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 potential difference Solution: change in in potential per unit change in height = 6 / 10 let  V be the change in potential for a change in height of 2.5 m change in in potential per unit change in height =  V / 2.5 Since gravitational field is constant uniform, change in potential per unit change in height is constant 6/10 =  V / 2.5  V = 1.5 work done to bring 2 kg of mass up 2.5 m = m  V = 2 x 1.5 = 3.0 J Answer: B 11 Essential Question(s):  What is the period of a geostationary satellite? Solution: v = rω= r (2/T) = (36000000+6400000) (2/ 24x60x60) = 3100 m s-1 Answer: D 12 Essential Question(s):  What is the relation between rms speed and any other properties of a gas? Solution: ݌= 1 3 ݉ܰ ܸ〈ܿଶ〉 = 1 3 (݈ܽݐ݋ܶ ݏݏܽ݉) ܸ〈ܿଶ〉 = 1 3 (݋݊.݂݋ ݏ݈݁݋݉ ݔ ݎ݈ܽ݋݉ ݏݏܽ݉) ܸ〈ܿଶ〉 1.0 ݔ 10ହ = 1 3 ൬5000 ݔ 0.029 10 ݔ 3.0 ݔ 4.0൰ 〈ܿଶ〉 ݎ.݉ .ݏ .݀݁݁݌ݏ = ඥ〈ܿଶ〉 = 498 = 500 ݉ ݏିଵ Answer: C 13 Essential Question(s): o What is meant by specific latent heat and specific heat capacity? Solution:

4 Q = heat to raise temp to 100 + heat to convert water to steam = mc + mL = 5 x 4190 x (100-30) + 5 x 2260 = 1.28 x 107 J Answer: D 14 Essential Question(s):  What is the internal energy of an ideal gas? Solution: Given Q = 0 and W = 0, thus ∆U = 0. So no change in internal energy; internal energy remains at the initial value of 3.0 kJ. Recall also that internal energy is equal to the sum of the KE and PE associated with the microscopic molecules of the gas, and independent of the bulk changes in KE and PE of the system as a whole. Answer: B 15 Essential Question(s):  How to relate PE to KE of a SHM Solution: At t = 0, and t = 2 s, PE is max, KE is zero, v = 0 At t = 1 s, PE = 0, KE is max, v is max Answer: C 16 Essential Question(s):  How is determine phase difference from path difference? Solution: ∆߮ 2ߨ= ∆ݔ ߣ= ∆ݔ ቀ݂ܿ ൗ ቁ ∆߮ 2ߨ= 1.5 ݔ 10ି଺ 3.00 ݔ 10଼ 5.0 ݔ 10ଵସൗ ∆߮= 1.0 ߨ ݀ܽݎ Answer: D 17 Essential Question(s):  How does the direction of polarization and intensity change with angle of between polarization axle of polarizing filter and the direction of polarization of incident light? Solution: At 45o After first polarizing filter Intensity = Io cos245 = Io cos245 = Io /2 direction of polarisation: 45 degree to vertical After second polarizing filter Intensity = (Io / 2 ) cos245 = Io /4 when direction of polarisation: vertical At 90o After first polarizing filter

5 Intensity = Io cos290 = 0 direction of polarisation: NA After second polarizing filter Intensity = 0 direction of polarisation: NA At 135o After first polarizing filter Intensity = Io cos2(135 -90) = Io cos245 = Io /2 direction of polarisation: 45 degree to vertical After second polarizing filter Intensity = (Io / 2 ) cos245 = Io /4 when direction of polarisation: vertical At any angle  less than 90 After first polarizing filter Intensity = Io cos2() = Io cos2 direction of polarisation:  degree to vertical After second polarizing filter Intensity = (Io cos2 ) cos2(90-) Intensity = (Io cos2 ) (sin2) =Io (1 – cos2 )/2 direction of polarisation: (90 -  ) degree to vertical Answer: B 18 Essential Question(s):  What is the Rayleigh’s criteria for resolving two point sources through a circular aperture? Solution: Rayleigh’s criteria min  /b = 620 x 10-9 / 0.50 = 1240 x 10-9 rad Distance between two stars, D  L/ To just resolve the two stars,  = min D  L/min = 5.7 x 1018 x 1240 x 10-9 = 7.1 x 1010m Answer: D 19 Essential Question(s):  How to determine the direction of electric force on a charged particle placed in an electric field? Solution: Definition of electric field strength Answer: D 20 Essential Question(s):  What is the Coulomb’s law on the force between two charged particles?

6 Solution: F1 = Q1 Q2 /4 r2 = 2 N F2 = Q3 Q2 /4 (2r)2 = 2Q1 Q2 /4 (2r)2 = {Q1 Q2 /(4 r)}/2= 2/2 = 1 N Answer: C 21 Essential Question(s):  How to determine the resistance of a uniform wire given its length, cross-section area and resistivity?  For components in parallel, how to determine their effective resistance? Solution: gradient of graph G = resistance per unit length Each wire R = resistance per unit length x L = GL 4 identical wires in parallel 1 ܴ௘௙௙ = 1 ܮܩ+ 1 ܮܩ+ 1 ܮܩ+ 1 ܮܩ Reff = ¼ (GL) Answer: B 22 Essential Question(s): 1. what is the same in a series circuit of two components? Solution: Current is the same sum of p.d. across the two components = 3.0 V By inspecting the graphs, when I = 0.1 A , V = 1.0 + 2.0 = 3.0 V Answer: A 23 Essential Question(s):  What determines the brightness of a bulb in a circuit? Solution: Brightness is determined by the power dissipated in a bulb = I2 R or V2 / R Assume the resistance of the bulb is unchanged, power is in