2024 DHS Prelim H1 Phy P1_Ans

2024 DHS H1 Physics Prelim Paper 1 Suggested Solutions 1 2 3 4 5 6 7 8 9 10 D A A B B C A C D A 11 12 13 14 15 16 17 18 19 20 B C A C D A B B D C 21 22 23 24 25 26 27 28 29 30 B D D A D C D A C B Worked Solutions & Explanations: 1 D Units of v = m s-1 Units of ඥߣ݃= ඥm s-2 × m = ඥm2 s-2 = m s-1 2 A For a pair of forces of magnitude X and Y, Minimum Fnet = |X – Y| = 6 – 4 = 2 N Maximum Fnet = |X + Y| = 6 + 4 = 10 N Hence, the resultant force cannot be 1 N. 3 A average length measured = 891.5 mm difference from true value = 895 mm – 891.5 mm = 3.5 mm (results are not accurate to within 1 mm) ∆L = Lmax – Lavg = 0.5 mm (results are precise to within 1mm)

4 B vy2 = uy2 + 2as At max height, vertical component of velocity = 0 vy2 = 0 + 2(9.81)(13.0) = 255.06 m s-1 horizontal component of velocity is a constant, vx = 22 cos 30° v = ටvy 2 + vx 2 = ඥ225.06+222 cos2 30° = 24.9 m s-1 5 D The total area under the acceleration-time graph is the change in velocity. The maximum magnitude of area occurs at point K. Hence magnitude of the change in velocity (and hence speed) is the largest at point K. The total area under the graph from t = 0 s onwards is also strictly negative. This indicates that the velocity of the object is strictly negative. (car is always moving in the opposite direction to the defined positive from t = 0 to point M). Hence, the maximum displacement is at M. 6 C The 6.0 kg mass will accelerate upwards (F net upwards), while the 10 kg mass will accelerate downwards (Fnet downwards) By considering the free body diagrams of the two masses, From the 6.0 kg mass: T – 6g = 6a -------------(1) From the 10 kg mass: 10g – T = 10a ------------------(2) (1) + (2): 4g = 16a a =0.25 g Therefore, T = 6g + 6(0.25g) = 7.5g = 74 N

7 A By Newton’s Second Law, the rate of change in momentum is directly proportional to the resultant force. Hence, the resultant force can be determined from the gradient of a momentum-time graph of a body. At terminal velocity, the momentum of the object is a constant (due to constant v) and non- zero. (graph should be a horizontal line with a non-zero value) After the time of impact P, the deceleration is a constant and hence net force is a constant. The momentum -time graph should therefore have a constant negative gradient (linear graph sloping downwards) after P. 8 C During a collision, momentum is always conserved. However kinetic energy is only conserved when the collision is elastic. 9 A For an object to be in equilibrium, the three forces must be concurrent (i.e. the line of actions of the 3 forces must intersect at a point). Only option A fits this description. 10 A F = kx F = (500) [(90−60) × 10-3] = 15 N 11 B For a pair of equal and opposite forces, torque = F × perpendicular distance between the two forces = F (L sinߠ) 12 C Power input = rate of loss of GPE = m t gh = (6.0 × 1000)(9.81)(80) = 4.70