DHS 2023 Prelim Physics (Paper 3 Solutions)

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 1 Answers to Prelim Exam H2 Physics Paper 3 1 (a) The principle of conservation of momentum states that the total momentum of a system of bodies is constant provided no resultant external force acts on the system. B1 (b) The total momentum of the two-nuclei system is non-zero. Thus, if one of the nuclei is at rest, then the other nucle us must possess a non-zero momentum (i.e. it is moving). Hence the two nuclei cannot be at rest simultaneously. B1 B1 (c) Taking right to be positive, By conservation of momentum, 3 2 3 2 32 td td mv mv mv mv v v v − = + =+ relative speed of approach = relative speed of separation 2 dtv v v=− Solving simultaneously for vt and vd, 1.4 0.6 (i.e. tritium nucleus is travelling leftward) d t vv vv = =− C1 C1 A1 for both correct answers Marker’s comments: There were answers that did not treat momentum as a vector. Answers should never be left in fraction form. (d) Let Favg be the average force exerted by the deuterium nucleus on the tritium nucleus, 3 ( 0.6 ) 4.8 (negative sign indicates it points to the left) tritium avg pF t m v v t mv t = −−= =− Magnitude of the average force = 4.8mv t C1 A1 Marker’s comments: The magnitude of the average force should be the same for either deuterium or tritium nucleus. There were mistakes in using the wrong mass or wrong change in velocity. The final numerical value should be shown and not left in fraction form. after collision vt vd

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 2 2 (a) Loss in GPE 70.0(9.81)(1.50) 1030 J mgh= == A1 (b) 2 2 1kinetic energy of man 2 1 (70.0)(2.00) 140 J2 mv= == A1 (c)(i) The perpendicular distance between the line of action of the weight and the pivot is zero. B1 (c)(ii) The man will fall a further distance equal to the extension of the spring, x before coming to a stop. Assuming no loss in energy to the surroundings, By conservation of energy, loss in k.e. + loss in g.p.e. = gain in e.p.e (1030 + 140) + 70x = 1 2 kx2 (1030 + 140) + 70x = 1 2(10000)x2 Solving: x = 0.556 m 1 sin 2.40 0.557sin 2.40 13.4 (to 3 s.f.) x  − = =  = M1 M1 M1 M1 A0 Marker’s comments: The board is NOT in rotational equilibrium, i.e. sum of clockwise moments is NOT equal to sum of anti-clockwise moments. (c)(iii) F = kx = (10000) (2.40 sin 13.4o) = 5560 N A1 (c)(iv) The board is in translational equilibrium so net force acting on the diving board is zero. N = Wboard + Wman + Wspring = 300 + 70 (9.81) + 5560 = 6550 N (3 s.f.)