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 step by step, we will use the principles of conservation of momentum and kinetic energy.

### Step 1: Understand the problem
We have a bomb that explodes into two parts with masses \( m_1 = 1 \, \text{kg} \) and \( m_2 = 3 \, \text{kg} \). The total kinetic energy after the explosion is given as \( 2400 \, \text{J} \). We need to find the kinetic energy of the smaller part (1 kg).

### Step 2: Conservation of Momentum
Since the bomb was initially stationary, the total initial momentum is zero. According to the conservation of momentum:
\[ m_1 v_1 + m_2 v_2 = 0 \]
Where \( v_1 \) is the velocity of the smaller mass (1 kg) and \( v_2 \) is the velocity of the larger mass (3 kg). Rearranging gives:
\[ v_1 = -\frac{m_2}{m_1} v_2 = -\frac{3}{1} v_2 = -3 v_2 \]

### Step 3: Kinetic Energy Equation
The total kinetic energy (KE) after the explosion is given by:
\[ KE_{\text{total}} = KE_1 + KE_2 \]
Where:
- \( KE_1 = \frac{1}{2} m_1 v_1^2 \)
- \( KE_2 = \frac{1}{2} m_2 v_2^2 \)

Substituting the values:
\[ KE_{\text{total}} = \frac{1}{2} (1) v_1^2 + \frac{1}{2} (3) v_2^2 = 2400 \, \text{J} \]

### Step 4: Substitute \( v_1 \) in terms of \( v_2 \)
Using \( v_1 = -3 v_2 \):
\[ KE_1 = \frac{1}{2} (1) (-3 v_2)^2 = \frac{1}{2} (1) (9 v_2^2) = \frac{9}{2} v_2^2 \]
Substituting this back into the kinetic energy equation:
\[ \frac{9}{2} v_2^2 + \frac{3}{2} v_2^2 = 2400 \]
Combining the terms:
\[ \frac{12}{2} v_2^2 = 2400 \]
\[ 6 v_2^2 = 2400 \]
\[ v_2^2 = 400 \]
\[ v_2 = 20 \, \text{m/s} \]

### Step 5: Find \( v_1 \)
Now substituting \( v_2 \) back to find \( v_1 \):
\[ v_1 = -3 v_2 = -3 \times 20 = -60 \, \text{m/s} \]

### Step 6: Calculate the Kinetic Energy of the Smaller Part
Now we can find the kinetic energy of the smaller part (1 kg):
\[ KE_1 = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} (1) (-60)^2 \]
\[ KE_1 = \frac{1}{2} (1) (3600) = 1800 \, \text{J} \]

### Final Answer
The kinetic energy of the smaller part is \( 1800 \, \text{J} \).

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