A stationary bomb explode into two parts of masses 3kg and 1kg. The total KE of the two parts after explosioons is 2400J. The KE of the smaller part is

Options

A. 600J
B. 1800J
C. 1200J
D. 2160J

Explanation

To solve the problem, we need to find the kinetic energy of the smaller part (1 kg) after the explosion of the bomb. We know the following:

- Mass of the first part (m1) = 3 kg
- Mass of the second part (m2) = 1 kg
- Total kinetic energy (KE_total) after the explosion = 2400 J

### Step-by-step Solution:

1. **Conservation of Momentum**:
Since the bomb was initially stationary, the total momentum before the explosion was zero. Therefore, the total momentum after the explosion must also be zero.
\[ m_1 v_1 + m_2 v_2 = 0 \]
This implies:
\[ 3v_1 + 1v_2 = 0 \quad \Rightarrow \quad v_2 = -3v_1 \]

2. **Kinetic Energy Expression**:
The total kinetic energy after the explosion can be expressed as:
\[ KE_{total} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Substituting \(v_2 = -3v_1\):
\[ KE_{total} = \frac{1}{2} (3) v_1^2 + \frac{1}{2} (1)(-3v_1)^2 \]
Simplifying this:
\[ KE_{total} = \frac{3}{2} v_1^2 + \frac{1}{2} (9 v_1^2) = \frac{3}{2} v_1^2 + \frac{9}{2} v_1^2 = \frac{12}{2} v_1^2 = 6 v_1^2 \]

3. **Setting Up the Equation**:
Now we know that the total kinetic energy is 2400 J:
\[ 6 v_1^2 = 2400 \]
Solving for \(v_1^2\):
\[ v_1^2 = \frac{2400}{6} = 400 \]

4. **Finding Kinetic Energy of the Smaller Part**:
Now we can find the kinetic energy of the smaller part (1 kg):
\[ KE_{m2} = \frac{1}{2} m_2 v_2^2 = \frac{1}{2} (1)(-3v_1)^2 = \frac{1}{2} (1)(9v_1^2) = \frac{9}{2} v_1^2 \]
Substituting \(v_1^2 = 400\):
\[ KE_{m2} = \frac{9}{2} (400) = 1800 \, \text{J} \]

5. **Final Answer**:
The kinetic energy of the smaller part (1 kg) after the explosion is:
\[ \boxed{1800 \, \text{J}} \]