CJC 2022 A level H2 Physics Answers

1 H2Physics 9749– 2022 A Level Exam Paper 1 1 Essential Question(s): o How to use a vector diagram to find vector subtraction? Solution: X – Y = X + (-Y) When angle = 0, Y is opposite to X, |X – Y| = |X| + |Y| maximum When angle = 90, Y is perpendicular to X, | X – Y| =  (|X|2+ |Y|2) less than |X| + |Y| When angle = 180, Y is along X, | X – Y| = |Y| - |X| minimum Answer: A Examiner’s comments: Option B was frequently chosen, indicating that candidates were subtracting magnitudes and not thinking about the orientation of the vector. 2 Essential Question(s): o How to find the range of a projectile motion? Solution: Let  be the angle and u be the initial velocity Horizontal motion sx = uxt a = 0 100 = u cos  x 2 ucos  = 50 ………(1) Vertical motion sy= uyt + ½ a t2 a = g 0 = u sin  x 2 + ½ (-g) 22 u sin  = g ………………..(2) X Y X - Y X - Y X Y X - Y X Y X - Y X - Y X - Y

2 (2)/(1) : tan  = g / 50 = 9.81/50  = 11 Answer: C Examiner’s comments: 3 Essential Question(s):  What are the two mathematical relations that are the consequent of an elastic collision? Solution: Let v be the velocity of Q after collision along the initial velocity of P relative velocity of approach = relative velocity of separation (2 – 0) = v – (-0.5) v = 1.5 m s-1 Conservation of momentum mp (2) = - mp(0.5) + mq (1.5) mp / mq = 1.5/2.5 = 3/5 Answer: B Examiner’s comments: 4 Essential Question(s):  What are the conditions for a body to be in equilibrium? Solution: Weight = mg Upthrust = Vg = ½ (4/3 r3) g = ½ (4/3 x  x (0.5)3)(1030) g = 269.7g Body in equilibrium Weight + tension = upthrust tension = upthrust - weight = 269.7g – 200g = 69.7 x 9.81= 684 N ( Answer: B Examiner’s comments: 5 Essential Question(s):  What are the conditions for a body to be in equilibrium? Solution: Let the tension in string be T and the weight of the mass be W. Vertical resultant force = 0 Tcos 36 = Weight = W ………(1) Horizontal resultant force = 0

3 Tsin36 = tension in spring = ke = 25 x 0.060 = 1.5 N…….(2) (2)/ (1): tan 36 = 1.5 / W W = 1.5 / tan 36 = 2.064 = 2.1 N Answer: D Examiner’s comments: 6 Essential Question(s):  What is the gravitational field strength on earth surface given the radius and density of the Earth?  Solution: Gravitational field strength g = GM/r2 = G M/r2 = G (4/3  r3) / r2 = G (4/3  r  = g / G (4/3  r) = 9.81 / {6.67 x 10-11 x 4/3  x 6.7 x 106) = G (4/3  r3) / r2 = 5512 kg m-3 Answer: C Examiner’s comments: 7 Essential Quest