Two stones of masses m_1 and m_2 (such that m | vedclass

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(C) Let the first stone of mass $m_1$ be dropped at instant $t=0$.At time $t$,its velocity $v_1$ and displacement $s_1$ are:$v_1 = -gt$ and $s_1 = -\f

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$\Delta v$ remains constant with time and $\Delta s$ decreases with time

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(C) Let the first stone of mass $m_1$ be dropped at instant $t=0$.At time $t$,its velocity $v_1$ and displacement $s_1$ are:$v_1 = -gt$ and $s_1 = -\frac{1}{2}gt^2$.Since the second stone of mass $m_2$ is dropped $\Delta t$ time later,its velocity $v_2$ and displacement $s_2$ at instant $t$ are:$v_2 = -g(t - \Delta t)$ and $s_2 = -\frac{1}{2}g(t - \Delta t)^2$.The difference in speeds is $\Delta v = |v_1 - v_2| = |-gt - (-g(t - \Delta t))| = |-g\Delta t| = g\Delta t$.Since $g$ and $\Delta t$ are constants,$\Delta v$ remains constant with time.The mutual separation $\Delta s$ is given by $|s_1 - s_2|$:$\Delta s = |-\frac{1}{2}gt^2 - (-\frac{1}{2}g(t - \Delta t)^2)| = |\frac{1}{2}g((t - \Delta t)^2 - t^2)| = |\frac{1}{2}g(t^2 + \Delta t^2 - 2t\Delta t - t^2)| = |\frac{1}{2}g(\Delta t^2 - 2t\Delta t)|$.As time $t$ increases,the term $2t\Delta t$ increases,causing the magnitude of the separation $\Delta s$ to decrease.

Two stones of masses $m_1$ and $m_2$ (such that $m_1 > m_2$) are dropped $\Delta t$ time apart from the same height towards the ground. At a later time $t$,the difference in their speed is $\Delta v$ and their mutual separation is $\Delta s$. While both stones are in flight:

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