Work and energy during a head-on collision

When two balls collide head-on in an elastic collision, the work done on each ball by the impact force is equal to the change in kinetic energy of each ball. A calculation of the work done during the collision is presented as an interesting extension of the work-energy principle, including an example where one of the balls is infinitely heavy.


Introduction
When two balls collide head-on in a perfectly elastic collision, the outcome can be determined using conservation of momentum and kinetic energy.That problem is a standard exercise in elementary mechanics.The change in the velocity and kinetic energy of each ball during the collision itself can be evaluated in terms of the force acting on each ball.The latter problem is considered in the present paper, together with a calculation of the work done on each ball.The change in the velocity of each ball during a collision has been calculated previously [1,2].However, the work done during a two ball collision has not previously been described and is the main concern of the present paper.The work done provides additional insights into how the energy of one ball is transferred to the other ball via elastic compression and expansion of each ball.The work done includes changes in both the kinetic energy and the stored elastic energy, even if one of the balls has infinite mass and remains at rest.Calculations involving work done can be confusing for students [3][4][5][6] but an elastic collision between two balls is relatively simple since it involves changes only in kinetic and potential (elastic) energy.As an example, suppose that a ball of mass m 1 collides head-on with another ball of mass m 2 , as shown in figure 1.
Ball 1 is incident at speed V 0 and ball 2 is initially at rest.Conservation of momentum indicates that where V 1 is the speed of ball 1 after the collision and V 2 is the speed of ball 2 after the collision.If the collision is perfectly elastic then since the relative speed of the two balls after the collision is equal to the relative speed before

R Cross
Figure 1.A head-on collision between two balls.
the collision.Substitution of equation ( 2) into equation (1) gives and For example, if Equal and opposite forces act on each ball during the collision and the work done by each force is equal to the change in kinetic energy of each ball.It is not difficult to show that the decrease in kinetic energy of ball 1 is equal to the increase in kinetic energy of ball 2. The work done on each ball is therefore equal and opposite.However, these relations do not apply during the collision itself, since work is done to compress each ball during the collision.The work done during the collision can be calculated by estimating the force that acts on each ball.For simplicity, it could be assumed that the force remains constant in time, but in reality the force increases to a maximum and then decreases to zero at the end of the collision.

Impact force
The displacement of each ball varies with time during the collision but can be calculated if the ball stiffnesses are known.If both balls are very stiff then the collision time will be very short and the displacement of the balls during the collision will be very small.If one or both balls are soft then the collision time will be longer.The effect of ball stiffness can be modelled by a spring connecting the two balls, as shown in figure 2. The stiffness of the spring can vary with time, but to simplify the problem it is assumed that the spring constant, k, remains constant.
If ball 1 moves to the right through a distance x 1 and ball 2 moves to the right through a distance x 2 then the force exerted on each ball is and as described in [1].A numerical solution of equations ( 5) and ( 6) is shown in figure 3 for a case where m 1 = 0.05 kg, m 2 = 0.1 kg, k = 2 × 10 6 N m −1 and where ball 1 is incident at V 0 = 1.0 m s −1 on ball 2 at rest.During the collision, ball 1 slows down and then reverses direction, while ball 2 increases in speed during the whole collision.Solutions of equations ( 5) and ( 6) for other values of m 1 , m 2 and k can be found using the supplementary excel file 2-Ball Collisions.xlsx.The duration of the collision in figure 3 is 0.406 ms, after which the balls separate at speeds V 1 = −1/3 m s −1 and V 2 = 2/3 m s −1 as calculated above.The impact force, F, during the collision is shown in figure 4 and is given by F = 258 sin(7738t).
Figure 5 shows F vs x 1 and F vs x 2 .The two curves are different since ball 1 slowed down then reversed direction during the collision so x 1 increased at first then decreased.The work done by F acting on ball 1 is given by the integral ´F dx 1 over the duration of the collision and is equal to the decrease in kinetic energy of ball 1, ).The work done on ball 2 is given  by ´F dx 2 and is equal to the increase in kinetic energy of ball 2, 1 2 m 2 V 2 2 .Both integrals are equal, which means that the area under the F vs x 1 curve is the same as the area under the F vs x 2 curve and the decrease in kinetic energy of ball 1 is equal to the increase in kinetic energy of ball 2.

During the collision,
where v 1 and v 2 vary with time as shown in figure 3(b), and the total energy is given by since some of the initial kinetic energy is converted to elastic energy stored in the spring.The three components of the total energy vary with time and are shown in figure 6. Ball 1 comes to a stop during the collision, so its kinetic energy decreases to zero before the ball reverses direction.By the end of the collision the spring returns to its original and PE is the elastic energy stored in the spring.The total energy remains constant.
uncompressed length so the stored elastic energy decreases back to zero.
Since F acts to the right on the left end of the spring and acts to the left on the right end of the spring, the work done to compress the spring at any given time during the collision is given by Ball 2 moves to the right during the collision so it allows that end of the spring to expand, while ball 1 acts to compress the spring at the start of the collision.At the end of the collision, x 1 = x 2 so the work done on the spring at that time is zero since it is no longer compressed.That is, the spring returns to its original length at the end of the collision.The same basic physics is involved in all collisions of this type, regardless of the masses m 1 and m 2 and regardless of the stiffness of each mass.The overall decrease in kinetic energy of m 1 by the end of the collision is which is the same as the overall increase in kinetic energy of m 2 if the collision is perfectly elastic.
The net work done on m 1 is therefore the same as the net work done on m 2 , regardless of the values of m 1 and m 2 .However, the work done on m 1 during the collision is not the same as the work done on m 2 since work is also done to compress the spring.

Case where m 2 = ∞
A collision between a ball and the floor is one where m 2 is very large.Then V 2 will be very small, and V 1 will be close to −V 0 if the collision is perfectly elastic, as given by equations ( 3) and (4).In the limit where m 2 = ∞, no net work is done on m 1 or m 2 by the end of the collision, but that is not the case during the collision.If m 2 remains at rest, then the decrease in kinetic energy of m 1 during the collision is equal to the increase in potential energy stored in the spring, and vice-versa when the spring starts to expand.Even if m 2 = ∞, the force exerted by the spring on m 1 and m 2 will be similar to that shown in figure 4 if the ball and/or m 2 has finite stiffness.Then the work done by the spring on ball 1 is ∫ F dx 1 , which is equal and opposite the work done by ball 1 on the spring.The work done by the force exerted by m 2 on the other end of the spring is zero since x 2 remains zero during the whole collision.

Conclusion
The work-energy principle is easy to apply to a single particle, but gets more complicated when the work done involves changes in additional forms of energy other than translational kinetic energy.A head-on perfectly elastic collision between two balls involves a change in the stored elastic energy as well as a change in the kinetic energy of each ball, and provides a good example of the work done when these two different forms of energy are involved.A calculation of the work done by one ball on the other during a collision reveals that some of the work involves a change in kinetic energy of each ball and some involves a change in the stored elastic energy, unlike the work done on a single particle.The physics of the problem may be too complicated for high school students but would be a useful exercise in an introductory mechanics course for undergraduate students.

Figure 2 .
Figure 2. The two balls compress during the collision, as represented by the spring.

Figure 4 .
Figure 4. F vs t during the collision in figure 3.

Figure 5 .
Figure 5. F vs x 1 and F vs x 2 during the collision in figure 3.

Figure 6 .
Figure 6.Components of the total energy whereKE 1 = 1 2 m 1 v 2 1 , KE 2 = 1 2 m 2 v 2 2and PE is the elastic energy stored in the spring.The total energy remains constant.