Physics Chapter 4
Problems (Chapter 3)
Review Example problems #1 - 12 by yourself.
Problem 3 (page 96): A web page designer creates an animation in which a dot on a computer screen has a position of r = [4 cm + (2.5 cm/s2)t2]i + (5 cm/s)t j. a) Find the magnitude and direction of the dot’s average velocity between t = 0 and t = 2 s. b) Find the magnitude and direction of the instantaneous velocity at t = 0, t = 1 s, nd t = 2 s. c) Sketch the dot’s trajectory from t = 0 to t = 2 s, and show the velocities calculated in part (b).
(a) Identify and Set Up: From [pic] we can calculate x and y for any t. Then use Eq. (3.2), in component form. Execute: [pic] At [pic] [pic] At [pic] [pic] [pic] [pic]
| |[pic] …show more content…
| |[pic] |
| | | |[pic] |
| | | |[pic] |
|Figure 3.3a | | |
Evaluate: Both x and y increase, so [pic] is in the 1st quadrant. (b) Identify and Set Up: Calculate [pic] by taking the time derivative of [pic] Execute: [pic] [pic] [pic] [pic] [pic] and [pic] [pic] [pic] [pic] [pic] and [pic] [pic] [pic] [pic] [pic] and [pic] (c) The trajectory is a graph of y versus x. [pic] [pic] For values of t between 0 and 2.0 s, calculate x and y and plot y versus x.
|[pic] |
|Figure 3.3b |
Evaluate: The sketch shows that the instantaneous velocity at any t is tangent to the trajectory.
Problem 5 (page 96): A jet plane is flying at a constant altitude. At time t1 = 0 it has components of velocity vx = 90 m/s, vy = 110 m/s. At time t2 = 30 s the components are vx = -170 m/s, vy = 40 m/s. a) Sketch the velocity vectors at t1 and t2. b) For this time interval calculate the components of the average acceleration, c) The magnitude and direction of the average acceleration.
Identify and Set Up: Use Eq. (3.8) in component form to calculate [pic] and [pic] Execute: (a) The velocity vectors at [pic] and [pic] are shown in Figure 3.5a.
|[pic] |
|Figure 3.5a |
(b) [pic] [pic]
|(c)|[pic] | |[pic] |
| | | |[pic] |
| | | |[pic] |
|Figure 3.5b | | |
Evaluate: The changes in [pic] and [pic] are both in the negative x or y direction, so both components of [pic] are in the 3rd quadrant.
Problem 16 (page 97): On level ground a shell is fired with an initial velocity of 50 m/s at 60o above the horizontal and feels no appreciable air resistance. a) Find the horizontal and vertical components of the shell’s initial velocity. b) How long does it take the shell to reach its highest point? c) Find its maximum height above the ground. d) How far from its firing point does the shell land? e) At its highest point, find the horizontal and vertical components of its acceleration and velocity.
Identify: The shell moves in projectile motion. Set Up: Let [pic] be horizontal, along the direction of the shell’s motion, and let [pic] be upward. [pic] [pic]. Execute: (a) [pic] [pic] (b) At the maximum height [pic] [pic]gives [pic] (c) [pic]gives [pic] (d) The total time in the air is twice the time to the maximum height, so [pic] (e) At the maximum height, [pic]and [pic] At all points in the motion, [pic]and [pic] Evaluate: The equation for the horizontal range R derived in the text is [pic] This gives [pic] which agrees with our result in part (d).
Problem 26 (page 97): A model of a helicopter rotor has four blades, each 3.4 m long from the central shaft to the blade tip. The model is rotated in a wind tunnel at 550 rev/min. a) What is the linear speed of the blade tip, in m/s? b) What is the radial acceleration of the blade tip expressed as a multiple of the acceleration of gravity, g?
Identify: Each blade tip moves in a circle of radius [pic] and therefore has radial acceleration [pic] Set Up: [pic] corresponding to a period of [pic] Execute: (a) [pic] (b) [pic] Evaluate: [pic] gives the same results for [pic] as in part (b).
Problem 29 (page 98): A Ferris wheel with radius 14 m is turning about a horizontal axis through its center (see Fig. E3.29). The linear speed of a passenger on the rim is constant and equal to 7 m/s. What are the magnitude and direction of the passenger’s acceleration as she passes through (a) the lowest point in her circular motion? (b) the highest point in her circular motion? (c) How much time does it take the Ferris wheel to make one revolution? [pic] Identify: Uniform circular motion. Set Up: Since the magnitude of [pic] is constant, [pic] and the resultant acceleration is equal to the radial component. At each point in the motion the radial component of the acceleration is directed in toward the center of the circular path and its magnitude is given by [pic] Execute: (a) [pic] upward. (b) The radial acceleration has the same magnitude as in part (a), but now the direction toward the center of the circle is downward.
