2023 VJC Prelim P2 AS

1 VICTORIA JUNIOR COLLEGE SUGGESTED SOLUTIONS TO 2023 PHYSICS H1 PRELIM EXAM PAPER 2 Section A Q1(a) The extension of a spring or wire is proportional to the external force applied provided the proportional limit is not exceeded. [1] Q1(b) 2 1( )F k x x  where x is the length of the spring. [1] = 20(0.200-0.100) = 2.0 N [1] 2 1( ) 20(0.001 0.001) 0.04F k x x       N [1] Hence, (2.00 0.04 )F   N [1] Q1(c) Let 2 1x x x  . So 21 2U kx . [1] 2 1 2 1 2 x xU U x x     [1] 0.1 0.12 0.04020.0 10.0    [1] Q2(a) For the lower spring, the force acting on it is the weight of the lower mass. By Hooke’s Law, Mg = ke2 or cm2 0.200(9.81) 8.1824.0e   [1] For the upper spring, 1 2Mge k [1] cm16.4 [1] Q2(a)(ii) The total elastic PE stored in the system is 2 2 1 2 1 1 2 2 totE ke ke  [1] J2 21(24.0)(0.0818 0.164 ) 0.4032 totE    [1] Q2(a)(iii) When the system is cut at P, the tension in the upper spring is 2Mg while the weight of the upper mass is Mg. [1] By N2L, 2Mg – Mg = Ma [1]

2 Hence the upward acceleration of the upper mass is 9.81 m s-2 upwards. [1] Q2(a)(iv) The student’s use of v = u + at is wrong. [1] This is because the acceleration of the mass is not constant. [1] Q3(a) Tangential speed of acrobat B is v r  [1] (2.0 1.1) 1.29 3.999     4.00 m s-1 [1] Q3(b) The net vertical forces on A towards the centre of the circle provide the centripetal force for circular motion of acrobat A. 2mvmg N r  When acrobat A just loses contact with the cage, contact force N = 0 2 0 mvmg r  [1] 2vg r 2 9.81 2.9 v (c.g. of A is 2.9 m from the centre) [1] 5.333v   5.33 m s-1 [1] Q3(c) The net vertical forces on A provide the centripetal force for its circular motion. 2mvmg N r  When acrobat A is just in contact with the top of cage, contact force N = 0 2 0 mvmg r  [1] 2vg r 2 9.81 3.1 v (c.g. of A is 3.1 m from the centre) 5.5146 5.51v   m s-1 Minimum velocity is 5.51 m s-1 [1]

3 Q3(d) [2] [ 1 mark for correct forces; 1 mark for identification of forces] Q4(a) 2V r   I Rearranging, 2V r   I or 2 2 rV   I [1] Q4(b) A To centre of circular motion Normal contact force weight Friction between feet and floor Current / A 0.20 0.40 0.60 0.80 1.00 1.20 1.40 Voltage / V 0 1.0 5.0 4.0 3.0 2.0 X X X X X X (0.20, 4.5) (1.40, 1.0)

4 2 2 rV            I + [Apply knowledge of y = mx + c] Thus, we can see that the gradient of this graph is equal to 2 r    . [1] Take gradient points (0.20, 4.5) and (1.40, 1.0), (1m for gradient points correctly and clearly indicated and line clearly drawn) 2 1 2 12 y yr x x       4.5 1.0 2.9172 0.20 1.40 r         [1] 5.8333 5r   .83 Ω [1] Q4(c) Using point (1.40, 1.0) and 5.8333 r   substitute into 2 2 rV            I + 5.83331.0 (1.40) 2 2            + 10.166  = 10.2 V [1] Q4(d) When kept at 5.0 , the equation now becomes (5.0)V r   I I Rearranging, (5.0 )V r   I (5.0 )V r    I . Comparing to the original equation of line, 2 2 rV            I + We see that the gradient will be steeper, more negative. [1] The y-intercept will be doubled [1] [Note: No marks if student guesses without suitable explanation. Do not accept y-intercept is increased, as opposed to doubled. Not specific enough] Q4(e) By potential divider principle. R total RV R   To increase the reading of V, the total resistance of the circuit must fall. [1] To reduce the resistance of LDR, it should be brighter. [1] To reduce the resistance of thermistor, temperature should be higher. [1]

