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 4 4 r Q r QE    --- (1) Electric field strength at surface of sphere B,  2 Bo A 2 Bo B B 12.0 4 4 r Q r QE    --- (2) (1) / (2)   676.0170 115 12.04 1 12.04 1 B A 2 B A 2 B B 0 2 A B 2 A A 0                      E E r Q r Q r Q r Q  

4               2 B A 2 B B 2 A B 2 A A 12.0 676.0 12.0 r Q r Q r Q r Q                       2 B 2 B B2 B 2 A A 12.0 1676.0 12.0 676.01 rr Q rr Q       66.3 6.012 676.0 4.1 1 4.112 1 6.0 676.0 12.0 676.01 12.0 1676.0 22 22 2 B 2 A 2 B 2 B B A            rr rr Q Q 4(b)(iii) Diagram Deduct 1 mark if any of the following is not shown.  Correct field line direction and shape.  At least more field lines radiating out of sphere A than sphere B.  Location of neutral point nearer to sphere B. 4(c)(i) The field strength is negative of the potential gradient, i.e. dx dVE  (not x V ). 4(c)(ii)     8cmx 2 cmx dxEV Hence, change in potential from x = 2.0 cm to x = 8.0 cm, ∆V = area under E−x graph (from x = 2.0 cm to x = 8.0 cm) Counting the number of squares, estimated about 3 (1 cm  25  10 6 N C−1) squares or 75 (2 mm  5  106 N C−1) squares ∆V = V 105.7)102501.0(3 56  (accept any logical estimation of area under E−x graph) Magnitude of W.D. by external force, W = q  ∆V W.D. = 0.20  7.5 x 105 = 1.5  105 J (accept ± 10 % deviation)

5 5(a)(i) 5(a)(ii) When p-type and n-type materials are placed together, free electrons, from n-type material, diffuse across junction to fill up holes in the p-type material producing negative ions in p- type material leaving positively charges ions in n-type material. This process continues until an electric field is set up to prevent any further diffusion of electron through the p-n junction. This leads to the formation of a layer depleted of any mobile charges at the junction and this layer is called the depletion region. 5(a)(iii) 5(b) The energy between valence band and conduction band is narrow at 1 eV. At 0K, there are no electrons in the conduction band and the valence band is fully filled. At temperatures > 0K, a significant number of electrons become thermally excited and move into the conduction band, leaving holes behind in the valence band. As temperature rises, more electrons-holes pairs are produced resulting in more charge carriers and thus reducing the resistance. 6(a)(i) ln ln B TRA e BRA T   Taking temperatures at o50 C(323.15K)and o80 C(353.15K) , R= 110 Ω and 50 Ω respectively. Fig. 5.1 p n p n Fig. 5.1

6 ln110 ln 323.15 BA … (1) ln50 ln 353.15 BA … (2) (1)  (2), ln110 ln50 323.15 353.15 BB  3 2 3.0 10 K 1.02 10 B A    6(a)(ii) 6(b)(i) At 30.0 oC, the resistance of X is approximately 188 Ω. By potential divider principle, 40 61 . 0 5 V40 188V   6(b)(ii) The voltmeter reading should increase. From Fig. 6.1, as the temperature of the water is raised, the resistance of Device X decreases. Using the potential divider principle, the p.d. across the 40 Ω will increase. 6(c)(i) On the positive cycle, the diode is forward biased. The diode conducts. On the negative cycle, the diode is reversed biased. The diode does not conduct. Hence, the AC input is V I

7 half-wave rectified. 6(c)(ii) 2 1163 V22 rms TV T    6(c)(iii) 1. LED 1 will be flashing but appears lighted up throughout due to the high frequency of flashing. 2. The LEDs will light up alternately, but the human eye will not be able to differentiate the rapid flashing, resulting in both LEDs being seemed to be lighted up at the same time. 3. Neither LED will light up.

8 7. Suggested solution: Diagram (equipment to be used is shown in the diagram) Fig. 7.1 Aim: To investigate the relationship between the thickness of the dielectric material in a capacitor and the amount of charge stored. Independent variable: thickness of the dielectric material Dependent variable: amount of charge stored in the capacitor, by determining the area under the current-time graph as the capacitor discharges. Controlled variables : - the e.m.f. of the battery used to charge the capacitors, - type of dielectric material Procedure: (a) Select a capacitor with a certain dielectric material of thickness t and connect the capacitor to the circuit shown in Fig. 7.1. (b) Close switch 1, leaving switch 2 open to charge the capacitor. The capacitor is fully charged when the voltmeter connected acro ss the capacitor reaches a maximum reading. (c) With the circuit connected, record the p.d. V of the fully charged capacitor. (d) After the capacitor is fully charged, open switch 1 to di