A 12 kg bomb at rest explodes into two pieces of 4 kg and 8 kg. If the momentum of 4 kg piece is 20 Ns, the kinetic energy of the 8 kg piece is

Options

A. 25 J
B. 20 J
C. 50 J
D. 40 J

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the conservation of momentum
The bomb is initially at rest, so the total momentum before the explosion is zero. After the explosion, the momentum of the two pieces must also sum to zero.

### Step 2: Set up the momentum equation
Let:
- \( m_1 = 4 \, \text{kg} \) (mass of the first piece)
- \( m_2 = 8 \, \text{kg} \) (mass of the second piece)
- \( v_1 \) = velocity of the 4 kg piece
- \( v_2 \) = velocity of the 8 kg piece

According to the problem, the momentum of the 4 kg piece is given as \( 20 \, \text{Ns} \). Therefore, we can write:
\[ m_1 v_1 = 20 \, \text{Ns} \]
\[ 4 v_1 = 20 \]
From this, we can find \( v_1 \):
\[ v_1 = \frac{20}{4} = 5 \, \text{m/s} \]

### Step 3: Use conservation of momentum
Since the total momentum must be zero:
\[ m_1 v_1 + m_2 v_2 = 0 \]
Substituting the values we have:
\[ 4 \cdot 5 + 8 v_2 = 0 \]
\[ 20 + 8 v_2 = 0 \]
Solving for \( v_2 \):
\[ 8 v_2 = -20 \]
\[ v_2 = -\frac{20}{8} = -2.5 \, \text{m/s} \]

### Step 4: Calculate the kinetic energy of the 8 kg piece
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
For the 8 kg piece:
\[ KE_2 = \frac{1}{2} \cdot 8 \cdot (-2.5)^2 \]
Calculating \( (-2.5)^2 \):
\[ (-2.5)^2 = 6.25 \]
Now substituting back into the kinetic energy formula:
\[ KE_2 = \frac{1}{2} \cdot 8 \cdot 6.25 \]
\[ KE_2 = 4 \cdot 6.25 = 25 \, \text{J} \]

### Final Answer
The kinetic energy of the 8 kg piece is \( 25 \, \text{J} \).

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