1. **State the problem:** Particle A (mass $m_A=0.3$ kg) moves at velocity $u_A=5$ m/s and collides with particle B (mass $m_B=0.2$ kg) initially at rest ($u_B=0$ m/s). The coefficient of restitution $e=0.5$. Find the velocities $v_A$ and $v_B$ after collision. 2. **Relevant formulas:** - Conservation of momentum: $$m_A u_A + m_B u_B = m_A v_A + m_B v_B$$ - Coefficient of restitution: $$e = \frac{v_B - v_A}{u_A - u_B}$$ 3. **Apply conservation of momentum:** $$0.3 \times 5 + 0.2 \times 0 = 0.3 v_A + 0.2 v_B$$ $$1.5 = 0.3 v_A + 0.2 v_B \quad (1)$$ 4. **Apply coefficient of restitution formula:** $$0.5 = \frac{v_B - v_A}{5 - 0} = \frac{v_B - v_A}{5}$$ Multiply both sides by 5: $$v_B - v_A = 2.5 \quad (2)$$ 5. **Solve equations (1) and (2):** From (2): $$v_B = v_A + 2.5$$ Substitute into (1): $$1.5 = 0.3 v_A + 0.2 (v_A + 2.5) = 0.3 v_A + 0.2 v_A + 0.5 = 0.5 v_A + 0.5$$ Subtract 0.5 from both sides: $$1.5 - 0.5 = 0.5 v_A$$ $$1.0 = 0.5 v_A$$ Divide both sides by 0.5: $$v_A = 2.0 \text{ m/s}$$ 6. **Find $v_B$:** $$v_B = v_A + 2.5 = 2.0 + 2.5 = 4.5 \text{ m/s}$$ **Final answer:** - Velocity of particle A after collision: $v_A = 2.0$ m/s - Velocity of particle B after collision: $v_B = 4.5$ m/s