DHS 2022 Prelim P3 Guide

1 2022 DHS H2 Physics Prelim Paper 3 Suggested Solutions Section A 1 (a) (i) (50 to 200) x 10-3 kg A1 (ii) diameter of tennis ball is about 6 – 7 cm volume = 4 3r3 from 110 to 180 cm3 A1 (b) units of Q: As C1 C = Q2 2E units of C = A2s2 kgm2s−2 = kg–1 m–2 A2 s4 A1 2 (a) (i) 1. equal B1 2. density of ice is less than density of water B1 explanation: 1. at equilibrium, weight of ice = upthrust on ice miceg = mdisplaced waterg mice = mdisplaced water 2. Viceice = Vdisplaced waterwater since Vice  Vdisplaced water, then ice < water (ii) mass of ice becomes equal mass of water after melted/ mass of melted ice equal mass of displaced water B1 volume of melted ice equals volume of water displaced/ melted ice fills the space of water displaced by ice B1 so level does not change A1 (b) (i) Upthrust is equal in magnitude and opposite in direction to the weight of the fluid displaced by a submerged or floating object. B1 (ii) Upthrust = weight of water displaced by the anchor = density of water x volume of anchor x g = 1030 x 0.50 x 9.81 C1 = 5050 N A1 (iii) Let volume of air be V Upthrust on lifting bag + upthrust on anchor – weight of anchor = ma (1030)V (9.81) + 5050 – 7800(0.50)(9.81) = 7800(0.50)(2.50) C1 V = 4.25 m3 A1

2 3 (a) (i) 5.00 cm A1 (ii)  = 2 T C1 = 2 4.0 = 1.6 rad s−1 A1 (iii) v0 =  x0 C1 = (1.6)(5.0) = 8.0 cm s−1 A1 (b) Any three points below, 1 mark each B3 • initial pull was to the right • distance from X to trolley (at equilibrium) is 20 cm • initial motion undamped • motion becomes damped at/from 12 s • damping is light • maximum speed at 1 s, 3 s, etc. / stationary at 2 s, 4 s, etc. (c) sketch closed loop encircle (20,0) – see below B1 minimum L shown as 15 cm and maximum L shown as 25 cm and minimum v shown as –8.0 cm s–1 and maximum v shown as 8.0 cm s–1 B1

3 4 (a) (i) Fundamental mode of vibration,  = 4L = 0.680 or L = 0.17 m (not formed) 1st overtone:  = 4L 3 = 0.680 or L = 0.51 m (formed) A1 2nd overtone:  = 4L 5 = 0.680 or L = 0.85 m (formed) A1 (ii) (b) waves pass through/enter the slits in the grating B1 waves spread out after passing through/entering the slits B1 (c) (i) d sin = n C1 or sin = n d  Hence G = n d or d = 4 G A1 (ii) straight line from 400 mm to 700 mm that is always below given line M1 straight line has smaller gradient than given line and is 5 small squares high at wavelength 700 nm A1 A at open end, N at closed end; An A and N in between, equally spaced (by eye) A1 (ii)

4 5 (a) The velocity v may be resolved into two components – one parallel to the magnetic field B, which is v cos , and the other perpendicular to B, which is v sin . B0 v sin results in a circular motion of the proton in the plane perpendicular to B. B1 Using Fleming’s Left-hand Rule, the centripetal force is directed into the page initially. B1 v cos causes the proton / c ircle to move at constant speed in the same direction as B. B1 Consequently, the proton moves in a helical path. B1 (b) (i) A1 (ii) Using sinF Bqv = , we get ( )( ) ( ) F to sf − − − − =    =  = 4 19 5 0 18 18 3.8 10 1.6 10 1 10 sin55 4.98 10 N 5.0 10 N 2 (iii) central axis coil v = 1.0  105 m s−1 P 55o Fig. 5.2 C1 A1 magnetic flux density distance coil 1 coil 2 mid-plane shape; symmetry about mid-plane A1

