Momentum Lab Report

INTRODUCTION

During collisions involving two bodies, equal and opposite forces are set up between them. These impact forces influence the subsequent motion of the bodies. Momentum of the system (consisting of both bodies) is preserved if both bodies are free to move in space. This is because there is no external forces act on the system.

The forces acting between the bodies during the small interval of time when they are in contact cause changes in the velocities of each separate body. An exact determination of these forces is not practical but the presence of the forces can be allowed for by using a property known as the coefficient of restitution. The coefficient of restitution is the ratio of speeds of a falling object, from when it hits a given surface to when it leaves the surface. In laymen's terms, the coefficient of restitution is a measure of bounciness. It basically is a property of collisions and depends upon the materials that are colliding.

In this experiment, the coefficient of restitution between two balls, (a glass marble and a steel ball bearing) and the apparatus it is colliding with will be determined.

AIMS

To determine the coefficient of restitution between two balls, (a glass marble and a steel ball bearing) and the apparatus it is colliding with.

THEORY

When two bodies collide, equal and opposite forces act on each body and will cause a motion. If there is no external force exerted to the system, then momentum will be conserved. Momentum is defined as (kg.m/s) and is a vector in the direction of v.

(Newton’s Second Law) … equation (1)

Impulse is defined I =F dt as which has useful applications in solving problems for forces when very short times are involved, such as during collisions.

By taking the equation (1) for an integration, it shows that the impulse due to a force over a given time period is equal to the change in momentum as shown in equation (2).

….. equation (2)

Figure 1

Consider the collision of the bodies in figure 1, there are equal and opposite forces (Newton’s Third Law) acting on each body for the same amount of time during the collision. This condition can be represented mathematically as in equation (3).

…. Equation (3)

Conservation of momentum tells that G1 + G2 = 0 during a collision. But, this is unlikely to happen in real case because there is always some energy loss, whether through deformation, heat or sound. The coefficient of restitution is used to described this loss. This coefficient is constant and depends only on the materials from which the bodies are made, provided that the relative velocity of the bodies coming into contact is not too low,. The coefficient of restitution, e, is defined as

After carrying out a simple substitution and simplification, the coefficient of restitution can be given by:

e=|v'Bn-v'An|vAn-vBn

For a perfectly elastic collision, the coefficient of restitution is 1.0 where all the kinetic energy is conserved. For a perfect plastic collision, the coefficient of restitution is 0 in which the loss in kinetic energy is a maximum. All collisions lie somewhere between these extremes. ( 0 < e < 1)

APPARATUS

Marble, steel ball, white paper, carbon paper, cylindrical pipe and ruler.

PROCEDURE

Figure 2

1. The apparatus is set up as shown above in Figure 2.

2. An inclined steel block is set to 35 ̊, a white paper is placed on the board and covered with carbon paper.

3. The height of the steel pipe from the point of impact is measured by using a ruler.

4. Drop the marble through the pipe and record the distance to the point of first impact.

5. Repeat step 4 at least three times.

6. Repeat steps 4 and 5 for the steel ball.

7. Set the steel block to 10 degrees and repeat steps 4-6.

8. Set the steel block to 22.5 degrees and repeat steps 4-6

DATA PROCESSING

Sample Calculations

The velocity of the ball bearing changes during impact, these calculations along with the measurements of distance travelled, the angle of the slider and the height dropped can determine the coefficient of restitution.

By using x/y coordinates from curvilinear motion, the time taken to travel distance d can be determined

t=2v'yg

For that interval period, the horizontal distance travelled is given by the following formula,

The initial velocities v’x and v’y can be broken down into normal and tangential components.

v'x =v'nsinø+v'tcosø v'y =v'ncosø-v'tsinø

The coefficient of restitution is in the direction of the contact forces, the normal direction. e=|v'Bn-v'An|vAn-vBn (where A is the ball and B is the surface)
(where A is the ball and B is the surface)

Due to the assumption of frictionless surface:

v't = vt

Combining these equations the formula for e occurs as,

d=2hsin(2ø)(ecos2ø-sin2ø)(1+e)

This equation gives two values for e, and the negative one can be disregarded, leaving the solution for the coefficient of restitution.

