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In the figure shown, coefficient of rest...

In the figure shown, coefficient of restitution between A and B is `e=1/2`, then

A

velocity of B after collision is `(v)/(2)`

B

impulse on one of the balls during collision is `(3)/(4)`mv

C

loss of kinetic energy in collision is `(3)/(16) mv^(2)`

D

loss of kinetic energy in the collision is `(1)/(4) mv^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B, C

By momentum conservation in horizontal direction `V = V_(1) + V_(2)` ….(i)
and `e = (V_(2) - V_(1))/(V) = (1)/(2)` ….(ii)
By (i) and (ii) `V_(2) = (3V)/(4)`
So impulse on `B = m ((3V)/(4))`
and loss in `K.E. = (3)/(16) mV^(2)`
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