Step 1: Application of the Law of Conservation of Momentum
The initial momentum of the stationary body is zero. By the law of conservation of momentum, the total momentum after the explosion must also be zero.
The masses are in the ratio \(1:1:2\). Let the masses be \(m_1 = m\), \(m_2 = m\), and \(m_3 = 2m\).
\[\text{Final total momentum} = \vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0\]
\[ m_1 \vec{v}_1 + m_2 \vec{v}_2 + m_3 \vec{v}_3 = 0 \] Where:
- \(\vec{v}_1 = 30 \, \text{m/s}\) (velocity of the first piece)
- \(\vec{v}_2 = 40 \, \text{m/s}\) (velocity of the second piece)
- \(\vec{v}_3\) is the velocity of the third piece.
Step 2: Momentum Component Resolution Given that the pieces are ejected perpendicularly, momentum can be resolved into horizontal and vertical components.
Horizontal component: \[ m \cdot 30 + m \cdot 0 + 2m \cdot v_{3x} = 0 \quad \Rightarrow \quad 30 + 2v_{3x} = 0 \quad \Rightarrow \quad v_{3x} = -15 \, \text{m/s} \]
Vertical component: \[ m \cdot 0 + m \cdot 40 + 2m \cdot v_{3y} = 0 \quad \Rightarrow \quad 40 + 2v_{3y} = 0 \quad \Rightarrow \quad v_{3y} = -20 \, \text{m/s} \]
Step 3: Calculation of the Third Piece's Velocity The resultant 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 \(25 \, \text{m/s}\).