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Question

A particle of mass m travelling with velocity v and kinetic energy E collides elastically to another particle of mass nm, at rest. What is the fraction of the total energy retained by the particle of mass m?

A
(n+1n)2
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B
(n+1n1)2
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C
(n1n+1)2
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D
none of these
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Solution

The correct option is A (n1n+1)2
Given : m1=m m2=nm
Let the velocities of A and B after the collision be v1 and v2 respectively.
initial velocity of A before collision is v and that of B is zero.
Initial kinetic energy of the system E=12mv2

Using v1=(m1em2)u1+(1+e)m2u2m1+m2
v1=(m1×nm)v+(1+1)nm(0)m+nm v1=(1n)(1+n)v
Thus Kinetic energy of A after the collision EA=12(m)v21

EA=12(m)(1n)2v2(1+n)2 EA=12mv2(1n1+n)2

Fraction of total kinetic energy retained by A EAE=(n1n+1)2

482272_156406_ans.png

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