By substituting the data obtained from the experiment, for the marble with the incline set to 10˚.
H=0.359m Ø=10˚ D=0.1463 m

0.1463 = 2 × 0.359 × sin(10)[ ecos2(10)-sin2(10)](1+e)

1.173410 = [e2.cos2(10) + (cos2(10) – sin2(10))e – sin2(10)]

= (0.969846e2 + 0.939692e – 0.030154)

0.969846e2 + 0.939629e - 1.203564 = 0

This can be solved using quadratic formula :

e=-b±b2-4ac2aA=πr2

e = 0.45366

RESULTS

All the coefficient of restitution results in the calculations as performed previously are presented in the table 1 below and the trajectories of both marble and steel ball at various angle are plotted to show the comparison are presented in the appendix(refer to page ) .

| Angle | Trial 1(cm) | Trial 2(cm) | Trial 3
(cm) | Average distance (cm) | Averagee | Marble | 10 ˚ | 15.8 | 14.5 | 13.6 | 14.63 | 0.45366 | | 22.5 ˚ | 20.7 | 20 | 19.3 | 20 | 0.48282 | | 30 ˚ | 24.3 | 24.9 | 27.6 | 25.6 | 0.80378 | Steel | 10 ˚ | 7.4 | 9.1 | 8.6 | 8.37 | 0.30118 | | 22.5 ˚ | 15.2 | 15.4 | 14 | 14.87 | 0.41421 | | 30 ˚ | 18.4 | 18.8 | 18.2 | 18.47 | 0.72658 |

DISCUSSION

i) There are basically two factors that affect the coefficient of restitution in a collision which are the type of material of the collided bodies are made of and the temperature. However, since this experiment is made in a laboratory where the temperature is always constant at room temperature, the only factor that affects the coefficient of restitution between two bodies in the collision is the types of the two materials are made of. If two materials remain constant then the coefficient of restitution will be a constant.

ii) The coefficient of restitution changes depending on the angle of the sheet metal slider. This is an error in the experimental set up, however it is unavoidable, since in reality there is a loss of energy to the environment in the form of heat, sound or friction, then the coefficient of restitution will change as the amount of energy lost varies.

iii) There are a few errors associated with this laboratory. The board that the marbles and ball bearing landed on did not seem to be long enough, as the ball bearings seemed to drop short when the steel block was set to 35˚.

Besides that, another error happened when releasing the ball from a hand the ball rotates slightly, causing the ball to lose some of its translational energy which in turn causes the ball to not bounce as high as it would in a perfect world.

Further errors occurred with measurement of both the exact height of the tube, and of each mark where the ball bearing or marble had landed.
As mentioned above there were also inherent errors in the experiment because the coefficient of restitution will not always be the same due to different amounts of energy being lost to the environment and hence different changes in the impulse during the collision.

iv) The experiment could be improved if it were possible to more accurately measure exactly how far the marbles and ball bearings had travelled. Another improvement would be a regulated height at which the marble or ball bearing was dropped, because whilst it can roughly be dropped from the same height each time it is not exact. If the board were also lengthened it would eradicate the error of the ball bearing dropping short when the slider was set to 35˚. A good release of the ball possibly without rotation would also help in improving the result in order to reduce loss of translational kinetic energy of the ball.

CONCLUSION

This experiment is effectively determined the coefficient of restitution for the collisions of a ball bearing and the metal slider, as well as a marble and the metal slider. The experiment revealed the coefficient to be equal, as expected, except in the case of when the slider was set to 35˚ and this can be explained by changes in the energy lost to the environment as well as measurement and analytical errors on the part of the experimenters.

REFERENCES

* MMAN1300 Impulse Momentum Lab Assignment
(UNSW Moodle)

* Physics for Scientists and Engineers with Modern
( Author : Raymond A. Serway, John W. Jewett )

APPENDIX

The trajectory graphs for the marble: At angle of launch of 10 ̊ :

Graph 1
At angle of launch of 22.5 ̊ :

Graph 2
At angle of launch of 35 ̊ :

Graph 3

The trajectory graphs for the steel ball

At angle of launch of 10 ̊ :

Graph 4

At angle of launch of 22.5 ̊ :

Graph 5
At angle of launch of 35 ̊

Graph 5