2021 GCE A Level H3 Physics 9814 Suggested Solutions

1 H3 9814 Physics (2023) Practice Papers 2021 A-Level H3 Physics Suggested Solutions 1 (a) ( ) 2 2 1 2 1100 9.812 4.5152 s 4.5 s (shown) s ut at t t = + = = ≈ [1] (b) (i) ( ) ( )( ) 2 2 1 2 1 9.81 2 4.51522 400 m s ut at s = + =  = [1] (ii) Objects can be launched upwards from the bottom (with a speed of 44.3 m s–1) instead of being dropped at the top from rest. [1] (c) (i) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) 22 2 1 2 2 12 12 12 1 2 2 Before safety net, 0 2 9.81 During safety net, 0 2 4 9.81 2 9.81 2 4 9.81 0 4 0 1 100 2 Solving simultaneous equations, 80 180 9.812 4.0386 s 4.0 s 4.0 s is v u as vx vx xx xx xx x t t = + = + = −  −=  −= += = ∴= = ≈   about 0.5 s less than the previous time of 4.5 s. (shown) [1] [1] [1] [1] (ii) 2 From previous part's simultaneous equations, 20 mx = [1]

2 H3 9814 Physics (2023) Practice Papers 2 (a) ( ) ( ) ( ) 12 12 21 2 2 2 By conservation of momentum, 44 44 1 Since collision is elastic, 2 Solving simultaneous equations, 8 5 Fraction of KE transferred 812 5 0.64 64%1 42 mv mv mv v vv vv v vv mv mv = + = + = − =  = = ≡   [1] [1] [1] (b) (i) 4, , 45 4 5 41 55 440 55 C C mC mC mv mv vv u vvv u vv = = = −= = −= − [3] (ii) 11 max 1 15sin sin 14.5 (3 s.f.)4 45 v v θ −−  = = = °  [1] [2] 4 5 v 1 5 v 4 5 v 1 5 v 4 5 v 4 5 v 1 5 v 1v maxθ The maximum angle is when the COM velocity, the velocity of the mass in the COM frame, and the velocity of the mass in the lab frame form a right- angled triangle as shown. The magnitudes of the velocities of each mass in the COM frame do not change before and after the collision, as dictated by kinetic energy conservation (elastic collision).

3 H3 9814 Physics (2023) Practice Papers 3. (a) 2 2 2 2 2 22 2 2 2 11KE of mass is 22 Since angular momentum 1KE 2 2 The total energy of the system is GP E + KE 2 T pm mv m LL pr p r pL m mr GMm LE r mr = = =⇒= ∴= = = = −+ [1] [1] [1] (b) (i) For a body moving in an elliptical orbit, its total energy is given by 22 2 T 1 dr d GMmE= m r dt dt r θ  +−   22 22 22 T T T eff 11 22 22 dr d GMmm =E m rdt dt r L GMm L GMmE E EUmr r mr r θ  ∴ −+      = −+ = − − = −   22 22 22 2 2 2 22 2 At the apses, 0, furthermore, both total energy and angular momentum are constant 22 1 1 11 222 1 2 T T pa pa p a pap a pa dr dt EL GMm L GMm LE rr mr mr L L GMm GMm L GMmr r m rrmr mr r r L m = ∴ = −+= −+   ∴ − = − ⇒ −= −    ⇒ ( ) ( ) ( ) 2 2 2 2 111 11 2 11 The total energy 2 1 pa pa pa pa a p pa pa T pp p a pp pa pp ap ap a GMmr rL GMmGMmrrrr rr m r r rr GMmr rGMm L GMmE rr mr r rr rrGMm GMm rr rr rr GMm rr    − += −⇒= =     +   + ∴ = −+= −+ +      = −− = −   ++   =− +( ) 2p GMm a=− [1] [1] [1] [1]

4 H3 9814 Physics (2023) Practice Papers (ii) 2 11 32 12 12 1 Total energy of the system is GPE + KE 1 + 22 112 2 112 6.67 10 5.4 10 2.8 10 2 3.8 10 130 km s TE GMm GMmmvra v GM ra − − = ∴− =− ∴= −  =× × ×× − × ×× = [1] [1] [1] 4 (a) oN VdBk L µ ρ= = I l oNk Vd L µρ= × Il Substitute L Nd= and 2 4 V VA V dI R π ρρ= = = ll into the above equation: 2 7 4 4 9.87 10 o o Vdk Vd d µρπ ρ µπ − =×× = = × l l The unit of k is the same as that of µo, which is T m A−1. [1] [1] [1] [1] [1] (b) (i) 3 7 8 3.0 0.25 109.87 10 1.7 10 2.9 VdBk ρ − − − = ××=×× ×× = 0.015 T l [1] [1] (ii) The formula oNB L µ= I is obtained by assuming that the solenoid is infinitely long and hence has infinite number of turns. Real solenoids are of finite length and finite number of turns. Fewer turns mean weaker magnetic field. [1] 5 (a) Heating a gas at constant volume: Using the first law of thermodynamics, U W Q pV Q∆= +=∆+ , since 0QU V∴= ∆ ∆= 3 32 2 V Nk TQUC NkTT T ∆∆= = = =∆∆ ∆ [1] [1]

