(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 force on helicopter is zero, that is, lift force = weight of helicopter. B1 (ii) volume of air displaced downwards V = cross-section area of air x distance moved by the air in 5 seconds = (5.02)(12)(5.0) mass of air displaced downwards in 5 seconds = volume of air x density of air = (5.02)(12)(5.0)(1.3) C1 = 6100 kg (2 s.f.) A0

3 MC: As it is a ‘show’ type of question, answers need to be written in full, before any substitution of values. Any symbols used must be defined. (iii) By Newton’s second law, F = rate of change of momentum = 6000 5.0 (12) C1 = 1.4 × 104 N A1 MC: Some students mistakenly took the mass of air propelled downwards as the mass of the helicopter. 3 (a) The density of fluid , the cross-section area A of tube, the total mass M of tube and sand and g is constant, B1 hence acceleration a is proportional to displacement x from the equilibrium position. B1 The ne gative sign indicates that acceleration a is in opposite direction to displacement x from the equilibrium position. B1 These indicate that the motion of the tube is simple harmonic where a = −2x. A0 (b) 2 = Ag M = (1.2 x 103)(5.3 x 10-4)9.81 130 x 10-3 = 48 rad2 s-2 C1 f =  2 = 1.1 Hz A1 MC: There were some careless mistakes in converting cm 2 to SI units. In addition, some students mistakenly thought  = 48 rad s-1.

4 4 (a) The re-drawn Fig. 4.1 is as follows i.e. R1, R3 and (R2, R4 in series) are in parallel across X and Y. Fig. 4 The effective resistance is ( 1 R + 1 R + 1 2R ) -1 = 0.4R M1 0.4R = 2.4 R = 6.0  A0 MC: There are different ways to view the re -drawn Fig. 4 to solve the (b)(i) to (iii). Below is the working that should be clear enough for you to understand. (b) (i) Total resistance = 2.4+0.6 = 3.0  M1 Current in A1 = E R = 1.5 3.0 = 0.50 A A1 A2 A1 R2 R4 R3 R1 Z Y X 0.6  A1 Y X 2.4  1.5 V 0.50 A

5 (ii) Terminal potential difference between X and Y = 2.4 3.0 x 1.5 = 1.2 V M1 Current through resistor R1 = V R = 1.2 6.0 = 0.20 A B1 Thus, current in A2 = 0.50 – 0.20 = 0.30 A A1 (iii) Potential difference across R4 = 6.0 6.0 + 6.0 = x 1.2 = 0.60 V M1 Potential at Z = 0 – 0.60 = −0.60 V A1 (iv) As reading is zero, the potential difference across S is 0.6 V. C1 The potential difference across S = S S+0.60 x 1.5 = 0.60 V Hence S = 0.4 Ω A1 5(a)(i) The magnetic force acting on the charged particle provides the centripetal force for the charged particle to move in uniform circular motion. 2 2 cBqv ma mv T mT Bq   ==  = sin The circular motion is horizontal, so the net vertical force is zero. mg qE= Combining both equations, 2 2 (Shown) mT Bq E Bg   = = B1 C1 C1 A0 weight electric force magnetic force

6 CKW MC • Some students stated that both the magnetic force and electric force (or sometimes even just the electric force) provide the centripetal force, which was a misconception! Though not credited, an initial free body diagram to identify the correct forces acting on the charged particle in the problem will be a useful starting approach. • The equation based on vertical equilibrium i.e. (1) could not be derived or was not shown separately by quite a number of students. • Some students wrote F = ma = mg for the vertical direction, which is conceptually incorrect application of Newton’s 2nd law, since there is no acceleration in the vertical direction i.e. a = 0. Thus, mg = qE means that the magnitude of the weight of the particle, mg is equal to the magnitude of the electric force, qE on the particle 5(a)(ii) From (i), T = 2E Bg = 2(150) (0.50)(9.81) = 192 s A1 CKW MC This part is well done, except for a few students who make careless mistakes. 5(b) When the electric field is removed, the only vertical force acting on the charged particle is its weight. Hence, the particle falls with uniform acceleration g. The magnetic force still provides the centripetal force for the charged particle to movie in uniform horizontal circular motion, resulting in the charged particle moving in a helical path in which the distance between adjacent loops increases as the particle falls. Using Fleming’s Left-hand Rule, the charged particle will continue to move in a clockwise circle when viewed from the top of its helical path. B1 B1 B1 CKW MC • Many students use inappropriate terms such as “s