A bomb of mass 30 kg at rest explodes in to two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is `6ms^(-1)`. The kinetic energy of the outer mass is

Options

A. 324 J
B. 486 J
C. 256 J
D. 5245 J

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a bomb of mass 30 kg that explodes into two pieces: one piece has a mass of 18 kg and the other has a mass of 12 kg. The 18 kg piece moves with a velocity of 6 m/s. We need to find the kinetic energy of the 12 kg piece.

### Step 2: Apply the conservation of momentum
Since the bomb was initially at rest, the total initial momentum is zero. According to the law of conservation of momentum, the total momentum before the explosion must equal the total momentum after the explosion.

Let:
- \( m_1 = 18 \, \text{kg} \) (mass of the first piece)
- \( v_1 = 6 \, \text{m/s} \) (velocity of the first piece)
- \( m_2 = 12 \, \text{kg} \) (mass of the second piece)
- \( v_2 \) = velocity of the second piece (unknown)

The equation for conservation of momentum is:
\[ m_1 v_1 + m_2 v_2 = 0 \]

Substituting the known values:
\[ 18 \times 6 + 12 \times v_2 = 0 \]

### Step 3: Solve for \( v_2 \)
Now we can solve for \( v_2 \):
\[ 108 + 12 v_2 = 0 \]
\[ 12 v_2 = -108 \]
\[ v_2 = -9 \, \text{m/s} \]

### Step 4: Calculate the kinetic energy of the second piece
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]

For the second piece (12 kg mass):
\[ KE_2 = \frac{1}{2} \times 12 \times (-9)^2 \]
\[ KE_2 = \frac{1}{2} \times 12 \times 81 \]
\[ KE_2 = 6 \times 81 \]
\[ KE_2 = 486 \, \text{Joules} \]

### Final Answer
The kinetic energy of the 12 kg mass is **486 Joules**.

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