A bullet of mass 50g is fired from a gun of mass 2kg. If the total kinetic energy produced is 2050J, the kinetic energy of the bullet and the gun respectively are

Options

A. 200 J, 5J
B. 2000J,50J
C. 5J,200J
D. 50J,2000J

Answer

To solve the problem, we need to find the kinetic energy of the bullet and the gun after the bullet is fired. We will use the principle of conservation of momentum and the relationship between kinetic energy and momentum.

### Step-by-Step Solution:

1. **Identify the masses**:
- Mass of the bullet, \( m_1 = 50 \, \text{g} = 0.05 \, \text{kg} \)
- Mass of the gun, \( m_2 = 2 \, \text{kg} \)

2. **Use the conservation of momentum**:
- Initially, both the gun and the bullet are at rest, so the initial momentum is \( 0 \).
- After firing, let the velocity of the bullet be \( v_1 \) and the velocity of the gun be \( v_2 \).
- According to the conservation of momentum:
\[ m_1 v_1 + m_2 (-v_2) = 0 \]
- This implies:
\[ m_1 v_1 = m_2 v_2 \]

3. **Relate kinetic energies**:
- The kinetic energy of the bullet \( K_1 \) and the gun \( K_2 \) can be expressed as:
\[ K_1 = \frac{1}{2} m_1 v_1^2 \]
\[ K_2 = \frac{1}{2} m_2 v_2^2 \]

4. **Express \( v_2 \) in terms of \( v_1 \)**:
- From the momentum equation, we can express \( v_2 \):
\[ v_2 = \frac{m_1}{m_2} v_1 \]

5. **Substitute \( v_2 \) into the kinetic energy equation for the gun**:
- Substitute \( v_2 \) into \( K_2 \):
\[ K_2 = \frac{1}{2} m_2 \left( \frac{m_1}{m_2} v_1 \right)^2 = \frac{1}{2} m_2 \frac{m_1^2}{m_2^2} v_1^2 = \frac{m_1^2}{2 m_2} v_1^2 \]

6. **Total kinetic energy**:
- The total kinetic energy produced is given as \( K_1 + K_2 = 2050 \, \text{J} \).
- Substitute \( K_1 \) and \( K_2 \):
\[ \frac{1}{2} m_1 v_1^2 + \frac{m_1^2}{2 m_2} v_1^2 = 2050 \]
- Factor out \( \frac{1}{2} v_1^2 \):
\[ \frac{1}{2} v_1^2 \left( m_1 + \frac{m_1^2}{m_2} \right) = 2050 \]

7. **Solve for \( v_1^2 \)**:
- Rearranging gives:
\[ v_1^2 = \frac{2050 \cdot 2}{m_1 + \frac{m_1^2}{m_2}} \]
- Substitute \( m_1 = 0.05 \, \text{kg} \) and \( m_2 = 2 \, \text{kg} \):
\[ v_1^2 = \frac{4100}{0.05 + \frac{(0.05)^2}{2}} = \frac{4100}{0.05 + 0.00125} = \frac{4100}{0.05125} \]

8. **Calculate \( K_1 \) and \( K_2 \)**:
- Calculate \( K_1 \) and \( K_2 \) using the ratios:
\[ \frac{K_1}{K_2} = \frac{m_2}{m_1} = \frac{2}{0.05} = 40 \]
- Let \( K_2 = x \) and \( K_1 = 40x \):
\[ 40x + x = 2050 \implies 41x = 2050 \implies x = \frac{2050}{41} = 50 \]
- Thus, \( K_2 = 50 \, \text{J} \) and \( K_1 = 2050 - 50 = 2000 \, \text{J} \).

### Final Answers:
- Kinetic energy of the bullet \( K_1 = 2000 \, \text{J} \)
- Kinetic energy of the gun \( K_2 = 50 \, \text{J} \)