Two persons of mass m_1 and m_2 are standing | vedclass

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

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None of these

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(D) The system consists of the trolley and the two persons. Since there is no external horizontal force acting on the system,the net external horizontal force is zero $(F_{ext, x} = 0)$.According to the law of conservation of linear momentum,the velocity of the centre of mass of the system remains constant. Since the system was initially at rest,the velocity of the centre of mass remains zero.Let the horizontal velocity of the person of mass $m_1$ be $v$ towards the left and the person of mass $m_2$ be $v$ towards the right. Let the velocity of the trolley be $V$ towards the right.Applying the conservation of linear momentum in the horizontal direction:$m_1(-v) + m_2(v) + MV = 0$$MV = m_1v - m_2v$$V = \frac{(m_1 - m_2)v}{M}$If $m_1 > m_2$,then $V > 0$,meaning the trolley moves towards the right (towards end $B$).If $m_2 > m_1$,then $V Thus,the trolley moves towards the end where the person with the lighter mass was standing,which is equivalent to saying it moves away from the heavier mass. Therefore,none of the given options 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 both the persons jump simultaneously with the same speed,then:

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