2020 RI Prelims H2 Phy Paper 1 Soln

© Raffles Institution 9749/01   2020 Raffles Institution Preliminary Examinations – H2 Physics Paper 1 Suggested Solutions 1 B    12 12 22 33 1 1222 2 2 33 122 J s N m kgunits of m s kg m s m s kg m s m kg m s mm G c               2 C  1212 2 3 4 14 3 2 2Pa b c d a b c d   32 42 134% 2% 6% 2 5%42 1% 2% 9% 10% 22% Pab c d Pa b cd         3 B The gradient of the displacement-time graph gives the velocity. Car M is accelerating as it has an increasing velocity (since gradient increases). Car N is not accelerating as it has a const ant velocity (since gradient is constant). Statement is incorrect. 4 D p mu Since horizontal velocity u remains unchanged, cosuv  . horizontal direction: Since horizontal velocity remains unchanged, initial momentum = final momentum change in momentum = 0 vertical direction: initial momentum = 0 final momentum sinmv  (downwards) change in momentum sin 0 sin tan tancos umv m mu p       (downwards) 5 D Frictional force on the boxes by the belt, in the forward direction, is needed to change their horizontal state of motion from rest to the final speed of 0.80 m s–1. By Newton’s third law, these boxes exert an equal and opposite force (in the backward direction) on the belt which slows it down. An external force that is equal and opposite (in the forward direction) to this force needs to be applied to the belt to keep it moving at constant speed. horizontal force applied,   0.80 2.5 2.0 NdmFv dt 

2 © Raffles Institution 9749/01   6 C Since the pyramid is floating in the water, this means the upthrust on the pyramid is equal and opposite to the weight of the pyrami d. (Hence option B is correct since the pyramid remains floating even when a liquid of higher density is added.) Upthrust is given by the weight of the water displaced by the submerged part of the pyramid. When the pyramid is inverted and floating, it w ill sink deeper in order to displace the same volume of water to achieve the same upthrust. This means the pressure at the submerged apex is higher than that at the submerged base when the pyramid is upright. Hence the pressure difference between the surface of the water and the lowest point of the pyramid is greater when it is inverted. (This explains why option C is incorrect and why option A is correct.) Option D is correct as proven below: ('w' for water, 'o' for object) constant for water and object of the same density ww oo o w wo Vg Vg V V      7 C decrease in G.P.E. increase in K.E.  energy lost to surroundings energy lost to surroundings  decrease in G.P.E. – increase in K.E.     22 22 11 22 110.80 9.81 10 4.0 1.222 72.7 73 J mg h mv mu        8 C Since body is moving up the track with constant velocity, there is no change in its K.E. power supplied = rate of increase in G.P.E. + rate at which work is done against the resistive forces      70 9.81 20sin30 120 20 9267 9300 W mg h Fd tt     9 A linear speed  12 200total distance 2 or 20.9 21 m stotal time 2 30 rr T       average velocity  12 200total displacement 2 13.3 13 m stotal time 30 30 r     10 B By Newton’s third law, force on bridge by car = force on car by bridge (i.e. normal contact force N) Since speed is constant, for each bridge, the car experiences the same centripetal force. Hence the normal contact force is larges t when the car is at the highest point of each curvature. Compare the normal contact force at the highest point of each curvature.

3 © Raffles Institution 9749/01   For P and Q: 22 vm vWNm NW rr    P has a smaller radius, hence NP < NQ For R: 22 RR vm vNW m NW rr   For S: NS = W Therefore NP < NQ < NS < NR 11 A Gravitational force provides the centripetal force for the satellite. 2 2 2 2 GMm mv rr GM vr v    2 11 2 1 1 16 4 25 5 v v vv v v          12 D The minimum kinetic energy that the object has just before it hits the surface of Planet Y at R is when the object just reaches Q i.e. K.E. is zero at Q. By the conservation of energy, for the object moving from Q to R, increase in K.E. = decrease in G.P.E.   , , 0 5.0 1.2 62.0 304 MJ kR Q R kR Em E       13 D The pressure of the gas on the walls is due to the force per unit area the gas molecules exert on the walls when they collide with the walls. This force is due to the rate of transfer of momentum to the wall which increases with increased frequency of collisions when the gas is compressed to a smaller volume. Hence pressure increases. 14 D For the isothermal process: U1 = Q1 + W1  0 = Q + W1  W1 = Q For the adiabatic process: U2 = Q2 + W2  U2 = 0 + W2  U2 = W2 Compare area under the p-V curves for each process, |W2| > |W1|  W2 > W1 = Q  U2 = W2 > Q 15 D The total distance travelled in one period T is 04x . So average speed is 00 0 0442 2 2 x xx vv T       16 C The frequency of forced oscillations will follow that of the driving force.

4 © Raffles Institution 9749/01   17 A Air molecules immediately to the left of P have positive displacement i.e. they are moving to the right towards P. Air molecules to t he right of P have negative displacement i.e. they are moving to the left towards P. Hence, Point P is a point of compression. 18 C Using Malus’ Law 2 10 22 2 2 0 21 0 cos cos (90 ) cos sin sin 2 4          II III I      When 20 and 90 , 0    I . When Q is being rotated through 90, I2 increases then decreases. 19 D At resonance, a stationary wave is formed in the tube. If f is the fundamental frequency of the air column in the tube, for resonance to occur, frequency of the tuning fork can be f, 3f, (2n–1)f for n 1, 2, 3… (i.e. odd harmonics). 20 A In order to increase the resolving power of the instrument, the limiting angle of resolution min of the aperture should be made smaller. From Rayleigh criterion, min b   , decreasing  (i.e. increasing f) and increasing the aperture size b would decrease min. 21 B Graph of option B satisfies the following points: The negative of the potential gradient gives the electric field strength ( dVE dr ). The closer the electric field lines, the larger the electric field strength E. Field lines point from a region of higher potential to a region of lower potential. 22 B EqFE qm a a m  2 211 22 Eq Lya t mu   where L is the length of the plate and u is the initial horizontal speed  2 0.5 .. 0.5 2 0.5040 . 5 0.50 p qqyy KEmu ey K yK e yy        OR Since the mass of the alpha particle is four times the mass of the proton and the alpha particle’s K.E. is also four times the proton’s K.E., this implies that the proton and the alpha particle had the same initial speed. 2 2 0.504 0.50 p pp qqyy mmu my e ym e yy        P Q R I0 I1 I2 

5 © Raffles Institution 9749/01   23 A Same current flows through both lamps in series. Hence, P  R. 1 32 4 24 6 V4 XX X YY Y X X X PRV PRV V V V     24 A Since 3 VPQV  and 6 VPSV  , potential at Q is 3 V higher than the potential at S. The p.d. across QS is 3 V. Since 5 VQRV  and potential at Q is higher than the potential at S by 3 V, the p.d. across SR is 2 V. 25 A magnetic flux density at centre of coil Q, 00 22 QQ Q NB rr  II magnetic flux density at a distance 2r from wire P at centre of coil Q,  00 22 2 PP PB dr    II For the resultant magnetic field to be zero at the centre of coil Q, BQ and BP a the centre must be equal in magnitude and opp