The acceleration at this point in the motion is [pic] downward. (c) Set Up: The time to make one rotation is the period T, and the speed v is the distance for one revolution divided by T. Execute: [pic] so [pic] Evaluate: The radial acceleration is constant in magnitude since v is constant and is at every point in the motion directed toward the center of the circular path. The acceleration is perpendicular to [pic] and is nonzero because the direction of [pic] changes.
Problem 63 (page 100): A grasshopper leaps into the air from the edge of a vertical cliff, as shown in Fig. P3.63. Use information from the Figure to find:
(a) The initial speed of the grasshopper
(b) The height of the cliff. [pic]
Identify: From the figure in the text, we can read off the maximum height and maximum horizontal distance reached by the grasshopper. Knowing its acceleration is g downward, we can find its initial speed and the height of the cliff (the target
variables). Set Up: Use coordinates with the origin at the ground and [pic] upward. [pic] [pic] The constant-acceleration kinematics formulas [pic] and [pic] apply. Execute: (a) [pic] when [pic] [pic] gives [pic] [pic] so [pic] (b) Use the horizontal motion to find the time in the air. The grasshopper travels horizontally [pic] [pic] gives [pic] Find the vertical displacement of the grasshopper at [pic] [pic] The height of the cliff is 4.66 m. Evaluate: The grasshopper’s maximum height (6.74 cm) is physically reasonable, so its takeoff speed of 1.50 m/s must also be reasonable. Note that the equation [pic] does not apply here since the launch point is not at the same level as the landing point.
Problem 67 (page 101): An engineering student did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (see Fig. P3.67). The takeoff ramp was inclined at 53o, the river was 40 m wide, and the far bank was 15 m lower than the top of the ramp. The river itself was 100 m below the ramp. You can ignore air resistance. a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? b) If his speed was only half the value found in part (a), where did he land? [pic]
(a) Identify: Projectile motion.
|[pic] | |Take the origin of coordinates at the top of the|
| | |ramp and take [pic] to be upward. |
| | |The problem specifies that the object is |
| | |displaced 40.0 m to the right when it is 15.0 m |
| | |below the origin. |
|Figure 3.67 | | |
We don’t know t, the time in the air, and we don’t know [pic] Write down the equations for the horizontal and vertical displacements. Combine these two equations to eliminate one unknown. Set Up: y-component: [pic] [pic] [pic] [pic] Execute: [pic] Set Up: x-component: [pic] [pic] [pic] [pic] Execute: [pic] The second equation says [pic] Use this to replace [pic] in the first equation: [pic] [pic] Now that we have t we can use the x-component equation to solve for [pic] [pic] Evaluate: Using these values of [pic] and t in the [pic] equation verifies that [pic] (b) Identify: [pic] This is less than the speed required to make it to the other side, so he lands in the river. Use the vertical motion to find the time it takes him to reach the water: Set Up: [pic] [pic] [pic] [pic] gives [pic] Execute: [pic] and [pic] [pic] so [pic] The horizontal distance he travels in this time is [pic] He lands in the river a horizontal distance of 28.4 m from his launch point. Evaluate: He has half the minimum speed and makes it only about halfway across.
Problem 69 (page 101): (NOT INCLUDED IN EXM_1): A 5500 kg cart carrying a vertical rocket launcher moves to the right at a constant speed of 30 m/s along a horizontal track. It launches a 45 kg rocket vertically upward with an initial speed of 40 m/s relative to the cart. a) How high will the rocket go? b) Where, relative to the cart, will the rocket land? c) How far does the cart move while the rocket is in the air? d) At what angle, relative to the horizontal, is the rocket traveling just as it leaves the cart, as measured by an observer at rest on the ground? e) Sketch the rocket’s trajectory as seen by an observer (i) stationary on the cart and (ii) stationary on the ground.
Identify and Set Up: Take [pic] to be upward. The rocket moves with projectile motion, with [pic] and [pic] relative to the ground. The vertical motion of the rocket is unaffected by its horizontal velocity. Execute: (a) [pic] (at maximum height), [pic] [pic] [pic] [pic] gives [pic] (b) Both the cart and the rocket have the same constant horizontal velocity, so both travel the same horizontal distance while the rocket is in the air and the rocket lands in the cart. (c) Use the vertical motion of the rocket to find the time it is in the air. [pic] [pic] [pic] [pic] [pic] gives [pic] Then [pic] (d) Relative to the ground the rocket has initial velocity components [pic] and [pic] so it is traveling at [pic] above the horizontal.
|(e) (i) |[pic] |
|Figure 3.69a | | |
Relative to the cart, the rocket travels straight up and then straight down.
| (ii) |[pic] |
|Figure 3.69b | | |
Relative to the ground the rocket travels in a parabola. Evaluate: Both the cart and rocket have the same constant horizontal velocity. The rocket lands in the cart.
