Question

A body of mass \(M\) at rest explodes into three pieces, in the ratio of masses 1:1:2. Two smaller pieces fly off perpendicular to each other with velocities of 30 m/s and 40 m/s respectively. The velocity of the third piece will be:

• A. 15 m/s
• B. 25 m/s
• C. 35 m/s
• D. 50 m/s

Hint:

- When solving explosion problems, apply the conservation of momentum in both the horizontal and vertical directions.
- Use vector addition to find the resultant velocity.
- The negative signs indicate opposite directions but do not affect magnitude.

Answer 1

The Correct Option is B

Step 1: Applying the Law of Conservation of Momentum 
Since the body is initially at rest, the total momentum before the explosion is zero. The total momentum after the explosion must also be zero to satisfy the law of conservation of momentum. 
The masses of the pieces are in the ratio \(1:1:2\), so we can denote the masses of the pieces as \(m_1 = m\), \(m_2 = m\), and \(m_3 = 2m\). 

\( \text{Total momentum before explosion} = 0 \)

\( \text{Total momentum after explosion} = \text{momentum of piece 1} + \text{momentum of piece 2} + \text{momentum of piece 3} = 0 \)

\[m_1 v_1 + m_2 v_2 + m_3 v_3 = 0\]

 Where: 
- \(v_1 = 30 \, \text{m/s}\) (velocity of first piece) 
- \(v_2 = 40 \, \text{m/s}\) (velocity of second piece) 
- \(v_3\) is the velocity of the third piece. 

Step 2: Resolving Momentum Components Since the pieces fly off perpendicular to each other, we can resolve the momentum into two components: horizontal and vertical. 
In the horizontal direction: \[ m \cdot 30 + m \cdot 0 + 2m \cdot v_{3x} = 0 \quad \Rightarrow \quad 30 + 2v_{3x} = 0 \quad \Rightarrow \quad v_{3x} = -15 \, \text{m/s} \] 
In the vertical direction: \[ m \cdot 0 + m \cdot 40 + 2m \cdot v_{3y} = 0 \quad \Rightarrow \quad 40 + 2v_{3y} = 0 \quad \Rightarrow \quad v_{3y} = -20 \, \text{m/s} \] 
Step 3: Calculating the Velocity of the Third Piece The total velocity of the third piece is the vector sum of the horizontal and vertical components of velocity: \[ v_3 = \sqrt{v_{3x}^2 + v_{3y}^2} = \sqrt{(-15)^2 + (-20)^2} = \sqrt{225 + 400} = \sqrt{625} = 25 \, \text{m/s} \] Final Answer: The velocity of the third piece is \(25 \, \text{m/s}\).