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A body of mass 1kg initially at rest exp...

A body of mass `1kg` initially at rest explodes and breaks into threee fragments of masses in the ratio `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 ? .

A

`(10)/(sqrt(2))m//s`

B

`10sqrt(2)m//s`

C

`20sqrt(2)m//s`

D

`30sqrt(2)m//s`

Text Solution

AI Generated Solution

The correct Answer is:
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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