Two particles of mass $m_{1}$ and $m_{2}$ in projectile motion have velocities $_{1} <_{2}$ respectively at tine $t = 0$ . They collide at time $t_{0}$ . Their velocities become $\to v_{1}^{'}$ and $\to v_{2}^{'}$ at time $2 t_{0^{'}}$ while still moving in air. The value of $∣ ∣ (m_{1} \to v_{1}^{'} + m_{2} \to v_{2}^{'}) - (m_{1}_{1} + m_{2}_{2}) ∣ ∣$ is

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Solution

Consider $m_{1}$ and $m_{2}$ together as system.
Gravitational force, $(m_{1} + m_{2}) g$ , is the only external force acting on the system Apply Newton's second law,

$∣ ∣ {\to F}_{e x t} ∣ ∣ = (m_{1} + m_{2}) g = \frac{| Δ \to p |}{Δ t} = \frac{∣ ∣ ∣ (m_{1} \to v_{1}^{'} + m_{2} \to v_{2}^{'}) - (m_{1} \to v_{1} + m_{2} \to v_{2}) ∣ ∣ ∣}{2 t_{o} - 0}$

So, we get $∣ ∣ ∣ (m_{1} \to v_{1}^{'} + m_{2} \to v_{2}^{'}) - (m_{1} \to v_{1} + m_{2} \to v_{2}) ∣ ∣ ∣ = 2 (m_{1} + m_{2}) g t_{o}$

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