A body of mass `1kg` initially at rest explodes and breaks into threee fragments of masses in the ration breaks into three fragments of masses in the ration `1: 1: 3`. The two pieces of equal masses fly off perpendicular to each other with a speed of `30m//s` each What is the velocity of heavier fragments ? .

Explanation

To solve the problem, we will apply the principle of conservation of momentum. Here are the steps to find the velocity of the heavier fragment:

### Step-by-Step Solution:

1. **Identify the Masses of the Fragments:**
The total mass of the body is 1 kg, and it breaks into three fragments in the ratio 1:1:3.
- Let the masses of the fragments be \( m_1 = \frac{1}{5} \) kg, \( m_2 = \frac{1}{5} \) kg, and \( m_3 = \frac{3}{5} \) kg.

2. **Determine the Initial Momentum:**
Since the body is initially at rest, the initial momentum \( P_{\text{initial}} \) is:
\[ P_{\text{initial}} = 0 \]

3. **Calculate the Momentum of the Two Equal Mass Fragments:**
The two fragments \( m_1 \) and \( m_2 \) move off perpendicular to each other with a speed of \( 30 \, \text{m/s} \).
- The momentum of \( m_1 \):
\[ P_1 = m_1 \cdot v_1 = \frac{1}{5} \cdot 30 = 6 \, \text{kg m/s} \]
- The momentum of \( m_2 \):
\[ P_2 = m_2 \cdot v_2 = \frac{1}{5} \cdot 30 = 6 \, \text{kg m/s} \]

4. **Determine the Resultant Momentum of the Two Fragments:**
Since \( P_1 \) and \( P_2 \) are perpendicular, we can find the resultant momentum \( P_{R} \) using the Pythagorean theorem:
\[ P_{R} = \sqrt{P_1^2 + P_2^2} = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \, \text{kg m/s} \]

5. **Apply Conservation of Momentum:**
The total momentum after the explosion must equal the total momentum before the explosion. Therefore, the momentum of the heavier fragment \( m_3 \) must balance the resultant momentum of the other two fragments:
\[ P_{3} = -P_{R} = -6\sqrt{2} \, \text{kg m/s} \]

6. **Calculate the Velocity of the Heavier Fragment:**
The momentum of the heavier fragment \( m_3 \) is given by:
\[ P_3 = m_3 \cdot v_3 \]
Substituting the mass of the heavier fragment:
\[ \frac{3}{5} \cdot v_3 = 6\sqrt{2} \]
Solving for \( v_3 \):
\[ v_3 = \frac{6\sqrt{2}}{\frac{3}{5}} = 6\sqrt{2} \cdot \frac{5}{3} = 10\sqrt{2} \, \text{m/s} \]

### Final Answer:
The velocity of the heavier fragment is \( 10\sqrt{2} \, \text{m/s} \).

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