A bullet of mass `m` hits a target of mass M hanging by a string and gets embeded in it. If the block rises to a height h as a result of this collision, the velocity of the bullet before collision is

Options

A. `v=sqrt(2gh)`
B. `v=sqrt(2gh)[1+m/M]`
C. `v=sqrt(2gh)[1+M/m]`
D. `v=sqrt(2gh)[1-m/M]`

Explanation

To solve the problem, we need to analyze the situation using the principles of conservation of momentum and the work-energy theorem.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A bullet of mass \( m \) strikes a target of mass \( M \) that is hanging by a string.
- The bullet embeds itself in the target, and together they rise to a height \( h \).

2. **Conservation of Momentum**:
- Before the collision, the momentum of the system is given by the bullet's momentum since the target is at rest.
- Initial momentum \( P_{\text{initial}} = mv \) (where \( v \) is the velocity of the bullet before the collision).
- After the collision, the bullet and the target move together with a combined mass of \( m + M \) and a velocity \( U \).
- By conservation of momentum:
\[ mv = (m + M)U \]
- Rearranging gives:
\[ U = \frac{mv}{m + M} \]

3. **Work-Energy Theorem**:
- The work done by gravity when the block rises to height \( h \) is equal to the change in kinetic energy.
- The work done against gravity is:
\[ W = - (m + M)gh \]
- The change in kinetic energy is:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 0 - \frac{1}{2}(m + M)U^2 \]
- Setting the work done equal to the change in kinetic energy:
\[ - (m + M)gh = -\frac{1}{2}(m + M)U^2 \]
- Simplifying gives:
\[ (m + M)gh = \frac{1}{2}(m + M)U^2 \]

4. **Solving for \( U \)**:
- Cancel \( (m + M) \) from both sides (assuming \( m + M \neq 0 \)):
\[ 2gh = U^2 \]
- Thus:
\[ U = \sqrt{2gh} \]

5. **Substituting \( U \) Back**:
- Substitute \( U \) back into the momentum equation:
\[ U = \frac{mv}{m + M} \]
- Therefore:
\[ \sqrt{2gh} = \frac{mv}{m + M} \]

6. **Solving for \( v \)**:
- Rearranging gives:
\[ v = \frac{(m + M)\sqrt{2gh}}{m} \]
- This can be simplified to:
\[ v = \sqrt{2gh} \left(1 + \frac{M}{m}\right) \]

### Final Result:
The velocity of the bullet before the collision is:
\[ v = \sqrt{2gh} \left(1 + \frac{M}{m}\right) \]