The kinetic energy of a body of mass 'm' moving with a speed 'v' is given by `(1)/(2)mv^(2)` . A bullet fired from gun moves with a speed of `50 m s^(-1)` strikes target. If 50 per cent of the kinetic energy is absorbed by the bullet as heat energy, calculate the change in temperature of the bullet. The specific heat capacity of the material of the bullet is `42 J kg ^(-1) ""^(@)C^(-1)`

Options

A. `11.35^(@)C`
B. `14.8^(@)C`
C. `22.5^(@)C`
D. `44^(@)C`

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Calculate the Kinetic Energy (KE) of the bullet
The formula for kinetic energy is given by:

\[ KE = \frac{1}{2} mv^2 \]

Where:
- \( m \) is the mass of the bullet (which we will keep as \( m \) for now since it will cancel out later),
- \( v \) is the speed of the bullet, which is given as \( 50 \, \text{m/s} \).

Substituting the value of \( v \):

\[ KE = \frac{1}{2} m (50)^2 = \frac{1}{2} m (2500) = 1250m \, \text{J} \]

### Step 2: Determine the heat energy absorbed by the bullet
According to the problem, 50% of the kinetic energy is absorbed as heat energy. Therefore, the heat energy \( Q \) absorbed by the bullet is:

\[ Q = 0.5 \times KE = 0.5 \times 1250m = 625m \, \text{J} \]

### Step 3: Relate heat energy to temperature change
The heat energy absorbed by the bullet can also be expressed using the specific heat capacity formula:

\[ Q = m \cdot S \cdot \Delta T \]

Where:
- \( S \) is the specific heat capacity of the bullet, given as \( 42 \, \text{J/(kg} \cdot {}^\circ C) \),
- \( \Delta T \) is the change in temperature.

### Step 4: Set the equations equal to each other
We can set the two expressions for \( Q \) equal to each other:

\[ 625m = m \cdot 42 \cdot \Delta T \]

### Step 5: Cancel \( m \) from both sides
Since \( m \) is present on both sides of the equation, we can cancel it out (assuming \( m \neq 0 \)):

\[ 625 = 42 \cdot \Delta T \]

### Step 6: Solve for \( \Delta T \)
Now, we can solve for \( \Delta T \):

\[ \Delta T = \frac{625}{42} \]

Calculating this gives:

\[ \Delta T \approx 14.88 \, {}^\circ C \]

### Final Answer
Thus, the change in temperature of the bullet is approximately:

\[ \Delta T \approx 14.88 \, {}^\circ C \]