5 6 (a) (i) The direction of induced e.m.f. produces effects that oppose the change in magnetic flux B1 (ii) As the current in the solenoid is switched on, the magnetic field in solenoid is increases. B1 This causes an increase in magnetic flux linkage through the small coil. B1 By Lenz’s law, the magnetic field in the coil point upwards to oppose the increase in flux linkage. B1 (iii) B1 (b) e.m.f. = N∆ / ∆t C1 = (75 × 1.4 × 10–3 × 7.0 × 10–4) / 0.12 C1 = 6.1 × 10–4 V A1 7 (a) (i) Q = (MRa − MRn −M)c2 = (226.0254 − 222.0176 −4.0026) x (3.00 x 108)2 = 7.7688 x 10−13 J C1 = (7.7688 x 10−13) / (106 x 1.6 x 10-19) M1 = 4.86 MeV A0 (ii) Conservation of momentum: MV + M V = 0 (1) Conservation of energy: (2) From (1): (3) Sub (3) into (2) (iii) 1. 4.00264.86 1 222.0176K =+  C1 4.77 MeVK = A1 2. The alpha particles carries away most of the energy (98%). B1  Q = 1 2 MV2 + 1 2 MV 2  V = − M M       V  Q = 1 2 M − M M       V       2 + 1 2 MV 2 = 1 2 MV 2 M M +1       = K M M +1       M1 M1 V V M1 A0

6 (b) alpha particle kinetic energy per mm = 4.77 25 = 0.191 MeV C1 energy to produce an ion pair = 0.191 x 1.6x10−13 5.0 x 103 C1 = 6.1 x 10−18 J A1 Section B 8 (a) (i) 2mu A1 (ii) 2L / u A1 (iii) force = N x change in momentum / time = N 2mu / (2L / u) = Nmu2 / L A1 (iv) pressure = force / area = (Nmu2 / L) / L2 = Nmu2 / L3 A1 (b) (i) Molecules has component of velocity in three directions. B1 or 2 2 2 2 X Y Zc c c c= + + or 2 2 2 2 X Y Zc c c c =  +   +   Since the molecules are in random motion, on average, 2 2 2 X Y Zc c c = =  B1 Thus 22 22 3 1 3 X X cc cc  =       From (a) (iv), p = Nmu2 / L3 or pV = Nmu2 21 3pV Nm c=   (ii) pV = NkT C1 NkT = 1 3Nm<c2> multiply 3 2 to both sides: 1 2m<c2> = 3 2kT and 1 2m<c2> = EK M1 so EK = 3 2kT A0 (c) 1 2 × 3.34 × 10–27 × <c2> = 3 2 × 1.38 × 10–23 × (25 + 273) C1 r.m.s. speed = 1.9 × 103 m s–1 A1 B1 A0 c2 = ux2 + uy2 + uz2 <c2> = <ux2> + <uy2> + <uz2> <ux2> = <uy2> = <uz2> <c2> = 3<ux2> <ux2> =

7 (d) internal energy = ΣEK of molecules + ΣEP of molecules B1 no forces between molecules, so potential energy of molecules is zero B1 (e) (i) increase in internal energy = Q + work done B1 constant volume so no work done B1 (ii) thermal energy per unit mass to cause a unit change in temperature B1 (iii) c = Q / NmΔT C1 = [N × (3/2)kΔT] / (NmΔT) M1 = 3k / 2m A0 (f) As the gas expands, it does work against the atmosphere/external pressure B1 for same temperature rise , more thermal energy needed, so larger specific heat capacity B1 9 (a) similarity: both are radial or both have inverse square variation with distance B1 difference: direction is always/only towards the mass or direction can be towards or away from charge B1 (b) gravitational force = mg = 1.67 x 10−27 (9.81) = 1.64 x 10−26 N A1 electric force = qE = qV d = 1.6x10-19