To solve the problem step by step, we will use the principles of conservation of momentum.
### Step 1: Determine the masses of the fragments
The total mass of the body is 1 kg. The fragments are in the ratio 1:1:3.
Let the masses of the fragments be:
- \( m_1 = x \)
- \( m_2 = x \)
- \( m_3 = 3x \)
From the equation:
\[ m_1 + m_2 + m_3 = 1 \, \text{kg} \]
\[ x + x + 3x = 1 \]
\[ 5x = 1 \]
\[ x = \frac{1}{5} \, \text{kg} \]
Thus, the masses are:
- \( m_1 = \frac{1}{5} \, \text{kg} = 0.2 \, \text{kg} \)
- \( m_2 = \frac{1}{5} \, \text{kg} = 0.2 \, \text{kg} \)
- \( m_3 = \frac{3}{5} \, \text{kg} = 0.6 \, \text{kg} \)
### Step 2: Analyze the motion of the fragments
The two lighter fragments (masses \( m_1 \) and \( m_2 \)) fly off perpendicular to each other with a speed of \( 30 \, \text{m/s} \).
Let:
- \( v_1 = 30 \, \text{m/s} \) (for \( m_1 \))
- \( v_2 = 30 \, \text{m/s} \) (for \( m_2 \))
### Step 3: Apply conservation of momentum
Since the body was initially at rest, the total initial momentum is zero. Therefore, the total final momentum must also be zero.
In the x-direction:
\[ 0 = m_1 v_1 + m_3 v_3 \]
In the y-direction:
\[ 0 = m_2 v_2 - m_1 v_1 \]
### Step 4: Set up the equations
For the x-direction:
\[ m_3 v_3 = - (m_1 v_1 + m_2 v_2) \]
For the y-direction:
\[ m_2 v_2 = m_1 v_1 \]
### Step 5: Substitute known values
From the y-direction equation:
\[ 0.2 \cdot 30 = 0.2 \cdot 30 \]
This equation holds true, confirming the momentum in the y-direction is balanced.
Now substitute \( m_1 \), \( m_2 \), and \( v_1 \) into the x-direction equation:
\[ 0.6 v_3 = - (0.2 \cdot 30 + 0.2 \cdot 30) \]
\[ 0.6 v_3 = - (6 + 6) \]
\[ 0.6 v_3 = -12 \]
### Step 6: Solve for \( v_3 \)
\[ v_3 = \frac{-12}{0.6} \]
\[ v_3 = -20 \, \text{m/s} \]
### Step 7: Interpret the result
The negative sign indicates that the direction of the heavier fragment \( m_3 \) is opposite to the resultant direction of the lighter fragments.
### Final Answer
The velocity of the heavier fragment \( m_3 \) is \( 20 \, \text{m/s} \) in the opposite direction.
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