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. (a) `96J`
B. (b) `384J`
C. (c) `192J`
D. (d) `768J`

Explanation

To solve the problem step by step, we will use the principle of conservation of momentum and the formula for kinetic energy.

### Step 1: Understand the problem
A bomb of mass 9 kg explodes into two pieces: one piece of mass 3 kg and another piece of mass 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
Before the explosion, the total momentum of the bomb is zero because it is at rest. After the explosion, the total momentum must still equal zero. The equation for momentum conservation can be written as:

\[ m_1 v_1 + m_2 v_2 = 0 \]

Where:
- \( m_1 = 3 \) kg (mass of the first piece)
- \( v_1 = 16 \) m/s (velocity of the first piece)
- \( m_2 = 6 \) kg (mass of the second piece)
- \( v_2 \) = velocity of the second piece (unknown)

### Step 3: Set up the equation
Substituting the known values into the momentum equation:

\[ 3 \times 16 + 6 \times v_2 = 0 \]

### Step 4: Solve for \( v_2 \)
Rearranging the equation gives:

\[ 48 + 6 v_2 = 0 \]

\[ 6 v_2 = -48 \]

\[ v_2 = -8 \text{ m/s} \]

The negative sign indicates that the 6 kg piece moves in the opposite direction to the 3 kg piece.

### Step 5: Calculate the kinetic energy of the 6 kg piece
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 piece:

\[ KE = \frac{1}{2} \times 6 \times (-8)^2 \]

Calculating this gives:

\[ KE = \frac{1}{2} \times 6 \times 64 \]

\[ KE = 3 \times 64 \]

\[ KE = 192 \text{ Joules} \]

### Conclusion
The kinetic energy of the 6 kg piece is 192 Joules.

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