Two persons of mass m_1 and m_2 are standing | vedclass

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(D) Initially,the system (trolley + two persons) is at rest,so the initial momentum of the system is zero. Since no external horizontal force acts on

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all the above

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(D) Initially,the system (trolley + two persons) is at rest,so the initial momentum of the system is zero. Since no external horizontal force acts on the system,the total momentum of the system must remain conserved (i.e.,zero). When the person of mass $m_2$ jumps towards the right with a relative velocity $u_{rel}$ with respect to the trolley,let the velocity of the trolley (and the person $m_1$) be $v$ towards the left. The velocity of the person $m_2$ with respect to the ground is $v_{m2} = u_{rel} - v$ towards the right. Applying the conservation of linear momentum: $0 = (m_1 + M)(-v) + m_2(u_{rel} - v)$. $0 = -m_1 v - Mv + m_2 u_{rel} - m_2 v$. $m_2 u_{rel} = (m_1 + m_2 + M)v$. $v = \frac{m_2 u_{rel}}{m_1 + m_2 + M}$. Since $v$ is towards the left,the trolley moves towards the left. Also,as there is no external force,the centre of mass of the system remains stationary. Thus,all the given statements are correct.

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. When only the person standing at $B$ jumps from the trolley towards the right while the person at $A$ keeps standing,then

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