CJC 2020 A level H2 Physics Answers

1 H2 Physics 9749 – 2020 A Level Exam Paper 1 1 Essential Question(s):  What is the relationship between V and r?  Which rule do we apply to determine percentage uncertainty for value obtained through multiplication/division? Solution: volume V = 4/3 r3 V / V = 3 r/r V / V x 100 % = = 3 r/r x 100% percentage uncertainty in volume = 3 percentage uncertainty in radius graph of percentage uncertainty in volume vs perc entage uncertainty in radius is a straight line graph passing through the origin with gradient = 3 Answer: B 2 Essential Question(s):  What are the four conditions that action-reaction force need to meet? Solution: The reaction force of the weight of the car is the gravitational force on the Earth by the car; the reaction force of the normal contact force on the car in the force on the ground by the car. Answer: D 3 Essential Question(s):  What is the unit of stress?  What is the unit of strain?  How do we determine the unit of a quantity represented as the area under a graph? Solution: stress= F/ A ; strain = e/L Units of stress = N m-2 Units of e = unit of L = m units of strain = unit of e / units of L = m / m strain has no unit units of area under stress-strain graph = units of stress x units of strain = N m-2 J = N m  N = J m-1 units of area under stress-strain graph = N m-2 = (J m-1) m-2 = J m-3 Answer: D

2 This question gave candidates a novel situation and asked them to use their knowledge of units to answer it. Those candidates who got this wrong tended to choose distractor C, probably because they confused newtons and joules. 4 Essential Question(s):  What does the area under the velocity – time graph represent? Solution: distance travelled by car X = P + Q + R distance travelled by car Y = S + Q + R Both cars were at the same position at t = 0, and again at t = T. Hence between t = 0 to t = T, both cars travelled the same distance. P + Q + R = S + Q + R P = S Answer: A 5 Essential Question(s):  What is relative velocity?  How do we determine the relative velocity on an object with respect to another? Solution: relative velocity of bicycle with respect to car vbc= vb– vc = vb + (- vc) vb = velocity of the bicycle vc = velocity of the car vbc = velocity of the bicycle relative to the car Answer: C Examiner’s comments: This was on relative velocities and proved dem anding with a significant number of candidates choosing distractor D, where the relative velocity of the bi cycle is shown pointing away from the car. 6 Essential Question(s):  What are the two conditions for a body to be in equilibrium? Solution: Option C is wrong because the direction of the torque is to be mentioned. The statement should be “The torque provided by the vertical forces is equal and opposite to the torque provided by the horizontal forces.” Option A, B and D are statements that follow the condition that the sum of forces is equal to zero for the object to be in equilibrium. -vc vb vbc

3 Answer: C 7 Essential Question(s):  What is the torque of a couple?  How do we determine the torque of a couple? Solution: The perpendicular distance of F from the centre of wheel, d = (d/2)sin Torque of F = 2 x F x d = (d/2)sin = 2 x F (d/2)sin = F d sin Answer: B 8 Essential Question(s):  How does the Free-Body Diagram of the basket and the balloon look like?  How to determine the upthrust of a hot air balloon? Solution: forces acting on the balloons are 1. Upthrust = V air outside g = 2800 (1.204)g acts upwards 2. Weight of air inside balloon = V air inside g = 2800 (0.898)g acts downwards 3. Weight of basket and balloon = mg = 700 g acts downwards 4. Total tension of the two rope = 2 T acts downwards , where T = tension in one rope the balloon is at rest, net force = 0 2800 (1.204)g = 2800 (0.898)g + 2T + 700g 2T = 2800 (1.204)g - 700g - 2800 (0.898)g = [2800 (1.204) - 700 - 2800 (0.898)] 9.81 = 1538 T = 769 N = 770 N Answer: A Examiner’s Comments: The incorrect answers were spread fairly evenly among the distractors, indicating that candidates forgot that there were two ropes (distractor B), confused the air densities (distractor C) or neglected to account for the weight of air inside the balloon (distractor D). 9 Essential Question(s):  How do we determine the mechanical power output of the car?  How do we determine the input power of the car?  What is the efficiency of a car in converti ng the thermal energy of fuel to useful work done on the car? Solution: power input to the car engine = thermal energy of fuel per second = 3.3 x 106 / 60 J s-1 power output from the car engine = useful work done of car per second = driving force x speed = 1.6 x 10 3 x 22 J s-1 efficiency of car engine = (power output from the car engine / power input to the car engine ) x 100

4 = (1.6 x 103 x 22) / [3.3 x 106 / 60] x 100 = 64 % Answer: D 10 Essential Question(s):  How do we determine the centripetal acce leration on an object in circular motion? Solution: centripetal acceleration of moon = r ω2 = r (2/T)2 = (3.85 x 108 ) x (2/ 27.3 x 24 x 60 x 60)2 = 2.73 x 10-3 m s-2 Answer: B 11 Essential Question(s):  What is the period of a minute hand of a clock?  How do we determine the angular velocity of an object in circular motion? Solution: angular velocity ω = 2/T = 2/(60x60) = 1.75 x 10-3 rad s-1 Answer: B Examiner’s Comments: Some candidates did not realise that the minute hand of an analogue clock takes 1 hour (3600 s) to complete one revolution. 12 Essential Question(s):  How do we determine the GPE of two masses?  How do we determine the change in GPE of two masses? Solution: change in GPE = final GPE – initial GPE = (-Gm 1m2/ 2r) - (-Gm1m2/ r) = Gm1m2/ 2r Answer: A 13 Essential Question(s): o How do we determine the gravitational field strength of a point mass at a distance away? Solution: gravitational field strength = GM / r 2 = 6.67 x 10-11 x 2.0 x 1031 / (150 x 106 x 103)2 = 5.9 x 10-2 N kg-1 Answer: A 14 Essential Question(s):  What is an ideal gas?  How is the graph related to the macroscopi c properties, p, V and T of an ideal gas?

5 Solution: Apply pV = nRT At L, (2.0 x 106 x 0.001) = nRT  T = 2 x 103(1/nR) At M, (2.0 x 106 x 0.003) = nRT  T = 6 x 103(1/nR) At N, (0.8 x 106 x 0.005) = nRT  T = 4 x 103(1/nR) temperature TM > TN > TL Answer: B Examiner’s Comments: Candidates who got it wrong tended to choose distractor C as they thought that line MN on the graph was showing an isothermal change, when the numbers show that in fact the temperature is decreasing. 15 Essential Question(s):  What is the 1 st Law of Thermodynamics  What do ΔU = ΔQ + W mean? Solution: Experiment 1, at constant volume, work done is zero, W = 0, hence U = Q + W = Q Option A is wrong Option B is correct Experiment 2, at constant pressure, as heat is supplied, part of the heat is used to do work against external environment and part become internal energy (Note that for a constant pressure process and fixed mass of ideal gas, as volume increases, product pV and hence T increases, which implies that internal energy U also increases! A quick way is to sketch the p-V graph for a constant p process – you’ll notice that the state moves to a point of higher T.) the volume increases, work is done by the gas, W is negative the temperature and internal energy increases, U is positive from U = Q + W since U is positive and W is negative Q must be positive and more than U D is incorrect Also since U is positive, (Q + W) must be positive, since Q is positive