1. **Problem statement:** We have two objects A and B connected by a light inextensible string over a smooth fixed pulley. Mass of A, $m_A = 4.5$ kg, mass of B, $m_B = 7.5$ kg. Initially, A is on a horizontal surface, B hangs 1.8 m above the surface. **a) Find the tension in the string when A is held stationary.** 2. Since A is held stationary on the surface, it does not move, so the tension $T$ in the string must balance the weight of B. 3. Weight of B is $W_B = m_B g = 7.5 \times 9.8 = 73.5$ N. 4. Therefore, the tension in the string is $T = 73.5$ N. --- **b) Object A is released.** **i) Calculate the magnitude of the acceleration of A.** 5. When released, both masses accelerate: A moves horizontally, B moves vertically. 6. Let acceleration be $a$ and tension be $T$. 7. For A (horizontal surface, no friction): $T = m_A a$. 8. For B (vertical motion): $m_B g - T = m_B a$. 9. Substitute $T$ from (7) into (8): $$m_B g - m_A a = m_B a$$ 10. Rearranged: $$m_B g = m_A a + m_B a = (m_A + m_B) a$$ 11. Solve for $a$: $$a = \frac{m_B g}{m_A + m_B} = \frac{7.5 \times 9.8}{4.5 + 7.5} = \frac{73.5}{12} = 6.125 \text{ m/s}^2$$ **ii) Find the time taken for the objects to be at the same height above the horizontal surface.** 12. Initially, B is 1.8 m above the surface, A is at 0 m. 13. Since the string is inextensible, the distance A moves horizontally equals the distance B moves vertically. 14. Let $s$ be the distance moved by A and B until they are at the same height. 15. At the moment they are at the same height, B has descended $s$ meters, so its height is $1.8 - s$. 16. A has moved $s$ meters horizontally, so its height remains 0. 17. For them to be at the same height: $$1.8 - s = 0 \implies s = 1.8 \text{ m}$$ 18. Use the equation of motion for A starting from rest: $$s = \frac{1}{2} a t^2$$ 19. Substitute $s=1.8$ m and $a=6.125$ m/s$^2$: $$1.8 = \frac{1}{2} \times 6.125 \times t^2$$ 20. Solve for $t^2$: $$t^2 = \frac{2 \times 1.8}{6.125} = \frac{3.6}{6.125} \approx 0.5878$$ 21. Therefore, $$t = \sqrt{0.5878} \approx 0.766 \text{ s}$$ **iii) Effect of air resistance on the time calculated in (ii).** 22. Air resistance opposes motion, reducing acceleration. 23. Reduced acceleration means it takes longer for the objects to reach the same height. 24. Therefore, the actual time taken will be greater than $0.766$ seconds due to air resistance.