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Question: Two similar bogies A and B of same mass M (empty bogie) move with constant velocities v<sub>A</sub> ...

Two similar bogies A and B of same mass M (empty bogie) move with constant velocities vA and vB towards each other on smooth parallel tracks. At an instant a boy of mass m from bogie A and a boy of same mass from bogie B exchange their position by jumping in a direction normal to the track, then bogie A stops while B keeps moving in the same direction with new velocity vB. The initial velocities of bogie A and B are given by

A

M−mmvB,M−mMvB\frac{M - m}{m}v_{B},\frac{M - m}{M}v_{B}

B

mvB(M−m),MvB(M−m)\frac{mv_{B}}{(M - m)},\frac{Mv_{B}}{(M - m)}

C

mvB(M+m),MvB(M+m)\frac{mv_{B}}{(M + m)},\frac{Mv_{B}}{(M + m)}

D

(M−m)vBm,(M−m)vBM\frac{(M - m)v_{B}}{m},\frac{(M - m)v_{B}}{M}

Answer

mvB(M−m),MvB(M−m)\frac{mv_{B}}{(M - m)},\frac{Mv_{B}}{(M - m)}

Explanation

Solution

Since bogie A stops after exchange of positions of boys, we get

(M + mu)uA – MuA – MuB= 0

(Q boy in A carries away momentum and boy in B brings in – ve momentum)

or MuA = MuB

For bogie B

(M + m)uB ……….. (i)

For bogie B

(M + m)uB – muB – muA = (M + m)vB

or MuB – MuA = (M + m)vB ………… (ii) From (i) and (ii)

uA = mvBM−m\frac{mv_{B}}{M - m} and uB = mvBM−m\frac{mv_{B}}{M - m}