2013_PJC_H2_Physics_P2 (Solutions)

1 Answers to 2013 JC2 Preliminary Examination Paper 2 (H2 Physics) Suggested Solutions: No. Solution 1(a) A head-on collision takes place along the line joining the centres of the colliding bodies. An elastic collision is one in which the kinetic energy is conserved. 1(b)(i) The total momentum of a system is constant, provided no external resultant force acts on it. 1(b)(ii) By conservation of linear momentum, mu1 = mv1  12mv2 ==> u1  v1 = 12v2 ……….(1) velocity of approach = -velocity of separation u1  0 = v2  v1 u1  v1 = v2 ………(2) (1)  (2) 2u1 = 13 v2 ==> v2 = 2 13 u1 From (2) v1 = v2  u1 = 13 2 u1  u1 =  13 11 u1 Thus ratio 1 1 u v =  13 11 1(c) Required fraction 22 11 21 2 11 1 ()2 1[ ]1 2 mu v v umu    28.0 ]13 11[1 2   1(d) Carbon-12 atom is more massive and slows down the neutrons while the neutron has the same mass and will not slow down the incoming neutron.

2 2(a) Work done is W = Fs , where s is the displacement in the direction of the force. To move the mass through a vertical height h without acceleration, an external force is needed to overcome the weight of the mass. Thus, F ext = mg Work done = Fext s = mgh = change in gravitational potential energy. 2(b)(i) Applying Hooke’s law, F = ke 150 500 N m0.100 Fk e    Elastic potential energy stored in both strands 22 212 (500)(0.350 0.200) =11.3 J 2 ke ke   2(b)(ii) From conservation of energy, gained in G.P.E. = lost in E.P.E. mgh = 11.3 h = 19.1 m 3(a)(i) x = rsinωt 3(a)(ii) Differentiating x with respect to time t, we have v = ωrcosωt Differentiating v with respect to time t, we have a = ω 2rsinωt Replacing rsinωt by x, we have a = ω2x Hence the shadow on the screen undergoes simple harmonic motion. s = h mass m Fext Fext Direction of gravitational force due to uniform field

3 3(b) v = ω√[xo 2x2] = 4.0√[0.1502 0.0752] = 0.52 m s1 3(c) a = ω2xo = 4.02  0.15 = 2.4 m s2 4(a) Zero electric field strengths in sphere A (between x = 0 and x = 1.4 cm) and in sphere B (between x = 11.4 and x = 12.0 cm) 4(b)(i) The charges on the spheres are both positive because the field strength is zero at a point between the spheres or the electric fields are in opposite directions. 4(b)(ii) At x = 0.08 m, the electric field strength due to sphere A cancels out the electric field strength due to sphere B. Electric field strength due to sphere A 2 o A A 4 x QE   --- (1) Electric field strength due to sphere B  2 o B B 12.0 4 x QE    --- (2) EA = EB  2 o B 2 o A 12.0 4 4 x Q x Q      2 o B 2 o A 04.0 408.0 4  QQ  404.0 08.0 2 B A     Q Q Allow estimation from graph, 7.8 cm < x < 8.2 cm Alternatively: Electric field strength at surface of sphere A,  2 Ao B 2 Ao A A 12.0