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) Part