(DHS) 2024 Prelim Phy P2 Ans_final

1 2024 DHS H2 Physics Prelim Paper 2 Suggested Solutions 1 (a) The bomb is travelling at constant speed in the horizontal direction B1 and is accelerating uniformly in the perpendicular (vertical) direction at a rate of 9.81 m s-2 (acceleration of free fall) B1 MC: Generally well answered. Most candidates were able to link their explanations to the definition of projectile motion and hence were able to attain full or minimally partial credit. Most who attained partial credit typically only stated that there is an acc eleration in the vertical direction but did not explicitly mention that it is constant. (b) 100 = 4.91 t2 C1 t = 4.5 s A1 MC: Generally well answered. Some candidate forgot to square in their working. (c) (i) 2 straight line graphs to show intersect at t = 4.5 s A1 and when t = 0, xbomb = 0 and xtruck = 125 m A1 No indication of values or graph labels: deduct 1 mark MC: Decent attempt. Most candidates were able to obtain the graph above. Common errors included having both graphs start at the origin. (ii) v = gradient of truck graph = [(72 x 4.5) – 125] / (4.5) C1 = 44 m s-1 A1 MC: Generally well answered. Ecf was should their answers in (b) or (c)(i) be wrong. x/m t/s 125 4.5 T B

2 2 (a) (i) It states that the total momentum of a system of bodies is constant provided no external resultant force acts on the system. B1 MC: This principle applies not only to collision problems e.g. applies during alpha decay as well. In addition, students must explain the meaning of conserve in their answers. It is also not necessary to mention that the system is isolated as no external force would imply such as. (ii) By conservation of momentum, C1 (2.4×3.0) – (1.2×2.0) = 3.6v v = 1.3 m s−1 A1 (iii) Initial total kinetic energy calculated correctly (10.8 + 2.4 = 13.2 J) and Final total kinetic energy calculated correctly (3.2 J) M1 Since total kinetic energy of system is not constant, the collision is inelastic. B1 OR relative speed of approach = 5.0 m s −1, relative speed of separation = 0 m s−1 M1 Since relative speed of approach is not equal to relative speed of separation, the collision is inelastic. B1 MC: As it is a ‘show’ type of question, answers need to be written in full with no short form used, i.e. KE is not accepted but have to spell out as kinetic energy. (b) (i) By Newton’s second law, there is a net downward force by the rotor on the air to increase the momentum of air. B1 By Newton’s third law, the air exerts a force of equal magnitude but upward direction (i.e. lift force) on the rotor. B1 Since the helicopter is hovering, by Newton’s first law, the net fo