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) mv^(2) xx (m)/((m + M))`
B. `(1)/(2) mv^(2) xx (M)/((m + M))`
C. `(1)/(2) mv^(2) xx ((M + m))/(M)`
D. `(1)/(2) mv^(2) ((M + m))/(m)`

Explanation

To find the kinetic energy of the composite block after a bullet of mass `m` strikes a block of mass `M` at rest and gets embedded into it, we can follow these steps:

### Step 1: Understand the system
- We have a bullet of mass `m` moving with velocity `v`.
- A block of mass `M` is at rest.
- After the collision, the bullet gets embedded in the block.

### Step 2: Apply the conservation of momentum
- The initial momentum of the system is the momentum of the bullet since the block is at rest.
- Initial momentum, \( p_i = mv + M \cdot 0 = mv \).
- After the collision, the bullet and block move together with a common velocity \( V \).
- The final momentum, \( p_f = (m + M)V \).

By conservation of momentum, we have:
\[ mv = (m + M)V \]

### Step 3: Solve for the final velocity \( V \)
- Rearranging the equation gives:
\[ V = \frac{mv}{m + M} \]

### Step 4: Calculate the kinetic energy of the composite system
- The kinetic energy \( KE \) of the composite block (bullet + block) after the collision is given by:
\[ KE = \frac{1}{2} (m + M) V^2 \]

### Step 5: Substitute \( V \) into the kinetic energy equation
- Substitute \( V = \frac{mv}{m + M} \) into the kinetic energy formula:
\[ KE = \frac{1}{2} (m + M) \left(\frac{mv}{m + M}\right)^2 \]

### Step 6: Simplify the kinetic energy expression
- Expanding this gives:
\[ KE = \frac{1}{2} (m + M) \cdot \frac{m^2 v^2}{(m + M)^2} \]
- Simplifying further:
\[ KE = \frac{1}{2} \cdot \frac{m^2 v^2}{m + M} \]

### Final Answer
Thus, the kinetic energy of the composite block after the collision is:
\[ KE = \frac{m^2 v^2}{2(m + M)} \]

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