Two particles of masses m_1 and m_2 in projec | vedclass

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(C) The initial momentum of the two-particle system at $t = 0$ is $\vec{P}_i = m_1\vec{v}_1 + m_2\vec{v}_2$.Collision between the two particles is an

Vedclass · Accepted Answer

$2(m_1 + m_2)gt_0$

Vedclass · Answer

(C) The initial momentum of the two-particle system at $t = 0$ is $\vec{P}_i = m_1\vec{v}_1 + m_2\vec{v}_2$.Collision between the two particles is an internal interaction and does not affect the total momentum of the system.The only external force acting on the system is gravity,which is $\vec{F}_{ext} = (m_1 + m_2)g$ acting downwards.According to the impulse-momentum theorem,the change in momentum $\Delta \vec{P}$ is equal to the impulse of the external force: $\Delta \vec{P} = \int_{0}^{2t_0} \vec{F}_{ext} dt$.Since the force is constant,$\Delta \vec{P} = (m_1 + m_2)g 	imes (2t_0 - 0) = 2(m_1 + m_2)gt_0$.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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