1. **Problem Statement:** I) Determine the time interval over which blocks X and Y are in contact. II) Determine the magnitude of the force exerted by block X on block Y using the momentum-time graph. III) Find the velocity $v$ of object Y after collision given masses, velocities, and angles. --- 2. **Relevant Formulas and Principles:** - Momentum $p = mv$ where $m$ is mass and $v$ is velocity. - Impulse-momentum theorem: $$F\Delta t = \Delta p$$ where $F$ is force, $\Delta t$ is contact time, and $\Delta p$ is change in momentum. - Conservation of linear momentum in both $x$ and $y$ directions: $$m_X u_X = m_X v_X \cos(\theta_X) + m_Y v_Y \cos(\theta_Y)$$ $$0 = m_X v_X \sin(\theta_X) - m_Y v_Y \sin(\theta_Y)$$ where $u_X$ is initial velocity of X, $v_X$, $v_Y$ are velocities after collision, and $\theta_X$, $\theta_Y$ are angles. --- 3. **Step I: Time interval of contact** From the graph, block Y's momentum starts changing at $t=60$ ms and stops changing at $t=100$ ms. Therefore, contact time $\Delta t = 100 - 60 = 40$ ms. --- 4. **Step II: Force exerted by block X on block Y** - Change in momentum of block Y during contact: $$\Delta p = 0.50 - 0 = 0.50\ \text{kg m/s}$$ - Contact time in seconds: $$\Delta t = 40\ \text{ms} = 40 \times 10^{-3} = 0.04\ \text{s}$$ - Using impulse-momentum theorem: $$F = \frac{\Delta p}{\Delta t} = \frac{0.50}{0.04} = 12.5\ \text{N}$$ Rounded to 13 N. --- 5. **Step III: Velocity $v$ of object Y after collision** Given: - $m_X = 0.3$ kg, $u_X = 3$ m/s - $m_Y = 0.5$ kg, $u_Y = 0$ m/s - After collision: - $v_X = 2$ m/s at $60^\circ$ - $v_Y = v$ at $41^\circ$ Apply conservation of momentum along $x$-axis: $$m_X u_X = m_X v_X \cos 60^\circ + m_Y v_Y \cos 41^\circ$$ $$0.3 \times 3 = 0.3 \times 2 \times 0.5 + 0.5 \times v \times 0.7547$$ $$0.9 = 0.3 + 0.3774 v$$ $$0.9 - 0.3 = 0.3774 v$$ $$0.6 = 0.3774 v$$ $$v = \frac{0.6}{0.3774} \approx 1.59\ \text{m/s}$$ Apply conservation of momentum along $y$-axis: $$0 = m_X v_X \sin 60^\circ - m_Y v_Y \sin 41^\circ$$ $$0 = 0.3 \times 2 \times 0.8660 - 0.5 \times v \times 0.6561$$ $$0 = 0.5196 - 0.328 v$$ $$0.328 v = 0.5196$$ $$v = \frac{0.5196}{0.328} \approx 1.58\ \text{m/s}$$ Both components agree, so final velocity magnitude of Y is approximately $1.6$ m/s. --- **Final Answers:** I) Contact time $= 40$ ms II) Force exerted by X on Y $\approx 13$ N III) Velocity of Y after collision $v \approx 1.6$ m/s