A shell of mass `20 kg` at rest explodes into two fragments whose masses are in the ratio `2 : 3`. The smaller fragment moves with a velocity of `6 m//s`. The kinetic energy of the larger fragment is

Options

A. 96 J
B. 216 J
C. 144 J
D. 360 J

Answer

To solve the problem step by step, we will follow the logic of conservation of momentum and kinetic energy calculations.

### Step 1: Determine the Masses of the Fragments
Given that the total mass of the shell is 20 kg and the masses of the fragments are in the ratio 2:3, we can denote the masses as:
- Let the mass of the smaller fragment (m1) = 2x
- Let the mass of the larger fragment (m2) = 3x

From the problem, we know:
\[ m1 + m2 = 20 \]
\[ 2x + 3x = 20 \]
\[ 5x = 20 \]
\[ x = 4 \]

Now, substituting back to find the masses:
- \( m1 = 2x = 2 \times 4 = 8 \, \text{kg} \)
- \( m2 = 3x = 3 \times 4 = 12 \, \text{kg} \)

### Step 2: Use Conservation of Momentum
The initial momentum of the system is zero since the shell is at rest. After the explosion, the momentum must still equal zero:
\[ m1 \cdot v1 + m2 \cdot v2 = 0 \]
Where:
- \( v1 = 6 \, \text{m/s} \) (velocity of the smaller fragment)
- \( v2 \) is the velocity of the larger fragment.

Substituting the known values:
\[ 8 \cdot 6 + 12 \cdot v2 = 0 \]
\[ 48 + 12v2 = 0 \]
\[ 12v2 = -48 \]
\[ v2 = -4 \, \text{m/s} \]

The negative sign indicates that the larger fragment moves in the opposite direction.

### Step 3: Calculate the Kinetic Energy of the Larger Fragment
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
For the larger fragment (mass = 12 kg and velocity = 4 m/s):
\[ KE = \frac{1}{2} \cdot 12 \cdot (-4)^2 \]
\[ KE = \frac{1}{2} \cdot 12 \cdot 16 \]
\[ KE = 6 \cdot 16 \]
\[ KE = 96 \, \text{Joules} \]

### Final Answer
The kinetic energy of the larger fragment is **96 Joules**.

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