2023 VJC Prelim P1 AS

1 VICTORIA JUNIOR COLLEGE SUGGESTED SOLUTIONS TO 2023 PHYSICS H1 PRELIM EXAM PAPER 1 Q1 C Q7 A Q13 C Q19 D Q25 B Q2 C Q8 D Q14 D Q20 B Q26 A Q3 B Q9 C Q15 B Q21 B Q27 A Q4 A Q10 C Q16 A Q22 C Q28 B Q5 D Q11 D Q17 C Q23 D Q29 C Q6 D Q12 B Q18 B Q24 C Q30 D Q1 Check for homogeneity of equations. An inhomogeneous equation is an incorrect equation. Let base units be represented by BU. A: LHS: BU = kg m2 s-2 RHS: BU = (kg m s-1)(m s-1) = kg m2 s-2 LHS = RHS  homogeneous equation (wrong answer) B: LHS : BU = (kg m2 s-2)2 = kg2 m4 s-4 RHS: BU of p2c2 = (kg m s-1)2 (m s-1)2 = kg2 m4 s-4 BU of m2c4 = kg2 (m s-1)4 = kg2 m4 s-4 All terms on both sides of the equal sign have the same units  equation is homogeneous (wrong answer) C: LHS: BU = kg m2 s-2 RHS: BU =(kg m s-1) 2(m s-1) = kg m3 s-3  inhomogeneous equation (correct answer) D: LHS: BU = kg m2 s-2 RHS: BU =   2  -1 2 -2kg m s kg m skg homogeneous equation (wrong answer) Ans: C Q2 2 26 AD B A D CB C B A D C         20.1 0.01 0.052 7.7 10 7.7%2.0 3.45 4.10 B xB       Ans: C Q3 The second student’s measurements are closer to one another and are therefore more precise compared to the measurements of the first student. Ans: B

2 Q4 Area under v-t graph is the displacement. For graph P, the area is always positive, meaning that there is positive displacement. For graph Q, the area during the negative velocities and the area during positive velocities are equal. Hence the total displacement is zero. Displacements for graphs P and Q are thus different. The magnitudes of the total area under the graphs P and Q are different. Thus the total distances travelled for P and Q is different. Ans: A Q5 2 2 2 2 2 for vertical motion 3 2( 9.81)( 2.4) taking upward as posi tive direction 56.09 y y y y yv u a s v        2 -2 m s Horizontal velocity, 5.0x xv u  -1 ms Speed of toy pilot,   2 2 2 1 56.09 5.0 9.0msy xv v v      Ans: D Q6 Here the vertical forces are gravitational and magnetic forces. The horizontal forces are the electric force and air resistance. The electrically charged oil drop is moving horizontally at constant velocity. The oil drop is in dynamic equilibrium and the resultant force is zero in all directions. Vertically, the gravitational force is acting downwards. Therefore, the magnetic force must be acting upwards to ensure equilibrium. The electric force cannot act vertically in this case because the electric field is horizontal, and air resistance should be acting in the direction opposite to the motion, which is horizontal in this case. Ans: D Q7 By N2L and taking the loads-pulley system as a whole, 8.0g – 2.0g – f = (8.0 + 2.0)a 6.0 x 9.81 – 3.0 = 10.0a  a = 5.6 m s-2 Ans: A

3 Q8 From Newton’s 3rd Law, force of P on Q (action force) is equal and opposite to force Q on P (reaction force). Ans: D Q9 Ans: C Q10 Net moment about P = 0 (rotational equilibrium) T1 × x = T2 × (1 – x) k1 (e) x = k2 (e) (1 – x), since both springs have the same extension e. Also, k1 = 2k2 (given), so 2x = 1 – x x = 1/3 = 0.33 m Ans: C Q11 Action-reaction forces are equal in magnitude, are oppositely directed, act on two different bodies and are of the same type. Ans: D Q12 For the mass of 500 g to move downwards and the mass of 200 g to move upwards, the friction F in the pulley must act opposite to the motion of the masses. For the masses to move at constant speed, net force = 0. Considering the pulley and masses as a whole, Mg – F – mg = 0 or F = (M – m)g F = (0.500 – 0.200) x 9.81 = 2.94 N Speed of rope = v = r ω = 0.035 x 2.4 = 0.084 m s-1 Hence, rate of work done = P = F v = 2.94 x 0.084 = 0.25 W Since the force exerted by the motor on the rope is opposite in direction to the velocity of the rope, the rate of work done = - 0.25 W Ans: B Q13 From graph, F = kx or 4.5 = k (0.200 – 0.050)  k = 30.0 N m-1 By the Law of Energy Conservation, Loss in GPE of load = Gain in elastic PE of spring mgx = (1/2)kx2 or 0.150 × 9.81 (x) = (1/2) (30.0) x2 x (15.0 x – 1.47) = 0 or x = 0.0981 m = 9.81 cm So, maximum length attained = 5.0 + 9.81 = 14.8 cm Ans: C

