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Question

A ball of mass m moving with speed u undergoes a head on perfect elastic collision with mass nm initially at rest. The fraction of the energy transferred to the second ball by the first ball is


A
n(1+n)
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B
n(1+n)2
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C
2n(1+n)2
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D
4n(1+n)2
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Solution

The correct option is D 4n(1+n)2

Using momentum conservation,
mu+0=mv1+nmv2            ... (1)
For elastic collision,
Velocity of approach = velocity of separation
(u0)=(v2v1)             ... (2)
From both the above equations, we get
v2=2u1+nv2u=2(1+n)
Hence, fraction of incident energy transfered.
=12nmv2212mu2=n(v2u)2
=4n(1+n)2

Physics

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