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Two particles of masses $m_1$ and $m_2$ in projectile motion have velocities $\vec u_1$ and $\vec u_2$ respectively at time $t=0$. They collide at time $t_0$. Their velocities become $\vec v_1$ and $\vec v_2$ at time $2t_0$ while still moving in air. The value of $$\big[(m_1\vec v_1+m_2\vec v_2)-(m_1\vec u_1+m_2\vec u_2)\big]$$ is: A. Zero B. $(m_1+m_2)gt_0$ C. $2(m_1+m_2)gt_0$ D. $\tfrac{1}{2}(m_1+m_2)gt_0$

1. Zero
2. (m_1+m_2)gt_0
3. 2(m_1+m_2)gt_0
4. 1/2(m_1+m_2)gt_0

Correct answer: 2(m_1+m_2)gt_0

Solution

The collision force is internal to the two-particle system, so it does not change the total momentum of the system. The only external force during flight is gravity, so the total change in momentum over time $2t_0$ is $(m_1+m_2)g(2t_0) = 2(m_1+m_2)gt_0$.

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