2023 RI H2 Physics Prelims P1 Answers

1 © Raffles Institution [Turn over 2023 Preliminary Examinations H2 Physics Paper 1 Solutions 1 B When  is zero, F2 = F (largest), F1 = 0 (smallest). When  is 90, F2 = 0 (smallest), F1 = F (largest). 2 C 22 2yyv u gy=− At the highest point, vy = 0. Hence, 202 yu gy=− 2yu gy= Since the maximum height reached is the same for all three paths, uy is the same too. Option A: Sy = uyt + at2, since all three paths end on ground level, time of flight is the same. Path Z has the longest range. Sx = uxt, path Z has the highest horizontal component and the largest initial speed 22() xyu u u=+ . Option B: Sy = uyt + at2, since all three paths end on ground level, time of flight is the same. Option D: Path X has the shortest range. Sx = uxt, path X has the lowest horizontal component. 3 A By Newton’s second law, taking the direction of F1 as positive, 12 12 netF ma F F ma FFa m = −= −= Since object was at rest, F2 is initially zero ( ) 2 2Fv and acceleration is maximum initially. Acceleration then decreases as F2 increases with increasing speed as object accelerates. Acceleration becomes zero eventually when F1 = F2 and object travels at constant (maximum) speed henceforth. 4 A At equilibrium, the lines of action of three forces must meet at a point. The arrows in the vector triangle must form a closed loop. 5 C ( )( )( ) ( )( )( ) 3 3 0.120 750 9.81 0.060 1000 9.81 1.4715 10 1.47 10 Pa o o w wP h g h g=+ =+ = = 6 D Work done is area under the force extension graph. This would be the work done to stretch the spring from x1 to x2. ceiling wall string Q bar P weight tension Force of wall on bar weight tension Force of wall on bar

2 © Raffles Institution 7 C ( ) ( ) ( ) P PQ Q PQ total gain in energy total loss in energ y gain in GPE gain in KE loss in GPE W.D. b y friction 1.5sin30 0 gain in KE 1.5 0 1.5PQm g m g f = + = + − + = − − ( ) ( ) ( ) ( )( )( ) ( )( ) ( )( )( ) PQgain in KE 1.5 1.5 1.5sin30 4.0 9.81 1.5 2.5 1.5 3.0 9.81 1.5sin30 33.0375 33 J QPm g f m g= − −  = − −  = = 8 D 2 2 2 1.2 0.040 9.81 0.30 1 0.238862 mvT mg r mv mv += +  = = By conservation of energy, ( ) 22 2 2 11 22 10.23886 0.040 9.81 2 0.30 0.040 2 23.715 mv mgh mu u u += +    =   = 2 223.715 79.1 m s0.30 ua r −= = = 9 B Graph G1 for path from S to Q means that the two stars have different masses (since there is no point where resultant force is zero). Graph G 3 shows that star X has a smaller mass than star Y (the values of g at start is lesser than at the end). The point of g = 0 is closer to X. Since the spacecraft is accelerating, it is possible for the time duration from P to point of g = 0 to be approximately equal to the time duration from point of g = 0 to R. 10 C Given Q = – 2.0 J kg−1 R Q R PQ 1 PR 4 1 (2.0) 0.5J4    == = − = − Work done = m ( R – Q ) = (3.0) (−0.5 + 2.0) = 4.5 J Work done is positive, since direction of external force is same as direction of displacement from Q to R. 11 A The defining equation of simple harmonic motion is 2ax =− with a and x having opposite signs. Hence, the graphs are a − t and x − t.

