A bullet of mass m moving horizontally with a velocity v strikes a block of wood of mass M and gets embedded in the block. The block is suspended from the ceiling by a massless string. The height to which block rises is

Options

A. `v^2/(2g)(m/(M+m))^2`
B. `v^2/(2g)((M+m)/(m))^2`
C. `v^2/(2g)(m/M)^2`
D. `v^2/(2g)(M/m)^2`

Explanation

To solve the problem of a bullet embedding into a block of wood and determining the height to which the block rises, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**:
- We have a bullet of mass \( m \) moving with velocity \( v \).
- The bullet strikes a block of wood of mass \( M \) which is hanging from the ceiling by a massless string.
- After the collision, the bullet gets embedded in the block.

2. **Conservation of Momentum**:
- Before the collision, the momentum of the bullet is \( mv \) and the block is at rest (momentum = 0).
- After the collision, the combined mass of the bullet and block is \( (M + m) \) and let \( v_1 \) be their common velocity after the collision.
- According to the conservation of momentum:
\[ mv + 0 = (M + m)v_1 \]
- Rearranging gives:
\[ v_1 = \frac{mv}{M + m} \]

3. **Conservation of Energy**:
- At the moment just after the bullet embeds in the block, all kinetic energy is converted to potential energy at the maximum height \( h \).
- The total kinetic energy just after the collision is:
\[ KE = \frac{1}{2}(M + m)v_1^2 \]
- The potential energy at the height \( h \) is:
\[ PE = (M + m)gh \]
- Setting the kinetic energy equal to the potential energy gives:
\[ \frac{1}{2}(M + m)v_1^2 = (M + m)gh \]

4. **Solving for Height \( h \)**:
- We can cancel \( (M + m) \) from both sides (assuming \( M + m \neq 0 \)):
\[ \frac{1}{2}v_1^2 = gh \]
- Rearranging for \( h \):
\[ h = \frac{v_1^2}{2g} \]

5. **Substituting for \( v_1 \)**:
- Substitute \( v_1 = \frac{mv}{M + m} \) into the equation for \( h \):
\[ h = \frac{\left(\frac{mv}{M + m}\right)^2}{2g} \]
- Simplifying gives:
\[ h = \frac{m^2v^2}{(M + m)^2 \cdot 2g} \]

### Final Result:
The height to which the block rises is:
\[ h = \frac{m^2v^2}{2g(M + m)^2} \]