4 Q14 The spring is shortest when the carts are moving with the same velocity. By the Law of Conservation of Momentum, Initial total momentum = Total momentum at instant of common velocity (2m)u + m(-u) = (2m + m)V taking right as positive direction. V = (1/3)u = common velocity when spring is shortest. By the Law of Conservation of Energy, Gain in elastic PE = Loss in KE = initial total KE – final total KE = (1/2)(2m)u2 + (1/2)mu2 – (1/2)(3m)V2 = (3/2)mu2 – (3/2)m (u/3)2 = (3/2)mu2 (1 – 1/9) = (4/3)mu2 Ans: D Q15 Q16 v r  24.0 201.2 v r    rad s-1. Ans: A Q17 The resultant gravitational field on the 1 kg mass is zero. Therefore, 2 31 2 2 2 (15 ) GM MGM M x x   or 2 3 2 1 (15 ) 16 15 9 4 15 4 45 33 M x x M x x x x xx        Hence x = 6.4 m Ans: C N mg Friction, f θ To centre of path The centripetal force for circular motion is constituted by the sum of the component forces of N and f towards the centre of the circular path. Hence 2 sin cos mvN f r   Ans: B

5 Q18 a = g = GM/r2 is independent of the mass m of the falling body. M is the mass of the earth. Ans: B Q19 The p.d. across the 4 resistor, 1.5 4 6.0V R   I V Hence the current in the 13 resistor, 6.0 0.461513 V R  I A The total current flowing in the internal resistor = 0.4615 1.5 1.9615  A Using P V I , we get 2.1 1.0706 1.11.9615V    V Hence emf = total pd = 3.0 + 6.0 + 1.1 = 10.1 V Ans: D Q20 The resistance should start at zero (s = 0), increase in magnitude and then end at zero (s = L). Let resistance per unit length of the wire be k. The resistance between X and Y will be that of two resistors in parallel. Thus 1 1 2 1 1R R R        = 1 2 1 2 R R R R  or [ ( )] ( ) ( ) ks k L s ks L sR ks k L s L      The variation of R with s is therefore not a straight-line relationship. Ans: B Q21 Since the 10  resistor and 30 resistor are in parallel, they have the same p.d. 2 10 10 VP   , 2 30 30 VP   Hence 10 30 3P P    10 30 30 30 3 3 3 4total P P P P P        or 10 3 4P P   while 30 1 4P P  . Ans: B Q22 When connected in series, the same current runs through the resistor and the thermistor. From the graph, we can thus see that the p.d. across the resistor is around 1.2 V and the p.d. across the thermistor is around 3.1 V. This means that the power supply has an emf of 4.3 V.

6 If we now have the components in parallel with the supply, it means that the p.d. across each component is 4.3 V. Drawing a vertical line at 4.3 V, we see that the current through the resistor is around 0.355 A and the current through the thermistor is aro und 0.285 A. The total current drawn from the supply is the sum, hence current is 0.64 A Ans: C Q23 Ans: D Q24 The currents in the two wires are opposite and hence the two wires repel. The magnitude of the forces acting on each wire is the same by Newton’s third law. Ans: C Q25 Ans: B Q26 F = Eq = 3.0 x107x1.6 x10-19 = 4.8 x10-12 N Ans: A Q27 Binding energy = mass defect x c2 = (mass of nucleons – mass of nucleus ) x c2 = [(2 x 1.0087)+ (2 x 1.0073) – 4.0015 ] x 931 MeV = 28.4 MeV Ans: A

7 Q28 Let the parent and daughter nuclei be X and Y respectively. Then 0 1 1 A A Z ZX Y    Hence the atomic number represented by Z increases by 1 unit. Ans: B Q29 Ans: C Q30 In 4 half-lives, the number of nuclei remaining is 4 0 0 1 1 2 16N N     . Hence