5 Q5(a) Half-life refers to the time taken for the number of a radioactive substance to decrease to half its initial value. [1] Q5(b) Number of protons of Pu239 94 is 94. [1] Number of neutrons = 239-94 = 145 [1] Q5(c) The power of the source is P AE where E is the energy of the emitted alpha-particle and A is the activity. [1] Hence 13 2.5 8.2 10A x  = 3.05 x 1012 Bq [1] Assumptions: (1) the energy released is only via the alpha particles. (2) the activity is effectively constant because of the long half-life. [1] [choose one assumption for one mark] Q5(d) When 40% of Pu239 94 nuclei have decayed, only 0.60 of the original nuclei are left. Hence using where number of half-lives 0 1 2 1 2 n N t nN t       , we have or 10.60 0.7372 n n     [1] Hence 40.737 2.4 10 t x or t =1.77 x 104 yr [1] Q6(a)(i) pwater = 190  1000  9.81 = 1.86 x 106 Pa [1] Q6(a)(ii) Using Bernoulli’s equation, pA + (1/2)  vA2 + hA  g = pB + (1/2)  vB2 + hB  g (patm + pwater) + (1/2)  (0)2 + hA  g = patm + (1/2)  vB2 + hB  g [1] (pwater) + hA  g = (1/2)  vB2 + hB  g (pwater) + (hA – hB)  g = (1/2)  vB2 1.86  106 + 1100  1000  9.81 = (1/2)  1000 vB2 vB = 159.1 m s-1 [1] = 159 m s-1 Q6(a)(iii) The cylindrical volume of water that exits each pipe per second = V/t = (πr2) v [1] Hence, mass of water that exits each pipe per second = (πr2) vρ Hence, mass of water that exits 6 pipes per second = 6(πr2) vρ = 6 × π × 0.0812 × 159 × 1000 [1] = 1.97 × 104 kg s-1. [1]

6 Q6(a)(iv) Efficiency = electrical power output / work done by water per second [1] 0.92 = 183 MW / work done by water per second Work done by water per second = 199 MW [1] Q6(a)(v) = R  [1] v = 1.7 × 500 × 2 / 60 = 89.0 m s-1 [1] Q6(a)(vi) P = F v [1] 199 × 106 = F × 89.0 F = 2.24 × 106 N [1] Q6(b)(i) The shape of the bucket causes the direction of flow of the water to be reversed. [1] This maximises the change in momentum of the water and so increases the force exerted by the water on the bucket. [1] Q6(b)(ii) Possible answers:  The buckets are not able to capture the kinetic energy of the impacting water entirely.  Energy is wasted in overcoming any frictional or resistive forces present between the moving parts of the turbine or electric generator. [1] Section B Q7(a) Method 1: Using equations of motion and taking downwards to be positive:    2 2 2 2 (1.5) 2 9.81 0.65v u as    [1] Final speed = 3.9 m s-1 [1] OR Method 2: Using conservation of energy Loss in GPE = Gain in KE   2 21 2mgh m v u      2 29 2.81 0.65 1 . 1 5v  [1] Final speed = 3.9 m s-1 [1]

7 Q7(a)(ii)  Straight line from X to – 3.9 m s-1 [1]  Gradient straight line from t1 to t2 must be parallel to line in first segment [1]  Line at t1 may join the two straight lines but its width must be narrow. [1] Q7(a)(iii) The speed of the ball after rebound is less than the speed just before impact. The ground/Earth is assumed to be stationary, hence there is a loss in the kinetic energy of the system during the bounce. [1] The bounce is inelastic. [1] Q7(a)(iv)  Downward