AP EAPCET 07-Jul-2022 Shift 2 Paper

For an elastic collision of 2 masses $m_1$ and $m_2$ with initial velocities $u_1$ and $u_2$, final velocities of masses after collision are $v_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) u_1 + \frac{2 m_2 u_2}{(m_1+m_2)}$ and $v_2 = \left(\frac{m_2 - m_1}{m_1 + m_2}\right) u_2 + \frac{2 m_1 u_2}{m_1 + m_2}$ These are obtained as standard results by applying conservation of energy and momentum equations. Now, in given case at first bullet is embedded in mass $M_1$. Let speed of bullet + Block system be $v_1$ after collision. Then by conservation of momentum we have. $m v = (m + M_1) v_1$ or $v_1 = \frac{m v}{m + M_1}$ Note Here we are not using standard result as first collision is perfectly inelastic. Now, bullet $+M_1$ strikes $M_2$ and it is given in problem that this collision is elastic. Now, by formula we have, $v_2 = \left(\frac{(m + M_1) - M_2}{m + M_1 + M_2}\right) v_1 + 0$ and $v_3 = \left(\frac{M_2 - (m + M_1)}{M_2 + m + M_1} \times 0\right) + \left(\frac{2 (m + M_1) v_1}{m + M_1 + M_2}\right)$ $\Rightarrow v_3 = \frac{2 (m + M_1)}{m + M_1 + M_2} \cdot \frac{m v}{(m + M_1)}$ $= \frac{2 m v}{m + M_1 + M_2}$ Note You can remember these standard and useful results for finding final velocities in case of elastic collisions.