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

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the masses of the fragments
The total mass of the body is given as 5 kg, and the masses of the fragments are in the ratio 1:1:3. Let the mass of the first two fragments be \( m \) each, and the mass of the heaviest fragment be \( 3m \).

\[ m + m + 3m = 5 \, \text{kg} \]
\[ 5m = 5 \, \text{kg} \implies m = 1 \, \text{kg} \]

Thus, the masses of the fragments are:
- Fragment 1: \( 1 \, \text{kg} \)
- Fragment 2: \( 1 \, \text{kg} \)
- Fragment 3 (heaviest): \( 3 \, \text{kg} \)

### Step 2: Analyze the motion of the fragments
The two fragments with mass \( 1 \, \text{kg} \) each move in mutually perpendicular directions with a speed of \( 21 \, \text{m/s} \). We can assume:
- Fragment 1 moves along the x-axis: \( \vec{V_1} = 21 \hat{i} \, \text{m/s} \)
- Fragment 2 moves along the y-axis: \( \vec{V_2} = 21 \hat{j} \, \text{m/s} \)

### Step 3: Apply conservation of momentum
Since the body was initially at rest, the total momentum before the explosion was zero. Therefore, the total momentum after the explosion must also be zero.

\[ \text{Total momentum} = m_1 \vec{V_1} + m_2 \vec{V_2} + m_3 \vec{V_3} = 0 \]

Substituting the values:
\[ 1 \cdot (21 \hat{i}) + 1 \cdot (21 \hat{j}) + 3 \cdot \vec{V_3} = 0 \]
\[ 21 \hat{i} + 21 \hat{j} + 3 \vec{V_3} = 0 \]

### Step 4: Solve for the velocity of the heaviest fragment
Rearranging the equation gives:
\[ 3 \vec{V_3} = - (21 \hat{i} + 21 \hat{j}) \]
\[ \vec{V_3} = -7 \hat{i} - 7 \hat{j} \, \text{m/s} \]

### Step 5: Calculate the magnitude of the velocity of the heaviest fragment
The magnitude of \( \vec{V_3} \) is given by:
\[ |\vec{V_3}| = \sqrt{(-7)^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2} \, \text{m/s} \]
Calculating \( 7\sqrt{2} \):
\[ 7\sqrt{2} \approx 7 \times 1.414 \approx 9.899 \, \text{m/s} \]

### Final Answer
The velocity of the heaviest fragment is approximately \( 9.89 \, \text{m/s} \).

---