Two persons of mass m_1 and m_2 are standing | vedclass

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(B) The system consists of the trolley and two persons. Since there is no external horizontal force acting on the system,the momentum of the system is

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the final velocity of the trolley is in the direction of the person who jumps first

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(B) The system consists of the trolley and two persons. Since there is no external horizontal force acting on the system,the momentum of the system is conserved. Initially,the system is at rest,so the initial momentum is $0$. When the first person (mass $m$) jumps with relative velocity $u_{rel}$ towards the left,the trolley (mass $M+m$) moves to the right with velocity $v_1 = \frac{m u_{rel}}{M+m}$. When the second person (mass $m$) jumps with relative velocity $u_{rel}$ towards the right,the trolley (mass $M$) moves to the left with velocity $v_2 = \frac{m u_{rel}}{M}$. The final velocity of the trolley is the vector sum of these velocities. Option $A$ is correct because the external force is zero,so the centre of mass remains stationary. Option $B$ is incorrect because the final velocity of the trolley depends on the masses and the direction of the jumps,and it is not necessarily in the direction of the first person. Option $C$ is also incorrect as the expression for final velocity is not $\left( {\frac{{m{u_{rel}}}}{{M + m}} - \frac{{m{u_{rel}}}}{{M + 2m}}} \right)$.

Two persons of mass $m_1$ and $m_2$ are standing at the two ends $A$ and $B$ respectively,of a trolley of mass $M$ as shown. Choose the incorrect statement,if $m_1 = m_2 = m$ and both the persons jump one by one,then

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