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Question

A particle of mass m1 experienced a perfectly elastic collision with a stationary particle of mass m2. What fraction of the kinetic energy does the striking particle lose, if the collision is a head-on one?

A
η=4m1m2(m1+m2)2
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B
η=m1m2(m1+m2)2
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C
η=m1m24(m1+m2)2
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D
η=2m1m2(m1+m2)2
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Solution

The correct option is A η=4m1m2(m1+m2)2

In head-on collision both momentum and kinetic energy are conserved hence, by conservation of momentum

m1u1+m2u2=m1v1+m2v2

m1u1+m2(0)=m1v1+m2v2

m1u1=m1v1+m2v2 ............(1)

And conservation of kinetic energy gives

12m1(u1)2=12m1(v1)2+12m2(v2)2 ...........(2)

From equation (1) and (2)

m1(u1v1)=m2v2 ...........(3)

m1[(u1)2(v1)2]=m2v2 ...........(4)

Dividing equation (4) by (3) we get

u1+v1=v2

Using this value in (3)

m1(u1v1)=m2(u1+v1)

m1u1m2u2=m2v2+m1v1

u1(m1m2)=v1(m1+m2)

v1=u1(m1m2)(m1+m2)

Hence, fractional loss in KE of striking particle is ratio of change in KE to initial KE.

η=EiEfEi=1EfEi

η=1(v1u1)2

η=1((m1m2)(m1+m2))2

η=(m1+m2)2(m1m2)2m1+m2)2

η=4m1m2(m1+m2)2


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