# Coefficient of Restitution & Collisions

• wanchosen
In summary, the conversation is about a problem involving two spheres, A and B, in a smooth groove on a horizontal table. The first collision occurs at time T and the coefficient of restitution between the spheres is e. The velocities of A and B after the first collision are found to be Vb = V(1+e)/2 and Va = (1-e)/2, respectively. The second collision takes place at time t = T(2+e)/e. The conversation also includes a solution for the first part in Solution1.txt and a discussion about the distance traveled by A and B before the second collision. The solution is simplified by considering the relative velocity between the spheres, which is eV.

#### wanchosen

Hi there,

I have been struggling with this problem for a while. It's an A-Level Mechanics problem.

A smooth groove in the form of a circle of radius a is carved out of a horizontal table. Two small spheres, A and B, lie at rest in the groove at opposite ends of a diameter. At time t = 0, the sphere A is projected along the groove and the first collision occurs at time = T. Given that e is the coefficient of restitution between the spheres, find the velocities of A and B after the first collision. Hence, or otherwise, show that the second collision takes place at time

t = T(2+e)/e.

I have included my solution of the first part in Solution1.txt and can confirm the answers. It's the last part that I have been unable to do.

After collision 1, B is traveling faster than A and I am assuming they are both traveling in the same direction (I have tried assuming A rebounded but still can't get the answer!)

So distance traveled by A before the second collision will be

(Va is the velocity after the first collision, Sa is the distance traveled by A before 2nd impact.)

Sa = t x Va

The distance traveled by B before impact will be the circumference + Sa

Sb = 2πa + Sa

where Sb also equals

Sb = t x Vb

I've substituted the velocities for A and B but still come up with

t = 2T/e

Can somebody help?

#### Attachments

• Solution 1.txt
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wanchosen said:
A smooth groove in the form of a circle of radius a is carved out of a horizontal table. Two small spheres, A and B, lie at rest in the groove at opposite ends of a diameter. At time t = 0, the sphere A is projected along the groove and the first collision occurs at time = T. Given that e is the coefficient of restitution between the spheres, find the velocities of A and B after the first collision. Hence, or otherwise, show that the second collision takes place at time

t = T(2+e)/e.

I've substituted the velocities for A and B but still come up with

t = 2T/e

Hi wanchosen!

But 2T/e is right … add the original T, and you get T(2+e)/e.

btw, you're making this needlessly complicated …

Once you've obtained Vb = V(1+e)/2, Va = (1-e)/2,

all you need to know is Vb - Va, which is eV.

It's only the relative velocity that matters!

Hi tiny-tim,

Thank you so much for that. That makes sense now! It was much easier working with the relative velocity! I completely forgot about adding the original time T to travel to the first collision! Thanks again.