Two particles of masses m_1 and m_2 in projec | vedclass

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(C) The total momentum of the system at time $t = 0$ is $\vec{P}_i = m_1 \vec{v}_1 + m_2 \vec{v}_2$.The collision between the two particles is an inte

Vedclass · Accepted Answer

$2(m_1 + m_2)gt_0$

Vedclass · Answer

(C) The total momentum of the system at time $t = 0$ is $\vec{P}_i = m_1 \vec{v}_1 + m_2 \vec{v}_2$.The collision between the two particles is an internal interaction and does not change the total momentum of the system.The only external force acting on the system throughout the motion is the gravitational force,which is $\vec{F}_{ext} = (m_1 + m_2)g$ acting downwards.According to the impulse-momentum theorem,the change in momentum is equal to the impulse of the external force: $\Delta \vec{P} = \int_{t_1}^{t_2} \vec{F}_{ext} dt$.Here,the time interval is from $t = 0$ to $t = 2t_0$,so $\Delta t = 2t_0$.Since the force is constant,the impulse is $\vec{J} = \vec{F}_{ext} 	imes \Delta t = (m_1 + m_2)g 	imes 2t_0$.Therefore,the magnitude of the change in momentum is $|(m_1 \vec{v}_1' + m_2 \vec{v}_2') - (m_1 \vec{v}_1 + m_2 \vec{v}_2)| = 2(m_1 + m_2)gt_0$.

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