Two particles of masses $m_{1}$ and $m_{2}$ in projectile motion have velocities ${¯ ¯ ¯ v}_{1}$ and ${¯ ¯ ¯ v}_{2}$ respectively at time $t = 0$ . They strike at time $t_{0}$ and their velocities become ${^{'}}_{1}$ and ${^{'}}_{2}$ at time $2 t_{0}$ while still moving in air. The value of
$∣ ∣ (m_{1} {^{'}}_{1} + m_{2} {^{'}}_{2}) - (m_{1} {¯ ¯ ¯ v}_{1} + m_{2} {¯ ¯ ¯ v}_{2}) ∣ ∣$ is :

z e r o

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(m_{1} + m_{2}) g t_{0}

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2 (m_{1} + m_{2}) g t_{0}

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\frac{1}{2} (m_{1} + m_{2}) g t_{0}

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Solution

The correct option is D $2 (m_{1} + m_{2}) g t_{0}$
The quantity to be found out is nothing but the change in momentum of the system containing the masses $m_{1} a n d m_{2}$ . Since the only external force acting on the system is gravitational force(impulsive collision force is internal when we consider the two masses as one system) which is constant, we just have to multiply the force by total time taken. Mass of the system is $m_{1} + m_{2}$ and time taken is $2 t_{0}$ , by impulse momentum theorem change in momentum is $2 (m_{1} + m_{2}) g t_{0}$

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Two particles of masses $m_{1} a n d m_{2}$ in projectile motion have velocities ${\to v}_{1} a n d {\to v}_{2}$ respectively at time t = 0 and that is the moment when they collide. Their velocities become ${\to v}_{1}^{'} a n d {\to v}_{2}^{'}$ at time $2 t_{0}$ while still moving in air. The value of magnitude of change in total momentum would be