A bullet of mass `m` hits a target of mass M hanging by a string and gets embedded 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. (a) `v=sqrt(2gh)`
B. (b) `v=sqrt(2gh)[1+m/M]`
C. (c) `v=sqrt(2gh)[1+M/m]`
D. (d) `v=sqrt(2gh)[1-m/M]`

Explanation

To solve the problem, we need to find the velocity of the bullet before the collision, given that it gets embedded in a target and the combined mass rises to a height \( h \). We will use the principles of conservation of momentum and energy.

### Step-by-Step Solution:

1. **Understand the System**:
- A bullet of mass \( m \) strikes a target of mass \( M \) that is hanging from a string. After the collision, the bullet embeds into the target, and they move together.

2. **Conservation of Momentum**:
- Before the collision, the momentum of the system is just due to the bullet, which is moving with velocity \( v \). The target is initially at rest.
- After the collision, the bullet and target move together with a combined mass of \( (M + m) \) and a velocity \( u \).
- By the conservation of momentum:
\[ m \cdot v = (M + m) \cdot u \]
- Rearranging gives:
\[ u = \frac{m \cdot v}{M + m} \]

3. **Energy Considerations**:
- After the collision, the combined mass rises to a height \( h \). At the maximum height, all kinetic energy is converted into potential energy.
- The potential energy gained is:
\[ PE = (M + m) \cdot g \cdot h \]
- The kinetic energy just after the collision is:
\[ KE = \frac{1}{2} (M + m) \cdot u^2 \]
- Setting the kinetic energy equal to the potential energy gives:
\[ \frac{1}{2} (M + m) \cdot u^2 = (M + m) \cdot g \cdot h \]
- We can cancel \( (M + m) \) from both sides (assuming \( M + m \neq 0 \)):
\[ \frac{1}{2} u^2 = g \cdot h \]
- Solving for \( u \):
\[ u^2 = 2gh \quad \Rightarrow \quad u = \sqrt{2gh} \]

4. **Substituting Back**:
- Now substitute \( u \) back into the momentum equation:
\[ \sqrt{2gh} = \frac{m \cdot v}{M + m} \]
- Rearranging to solve for \( v \):
\[ v = \frac{(M + m) \cdot \sqrt{2gh}}{m} \]

5. **Final Expression**:
- Thus, the velocity of the bullet before the collision is:
\[ v = \left(1 + \frac{M}{m}\right) \sqrt{2gh} \]