Question

Two wooden blocks of mass \(M_1\) and \(M_2\) rest on a frictionless table. A bullet of mass \(m\) is fired at \(M_1\) with speed \(v\), which embeds in it, and the two together finally collide with \(M_2\). Find the velocity of \(M_2\) after collision.
[Ignore any energy loss and treat the problem to be one dimensional]

• A. \(\frac{2mv}{M_1+M_2+m}\)
• B. \(\frac{mv}{M_1+M_2+m}\)
• C. \(\frac{(M_1+M_2+m)v}{M_1+M_2+m}\)
• D. \(\frac{M_1+M_2}{M_1+M_2+m}v\)

Hint:

For a one-dimensional elastic collision where a mass \(m_1\) moving with speed \(u\) strikes a stationary mass \(m_2\), the velocity of \(m_2\) after collision is \[ v_2=\frac{2m_1u}{m_1+m_2}. \] First apply momentum conservation for embedding, then use elastic collision formula.

Answer 1

The Correct Option is A

Step 1: Bullet embeds in M1 — momentum conservation.
\( mv = (m + M_1)V_1 \Rightarrow V_1 = \frac{mv}{m+M_1} \)

Step 2: Elastic collision of \((m+M_1)\) with \(M_2\) — velocity of M2.
For elastic collision, velocity of target: \( V_2 = \frac{2(m+M_1)}{(m+M_1)+M_2}\cdot V_1 = \frac{2(m+M_1)}{m+M_1+M_2}\cdot\frac{mv}{m+M_1} = \frac{2mv}{m+M_1+M_2} \)

\[ \boxed{V_2 = \frac{2mv}{m+M_1+M_2}} \]