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Question

Consider a head-on collision between two particles of masses m1 and m2.The initial speeds of the particles are u1 and u2 in the same direction.The collision starts at t=0 and the particles interact for a time interval t.During the collision,the speed of the first particle varies as

v(t)=u1+tt(v1u1).

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Solution

Using law of consetrvation of momentum m1u1+m2u2=m1v(t)+m2v1

where v=speed of 2nd particle during collision.

m1u1+m2u2

=m1u1+m1.(tt)(v1u1)+m2v1

m2u2m2=m1m2(tt)(v1u1)+v1

v1=u2m1m2tt(v1u1)


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