In projectile motion,two particles of masses | vedclass

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(D) The external force acting on the system of two particles is the gravitational force: $F_{ext} = (m_1 + m_2)g$ (acting downwards).According to Newt

Vedclass · Accepted Answer

$2(m_1 + m_2) gt$

Vedclass · Answer

(D) The external force acting on the system of two particles is the gravitational force: $F_{ext} = (m_1 + m_2)g$ (acting downwards).According to Newton's second law,the rate of change of momentum is equal to the external force: $F_{ext} = \frac{\Delta P}{\Delta t}$.Here,$\Delta P = (m_1 \vec{V}_1^{\prime} + m_2 \vec{V}_2^{\prime}) - (m_1 \vec{V}_1 + m_2 \vec{V}_2)$ and $\Delta t = 2t - 0 = 2t$.Substituting these into the equation: $(m_1 + m_2)g = \frac{(m_1 \vec{V}_1^{\prime} + m_2 \vec{V}_2^{\prime}) - (m_1 \vec{V}_1 + m_2 \vec{V}_2)}{2t}$.Therefore,the change in momentum is: $[(m_1 \vec{V}_1^{\prime} + m_2 \vec{V}_2^{\prime}) - (m_1 \vec{V}_1 + m_2 \vec{V}_2)] = 2(m_1 + m_2)gt$.

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