CJC 2024 A level H2 Physics Answers

2024_A_Level_Solutions prepared by CJC 1 H2 Physics 9749 – 2024 A Level Exam Paper 1 1 Essential Question(s):  How to calculate the area of the circular orbit of the Moon around the Earth? Solution: The radius of the circular orbit is 4 × 108 m. The area of the circular orbit is π × (4 × 108)2 = 50 × 1016 m2 = 0.5 × 1018 m2 = 0.5 × (109 m)2 = 5 × 10−1 (Gm)2 Answer: A 2 Essential Question(s):  How to relate velocity (goal) to the a-t graph (given info)? Solution: Change in velocity = Area under a-t graph Note that the positive area (positive change in velocity) is lesser than the negative area (negative change in velocity). Answer: B 3 Essential Question(s):  What kinematics equation can be used to determine the time of flight of a projectile?  What do I know about v y or sy for a 2-D projectile motion where the object returns to the initial height? Solution: Resistance from the gases on the Moon is negligible. Take upwards as positive direction. Consider the vertical component of the initial speed u y = 15 sin 40° and the final speed of the vertical component vy = 0 when the golf reaches its maximum height. Using vy = uy + ayt, t = (vy – uy) ÷ ay = (0 – 15 sin 40°) ÷ (-1.62) = 5.95 s Time of flight of the golf ball = 2 × t = 11.9 s OR At maximum height, sy = 0 sy = uyt + ½ayt2 0 = (15 sin 40°)t + ½ (-1.62)t2 t = 0 (at the start) or t = 11.9 s (returns to the ground) Answer: C

2024_A_Level_Solutions prepared by CJC 2 4 Essential Question(s):  What is the force vector diagram that represents the equilibrium of forces acting on the cylinder? Solution: Smooth  No friction. Only contact force by the planes on the cylinder is the normal contact force. The angle between the planes X and Y is 90°. The force acting on the cylinder by plane X, FX, points to the centre of the cylinder at 30° from the vertical. The force acting on the cylinder by plane Y, FY, points to the centre of the cylinder at 60° from the vertical. The force acting on the cylinder by gravity, W, points vertically downward. For 3 non-parallel coplanar forces in equilibrium, their lines of action should intersect at a common point, and, these 3 forces when joined head-to-tail should form a closed polygon vector diagram that represents a zero net force. Answer: B 5 Essential Question(s):  What is the change in the setup?  What determines the extension of each spring?  What is the original force in each spring? What is the new force in each spring? Solution: The 3 springs are identical. From the diagram, the force acting on each spring is W/3 and the extension of each spring is x. When the middle spring is removed and a weight of 2W is hanging from the middle of the beam, the force acting on each of the spring is (2W) / 2 = W which is 3 times of W/3. Hence the extension of each spring should also be 3 times of x. OR Before: 3kx = W ---(1) Afte