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

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# Homework Help: Coefficient of resistution

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