HCI 2022 Prelim P2 Guide

1 2022 Preliminary Examination Paper 2 Suggested Solution Qn 1 Answer Marks (a) (i) When mass M is placed on the spring, the spring extends by a length of (L2 – L1). 21()mg k L L=− 21 1(0.0985)(9.81) 40.3 N m 0.037 0.013 mgk LL − = − == − C1 M1 (a) (ii) 21 21 () 0.2 0.2 + 0.1+ 98.5 3.7 - 1.3 LLkm k m L L − = + = − = 0.127 Thus 10.2 0.2 0.140.3( ) 5 N m 98.5 3.7 1.3 k k k k −+ = + = − = (Do not penalise if k is left to a value that is greater than 1 s.f. – this mark will be marked in (a)(iii)) M1 A1 (a)(iii) k = (40 ± 5) N m-1 k expressed to 1 s.f. and k expressed to the same place value as k. Allow ecf from (a)(ii) and (a)(iii). A1 (b)(i) Marking Guidance: B1 – 3 forces drawn with correct direction B1 – Relative lengths of forces show T+U = W [-1] – Missing labels/legend Note that first B1 mark must be obtained before the second B1 mark can be given. [Point of action is not marked here. But in general, students should know that the upthrust is acting at the centre of buoyancy – c.g. of the liquid displaced and weight is at the c.g. of the cube.] B2 (b)(ii) Initially (Fig. 1.2), P is inequilibrum, net force on it is zero. Weight = upthrust 07( . )W V g = M1 (b)(iii) After cube Q was connected (Fig. 1.3), at new equilibrium, net force on P is zero. Weight of P = Tension due to string + new upthrust Tension upthrust weight

2 'W T U=+ = T + 0.4Vg Solving the two equations above gives 0 4 304 0 7 7 .. . WT W V g W W= − = − = (Shown) M1 M1 (b)(iii) Consider the forces acting on cube Q. Resolving the weight of cube Q along the slope, and since it is in equilibrium, T = (3/7) W = W sin (θ)  Solving, θ = 25.4 ᵒ A1 [Total : 11 marks ]

3 Qn 2 Answer Marks (a) (i) For circular motion, centripetal force is provided by the gravitational force, 2 2 23 2 2 2 ) 4( J J JMmGm GM m mRR RTR MT GR  = = = M1 A1 (a) (ii) Since T2  R3, 2 3 2 3 0.676 3.18 2.62 Th Th Am Am Am TR T R T       ==             TAm = 0.506 Earth-days M1 A1 (b)(i) 2 JGMv TR R== OR Since the orbit is circular, centripetal force is provided by gravitational force, hence 2 2 JJGM m GMmv vRRR =  = A1 A1 (b)(ii) When the mass of the moon decreases, although the gravitational force and centripetal force required both decreases, the condition for orbit, that the gravitational force is equal to the centripetal force, is still true. Hence the moon will stay in orbit. B1 (c)(i) Read off graph with orbital period equal to one Jupiter-day = 0.417 Earth-days Orbital radius of geostationary orbit is 2.30 RJ. (Allow :  ½ smallest div) A1 (c)(ii) Able to continuously observe the same area on Jupiter for an extended period of time. B1 [Total : 8 marks ]

4 Qn 3 Answer Marks (a) Using v = fλ v = 3 1 1.4 4.0 10 −   = 350 m s-1 M1 A1 (b)(i) Particle R (As the particle is undergoing SHM, at the amplitude, the instantaneous velocity of the particle is zero). A1 (b)(ii) Particle Q . A1 (b)(iii) Particle Q. (As the Fig. 3.2 shows a particle that is initially at equilibrium and is moving up in the next instant. Particle P and Q are both at equilibrium at t = 0. To determine which particle is presented above, the subsequent displacement-position graph in the next instant needs to be used. As seen below, since the wave is travelling towards the left, it is clear that particle P will be displaced in the negative direction and particle Q in the positive direction. Hence particle Q is depicted in Fig. 3.2). A1 (c)(i) Using equation . 0.7360 3601.4 x   =  =   180= Accept working with radians. And answer as  radians M1 A1 Vertical lines represent equilibrium position of particle along the wave. Particle P is at the centre of compression and particle Q is at the centre of rarefaction | | | | 0.7 1.4 2.1 2.8 position / m displacement / nm 5.00 −5.00 P Q R Direction of wave progression is to the left.

