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(A) Step $1$: The bullet of mass $m$ strikes $M_1$ and gets embedded in it. This is a perfectly inelastic collision. By the law of conservation of lin

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$\frac{2 m v}{M_1+M_2+m}$

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(A) Step $1$: The bullet of mass $m$ strikes $M_1$ and gets embedded in it. This is a perfectly inelastic collision. By the law of conservation of linear momentum:$m v = (m + M_1) v_1$where $v_1$ is the velocity of the $(m + M_1)$ system after the first collision.$v_1 = \frac{m v}{m + M_1}$Step $2$: The system $(m + M_1)$ now moves with velocity $v_1$ and collides elastically with $M_2$,which is initially at rest $(u_2 = 0)$.For an elastic collision between two masses $m_A$ and $m_B$ where $m_B$ is initially at rest,the final velocity $v_B$ of mass $m_B$ is given by:$v_B = \frac{2 m_A}{m_A + m_B} u_A$Here,$m_A = (m + M_1)$,$m_B = M_2$,and $u_A = v_1$.Substituting these values:$v_2 = \frac{2(m + M_1)}{(m + M_1) + M_2} v_1$$v_2 = \frac{2(m + M_1)}{m + M_1 + M_2} \cdot \frac{m v}{m + M_1}$$v_2 = \frac{2 m v}{m + M_1 + M_2}$Thus,the velocity of $M_2$ after the collision is $\frac{2 m v}{M_1 + M_2 + m}$.

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$A$ small ball of mass $m$ starts at a point $A$ with speed $v_0$ and moves along a frictionless track $AB$ as shown. The track $BC$ has a coefficient of friction $\mu$. The ball comes to a stop at $C$ after traveling a distance $L$. The value of $L$ is:

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