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