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+1n−1)2
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C
(n−1n+1)2
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D
noneofthese
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Solution
The correct option is A(n−1n+1)2
Given : m1=mm2=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=(m1−em2)u1+(1+e)m2u2m1+m2
v1=(m−1×nm)v+(1+1)nm(0)m+nm⟹v1=(1−n)(1+n)v
Thus Kinetic energy of A after the collision E′A=12(m)v21
E′A=12(m)(1−n)2v2(1+n)2⟹E′A=12mv2(1−n1+n)2
Fraction of total kinetic energy retained by A E′AE=(n−1n+1)2