CJC 2021 A level H2 Physics Answers

1 H2 Physics 9749 – 2021 A Level Exam Paper 1 1 Essential Question(s):  What is the relationship between weight and mass?  What are the values of 1 cN and 1 dN in N? Solution: mass of a smartphone = 200 g weight = mg = 0.2 x 10 = 2 N = 200 x 10-2 N = 200 cN Answer: C 2 Essential Question(s):  What is the definition of the tesla? Solution: From F = BIL, B = F / I L 1 T = 1 N / 1 A x 1 m = 1 kg m s-2 / A m = 1 kg s-2 A-1 Answer: A 3 Essential Question(s):  What are the forces acting on the ball? Solution: forces are: weight acting downwards drag force acting to the left (opposite to velocity) Resultant force = vector sum of drag force and weight Answer: B 4 Essential Question(s):  What can you conclude about the velocity of the bodies undergo an elastic collision? Solution: for elastic collision, relative velocity of approach = relative velocity of separation Take vectors to the right as positive. v – 0 = 0.67v – final velocity of x final velocity of x = -0.33 v speed = 0.33 v Answer: B 5 Essential Question(s):  What is the Newton’s third law of motion?

2 Solution: W = force on brick by earth reaction of W = force on Earth by brick S = force on brick by floor reaction of S = force on floor by brick Answer: A 6 Essential Question(s):  How is the apparent loss in weight related to upthrust? Solution: apparent loss in weight = upthrust = Vg 100(9.81) – 100(9.81) x 0.75 = V (1000)(9.81) V = 25 x 10-6 m3 (This is the volume of fluid displaced) volume of ball = 2V = 2 x 25 x 10-6 = 50 x 10-6 m3 = 50 cm3 Answer: B 7 Essential Question(s):  How do we determine the elastic energy stored from extension-force graph? Solution: Elastic energy stored = area under force-extension graph = area bounded by the graph and VERTICAL axis in this case Initial elastic energy stored = ½ X0Wo Final elastic energy stored = ½ X1 W1 Extra elastic energy stored = ½ (X1 W1 - X0Wo) Answer: D 8 Essential Question(s):  What is the work done against a resistive force Solution: Constant speed  no energy is consumed to convert into KE Level road  no energy is consumed to convert into GPE WD against resistive force = resistive force x distance = 400 x 1000 = 400000 J Efficiency = energy output / energy input energy input = energy output / efficiency = 400000/ 0.16 = 250000 J amount of fuel used = 2500000/ 48 x 106 = 0.052 kg = 52 g Answer: A

3 9 Essential Question(s):  What is the relation between tangential speed and angular velocity of a circular motion? Solution: angular speed  = v / r = Be / m since r = mv / Be  will increase when B is increased while e and m are unchanged Answer: A 10 Essential Question(s):  What is the relation between gravitational field strength and potenti