A 20 kg bomb at rest explodes into two pieces each of 14 kg and 6 kg. If the speed of the 14 kg piece is `"2 m s"^(-1)`, the kinetic energy of the 6 kg piece is

Options

A. 60.3 J
B. 55.4 J
C. 50.5 J
D. 65.3 J

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a bomb of mass 20 kg that explodes into two pieces: one piece of mass 14 kg and another piece of mass 6 kg. The speed of the 14 kg piece after the explosion is given as 2 m/s. We need to find the kinetic energy of the 6 kg piece.

### Step 2: Apply the law of conservation of momentum
Since the bomb was initially at rest, the total initial momentum is zero. After the explosion, the momentum of the two pieces must also sum to zero.

Let:
- Mass of the first piece (m1) = 14 kg
- Speed of the first piece (v1) = 2 m/s
- Mass of the second piece (m2) = 6 kg
- Speed of the second piece (v2) = ?

Using the conservation of momentum:
\[ m_1 v_1 + m_2 v_2 = 0 \]

Substituting the known values:
\[ 14 \times 2 + 6 \times v_2 = 0 \]

### Step 3: Solve for v2
Rearranging the equation gives:
\[ 6 v_2 = -28 \]
\[ v_2 = -\frac{28}{6} \]
\[ v_2 = -4.67 \, \text{m/s} \]

The negative sign indicates that the 6 kg piece moves in the opposite direction to the 14 kg piece.

### Step 4: Calculate the kinetic energy of the 6 kg piece
The kinetic energy (KE) is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]

For the 6 kg piece:
\[ KE_2 = \frac{1}{2} \times 6 \times (-4.67)^2 \]

Calculating this:
\[ KE_2 = \frac{1}{2} \times 6 \times 21.81 \]
\[ KE_2 = 3 \times 21.81 \]
\[ KE_2 = 65.43 \, \text{J} \]

### Final Answer
The kinetic energy of the 6 kg piece is approximately **65.43 J**.

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