5 H3 9814 Physics (2023) Practice Papers Heating a gas at constant pressure: Using the first law of thermodynamics, UWQ∆= + 3 2 PNk T p V C T∆ = −∆ + ∆ since the gas expands, W has a negative value. 3 2 P pVNk C T ∆= −+∆ From ideal gas law, pV NkT pVp V Nk T Nk T = ∆∴∆ = ∆ ⇒ = ∆ 35 22 PC Nk Nk Nk∴=+ = [1] [1] [1] (c) In the first scenario, the heat supplied to the gas only increases its internal energy and there is no work done by the gas. However, in the second scenario, the heat supplied increases its internal energy by the same amount (for the same increase in temperature as before) and enables the gas to do work against the external pressure as well. Hence the heat supplied in the second scenario is greater and this explains why Cp is greater than CV. [1] 6 (a) (i) [1] (ii) As the electric force is upwards, which is opposite in direction as the downward electric field, the oil droplet has a negative charge. Hence, it gained electrons. [1] (iii) 3 3 3 3 4 3 4 32 1 6 1 (shown)6 Vr D D mV D π π π ρ πρ = =  = ∴= = [1] weight Electric force

6 H3 9814 Physics (2023) Practice Papers (iv) ( ) ( ) 3 36 17 1 6 1 960 0.5 106 6.2831 10 kg mD πρ π − − = = × = × ( ) 17 17 17 3 3 1 0.13 6.2831 10100 0.5 3.8327 10 4 10 kg (to 1 s.f.) mD mD Dmm D ρ ρ ρ ρ − − − ∆∆ ∆ = +  ∆∆ ∆= +    = +×   = × = × ( ) 176 4 10 kgm −∴=±× [1] [1] [1] (b) (i) [1] (ii) At terminal velocity, drag force = weight 2 2 3 3 3 3 3 3 3 3 1 6 18 D D D F Dv mg Fv D gDB v gD D F gD mg gD gD πη πη ρ η ρ πη η πηρ η πηρ η πρ = = = = =   = =   = [1] (iii) In equilibrium, weight = electric force drag force weight

7 H3 9814 Physics (2023) Practice Papers (where n is the number of electrons) (where d is the distance between plates) EWF mg neE Vne d neV mgd = = =  = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 3 2 3 2 2 3 2 3 2 2 3 2 3 2 2 3 3 2 3 2 2 3 2 1 2 3 3 35 2 2 11 22 11 -10 22 11 -10 22 6 6 6.0 10 1.8 10 18 6 960 9.81 1.8879 10 kg m s 1.89 10 kg m s (to 3 s.f.) neVk v neV gD B neV B gD mgd B gD D gd B gD dB g ρ η η ρ η ρ πρ η ρ πη ρ π −− − − = =   = = = = ×× = = × = × [1] [1] [1] (c) (i) Since 3 2kVv ne= , when a graph of V is plotted against v1.5, the gradient of the graph is k ne , which varies with n, the number of electrons in the oil droplet. Millikan analyzed several droplets with different number of electrons n in each, hence only for data with the same number of n will yield a linear fit, and not a linear fit for all the data. [1] [1] [1]

8 H3 9814 Physics (2023) Practice Papers (ii) [1 mark] for drawing the three lines of best-fit For the top line, 3 6 3 6 9 3.000 0.500 10Gradient 2.50 0.40 10 2.500 10 2.10 10 1.19 10 − − −= −  =   = × For the middle line, 3 6 3 6 9 3.250 1.000 10Gradient 5.50 1.70 10 2.250 10 3.80 10 0.592 10 − − −= −  =   = × (2.50, 3.000) (0.40, 0.500) (5.50, 3.250) (1.70, 1.000) (5.50, 2.150) (1.10, 0.400)

9 H3 9814 Physics (2023) Practice Papers For the bottom line, 3 6 3 6 9 2.150 0.400 10Gradient 5.50 1.10 10 1.750 10 4.40 10 0.398 10 − − −= −  =   = × [1 mark] for calculating the values of the three gradients. 9 9 Gradient of top line 1.19 10 Gradient of middle line 0.592 10 2.0101