Two similar bogies A and B of same mass M(empty bogie) move with constant velocities $v_{A}$ and $v_{B}$ towards each other on smooth parallel tracks. At an instant a boy of mass m from bogie A and a boy 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 $v_{B}$ . The initial velocities of bogie A and B are given by :

\frac{M - m}{m} v_{B}, \frac{M - m}{M} v_{B}

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

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\frac{m v_{B}}{(M + m)}, \frac{m v_{B}}{(M + m)}

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

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Solution

The correct option is B $\frac{m v_{B}}{(M - m)}, \frac{M v_{B}}{(M - m)}$
Initially:
Momentum of bogie A is given by $(M + m) v_{A}$
Momentum of bogie B is given by $(M + m) u_{B}$
Considering change in momentum of Bogie A:
$M v_{A} + m v_{A} - m u_{B} - m v_{A} = 0$
$v_{A} = \frac{m u_{B}}{M}$

Considering change in momentum of bogie B:
$(M + m) u_{B} - m u_{B} - m v_{A} = (M + m) v_{B}$
$M u_{B} - m v_{A} = (M + m) v_{B}$

Replacing value of $v_{A}$ obtained earlier,
$M u_{B} - m v_{A} = (M + m) v_{B}$
$M u_{B} - \frac{m^{2}}{M} u_{B} = (M + m) v_{B}$
$u_{B} = \frac{M}{M - m} v_{B}$
Using $v_{A} = \frac{m u_{B}}{M}$ , $v_{A} = \frac{m}{M - m} v_{B}$

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