A bullet of mass m is fired into a block of wood of mass M which hangs on the end of pendulum and gets embedded into it. When the bullet strikes with maximum rise R. Then, the velocity of the bullet is given by

Options

A. `(M)/(m + M)(sqrt(2gR)`
B. `(M+m)/(m) sqrt(2gR)`
C. `(M/m)sqrt(2gR)`
D. None of these

Explanation

To solve the problem of finding the velocity of the bullet when it strikes the block of wood and gets embedded into it, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**:
- We have a bullet of mass \( m \) that is fired into a block of wood of mass \( M \). The block is hanging from a pendulum.

2. **Conservation of Momentum**:
- When the bullet embeds itself into the block, we can use the principle of conservation of momentum. The momentum before the collision must equal the momentum after the collision.
- Let \( v \) be the velocity of the bullet just before the collision, and \( V \) be the velocity of the combined system (bullet + block) just after the collision.
- The equation for conservation of momentum is:
\[ mv = (M + m)V \]

3. **Maximum Rise of the Pendulum**:
- When the bullet embeds into the block, the system will rise to a maximum height \( R \). At this point, all the kinetic energy is converted into potential energy.
- The potential energy at height \( R \) is given by:
\[ PE = (M + m)gR \]
- The kinetic energy just after the collision is:
\[ KE = \frac{1}{2}(M + m)V^2 \]

4. **Equating Kinetic and Potential Energy**:
- At the maximum height, the kinetic energy is equal to the potential energy:
\[ \frac{1}{2}(M + m)V^2 = (M + m)gR \]
- We can simplify this equation by dividing both sides by \( (M + m) \):
\[ \frac{1}{2}V^2 = gR \]
- Rearranging gives:
\[ V^2 = 2gR \]
- Therefore, the velocity \( V \) just after the collision is:
\[ V = \sqrt{2gR} \]

5. **Substituting Back to Find Bullet's Velocity**:
- Now, substitute \( V \) back into the momentum conservation equation:
\[ mv = (M + m)\sqrt{2gR} \]
- Solving for \( v \) gives:
\[ v = \frac{(M + m)\sqrt{2gR}}{m} \]

### Final Answer:
The velocity of the bullet just before it strikes the block is:
\[ v = \frac{(M + m)\sqrt{2gR}}{m} \]