Review Example problems #1 - 12 by yourself.
Problem 3 (page 96): A web page designer creates an animation in which a dot on a computer screen has a position of r = [4 cm + (2.5 cm/s2)t2]i + (5 cm/s)t j. a) Find the magnitude and direction of the dot’s average velocity between t = 0 and t = 2 s. b) Find the magnitude and direction of the instantaneous velocity at t = 0, t = 1 s, nd t = 2 s. c) Sketch the dot’s trajectory from t = 0 to t = 2 s, and show the velocities calculated in part (b).
(a) Identify and Set Up: From [pic] we can calculate x and y for any t. Then use Eq. (3.2), in component form. Execute: [pic] At [pic] [pic] At [pic] [pic] [pic] [pic]
| |[pic] …show more content…
| |[pic] |
| | | |[pic] |
| | | |[pic] |
|Figure 3.3a | | |
Evaluate: Both x and y increase, so [pic] is in the 1st quadrant. (b) Identify and Set Up: Calculate [pic] by taking the time derivative of [pic] Execute: [pic] [pic] [pic] [pic] [pic] and [pic] [pic] [pic] [pic] [pic] and [pic] [pic] [pic] [pic] [pic] and [pic] (c) The trajectory is a graph of y versus x. [pic] [pic] For values of t between 0 and 2.0 s, calculate x and y and plot y versus x.
|[pic] |
|Figure 3.3b |
Evaluate: The sketch shows that the instantaneous velocity at any t is tangent to the trajectory.
Problem 5 (page 96): A jet plane is flying at a constant altitude. At time t1 = 0 it has components of velocity vx = 90 m/s, vy = 110 m/s. At time t2 = 30 s the components are vx = -170 m/s, vy = 40 m/s. a) Sketch the velocity vectors at t1 and t2. b) For this time interval calculate the components of the average acceleration, c) The magnitude and direction of the average acceleration.
Identify and Set Up: Use Eq. (3.8) in component form to calculate [pic] and [pic] Execute: (a) The velocity vectors at [pic] and [pic] are shown in Figure 3.5a.
|[pic] |
|Figure 3.5a |
(b) [pic] [pic]
|(c)|[pic] | |[pic] |
| | | |[pic] |
| | | |[pic] |
|Figure 3.5b | | |
Evaluate: The changes in [pic] and [pic] are both in the negative x or y direction, so both components of [pic] are in the 3rd quadrant.
Problem 16 (page 97): On level ground a shell is fired with an initial velocity of 50 m/s at 60o above the horizontal and feels no appreciable air resistance. a) Find the horizontal and vertical components of the shell’s initial velocity. b) How long does it take the shell to reach its highest point? c) Find its maximum height above the ground. d) How far from its firing point does the shell land? e) At its highest point, find the horizontal and vertical components of its acceleration and velocity.
Identify: The shell moves in projectile motion. Set Up: Let [pic] be horizontal, along the direction of the shell’s motion, and let [pic] be upward. [pic] [pic]. Execute: (a) [pic] [pic] (b) At the maximum height [pic] [pic]gives [pic] (c) [pic]gives [pic] (d) The total time in the air is twice the time to the maximum height, so [pic] (e) At the maximum height, [pic]and [pic] At all points in the motion, [pic]and [pic] Evaluate: The equation for the horizontal range R derived in the text is [pic] This gives [pic] which agrees with our result in part (d).
Problem 26 (page 97): A model of a helicopter rotor has four blades, each 3.4 m long from the central shaft to the blade tip. The model is rotated in a wind tunnel at 550 rev/min. a) What is the linear speed of the blade tip, in m/s? b) What is the radial acceleration of the blade tip expressed as a multiple of the acceleration of gravity, g?
Identify: Each blade tip moves in a circle of radius [pic] and therefore has radial acceleration [pic] Set Up: [pic] corresponding to a period of [pic] Execute: (a) [pic] (b) [pic] Evaluate: [pic] gives the same results for [pic] as in part (b).
Problem 29 (page 98): A Ferris wheel with radius 14 m is turning about a horizontal axis through its center (see Fig. E3.29). The linear speed of a passenger on the rim is constant and equal to 7 m/s. What are the magnitude and direction of the passenger’s acceleration as she passes through (a) the lowest point in her circular motion? (b) the highest point in her circular motion? (c) How much time does it take the Ferris wheel to make one revolution? [pic] Identify: Uniform circular motion. Set Up: Since the magnitude of [pic] is constant, [pic] and the resultant acceleration is equal to the radial component. At each point in the motion the radial component of the acceleration is directed in toward the center of the circular path and its magnitude is given by [pic] Execute: (a) [pic] upward. (b) The radial acceleration has the same magnitude as in part (a), but now the direction toward the center of the circle is downward.
