLER EFFECT

1. (a) 0.58 m (b) 569 Hz

2. 72.3 Hz

3. (a) Away (b) 0.32 m (c) 20.4 m/s

4. (a) 0.32 m (b) 997.4 Hz

MAGNETISM 1

1. (a) upwards (b) 56.9 mm

2. (a) 4.8 x 10-13 N (b) 50 m

3. 15.81 x 106 m/s

MAGNETISM 2

1. 6 x 10-4 T

2. (a) 4 x 10-6 T (b) 8 x 10-6 N/m

ELECTROMAGNETIC INDUCTION

1. (a) 4 x 10-3Wb (b) 0.02 V

2. 2 H

3. 1.25 x 10-3 J

4. 6 V

5. (a) 0.0625 Wb (b) 31.25 V (c) 1.56 H

6. Notes

7. (a) 9 mV (b) 0.144 J

8. (i) 54.4 V (b) 53.8 V (c) 0.117 s

INDUCTION
1. (a) 375 rad/s (b) 59.68 Hz

2. (a) 40.2 V

3. (a) 37.9 V (b) 3.16 A (c) I = 3.16 sin 200πt

AC CIRCUITS

1. Same as Q3 above

2. (i) 3026.55 (ii) 16.52 mA (iii) 7.590 (iv) Vc = 3.3V, VL = 9.9V, VR = 49.56V (out of phase)

3. (i) XC = 39.8, XL = 62.83 (ii) 30.50 (iii) 8.2 A (iv) 39.8 Hz

4. (i) 20 V (ii) 18 V (iv) 18 V (v) 53.05 Hz

PHOTOELECTRIC EFFECT

1. (i) 4.02 x 10-19 J . (ii) 6.08 x 1014 Hz

2. 1.25 eV

3. B

5. (i) 2.6 x 10-19 J (ii) 2.03 x 10-19 J

6. (i) 9.67 x 1014Hz (ii) Not possible

7. (a) 3.2 x 10-19 J (b) 3.42 x 10-19 J (c) 5.17 x 1014 Hz

ATOMIC SPECTRA

1. (a) Layman Series (b) 2.55 eV/4.1x 10-19J (c) 121.6 nm

2. 6.17 x 1014 Hz

3. 102 nm