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 both the bullet and the gun after the bullet is fired. We know the total kinetic energy produced is 2050 J, and we can 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, \( m_b = 50 \, \text{g} = 0.05 \, \text{kg} \)
- Mass of the gun, \( m_g = 2 \, \text{kg} \)

2. **Conservation of Momentum:**
- Initially, the system (gun + bullet) is at rest, so the initial momentum is 0.
- After firing, let \( P_b \) be the momentum of the bullet and \( P_g \) be the momentum of the gun.
- According to the conservation of momentum:
\[ P_b + P_g = 0 \implies P_b = -P_g \]

3. **Express the momenta:**
- Let \( P \) be the magnitude of the momentum of the bullet and the gun. Thus, we can write:
\[ P_b = P \quad \text{and} \quad P_g = -P \]

4. **Kinetic Energy Expressions:**
- The kinetic energy of the bullet \( KE_b \) can be expressed as:
\[ KE_b = \frac{P_b^2}{2m_b} = \frac{P^2}{2 \times 0.05} \]
- The kinetic energy of the gun \( KE_g \) can be expressed as:
\[ KE_g = \frac{P_g^2}{2m_g} = \frac{P^2}{2 \times 2} \]

5. **Total Kinetic Energy:**
- The total kinetic energy produced is given as:
\[ KE_b + KE_g = 2050 \, \text{J} \]
- Substituting the expressions for \( KE_b \) and \( KE_g \):
\[ \frac{P^2}{2 \times 0.05} + \frac{P^2}{2 \times 2} = 2050 \]

6. **Simplifying the equation:**
- This simplifies to:
\[ \frac{P^2}{0.1} + \frac{P^2}{4} = 2050 \]
- Finding a common denominator (which is 0.1):
\[ 10P^2 + 0.025P^2 = 2050 \]
- Therefore:
\[ 10.025P^2 = 2050 \]

7. **Solving for \( P^2 \):**
- Rearranging gives:
\[ P^2 = \frac{2050}{10.025} \approx 204.8 \, \text{J} \]

8. **Finding Kinetic Energies:**
- Now we can find the kinetic energy of the bullet:
\[ KE_b = \frac{P^2}{2 \times 0.05} = \frac{204.8}{0.1} = 2048 \, \text{J} \]
- And for the gun:
\[ KE_g = \frac{P^2}{2 \times 2} = \frac{204.8}{4} = 51.2 \, \text{J} \]

### Final Answer:
- Kinetic energy of the bullet: **2048 J**
- Kinetic energy of the gun: **51.2 J**