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Question

Two particles of masses m1 & m2 and velocities u1 and (αu1)(α0) make an elastic head on collision. If the initial kinetic energies of the two particles are equal and m1 comes to rest after collision, then

A
m2m1=9+22
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B
m2m1=322
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C
m1m2=322
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D
m1m2=9+2
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Solution

The correct option is C m1m2=322
Given,
Two particles of mass m1 & m2 with velocities u1 & αu1 respectively.
If the initial kinetic energies of the two particles are equal, and after the collision, m1 comes to rest.
12m1u21=12m2(αu1)2
m1m2=α2 ... (1)

By the conservation of linear momentum,
m1u1+m2(αu1)=0+m2v2
Here, v2= velocity of m2 particle after collision
Using eq. (1)
m2α2u1+m2αu1=m2v2
v2=u1α(α+1) ... (2)

Collision is elastic, hence the kinetic energy of the particles before and after the collision will be equal.
12m1u21+12m2(αu1)2=12m2v22
On putting v2 from equation (2)
m1u21+m2α2u21=m2u21α2(α+1)2
m1m2=α4+2α3
From equation (1),
α4+2α3=α2
α2+2α1=0
α=1+2 (taking +ve root)
m1m2=α2=(1+2)2
Hence, m1m2=322

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