A shell of mass m is at rest initially. It explodes into three fragments having masses in the ratio `2:2: 1.` The fragments having equal masses fly off along mutually perpendicular directions with speed v. What will be the speed of the third (lighter) fragment ?

Options

A. v
B. `sqrt(2v)`
C. `2sqrt(2v)`
D. `3sqrt(2v)`

Answer

To solve the problem of the shell that explodes into three fragments, we will use the principle of conservation of linear momentum. Here’s the step-by-step solution:

### Step 1: Identify the Masses of the Fragments
The total mass of the shell is \( m \). The fragments are in the ratio \( 2:2:1 \). Let the masses of the fragments be:
- Fragment 1: \( 2x \)
- Fragment 2: \( 2x \)
- Fragment 3: \( x \)

Since the total mass is \( m \), we have:
\[ 2x + 2x + x = m \implies 5x = m \implies x = \frac{m}{5} \]
Thus, the masses of the fragments are:
- Fragment 1: \( \frac{2m}{5} \)
- Fragment 2: \( \frac{2m}{5} \)
- Fragment 3: \( \frac{m}{5} \)

### Step 2: Determine the Directions and Speeds of the Fragments
The two fragments with mass \( \frac{2m}{5} \) fly off at speed \( v \) along mutually perpendicular directions. We can assume:
- Fragment 1 moves along the x-axis.
- Fragment 2 moves along the y-axis.

### Step 3: Calculate the Momentum of the Fragments
The momentum of each fragment can be calculated as follows:
- Momentum of Fragment 1 (mass \( \frac{2m}{5} \), speed \( v \)):
\[ p_1 = \frac{2m}{5} \cdot v \]
- Momentum of Fragment 2 (mass \( \frac{2m}{5} \), speed \( v \)):
\[ p_2 = \frac{2m}{5} \cdot v \]

### Step 4: Use Conservation of Momentum
Since the shell was initially at rest, the total initial momentum is zero. Therefore, the total momentum after the explosion must also be zero. The momentum of the third fragment (mass \( \frac{m}{5} \), speed \( v_3 \)) must balance the momentum of the first two fragments.

The total momentum in the x-direction and y-direction must be zero:
\[ p_{total} = p_1 + p_2 + p_3 = 0 \]

### Step 5: Set Up the Equations
In the x-direction:
\[ \frac{2m}{5} v + 0 + p_{3x} = 0 \implies p_{3x} = -\frac{2m}{5} v \]
In the y-direction:
\[ 0 + \frac{2m}{5} v + p_{3y} = 0 \implies p_{3y} = -\frac{2m}{5} v \]

### Step 6: Calculate the Speed of the Third Fragment
The momentum of the third fragment can be expressed as:
\[ p_3 = \frac{m}{5} v_3 \]
Using the Pythagorean theorem to combine the x and y components:
\[ \sqrt{p_{3x}^2 + p_{3y}^2} = p_3 \]
Substituting the values:
\[ \sqrt{\left(-\frac{2m}{5} v\right)^2 + \left(-\frac{2m}{5} v\right)^2} = \frac{m}{5} v_3 \]
\[ \sqrt{2 \left(\frac{2m}{5} v\right)^2} = \frac{m}{5} v_3 \]
\[ \sqrt{2} \cdot \frac{2m}{5} v = \frac{m}{5} v_3 \]
Cancelling \( \frac{m}{5} \) from both sides:
\[ \sqrt{2} \cdot 2v = v_3 \]
Thus, the speed of the third fragment is:
\[ v_3 = 2\sqrt{2} v \]

### Final Answer
The speed of the third (lighter) fragment is \( 2\sqrt{2} v \).