3 © Raffles Institution [Turn over 12 B Total energy E, ( ) 2 0 2 0 2 0 22 0 2 1 2 1 2 12 2 2 E mv mx mx T mx T    = = =  = ( ) ( ) ( ) 2 2 0 0 2 2 0 0 2 2 00 '' ' '' ' 12 18 ' 2 and ' 33 8.0 mJ xET E x T x TEE xT x x T T  =   =  = = = = 13 C ( ) ( ) ( ) ( ) II I I I II I I I      = = −   = = − = −=  − = = −    = = 2 0 2 012 1 2 0 2 01 2 2 1 00 0 0 0 0 0 0 cos 20 , and cos 20 cos 20 cos 20 0.25 cos20 cos 20 0.50 0.50cos 20 0.53209cos20 20 57.853 77.853 78 14 D A stationary wave forms between G and H. As such the phase difference between E and F is  as they are in adjacent loops. As the frequency and period of points E and F are the same, F has to travel faster and hence possesses a higher maximum speed and thus maximum kinetic energy as compared to E. 15 A ( )( ) ( )( ) 9 red red5 9 blue blue5 1 sin 1 700 10 20.495.0 10 1 sin 1 400 10 11.545.0 10 8.95 9.0    − − =   =  =   =   = =  16 A Constant pressure compression – in the p−T diagram it is still a horizontal line. pv = nRT. Since temperature decreases in a constant pressure compression, it is represented by a horizontal line from right to left. Constant temperature expansion – Temperature is constant and pressure decreases. It is represented by a downward vertical line in the p−T diagram. Constant volume process – From pV nRT= , since V is constant, the graph of p against T is a straight line passing through the origin. This is why option D is wrong.

4 © Raffles Institution 17 D Since temperature is constant, 0U= . Since the kettle is rated as 500 W, rate of electrical work done on the coil is 500 W. The heating coil continues to lose heat to the water. From the first law of thermodynamics, 500 WdQ dt =− . 18 A Taking towards the right as positive, The forces acting on the electric due to +Q and −Q are both towards the left. 19 B Since oil drop is stationary, weight of oil drop is equal to electric force. 19 13 50003.2 10 1.0 10 N0.016 Vmg q d −−= =   =  13 13 19 2 ' 4000 1.0 101.0 10 3.2 10 0.016 9.81 1.96 m s Vmg q mad a a − −− − −=  −   =  = OR 5000Vmg q q dd== 4000 4' 0.8 5 EF q mg mgd= = = mg − 0.8mg = ma a = 1.96 m s−2 20 B Since current is flowing through the specimen, current must be the same in each section. IX = IY = IZ Since I = nevA, as I, n and e are constant, v is inversely related to A. vX > vZ > vY 21 C p.d. across the 400  resistor = p.d. across the 600  resistor Since I V= R , Current through the 600  resistor = II = + 400 2 400 600 5 ( ) ( ) I I  = = =   2 2 120power dissipated across 120 1.25 1.3power dissipated across 600 2 6005 22 D Effective e.m.f. = 2.0 V p.d. across the 3.0  resistor = ( ) =+ 3.0 2.0 1.2 V2.0 3.0 p.d. between X and Y = 1.2 + 3.0 = 4.2 V (going from negative terminal to positive terminal of 3.0 V battery, potential increases by 3.0 V) 0 F r +Q −Q Fnet FX FY 3.0 V 2.0  3.0  5.0 V Y X I I I +3.0 V +1.2 V

5 © Raffles Institution [Turn over 23 A Since X and Z carry the same current of 3.0 A and are the same distance from Y, the magnitude of the forces Fon Z due to Y and Fon X due to Y are the same. 61on Z due to Y 0 0 5.0 3.0 8.0 10 N m22 Y Z z F L d d   −−=  =  = I I Since the IX and IY are in same directions, the two wires attract each other and the force per unit length acting on X due to Y is 8.0  10−6 N m−1 to the right. ( ) ( ) 6on X due to Z 0 0 3.0 1 3.03.0 8.0 102 2 2 2 2 5.0 Z X z F L d d   −=  =  =   I I = 2.4  10−6 N m−1 to the left Resultant force per unit length on wire X = (8.0 − 2.4)  10−6 = 5.6  10−6 N m−1 to the right 24 D When the magnitude of the magnetic flux density increases, the frame experiences an increase in magnetic flux linkage out of the plane of the paper. By Lenz’s law, an induced current will flow in the frame to create a magnetic field into the plane of the paper to oppose this change. Hence, this leads to a clockwise current flowing through the frame. Option A: Sliding the rod right or left will induce an e.m.f. across the rod. Since the rod is in the centre of the metal frame, it will produce currents in opposite directions in the left and right sections of the frame. Option B: Sliding the rod right or left will induce an e.m.f. across the rod. Since the rod