Let's solve the problem step by step using the **conservation of mechanical energy**. ### Step 1: **Determine the change in potential energy** Let: - \( m \) = mass of each body - \( g \) = gravitational acceleration - \( v_A \) = velocity of \( A \) as it crosses the vertical centerline #### **Initial Position:** - \( A \) is at a height above the centerline (let's call this \( h_A \)) - \( B \) is at its initial height (call this \( h_B \)) #### **Final Position:** - \( A \) is at the vertical centerline (lowest point) - \( B \) moves up to a new height Let's choose the vertical centerline as our reference (zero potential energy). #### **Calculate height differences:** For \( A \): - Initial vertical height above centerline: \( h_{A_i} = 200\,\text{mm} \cdot \sin{60^\circ} = 200 \cdot \frac{\sqrt{3}}{2} = 173.2\,\text{mm} \) - Final vertical height = 0 (since it's at the centerline) For \( B \): - Since the links are rigid, as \( A \) moves down, \( B \) moves up by the same vertical distance (\( \Delta h \)) So, initial potential energy: \[ PE_{initial} = mgh_{A_i} + m \cdot 0 \quad (\text{assuming } B \text{ is at reference height}) \] Final potential energy: \[ PE_{final} = m \cdot 0 + mgh_{B_f} \] Where \( h_{B_f} = 173.2\,\text{mm} \) (since B moves up by the same amount A moves down). So, the **change in potential energy**: \[ \Delta PE = mgh_{A_i} - mgh_{B_f} = mg(173.2) - mg(173.2) = 0 \] **But this result suggests no velocity, which is counterintuitive.** Let's reconsider the setup. --- ### Let's use **conservation of energy** more carefully: Both masses are moving, but only A starts with potential energy above the centerline. As A descends, B rises, but the actual path B takes is along the arc of its link, not simply vertical. But, since the links are rigid and there are no external forces (except gravity), the total mechanical energy of the system is conserved. #### **Initial total potential energy:** - \( A \) is at height \( h_{A_i} \) - \( B \) is at the bottom (reference) #### **Final total potential energy:** - \( A \) at centerline (height = 0) - \( B \) rises by a certain amount Let’s calculate the **vertical displacement of B** when A reaches the centerline. - Initial position of B: directly below O, 250 mm down - Final position: draw a circle of radius 250 mm with center O, and see where B moves as A reaches the bottom. But since the system is symmetrical, *as A descends by h, B ascends by h*. So the **net change in potential energy** is due to the redistribution between A and B: \[ \Delta PE = mg \cdot (h_{A_i} - h_{B_f}) \] But since \( h_{B_f} = h_{A_i} \), the net change is zero. **But, the two masses move together, so let's consider their kinetic energy.** #### **Kinetic Energy at centerline:** Let the velocity of A at centerline be \( v_A \) and of B be \( v_B \). Total kinetic energy: \[ KE = \frac{1}{2} m v_A^2 + \frac{1}{2} m v_B^2 \] --- ### **Constraint:** Since the links are rigid, - As A moves down, B moves up, and their velocities are related by the geometry. #### **Let’s use geometry:** When A is at the centerline, both links are vertical: - OA = 200 mm down from O (centerline) - OB = 250 mm down from O (B at centerline too) But initially: - A is at \( 200 \times \sin 60^\circ = 173.2 \) mm to the right of O, and \( 200 \times \cos 60^\circ = 100 \) mm below O. - B is at the bottom. So, as A moves from initial to centerline, its vertical displacement is \( 200 \times \cos 60^\circ = 100 \) mm. So, A falls a vertical distance of 100 mm. At the same time, B rises by an amount we can calculate from the geometry. But, for simplicity, assume both move by h. #### **Rewriting conservation of energy:** \[ \text{Initial PE} = \text{Final KE} \] \[ mg h = \frac{1}{2} m v_A^2 + \frac{1}{2} m v_B^2 \] #### **Find the relation between \( v_A \) and \( v_B \):** From rigid links: - The velocities are proportional to the lengths of the links: \[ \frac{v_A}{OA} = \frac{v_B}{OB} \] \[ v_B = \frac{OB}{OA} v_A = \frac{250}{200} v_A = 1.25 v_A \] #### **Plug back into energy equation:** \[ mg h = \frac{1}{2} m v_A^2 + \frac{1}{2} m (1.25 v_A)^2 \] \[ mg h = \frac{1}{2} m v_A^2 + \frac{1}{2} m (1.5625 v_A^2) \] \[ mg h = \frac{1}{2} m (v_A^2 + 1.5625 v_A^2) \] \[ mg h = \frac{1}{2} m (2.5625 v_A^2) \] \[ 2gh = 2.5625 v_A^2 \] \[ v_A^2 = \frac{2gh}{2.5625} \] \[ v_A = \sqrt{\frac{2gh}{2.5625}} \] #### **Plug in values:** - \( g = 9.81\, \text{m/s}^2 \) - \( h = 0.1\, \text{m} \) (since A falls 100 mm) \[ v_A = \sqrt{\frac{2 \times 9.81 \times 0.1}{2.5625}} \] \[ v_A = \sqrt{\frac{1.962}{2.5625}} \] \[ v_A = \sqrt{0.7658} \] \[ v_A \approx 0.875 \, \text{m/s} \] --- ## **Final Answer:** \[ \boxed{v_A \approx 0.88\ \text{m/s}} \] as it crosses the vertical centerline.