Two wooden blocks of mass $M_1$ and $M_2$ rest on a frictionless table. $A$ bullet of mass $m$ is fired at…

1. $\frac{2 m v}{M_1+M_2+m}$
2. $\frac{m v}{M_1+M_2+m}$
3. $\frac{\left(M_1+M_2+m\right) v}{M_1+M_2+m}$
4. $\frac{M_1+M_2}{M_1+M_2+m} v$

Solution

( ) 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 $v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right) u_1+\frac{2 m_2 u_2}{\left(m_1+m_2\right)}$ 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. $\begin{aligned} & m v=\left(m+M_1\right) v_1 \\ & \text { or } \quad v_1=\left(\frac{m v}{m+M_1}\right) \end{aligned}$ 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, $\begin{aligned} v_2 & =\left(\frac{\left(m+M_1\right)-M_2}{m+M_1+M_2}\right) v_1+0 \\ \text { and, } v_3 & =\left(\frac{M_2-\left(m+M_1\right)}{M_2+m+M_1} \times 0\right)+\left(\frac{2\left(m+M_1\right) v_1}{m+M_1+M_2}\right) \\ \Rightarrow \quad v_3 & =\frac{2\left(m+M_1\right)}{m+M_1+M_2} \cdot \frac{m v}{\left(m+M_1\right)} \\ & =\frac{2 m v}{m+M_1+M_2} \end{aligned}$ Note You can remember these standard and useful results for finding final velocities in case of elastic collisions.

Asked in: MHT CET Full Test 9

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