Physics 6 Tutorial 9  Coefficient of Restitution
These notes are for students studying Option C in the Welsh Board and EDUQAS syllabus.
When you drop a ball onto the ground, it bounces back, but the height to which it bounces is ALWAYS less than the height it fell from. (If you throw the ball at the ground in a temper, it will bounce higher and you may be disqualified... We are not considering this here.) You may well do a simple experiment in which you measure the bounce height of a ball.
If we measure the drop height, and the bounce height, we find that they are in a constant ratio, which we will look at later.
We measure the speed as the ball hits the floor and then as it rebounds. A good quality datalogger can be used for this. You need to place it at a point that is about the diameter of the ball above the floor.
Why do you think the ball should be placed at this point? 
We find that the speed just before the ball hits the floor and the speed after the ball bounces off the floor are related in this ratio:
The speed of separation is the difference between the velocities of objects A and B:
So we can say that:
Speed of separation = velocity of A  velocity of B
In this case, the relative velocity is the same as the measured speed. The velocity of the floor is 0 before and 0 after. We are ignoring the signs of direction and the the movement of the floor is considered to be negligible.
The constant is called the coefficient of restitution and is given the Physics code e. It has no units.
The coefficient of restitution of a ball is 0.675. What is the speed of the ball as it rebounds having hit the floor at 5.50 m s^{1}? 
The coefficient of restitution is less than 1. This is because energy is always lost as a ball hits a surface:
Sound is given off (you can hear it bounce);
Some is converted to heat;
The ball is deformed as it bounces, and work is done in deforming the ball;
When the ball bounces, elastic energy is converted back to kinetic. Some is lost due to hysteresis.
There are many factors that affect the coefficient of restitution. A cold squash ball has a low coefficient of restitution. It hardly bounces at all. When it is warmed up, the coefficient of restitution is much higher. You could investigate that by putting squash balls in water baths at different temperatures.
Some materials have a low coefficient of reinstitution. Others have a high value. Different surfaces can affect the coefficient of restitution. Balls bouncing on a carpet do not bounce as high as those bouncing from a hard floor.
A perfectly inelastic collision results in a coefficient of restitution of 0.
Sports such as bowls involve one ball hitting another. Usually one ball is stationary. In this next argument, we will think of two balls that strike each other in one dimension, i.e. the centres of mass are in line. We will ignore friction.
Consider two balls, A and B of masses m_{1} and m_{2} respectively. They come together and collide:
We can add up the momenta:
p _{before} = m_{1}u_{1} + m_{2}u_{2}
After the collision we see:
We again add up the momenta:
p _{after }= m_{1}v_{1} + m_{2}v_{2}
We know that momentum is conserved, so we can write:
m_{1}u_{1} + m_{2}u_{2 }= m_{1}v_{1} + m_{2}v_{2}
When we work on momentum calculations like this, we must use the velocities, as direction is critical. Left to right is considered positive.
The coefficient of restitution for the two balls is given by the expression:
When using the coefficient of restitution equation, we must use the speed (magnitude of velocity) without the direction.
Worked example Ball A has a mass of 2.5 kg and its initial velocity is 1.5 m s^{1} to the right. Ball B has a mass of 3.0 kg and its initial velocity is 2.0 m s^{1} to the left. They collide head on.
If the the coefficient of restitution is 0.75, what are the final velocities of balls A and B?

Answer Momentum before = (2.5 kg × +1.5 m s^{1}) + (3.0 kg × 2.0 m s^{1}) = 3.75 kg m s^{1} + 6.0 kg m s^{1} = 2.25 kg m s^{1}
Momentum after = (2.5 kg × v_{1} m s^{1}) + (3.0 kg × v_{2} m s^{1}) = 2.25 kg m s^{1}
2.5 v_{1} + 3.0 v_{2} = 2.25 kg m s^{1}
Now we use the coefficient of restitution: 0.75 = (v_{2}  v_{1}) ÷ (2.0  1.5)
v_{2}  v_{1} = 0.75 × (2.0  1.5) = 0.375
Only use the speeds (i.e. the magnitudes of the velocities) for the coefficient of restitution. Do not use the signs, or it will mess the calculation up.
Watch the signs; it is very easy to get a wrong sign somewhere.
So we have two simultaneous equations: v_{2}  v_{1} = 0.375 2.5 v_{1} + 3.0 v_{2} = 2.25
Substitute for v_{2}: v_{2} = 0.375 + v_{1}
2.5 v_{1} + 3.0 (0.375 + v_{1}) = 2.25 2.5 v_{1} + 3.0 v_{1} = 2.25  1.125 5.5 v_{1} = 3.375 v_{1} = 0.614 m s1
Now substitute for v_{1}: 2.5 × 0.614 + 3.0 v_{2} = 2.25 3.0 v_{2} = 2.25 + 1.535 = 0.715 v_{2} = 0.238 m s^{1}
Ball A travels at 0.61 m s^{1} to the left, while ball B travels at 0.24 m s1 to the left, like this:

If the balls stick together after the collision, the coefficient of restitution is 0.
Explain why the statement above is true. What kind of collision is this? 
In the example above, kinetic energy has been lost.
Using the values for masses and speeds in the example above, calculate the loss of kinetic energy. What proportion of the kinetic energy remains as kinetic energy? 
Consider a ball being dropped from a height H. It bounces off the floor, back to a height h.
When the ball is released, it will accelerate towards the ground at an acceleration g (= 9.81 m s^{2}). When the ball hits the ground, there is a change in momentum (impulse) leading to a force that accelerates the ball upwards (Newton II). Its final velocity is 0, when the height is h.
We know that a suitable equation of motion would be:
v^{2} = u^{2} + 2as
We can adapt that for our situation. We will use downwards as negative. The acceleration is g and the displacement downwards is H. We will call the speed before v _{drop}. So we can write:
(v _{drop})^{2} = 0 + 2×g × H
Therefore the two minus signs cancel out:
(v _{drop})^{2} = 2gH
This is important because you cannot get a square of a negative that is a real number.
When the ball bounces back, the final speed is zero, and the height reached is h. The displacement of h is upwards, so it's positive. We will call the the speed of the bounce v _{bounce}. So we write:
0^{2} = (v _{bounce})^{2} + 2× g × h
Therefore:
(v _{bounce})^{2} = 2gh
The minus signs helpfully cancel out.
We know that the coefficient of restitution, e, for a bouncing ball is:
Therefore:
And we can write:
And it doesn't take a genius to see that:
Therefore we can write a final expression for coefficient of restitution:
A ball has a coefficient of restitution of 0.75. It is dropped from a height of 1.0 m. (a) Show that the speed of the ball is about 4.4 m s^{1} as it is about to hit the floor. (b) Work out the speed of the bounce as the ball just leaves the floor. Give your answer to an appropriate number of significant figures. (c) Work out the height of the bounce. 
A lively ball has a high coefficient of restitution. A football bounces well on a hard surface, and reasonably well on a grass surface. A cricket ball on grass is as lively as a lump of lead.