A smooth sphere A of mass 4m is moving on a smooth horizontal plane with speed u. It(adsbygoogle = window.adsbygoogle || []).push({});

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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Coefficient of resistution

**Physics Forums | Science Articles, Homework Help, Discussion**