Step 1: Applying the Law of Conservation of Momentum
The initial momentum of the stationary body is zero. According to the law of conservation of momentum, the total momentum after the explosion must also be zero.
The masses of the fragments are in the ratio \(1:1:2\). Let these masses be \(m_1 = m\), \(m_2 = m\), 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 the first piece)
- \(v_2 = 40 \, \text{m/s}\) (velocity of the second piece)
- \(v_3\) is the velocity of the third piece.
Step 2: Resolving Momentum Components The fragments are emitted perpendicularly. Momentum can be resolved into horizontal and vertical components.
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 velocity of the third piece is the vector sum of its horizontal and vertical velocity components: \[ v_3 = \sqrt{v_{3x}^2 + v_{3y}^2} = \sqrt{(-15)^2 + (-20)^2} = \sqrt{225 + 400} = \sqrt{625} = 25 \, \text{m/s} \] Final Answer: The velocity of the third piece is \(25 \, \text{m/s}\).