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 collide at time $t_{0}$ . Their velocities become ${v^{'}}_{1}$ and ${v^{'}}_{2}$ at time $2 t_{0}$ while still moving in air. The value of $[(m_{1} {v^{'}}_{1} + m_{2} {v^{'}}_{2}) - (m_{1} v_{1} + m_{2} v_{2})]$

zero

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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 C $2 (m_{1} + m_{2}) g t_{0}$
External force = $F_{e x t} = \frac{Δ p}{Δ t} = ((m_{1} v_{1}^{^{'}} + m_{2} v_{2}^{^{'}}) - (m_{1} v_{1} + m_{2} v_{2})) / 2 t_{0} - 0$
$((m_{1} v_{1}^{^{'}} + m_{2} v_{2}^{^{'}}) - (m_{1} v_{1} + m_{2} v_{2}) = 2 t_{0} F_{e x t} = 2 t o (m_{1} + m_{2}) g$

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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