# Kinetic energy of a bouncing ball

## Homework Statement

I'm trying to find the coefficient of restitution (COR) and energy loss upon impact of a bouncing ball, and I'm trying to understand how these relate to each other.

## Homework Equations

COR=sqrt(2gh/2gH)
suvat equations (possibly)
1/2mv^2=mgH (using m=0.0585kg, g=9.8)
energy transfer efficiency = (useful output ÷ total input ) × 100

## The Attempt at a Solution

In my case of a ball released from 1.2m then bouncing to 0.64m, COR=sqrt(2gh/2gH) = 0.73 (fairly elastic). I know that COR is a ratio of the ball velocity before/after impact, therefore also of the kinetic energy before/after. Now doesn't this mean that the final velocity/kinetic energy is 73% of the original?

I'm struggling to verify this because I can't measure the kinetic energy after the bounce. The suvat equations assume constant acceleration or time so I don't think I can use those. Given that I know the final velocity upon bouncing using mgH=1/2mv^2, v = sqrt(23.52)m/s, and I know the bounce height (0.64m) is there another way?

mjc123
Homework Helper
I know that COR is a ratio of the ball velocity before/after impact, therefore also of the kinetic energy before/after. Now doesn't this mean that the final velocity/kinetic energy is 73% of the original?
Since kinetic energy is proportional to velocity squared, COR cannot be both the ratio of velocities and the ratio of kinetic energies. Which is it? Hint: what is it also the ratio of? see your equation above.

Since kinetic energy is proportional to velocity squared, COR cannot be both the ratio of velocities and the ratio of kinetic energies. Which is it? Hint: what is it also the ratio of? see your equation above.
Sorry putting it into words is pretty confusing for me. I know that it's sqrt(2gh/2gH) which is derived from mgh=1/2mv^2, so I guess COR is the ratio of the square root of the velocities, seeing as v=sqrt(2gh). Unless you mean it's the ratio of the square root of the heights?

haruspex
Homework Helper
Gold Member
The suvat equations assume constant acceleration
There is constant acceleration up to just before the bounce, and again from just after the bounce. You can use SUVAT for any interval that does not include the bounce itself.

There is constant acceleration up to just before the bounce, and again from just after the bounce. You can use SUVAT for any interval that does not include the bounce itself.
Thanks I was confused about that. so using v^2=u^2+2as for the bounce. u=3.541750979m/s, which is consistent with the COR=0.73

mjc123
Homework Helper
I know that it's sqrt(2gh/2gH) which is derived from mgh=1/2mv^2, so I guess COR is the ratio of the square root of the velocities, seeing as v=sqrt(2gh). Unless you mean it's the ratio of the square root of the heights?
No, it's the ratio of the velocities just before/after bounce. It is also the square root of the ratio of the potential energies at maximum height, and therefore the square root of the ratio of kinetic energies just before/after bounce.
u=3.541750979m/s
Do you think this is a reasonable number of significant figures to give in your answer?

Maybe I should start a new thread but I was also wondering about the conservation of energy upon collision between ball and floor. The ball will lose energy due to heat, sound and friction at least, but how does the type of surface influence these energy losses? I know hard surfaces tend to 'reflect' more energy than softer surfaces, but is this just because of the tight structure of solids, and how is it 'reflected'? Also is friction the biggest cause of energy loss in this type of collision?

haruspex