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 these steps:

### 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 of 2:3, we can denote the masses of the smaller fragment (m1) and the larger fragment (m2) as follows:

Let:
- m1 = 2x (mass of the smaller fragment)
- m2 = 3x (mass of the larger fragment)

Since the total mass is 20 kg:
\[ m1 + m2 = 20 \, \text{kg} \]
\[ 2x + 3x = 20 \]
\[ 5x = 20 \]
\[ x = 4 \]

Now, substituting back to find the masses:
- m1 = 2x = 2(4) = 8 kg
- m2 = 3x = 3(4) = 12 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 of the two fragments must also sum to zero.

Let the velocity of the smaller fragment (m1) be \( v_1 = 6 \, \text{m/s} \) (given). The larger fragment (m2) will have a velocity \( v_2 \) in the opposite direction.

Using the conservation of momentum:
\[ m_1 v_1 + m_2 (-v_2) = 0 \]
Substituting the known values:
\[ 8 \cdot 6 - 12 \cdot v_2 = 0 \]
\[ 48 - 12 v_2 = 0 \]
Solving for \( v_2 \):
\[ 12 v_2 = 48 \]
\[ v_2 = 4 \, \text{m/s} \]

### Step 3: Calculate the kinetic energy of the larger fragment
The kinetic energy (KE) of the larger fragment (m2) can be calculated using the formula:
\[ KE = \frac{1}{2} m v^2 \]
Substituting the mass of the larger fragment and its velocity:
\[ KE = \frac{1}{2} \cdot 12 \cdot (4)^2 \]
\[ KE = \frac{1}{2} \cdot 12 \cdot 16 \]
\[ KE = 6 \cdot 16 \]
\[ KE = 96 \, \text{Joules} \]

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

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