The acceleration at this point in the motion is [pic] downward. (c) Set Up: The time to make one rotation is the period T, and the speed v is the distance for one revolution divided by T. Execute: [pic] so [pic] Evaluate: The radial acceleration is constant in magnitude since v is constant and is at every point in the motion directed toward the center of the circular path. The acceleration is perpendicular to [pic] and is nonzero because the direction of [pic] changes.
Problem 63 (page 100): A grasshopper leaps into the air from the edge of a vertical cliff, as shown in Fig. P3.63. Use information from the Figure to find:
(a) The initial speed of the grasshopper
(b) The height of the cliff. [pic]
Identify: From the figure in the text, we can read off the maximum height and maximum horizontal distance reached by the grasshopper. Knowing its acceleration is g downward, we can find its initial speed and the height of the cliff (the target
variables). Set Up: Use coordinates with the origin at the ground and [pic] upward. [pic] [pic] The constant-acceleration kinematics formulas [pic] and [pic] apply. Execute: (a) [pic] when [pic] [pic] gives [pic] [pic] so [pic] (b) Use the horizontal motion to find the time in the air. The grasshopper travels horizontally [pic] [pic] gives [pic] Find the vertical displacement of the grasshopper at [pic] [pic] The height of the cliff is 4.66 m. Evaluate: The grasshopper’s maximum height (6.74 cm) is physically reasonable, so its takeoff speed of 1.50 m/s must also be reasonable. Note that the equation [pic] does not apply here since the launch point is not at the same level as the landing point.
Problem 67 (page 101): An engineering student did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (see Fig. P3.67). The takeoff ramp was inclined at 53o, the river was 40 m wide, and the far bank was 15 m lower than the top of the ramp. The river itself was 100 m below the ramp. You can ignore air resistance. a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? b) If his speed was only half the value found in part (a), where did he land? [pic]
(a) Identify: Projectile motion.
|[pic] | |Take the origin of coordinates at the top of the|
| | |ramp and take [pic] to be upward. |
| | |The problem specifies that the object is |
| | |displaced 40.0 m to the right when it is 15.0 m |
| | |below the origin. |
|Figure 3.67 | | |
We don’t know t, the time in the air, and we don’t know [pic] Write down the equations for the horizontal and vertical displacements. Combine these two equations to eliminate one unknown. Set Up: y-component: [pic] [pic] [pic] [pic] Execute: [pic] Set Up: x-component: [pic] [pic] [pic] [pic] Execute: [pic] The second equation says [pic] Use this to replace [pic] in the first equation: [pic] [pic] Now that we have t we can use the x-component equation to solve for [pic] [pic] Evaluate: Using these values of [pic] and t in the [pic] equation verifies that [pic] (b) Identify: [pic] This is less than the speed required to make it to the other side, so he lands in the river. Use the vertical motion to find the time it takes him to reach the water: Set Up: [pic] [pic] [pic] [pic] gives [pic] Execute: [pic] and [pic] [pic] so [pic] The horizontal distance he travels in this time is [pic] He lands in the river a horizontal distance of 28.4 m from his launch point. Evaluate: He has half the minimum speed and makes it only about halfway across.
Problem 69 (page 101): (NOT INCLUDED IN EXM_1): A 5500 kg cart carrying a vertical rocket launcher moves to the right at a constant speed of 30 m/s along a horizontal track. It launches a 45 kg rocket vertically upward with an initial speed of 40 m/s relative to the cart. a) How high will the rocket go? b) Where, relative to the cart, will the rocket land? c) How far does the cart move while the rocket is in the air? d) At what angle, relative to the horizontal, is the rocket traveling just as it leaves the cart, as measured by an observer at rest on the ground? e) Sketch the rocket’s trajectory as seen by an observer (i) stationary on the cart and (ii) stationary on the ground.
Identify and Set Up: Take [pic] to be upward. The rocket moves with projectile motion, with [pic] and [pic] relative to the ground. The vertical motion of the rocket is unaffected by its horizontal velocity. Execute: (a) [pic] (at maximum height), [pic] [pic] [pic] [pic] gives [pic] (b) Both the cart and the rocket have the same constant horizontal velocity, so both travel the same horizontal distance while the rocket is in the air and the rocket lands in the cart. (c) Use the vertical motion of the rocket to find the time it is in the air. [pic] [pic] [pic] [pic] [pic] gives [pic] Then [pic] (d) Relative to the ground the rocket has initial velocity components [pic] and [pic] so it is traveling at [pic] above the horizontal.
|(e) (i) |[pic] |
|Figure 3.69a | | |
Relative to the cart, the rocket travels straight up and then straight down.
| (ii) |[pic] |
|Figure 3.69b | | |
Relative to the ground the rocket travels in a parabola. Evaluate: Both the cart and rocket have the same constant horizontal velocity. The rocket lands in the cart.