A bomb of mass `9kg` explodes into two pieces of masses `3kg` and `6kg`. The velocity of mass `3kg` is `16ms^-1`. The kinetic energy of mass `6kg` is

Options

A. `96J`
B. `384J`
C. `192J`
D. `768J`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a bomb of mass 9 kg that explodes into two pieces: one piece has a mass of 3 kg and the other has a mass of 6 kg. The velocity of the 3 kg piece after the explosion is given as 16 m/s. We need to find the kinetic energy of the 6 kg piece.

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

Let:
- \( m_1 = 3 \, \text{kg} \) (mass of the first piece)
- \( v_1 = 16 \, \text{m/s} \) (velocity of the first piece)
- \( m_2 = 6 \, \text{kg} \) (mass of the second piece)
- \( v_2 \) = velocity of the second piece

Using the conservation of momentum:
\[ m_1 v_1 + m_2 v_2 = 0 \]
Substituting the known values:
\[ 3 \times 16 + 6 v_2 = 0 \]
\[ 48 + 6 v_2 = 0 \]
\[ 6 v_2 = -48 \]
\[ v_2 = -8 \, \text{m/s} \]

### Step 3: Calculate the kinetic energy of the 6 kg mass
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
Substituting the values for the 6 kg mass:
\[ KE_2 = \frac{1}{2} \times 6 \times (-8)^2 \]
Calculating \( (-8)^2 \):
\[ (-8)^2 = 64 \]
Now substituting back:
\[ KE_2 = \frac{1}{2} \times 6 \times 64 \]
\[ KE_2 = 3 \times 64 \]
\[ KE_2 = 192 \, \text{Joules} \]

### Final Answer
The kinetic energy of the 6 kg mass is **192 Joules**.

---