Two persons of mass m_1 and m_2 are standing | vedclass

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(D) Let the velocity of the trolley (along with the person at $B$) be $v$ towards the right after the person at $A$ jumps. The velocity of the person

Vedclass · Accepted Answer

all the above

Vedclass · Answer

(D) Let the velocity of the trolley (along with the person at $B$) be $v$ towards the right after the person at $A$ jumps. The velocity of the person at $A$ with respect to the ground is $v_1 = v - u_{rel}$ (towards the left). Since there is no external horizontal force on the system,the linear momentum of the system is conserved. Initial momentum $P_i = 0$. Final momentum $P_f = m_1(v - u_{rel}) + (m_2 + M)v = 0$. $m_1 v - m_1 u_{rel} + m_2 v + M v = 0$. $(m_1 + m_2 + M)v = m_1 u_{rel}$. $v = \frac{m_1 u_{rel}}{m_1 + m_2 + M}$. Since $v > 0$,the trolley moves towards the right. Also,since no external force acts on the system,the centre of mass 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 the person standing at $A$ jumps from the trolley towards the left with $u_{rel}$ with respect to the trolley,then:

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