5 (c)(ii) Distance travelled by wave in 1 ms = vt = (350)(1 x 10-3) = 0.35 m The graph should have shifted to the left by 0.35 m. Award mark as long as one full wavelength is drawn with displacement at 5.00 nm at initial position. C1 A1 (c)(iii) At t = 0, Particle R is at +5.00 nm displacement so Particle S should be at -5.00 nm displacement (phase difference of π rad). A1 [Total : 10 marks ] displacement / nm t / ms 5.00 −5.00 | | | | | | 2.0 4.0 6.0 8.0 10.0 12.0 Z | | | | 0.7 1.4 2.1 2.8 position / m displacement / nm 5.00 −5.00 P Q R Y

6 Qn 4 Answer Marks (a)(i) Power supplied by battery = Energy / time = (11.3 J) /(10 x 60 s) = 0.0188 W M1 A1 (a)(ii) From graph, resistance of thermistor = 3.1 kΩ Power dissipated through the resistor and thermistor = Power supplied by battery  0.0188 = E2 /(3100+1200) E = 8.99 V B1 M1 A1 (b) From graph, New resistance of thermistor = 2.0 kΩ Since the power delivered is the same, Total resistance before =total resistance after  3100+1200 = R +2000  R = 2 300 Ω B1 M1 A1 [Total : 8 marks ]

7 Qn 5 Answer Marks (a) (Using Fleming’s Left hand Rule), the magnetic force will always be perpendicular to the direction of motion of the charge. This force will provide the centripetal force for circular motion. Since the magnitude of the magnetic force is constant , the magnitude of the centripetal force is constant. The electron will describe a uniform circular path. Or since the force is always perpendicular to the motion, no work is done by the force on the system and there is no gain in kinetic energy of the electron and hence the electron is in uniform circular motion (or moving at constant speed). B1 – for explaining why the motion is that of a circular one. B1 – for explaining why the motion is uniform B1 B1 (b)(i) The magnetic force provides for the centripetal force, 2mvBqv r= 31 7 3 19 2 (9.11 10 )(8.5 10 ) (7.5 10 )(1.60 10 ) 6.45 10 m 6.5cm mvr Bq − −− − ==  =  = M1 A1 (b)(ii) B1- circular path curving downwards in B-field B1 - emerging at side PO (allow e.c.f.) B1 – straight path after leaving the B-field B3 (b)(iii) The magnetic force on electron provides the required centripetal force to keep the electron in circular motion. If m is the mass of the electron and ω is the angular velocity of the electron. 2 2 2    2   mr TT Be mBev mr  =  ==  Hence, T is independent of v and r. M1 (b)(iv) Since period T is independent of the speed and radius of the path taken, the period of both electrons are the same. Hence, the time spent by each electron in the magnetic field simply B1

8 depends on the fraction of the circular path traveled by each electron within the magnetic field. Electron e1 travels for less than ¼ of a period in the field, while electron e2 travels for ½ a period. Hence, electron e2 spends a longer time in the magnetic field. A1 [Total : 10 marks ] Qn 6 Answer Marks (a) The activity of a radioactive source is the rate at which a source of unstable nuclei decays or the number of disintegrations per unit time. A1 (b)(i) ( ) 23 6 6.02 102.40 10 235 N − =   15 6.148 10= M1 A1 (b)(ii) Decay constant 15 0.1919 6.148 10 A N  ==  17 17 1 3.1213 10 3.12 10 s (to 3 s.f.) − − − =  =  M1 A1 (c)(i) 42 42 0 19 20 1K Ca e antineutrino−→ + + beta particles are emitted. Accept electrons. Accept: beta with (anti-)neutrino A1 (c)(ii) Half-life = 12.5 h A1 (c)(iii) Correct Shape B1 sum