1. **State the problem:** 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 60º to the original direction, and Y moves at velocity $v$ at 41º to the original direction. Find $v$ using conservation of linear momentum. 2. **Relevant formula:** Momentum before collision = Momentum after collision. Since momentum is a vector, conserve momentum separately along x-axis and y-axis. 3. **Momentum along x-axis:** Before collision: $$p_{x,\text{initial}} = m_X u_X = 0.3 \times 3 = 0.9\,\text{kg m/s}$$ After collision: $$p_{x,\text{final}} = m_X v_{Xx} + m_Y v_{Yx} = 0.3 \times 2 \cos 60^\circ + 0.5 \times v \cos 41^\circ$$ 4. **Momentum along y-axis:** Before collision: $$p_{y,\text{initial}} = 0$$ After collision: $$p_{y,\text{final}} = m_X v_{Xy} + m_Y v_{Yy} = 0.3 \times 2 \sin (-60^\circ) + 0.5 \times v \sin 41^\circ$$ Note: Since X moves downward at 60º, $v_{Xy} = 2 \sin (-60^\circ)$ (negative direction). 5. **Set up equations:** Along x-axis: $$0.9 = 0.3 \times 2 \times \cos 60^\circ + 0.5 \times v \times \cos 41^\circ$$ Simplify: $$0.9 = 0.6 \times 0.5 + 0.5 v \times 0.7547$$ $$0.9 = 0.3 + 0.37735 v$$ Along y-axis: $$0 = 0.3 \times 2 \times \sin (-60^\circ) + 0.5 \times v \times \sin 41^\circ$$ $$0 = 0.6 \times (-0.8660) + 0.5 v \times 0.6561$$ $$0 = -0.5196 + 0.32805 v$$ 6. **Solve y-axis equation for $v$:** $$0.32805 v = 0.5196$$ $$v = \frac{0.5196}{0.32805}$$ $$v \approx 1.583\,\text{m/s}$$ 7. **Check x-axis equation with this $v$:** $$0.9 \stackrel{?}{=} 0.3 + 0.37735 \times 1.583$$ $$0.9 \stackrel{?}{=} 0.3 + 0.597$$ $$0.9 \approx 0.897$$ Close enough considering rounding. **Final answer:** $$v \approx 1.6\,\text{m/s}$$