A bullet fired from a gun. As a result the gun recoils the mass of the gun is 100 times the mass of the bullet. The ratio of the kinetic energies of the bullet and the gun will be

Options

A. `10^4:1`
B. 100:1
C. `1/100`
D. 1:1

Answer

To solve the problem, we need to find the ratio of the kinetic energies of the bullet and the gun after the bullet is fired.

### Step-by-Step Solution:

1. **Define Variables:**
- Let the mass of the bullet be \( m \).
- Therefore, the mass of the gun is \( 100m \) (since the gun's mass is 100 times that of the bullet).

2. **Conservation of Momentum:**
- When the bullet is fired, the total momentum before firing is zero (as both the gun and bullet are at rest).
- After firing, the momentum of the bullet \( p_b \) and the momentum of the gun \( p_g \) must balance out:
\[ m v_b + 100m v_g = 0 \]
- Here, \( v_b \) is the velocity of the bullet and \( v_g \) is the velocity of the gun (recoil velocity).
- Rearranging gives us:
\[ v_g = -\frac{v_b}{100} \]

3. **Kinetic Energy of the Bullet:**
- The kinetic energy \( KE_b \) of the bullet is given by:
\[ KE_b = \frac{1}{2} m v_b^2 \]

4. **Kinetic Energy of the Gun:**
- The kinetic energy \( KE_g \) of the gun is given by:
\[ KE_g = \frac{1}{2} (100m) v_g^2 \]
- Substituting \( v_g = -\frac{v_b}{100} \):
\[ KE_g = \frac{1}{2} (100m) \left(-\frac{v_b}{100}\right)^2 = \frac{1}{2} (100m) \frac{v_b^2}{10000} = \frac{m v_b^2}{200} \]

5. **Ratio of Kinetic Energies:**
- Now, we find the ratio of the kinetic energies of the bullet and the gun:
\[ \frac{KE_b}{KE_g} = \frac{\frac{1}{2} m v_b^2}{\frac{m v_b^2}{200}} = \frac{\frac{1}{2}}{\frac{1}{200}} = \frac{200}{2} = 100 \]

6. **Final Result:**
- Therefore, the ratio of the kinetic energies of the bullet to the gun is:
\[ \text{Ratio} = 100 : 1 \]