2023 RI Prelim P1 AS

1 © Raffles Institution [Turn over 2023 Preliminary Examination H1 Physics Paper 1 Solutions 1 D W W FdV Q t t   I I 2 2 3 1kg m sunits of V kg m s AA s m     2 C The average apple is between 70 and 100 grams. Weight = 0.100  9.81 = 0.98 N 3 B When  is zero, F2 = F (largest), F1 = 0 (smallest). When  is 90, F2 = 0 (smallest), F1 = F (largest). 4 C   2 1 1 1 2 2 2 2 1 2 1 2 1 2 1 (4 ) 82 1 (4 ) 82 Hence, 16 8 2 s a a s a a s s a a a a               2 1 2 1 2 At 8 s, 1 82 64 m t s s a a      5 B 2 2 2 2 2 2 2 2 2 2 2 v u as v u as ua s a ua s a                    Hence, the graph of v against s is a “square-root” graph shifted to the right by 2 2u a to the right and then reflected about the vertical line passing through 2 2s u a . 6 C 2 2 2y yv u gy  At the highest point, vy = 0. Hence, 20 2 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 2 2( ) x yu 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.

2 © Raffles Institution 7 A By Newton’s second law, taking the direction of F1 as positive, 1 2 1 2 netF ma F F ma F Fa m     Since object was at rest, F2 is initially zero   2 2F v 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. 8 A W is the force the Earth exerts on the brick and S is the force the floor exerts on the brick. Hence, by Newton’s third law, equal and opposite force to W is the force the brick exerts on the Earth, and equal and opposite force to S is the force the brick exerts on the floor. 9 C Let the mass of trolley be m and the angle the slope makes with the horizontal be . By Newton’s second law, taking the direction down the slope as positive, net sin sin F ma mg ma a g      where g is the acceleration of free fall. Acceleration is independent of mass. 10 D The initial momentum of the 2 kg sphere is 8 kg m s1 to the right. The initial momentum of the 3 kg sphere is 18 kg m s 1 to the left. The total initial momentum (before collision) is 10 kg m s1 to the left. Thus, both spheres cannot come to rest at the same time, otherwise principle of conservation of linear momentum will be violated. 11 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. 12 B Since the student’s head is only rotating, there is no resultant force acting on the head. The rotation of her head is produced by the couple acting on her head. ceiling wall string Q bar P weight tension Force of wall on bar weight tension Force of wall on bar

3 © Raffles Institution [Turn over 13 B                   clockwise moment about pivot anticlockwis e moment about pivot 2.0 2.0 5.0 is the cross-sectional area2.0 2.5      wood wood plastic plastic plastic wood wood plastic V g x V g x V x V x xA x AxA x     14 D Work done is area under the force extension graph. This would be the work done to stretch the spring from x1 to x2. 15 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. a gainst friction 1.5sin30 0 gain in KE 1.5 0 1.5P Qm 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 1.5 sin30 33.0375 33 J Q Pm g f m g          16 A            6 Work done against resistive forces 400 100 0 J 400 1000 JEnergy from fuel 0.16 400 1000 0.16Mass of fuel needed kg 45 10 55.6 g 56 g 17 B             Power Since force is constant, graph of vs is a straight line passing through the origin. Fp Fv F u at F a t = F t m F p t p t 18 D 2 2 2 1.2 0.040 9.81 0.30 1 0.238862 mvT mg r mv mv       By conservation of energy,   2 2 2 2 1 1 2 2 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   

4 © Raffles Institution 19 A Angular velocity and angular displacement are the same for both points since they are rotating on the same disc. 20 A 2 2 2 GMm mv rr GM vr   1 3 Q P P Q Q v r v r v v   21 B 2 A B 240 , 2.215 VP R R R      22 1 1 4 L LR R d A dd R               32.215 9.229 10 0.096240 A B B A d R d R      22 C p.d. across the 400  resistor = p.d. across the 600  resistor Since I V= R , Current through the 600  resistor = I I      400 2 400 600 5     I I          2 2 120power dissipated across 120 1.25 1.3power dissipated across 600 2 6005 23 D Effective e.m.f. = 2.0 V p.d. across the 3.0  resistor =    3.0 2.0 1.2V2.0 3.0 p.d. between X and Y = 1.2 + 3.0 = 4.2 V (going from negative terminal to positive terminal 24 B Use Fleming’s left hand rule to determine the direction of the forces. 25 B     1 1 3 3 2 2 1 1 0.5 cos 20 0.5 cos 20 0.5 7.5 10 cos 20 3.5 10 mN F B L F B L B L F            I I I 2 26 C B 2 F ma mv mv mBqv r r r Bq q       OLi Li O O Li 7 2 7 16 1 8 qr m r m q     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 27 A Each  decay causes the nucleus to lose 4 nucleons including 2 protons. After 8 decays, the nucleus has 241 – 8(4) = 209 nucleons left, and 94 – 8(2) = 78 protons left. After 5  decays, the nucleus has 78 + 5 = 83 protons, which matches 209 83Bi . 28 D Total energy of products = 939 + 940 = 1879 MeV Total mass of deuteron = 1876 MeV Hence, the deuteron needs to capture an energy of 1879 – 1876 = 3 MeV 29 D The mass of the radioisotope has no effect on safety concerns. Option A: Short half-life will be preferred so that the radioisotope is not radioactive in the body for too long. Option B: The daughter nucleus should not be harmful to the body. Option C: The intensity should not be too high to cause harm. 30 B This shows the random nature of radioactive decay. Option A: This is due to the law of decay, and not the spontaneous nature. Option C: This is the spontaneous nature in which the rate of decay is unaffected by external conditions. Option D: This is a fact not related to random nature.