2022 A Level H2 Physics P1 Soln

Suggested solutions to 2022 A-Level H2 Physics Paper 1 1 A 11 A 21 D 2 C 12 C 22 B 3 B 13 B 23 C 4 B 14 C 24 D 5 D 15 A 25 B 6 C 16 D 26 A 7 B 17 B 27 B 8 C 18 B 28 A 9 C 19 A 29 C 10 D 20 A 30 C Q Ans Working 1 A Therefore as Y is rotated through 180o, the magnitude of X – Y gets smaller. 2 C Vertical velocity:    2 2 1 1 2 10 (2) 9.81 2 2 9.81 m s y y y y s u t gt u u            Horizontal velocity: 1 100 (2) 50 m s x x x x s u t u u     Angle: 1 9.81tan 11.1 50 o         X Y X – Y X Y X – Y

3 B Let the mass of P be mP and the mass of Q be mQ. Since collision is elastic, relative speed of approach is equal to relative speed of separation,   1 2 0 1.5 m s " " 0.50 P Q Q P v v u u v v          By conservation of momentum,         0.502.0 0 2.5 1.5 1.5 0.602.5 P P Q P Q P Q m m m v m m m m        4 B Since the buoy is in equilibrium,         31 4 0.5 1030 9.81 200 9.812 3 683 N submerged U W T V g mg T T T                 P Q P Q 2.0 m s-1 v 0.50 m s -1 Upthrust U Weight W Tension T

5 D Forces acting on mass: Fs = kx = 25 (0.060) = 1.5 N Resolving forces horizontally T sin 36= 1.5 ------(1) Resolving forces vertically T cos 36 = W ------(2) (1)/(2): tan 36 = 1.5/W Hence, W =2.1 N 6 C     2 11 26 24 24 3 36 9.81 6.67 10 6.37 10 5.968 10 kg 5.968 10 5510 kg m4 6.37 103 Mg G r M M Mdensity Volume              7 B Work done against air resistance = loss in gravitational potential energy Height of pendulum at 2nd oscillations 0.10(2)0.60 0.491 m Hence, 0.600 0.491 0.109 m Loss in GPE 0.40(9.81)(0.109) 0.43 J h e h mg h           8 C 2 v r T     Angular velocity ω is independent of distance from centre of the disc. 9 C A and B are incorrect. The orbital period is independent of the mass of the satellite as it is given by Kepler’s Third Law: 2 3 4T r GM  Where M is the mass of the planet. D is incorrect. The geostationary satellite has the same angular velocity as the point on Earth below it, but it has much higher linear velocity than the point on Earth below it. For instance, the linear speed of a point on the Equator is about 465 m s-1 whereas the linear speed of a satellite in geostationary orbit is 3075 m s-1. 36° W T Fs

10 D The resultant gravitational potential at P is the scalar addition of the potentials of each mass at that point. The combined gravitational potential at point P is   4 4 2 2 10 M M G MGM d d GM d         11 A The relationship between the r.m.s. speed of gas molecules and the absolute (Kelvin) temperature is given by 21 3 2 2m c kT  Where m is the mass of a molecule, c is the RMS speed, k is the Boltzmann constant and T is the absolute temperature. Hence, dividing the equations gives     2. . 2 . . 1 1 . . 2 2