A bullet of mass m travelling with a speed v hits a block of mass M initially at rest and gets embedded in it. The combined system is free to move and there is no other force acting on the system. The heat generated in the process will be

Options

A. zero
B. `(mv^(2))/2`
C. `(Mmv^(2))/(2(M-m))`
D. `(mMv^(2))/(2(M+m))`

Answer

To solve the problem of finding the heat generated when a bullet of mass \( m \) traveling with speed \( v \) hits a block of mass \( M \) that is initially at rest and gets embedded in it, we can follow these steps:

### Step 1: Understand the System
We have a bullet of mass \( m \) moving with velocity \( v \) colliding with a stationary block of mass \( M \). After the collision, the bullet gets embedded in the block, and they move together with a common velocity \( V' \).

### Step 2: Apply Conservation of Momentum
Since there are no external forces acting on the system, we can apply the conservation of momentum principle:

\[ \text{Initial Momentum} = \text{Final Momentum} \]

The initial momentum before the collision is:

\[ p_{\text{initial}} = mv + 0 = mv \]

The final momentum after the collision (when both the bullet and block move together) is:

\[ p_{\text{final}} = (M + m)V' \]

Setting the initial momentum equal to the final momentum gives us:

\[ mv = (M + m)V' \]

### Step 3: Solve for the Final Velocity \( V' \)
From the equation above, we can solve for \( V' \):

\[ V' = \frac{mv}{M + m} \]

### Step 4: Calculate Initial and Final Kinetic Energy
Next, we need to calculate the initial and final kinetic energy of the system.

- **Initial Kinetic Energy** (\( KE_{\text{initial}} \)):
Only the bullet is moving initially, so:

\[ KE_{\text{initial}} = \frac{1}{2} mv^2 \]

- **Final Kinetic Energy** (\( KE_{\text{final}} \)):
After the collision, both the bullet and the block move together with velocity \( V' \):

\[ KE_{\text{final}} = \frac{1}{2} (M + m) (V')^2 \]

Substituting \( V' \) from Step 3:

\[ KE_{\text{final}} = \frac{1}{2} (M + m) \left(\frac{mv}{M + m}\right)^2 \]

### Step 5: Simplify the Final Kinetic Energy
Now, simplifying \( KE_{\text{final}} \):

\[ KE_{\text{final}} = \frac{1}{2} (M + m) \cdot \frac{m^2 v^2}{(M + m)^2} \]

This simplifies to:

\[ KE_{\text{final}} = \frac{1}{2} \cdot \frac{m^2 v^2}{M + m} \]

### Step 6: Calculate the Heat Generated
The heat generated (\( Q \)) in the process is equal to the change in kinetic energy:

\[ Q = KE_{\text{initial}} - KE_{\text{final}} \]

Substituting the values we calculated:

\[ Q = \frac{1}{2} mv^2 - \frac{1}{2} \cdot \frac{m^2 v^2}{M + m} \]

Factoring out \( \frac{1}{2} v^2 \):

\[ Q = \frac{1}{2} v^2 \left( m - \frac{m^2}{M + m} \right) \]

### Step 7: Simplify Further
To simplify the expression inside the parentheses:

\[ Q = \frac{1}{2} v^2 \left( \frac{m(M + m) - m^2}{M + m} \right) \]

This simplifies to:

\[ Q = \frac{1}{2} v^2 \left( \frac{mM}{M + m} \right) \]

### Final Answer
Thus, the heat generated in the process is:

\[ Q = \frac{1}{2} \frac{mM v^2}{M + m} \]