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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 travelling faster than A and I am assuming they are both travelling in the same direction (I have tried assuming A rebounded but still cant get the answer!)

So distance travelled by A before the second collision will be

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

Sa = t x Va

The distance travelled 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?

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 travelling faster than A and I am assuming they are both travelling in the same direction (I have tried assuming A rebounded but still cant get the answer!)

So distance travelled by A before the second collision will be

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

Sa = t x Va

The distance travelled 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?