A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of the heaviest fragment will be -

Options

A. 11.5 m/s
B. 14.0 m/s
C. 7.0 m/s
D. 9.89 m/s

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Determine the masses of the fragments
Given the mass of the body is 5 kg and the ratio of the masses of the fragments is 1:1:3, we can denote the masses as:
- Mass of fragment A = \( x \)
- Mass of fragment B = \( x \)
- Mass of fragment C = \( 3x \)

From the ratio, we have:
\[ x + x + 3x = 5 \quad \Rightarrow \quad 5x = 5 \quad \Rightarrow \quad x = 1 \text{ kg} \]

Thus, the masses of the fragments are:
- Mass of fragment A = 1 kg
- Mass of fragment B = 1 kg
- Mass of fragment C = 3 kg

### Step 2: Analyze the motion of the fragments
The two fragments with equal masses (1 kg each) are moving in mutually perpendicular directions with a speed of 21 m/s. We can assume:
- Fragment A moves along the x-axis (i-direction) with a velocity of \( 21 \, \text{m/s} \).
- Fragment B moves along the y-axis (j-direction) with a velocity of \( 21 \, \text{m/s} \).

### Step 3: Calculate the momentum of the fragments
The momentum of each fragment can be calculated as:
- Momentum of fragment A:
\[ p_A = m_A \cdot v_A = 1 \, \text{kg} \cdot 21 \, \text{m/s} = 21 \, \text{kg m/s} \, \text{i} \]
- Momentum of fragment B:
\[ p_B = m_B \cdot v_B = 1 \, \text{kg} \cdot 21 \, \text{m/s} = 21 \, \text{kg m/s} \, \text{j} \]

### Step 4: Apply conservation of momentum
Since the body was initially at rest, the total initial momentum was 0. Therefore, the momentum of the heaviest fragment (fragment C) must balance the momentum of fragments A and B:
\[ p_C + p_A + p_B = 0 \]
Let \( v_C \) be the velocity of fragment C. The momentum of fragment C is:
\[ p_C = m_C \cdot v_C = 3 \, \text{kg} \cdot v_C \]

Setting up the equation:
\[ 3v_C + 21 \, \text{i} + 21 \, \text{j} = 0 \]
This gives:
\[ 3v_C = -21 \, \text{i} - 21 \, \text{j} \]

### Step 5: Solve for the velocity of fragment C
Now, we can solve for \( v_C \):
\[ v_C = \frac{-21 \, \text{i} - 21 \, \text{j}}{3} = -7 \, \text{i} - 7 \, \text{j} \]

### Step 6: Calculate the magnitude of the velocity of fragment C
To find the magnitude of the velocity vector \( v_C \):
\[ |v_C| = \sqrt{(-7)^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2} \, \text{m/s} \]

### Final Answer
The magnitude of the velocity of the heaviest fragment is:
\[ |v_C| = 7\sqrt{2} \, \text{m/s} \approx 9.89 \, \text{m/s} \]