Two persons of mass $m_{1} a n d m_{2}$ are standing at the two ends A and B respectively, of a stationary trolley of mass M and negligible height kept on smooth horizontal surface as shown. When the person standing at A jumps from the trolley towards left with $u_{r e l}$ with respect to the trolley, then

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The trolley rebounds with velocity $\frac{m_{1} u_{r e l}}{m_{1} + m_{2} + M}$

The trolley moves towards right

The centre of mass of the system remains stationary

The trolley rebounds with velocity $\frac{m_{1} u_{r e l}}{m_{2} + M}$

answer is A, B, D.

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Detailed Solution

$F_{e x t}$ on the system $(m_{1} + m_{2} + M)$ in horizontal direction is zero.
Centre of mass of the system remains stationary.
Again ${\vec{V}}_{m_{1} g} = {\vec{V}}_{m_{1} t} + {\vec{V}}_{t g} = (- u_{r e l} + v) \hat{i}$
${\vec{V}}_{(m_{2} + M) g} = v \hat{i}$
Conserve linear momentum in horizontal direction
$m_{1} (- u_{r e l} + v) \hat{i} + (m_{2} + M) v \hat{i} = 0 ∴ v = (\frac{m_{1} u_{r e l}}{m_{1} + m_{2} + M}) \hat{i}$