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]
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]
[Ignore any energy loss and treat the problem to be one dimensional]
Show 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.
Updated On: Jun 26, 2026
- \(\frac{2mv}{M_1+M_2+m}\)
- \(\frac{mv}{M_1+M_2+m}\)
- \(\frac{(M_1+M_2+m)v}{M_1+M_2+m}\)
- \(\frac{M_1+M_2}{M_1+M_2+m}v\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Apply conservation of momentum for bullet and block \(M_1\).
Initially, bullet has momentum \[ mv. \] Block \(M_1\) is initially at rest.
After the bullet embeds in \(M_1\), their combined mass is \[ M_1+m. \] Let their common velocity be \(u\).
Using conservation of momentum, \[ mv=(M_1+m)u. \] Therefore, \[ u=\frac{mv}{M_1+m}. \]
Step 2: Collision of \((M_1+m)\) with \(M_2\).
Now the combined body of mass \[ M_1+m \] moving with velocity \[ u \] collides with block \(M_2\), which is initially at rest.
For one-dimensional elastic collision, when a body of mass \(m_1\) moving with velocity \(u\) collides with a stationary body of mass \(m_2\), the velocity of the second body after collision is \[ v_2=\frac{2m_1u}{m_1+m_2}. \] Here, \[ m_1=M_1+m \] and \[ m_2=M_2. \] So, \[ v_2=\frac{2(M_1+m)u}{M_1+m+M_2}. \]
Step 3: Substitute the value of \(u\).
Since \[ u=\frac{mv}{M_1+m}, \] we get \[ v_2=\frac{2(M_1+m)}{M_1+m+M_2}\cdot \frac{mv}{M_1+m}. \] Canceling \(M_1+m\), \[ v_2=\frac{2mv}{M_1+M_2+m}. \]
Step 4: Final conclusion.
Hence, the velocity of \(M_2\) after collision is \[ \boxed{\frac{2mv}{M_1+M_2+m}} \] Therefore, the correct option is \[ \boxed{(1)} \]
Initially, bullet has momentum \[ mv. \] Block \(M_1\) is initially at rest.
After the bullet embeds in \(M_1\), their combined mass is \[ M_1+m. \] Let their common velocity be \(u\).
Using conservation of momentum, \[ mv=(M_1+m)u. \] Therefore, \[ u=\frac{mv}{M_1+m}. \]
Step 2: Collision of \((M_1+m)\) with \(M_2\).
Now the combined body of mass \[ M_1+m \] moving with velocity \[ u \] collides with block \(M_2\), which is initially at rest.
For one-dimensional elastic collision, when a body of mass \(m_1\) moving with velocity \(u\) collides with a stationary body of mass \(m_2\), the velocity of the second body after collision is \[ v_2=\frac{2m_1u}{m_1+m_2}. \] Here, \[ m_1=M_1+m \] and \[ m_2=M_2. \] So, \[ v_2=\frac{2(M_1+m)u}{M_1+m+M_2}. \]
Step 3: Substitute the value of \(u\).
Since \[ u=\frac{mv}{M_1+m}, \] we get \[ v_2=\frac{2(M_1+m)}{M_1+m+M_2}\cdot \frac{mv}{M_1+m}. \] Canceling \(M_1+m\), \[ v_2=\frac{2mv}{M_1+M_2+m}. \]
Step 4: Final conclusion.
Hence, the velocity of \(M_2\) after collision is \[ \boxed{\frac{2mv}{M_1+M_2+m}} \] Therefore, the correct option is \[ \boxed{(1)} \]
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