Thread: (M2) Mechanics collision question

A small smooth ball strikes a smooth vertical wall at right angles.

Its kinetic energy after impact is one-half of its initial kinetic energy.

Find the coefficient of restitution between the ball and the wall.


This is what I have got so far:
Let speed of approach be u and speed of separation be v

KE = 1/2 mv^2

so therefore

1/2 m (2)^2 = 1/2 * 1/2 m (v) ^2

Then I just do not have a clue what to do from there.
If it helps the textbook tells me the answer is e = 1/√2, I just don't know how to get there...
Any help will be appreciated

Originally Posted by Lauwra

A small smooth ball strikes a smooth vertical wall at right angles.

Its kinetic energy after impact is one-half of its initial kinetic energy.

Find the coefficient of restitution between the ball and the wall.


This is what I have got so far:
Let speed of approach be u and speed of separation be v

KE = 1/2 mv^2

so therefore

1/2 m (2)^2 = 1/2 * 1/2 m (v) ^2

Then I just do not have a clue what to do from there.
If it helps the textbook tells me the answer is e = 1/√2, I just don't know how to get there...
Any help will be appreciated

Coefficient of restitution = (rebound speed)/(inbound speed)

where these are realy the components of the after and befor velocities normal to the plane of impact. So in this case:

Coefficient of restitution = sqrt(rebound KE)/sqrt(inbound KE) = sqrt[(rebound KE)/(inbound KE)] =sqrt(1/2)

RonL