A bullet of mass `m` moving with velocity `v` strikes a block of mass `M` at rest and gets embedded into it. The kinetic energy of the composite block will be

Options

A. `(1)/(2)m Mu^(2)`
B. `(1)/(2)(m+M)u^(2)`
C. `(m^2v^(2))/(2(m+M))`
D. `((m+M)/(2mM))u^(2)`

Answer

To solve the problem, we need to find the kinetic energy of the composite block formed when a bullet of mass \( m \) moving with velocity \( v \) strikes a block of mass \( M \) at rest and gets embedded into it.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Mass of the bullet, \( m \)
- Velocity of the bullet, \( v \)
- Mass of the block, \( M \) (initially at rest)

2. **Apply the Law of Conservation of Momentum:**
The total momentum before the collision must equal the total momentum after the collision.
\[ \text{Initial Momentum} = \text{Final Momentum} \]
Before the collision, the momentum is:
\[ p_{\text{initial}} = mv + M \cdot 0 = mv \]
After the collision, the bullet gets embedded in the block, and they move together with a common velocity \( U \):
\[ p_{\text{final}} = (m + M)U \]
Setting these equal gives:
\[ mv = (m + M)U \]

3. **Solve for the Final Velocity \( U \):**
Rearranging the equation to solve for \( U \):
\[ U = \frac{mv}{m + M} \]

4. **Calculate the Kinetic Energy of the Composite Block:**
The kinetic energy \( KE \) of the composite block is given by:
\[ KE = \frac{1}{2} (m + M) U^2 \]
Substituting \( U \) from the previous step:
\[ KE = \frac{1}{2} (m + M) \left(\frac{mv}{m + M}\right)^2 \]

5. **Simplify the Expression:**
\[ KE = \frac{1}{2} (m + M) \cdot \frac{m^2 v^2}{(m + M)^2} \]
\[ KE = \frac{1}{2} \cdot \frac{m^2 v^2}{m + M} \]

6. **Final Result:**
The kinetic energy of the composite block after the collision is:
\[ KE = \frac{m^2 v^2}{2(m + M)} \]