 #1
 19
 0
Hi there,
I have been struggling with this problem for a while. It's an ALevel 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 ALevel 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?
Attachments

1.7 KB Views: 258