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 Question(s): Solution: Height after 2 oscillations h = 0.60 e-0.10n = = 0.60 e-0.10(2) = 0.109 m Initial height = 0.60 m Loss in GPE = mg h = 0.4 x 9.81 (0.60 – 0.491) = 0.43 J Answer: B Examiner’s Comments: 8 Essential Question(s): Solution: All points on the disc rotate with the same angular velocity. Hence it is independent of the distance from the centre. Answer: C Examiner’s Comments: Some candidates chose option B, which means they did not think about the concept but tried to solve the question mathematically. 9 Essential Question(s):  What are the conditions for a geostationary satellite? Solution:

4 Option A is incorrect: Radius of orbit is independent of the mass of the satellite. Option B is incorrect: Period of the satellites is 24 hours. Option D is incorrect: The angular speed is the same at orbit and on Earth surface. Hence the linear speed (v = r) is proportional to the distance of orbit to the centre and the distance of surface to the centre of the Earth. Option C is a condition for geostationary satellite. Answer: C Examiner’s Comments: 10 Essential Questions:  Is gravitational potential a scalar or vector quantity? Solution: The sum of gravitational potential at P is the sum of individual potentials, P GM GM GM dd d 4 10 1 1 2 2                   Answer: D Examiner’s Comment(s): This question about gravitational potential at a point confused some candidates who subtracted the potentials instead of adding them and chose option B. Gravitational potential is a scalar. 11 Essential Questions:  What is the relation between r.m.s. speed of molecules and temperature in degree Celsius? Solution: r.m.s.speed  T final r.m.s. speed / initial r.m.s. speed = Tfinal / T initial = 160 + 273.15 / 80 +273.15 = 1.11 final r.m.s. speed = initial r.m.s. speed x 1.11 = 350 x 1.11 = 388 m s-1 Answer: A Examiner’s Comment(s): Some candidates found this question about r.m.s. speeds challenging. A proportion of candidates thought that the r.m.s. speed was proportional to the kelvin temperature to give option B. A similar number left the temperature in degrees Celsius and then square-rooted to give option C. 12 Essential Question(s):  What is the total kinetic energy of a gas in term of its volume and pressure? Solution: total kinetic energy of a gas = N (3/2kT) = 3/2 nRT = 3/2 PV

5 = 3/2 (1.0 x 105) (0.10) = 1500 J Answer: C Examiner’s comments: 13 Essential Question(s):  What is the heat absorb by a body of mass m, specific heat capacity c and rise in temperature ? Solution: Heat absorbed = mc  Loss in KE = ½ m v2 KE loss by pellet is converted to thermal energy in the pellet = 50% of ½ m v2 = 0.5 x ½ m v2 KE loss by pellet = gain in thermal energy in the pellet 0.5 x ½ m v2 = mc   = v2 /4c Answer: B Examiner’s comments: 14 Essential Question(s):  What is the general equation that represents the velocity of a body in s.h.m.? Solution: v = vo cos wt = xow cos wt from graph xo = 0.30 m period T = 5.0 s w = 2/T = 2 x / 5 = 1.26 v = 0.3 x 1.26 cos 1.3 t = 0.38 cos 1.3t Answer: C Examiner’s comments: 15 Essential Question(s):  What are some examples of resonance? Solution: Option A is incorrect. The volume control increases the signal ( voltage) in the loudspeaker which increases the amplitude of vibration of the speaker cone. This is Not a resonance effect. Option B is correct. The vibrating metal strip produces a sound wave which resonate in the organ pipe. Option C is correct. The engine produces vibration on the lorry body which cause a forced oscillation of the mirror which resonate with the energy vibration. Option D is correct. The swinging of the leg produces a period force on the child which undergoes forced oscillation which oscillate at resonance when the periodic force frequency matches the frequency of the child on the swing.

6 Answer: A Examiner’s comments: This question asked which was not an example of resonance. The most common incorrect answer was option D. 16 Essential Question(s):  How to determine the speed of a waves from the displacement-time and displacement- distance graph? Solution: From the displacement-time graph, period = 1.0/5 = 0.2 s From the displacement-distance graph, wavelength = 1.6 /2 = 0.80 m Speed = wavelength x frequency = wavelength / period = 0.8/0.2 = 4.0 m s-1 Answer: D Examiner’s comments: 17 Essential Question(s):  What is produced between the loudspeaker and the board? Solution: XY = distance between two adjacent antinodes = wavelength / 2 wavelength = 2 XY = 2 x 5.0 = 10.0 cm from the trace on screen, ½ cycle = 3 divisions = 3 x 0.050 = 0.150 ms hence ½ period = 0.150 ms period = 0.300 ms frequency = 1/period = 1/ 0.300 ms = 3.33 kHz Answer: B Exa