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(C) For a one-dimensional elastic collision between two masses $m_1$ and $m_2$, where $m_2$ is initially at rest, the velocity $v_2$ of the second mas

Vedclass · Accepted Answer

$\left(\frac{4}{3}\right)^{n-1} v_0$

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(C) For a one-dimensional elastic collision between two masses $m_1$ and $m_2$, where $m_2$ is initially at rest, the velocity $v_2$ of the second mass after the collision is given by $v_2 = \frac{2m_1}{m_1 + m_2} v_1$.1st collision: Mass $m_1 = m$ hits $m_2 = \frac{m}{2}$ with velocity $v_0$. The velocity of the second ball $v_1$ is:$v_1 = \frac{2m}{m + \frac{m}{2}} v_0 = \frac{2m}{\frac{3m}{2}} v_0 = \frac{4}{3} v_0$.2nd collision: Mass $m_2 = \frac{m}{2}$ hits $m_3 = \frac{m}{4}$ with velocity $v_1 = \frac{4}{3} v_0$. The velocity of the third ball $v_2$ is:$v_2 = \frac{2(\frac{m}{2})}{\frac{m}{2} + \frac{m}{4}} v_1 = \frac{m}{\frac{3m}{4}} v_1 = \frac{4}{3} v_1 = \left(\frac{4}{3}ight)^2 v_0$.Following this pattern, for the $n^{	ext{th}}$ ball, the velocity $v_{n-1}$ after $(n-1)$ collisions will be:$v_{n-1} = \left(\frac{4}{3}ight)^{n-1} v_0$.

There are $n$ elastic balls placed on a smooth horizontal plane. The masses of the balls are $m, \frac{m}{2}, \frac{m}{2^2}, \ldots, \frac{m}{2^{n-1}}$ respectively. If the first ball hits the second ball with velocity $v_0$, then the velocity of the $n^{\text{th}}$ ball will be,

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