# Homework Help: Coefficient of resistution

1. Dec 11, 2008

### nokia8650

A smooth sphere A of mass 4m is moving on a smooth horizontal plane with speed u. It
collides directly with a stationary smooth sphere B of mass 5m and with the same radius as A.

The coefficient of restitution between A and B is 1/2

(a) Show that after the collision the speed of B is 4 times greater than the speed of A.

Sphere B subsequently hits a smooth vertical wall at right angles. After rebounding from the
wall, B collides with A again and as a result of this collision, B comes to rest.

Given that the coefficient of restitution between B and the wall is e,

(b) find e.

For part (b), I recieve an answer of 1/14. This is due to the fact that I have equated the coefficient of resisution of the two balls in the following way for the final collision:

0.5 = v/(u+4eu)

where v = velovity of ball A after the final collision, and u = the velocity of ball A before the collision (and hence the velocity of B is -4eu).

This results in a result of e = 1/14 after solving simultaneously with a linear expression for the conservation of momentum.

However, the markscheme has done the above as: 0.5 = (0-v)/(u+4eu) - note the presence of the negative sign. This results in a result of 0.5 for the second coefficient of resistution. Can someone please explain the logic behind this?

Thanks

2. Dec 13, 2008

### nokia8650

Last edited by a moderator: May 3, 2017
3. Dec 13, 2008

### Staff: Mentor

I didn't look at the problem in detail, but this piece seems to make sense. By definition of coefficient of restitution (e):
V'a - V'b = e(Vb - Va)
v - 0 = e(u - -4e'u) = 0.5(u + 4e'u)

Where I use e' for the second COR.