1. **Problem statement:** An object X of mass 0.3 kg moves at 3 m/s horizontally and collides with a stationary object Y of mass 0.5 kg. After collision, X moves at 2 m/s at 60° to the original direction, and Y moves at velocity $v$ at 41° to the original direction. Find $v$. 2. **Relevant principle:** Use conservation of linear momentum separately along the x-axis (horizontal) and y-axis (vertical) because no external forces act. 3. **Momentum before collision:** - Object X: $p_{x,i} = m_X u_X = 0.3 \times 3 = 0.9$ kg·m/s (all in x-direction) - Object Y: stationary, so $p_{y,i} = 0$ 4. **Momentum after collision:** - Object X components: - $p_{x,X,f} = m_X v_X \cos 60^\circ = 0.3 \times 2 \times 0.5 = 0.3$ kg·m/s - $p_{y,X,f} = m_X v_X \sin 60^\circ = 0.3 \times 2 \times \frac{\sqrt{3}}{2} = 0.3 \times \sqrt{3} \approx 0.5196$ kg·m/s - Object Y components: - $p_{x,Y,f} = m_Y v \cos 41^\circ = 0.5 v \times 0.7547 = 0.37735 v$ - $p_{y,Y,f} = m_Y v \sin 41^\circ = 0.5 v \times 0.6561 = 0.32805 v$ 5. **Apply conservation of momentum along x-axis:** $$0.9 = 0.3 + 0.37735 v$$ $$0.9 - 0.3 = 0.37735 v$$ $$0.6 = 0.37735 v$$ $$v = \frac{0.6}{0.37735} \approx 1.59 \text{ m/s}$$ 6. **Apply conservation of momentum along y-axis:** Since initial y-momentum is zero, $$0 = 0.5196 - 0.32805 v$$ $$0.32805 v = 0.5196$$ $$v = \frac{0.5196}{0.32805} \approx 1.58 \text{ m/s}$$ 7. **Conclusion:** Both components give approximately $v \approx 1.6$ m/s, confirming the velocity magnitude of object Y after collision. **Final answer:** $$v \approx 1.6 \text{ m/s}$$