Two persons of mass m_1 and m_2 are standing | vedclass

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(A) Let the velocity of the trolley be $v$ in the direction of the person with mass $m_1$. Since the system is initially at rest,the total external fo

Vedclass · Accepted Answer

$\frac{{|{m_1} - {m_2}|{u_{rel}}}}{{{m_1} + {m_2} + M}}$

Vedclass · Answer

(A) Let the velocity of the trolley be $v$ in the direction of the person with mass $m_1$. Since the system is initially at rest,the total external force on the system (trolley + two persons) is zero. Thus,the linear momentum of the system is conserved. Let the velocity of the trolley be $v$ (towards the right). The velocity of person $m_1$ with respect to the ground is $v - u_{rel}$ (assuming they jump towards the left). The velocity of person $m_2$ with respect to the ground is $v + u_{rel}$ (assuming they jump towards the right). Applying the law of conservation of linear momentum: $0 = Mv + m_1(v - u_{rel}) + m_2(v + u_{rel})$ $0 = Mv + m_1v - m_1u_{rel} + m_2v + m_2u_{rel}$ $0 = (M + m_1 + m_2)v + (m_2 - m_1)u_{rel}$ $(M + m_1 + m_2)v = (m_1 - m_2)u_{rel}$ $v = \frac{(m_1 - m_2)u_{rel}}{M + m_1 + m_2}$ Taking the magnitude,the velocity of the trolley is $\frac{|m_1 - m_2|u_{rel}}{M + m_1 + m_2}$.

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 $u_{rel}$ with respect to the trolley,then the velocity of the trolley is

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