12 SMSS 2010 Prelim P1 P2 Ans

Secondary 4 Express Paper 1 Paper 2 1 (a) ]1[s010.0t ]1[t)0.800(2 140.0 t)vu(2 1s = += += (b) ]1[s/m8000a ]1[)010.0(a00.80 atuv 2= += += 1 D 11 A 21 B 31 C 2 C 12 C 22 A 32 A 3 C 13 A 23 D 33 A 4 B 14 C 24 B 34 D 5 B 15 B 25 C 35 B 6 C 16 D 26 A 36 D 7 D 17 D 27 D 37 C 8 B 18 C 28 A 38 A 9 C 19 C 29 A 39 C 10 B 20 C 30 D 40 D PHYSICS 5058/01 & 02 St. Margaret’s Secondary School Preliminary Examinations 2010

2 0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22 t/s speed/ m/s 80 leaves bow hits target (c) 2 marks for correct shape graph with clearly labelled values 1 mark for correct labelling of graph (leaves bow & hits target) 1 mark for correct calculation of distance travelled by arrow during flight (0.40 m for acc portion; 15.6 m for the 80 m/s portion) 2 (a) Upward forces at A and B. [1] (b) W = mg = 10.8 x 10 = 108 N [1] (c) ⊥ distance of W from B = (70.0 cos 45°) - 12.0 = 37.5 cm [1] Taking moments about the B, ]1[N5.67F ]1[5.371080.60F ACMCM A A = ×=× = F B = 108 – 67.5 = 40.5 N [1] 3 (a) P = (120 x 1000 x 10) [1] = 1200000 Pa [1] (b) Pressure increases with depth [1] so a thicker base is needed to withstand the greater force per unit area at the base. [1] (c) (i) loss in GPE = gain in KE mgh = ½ mv 2 10 × 100 = ½ × v2 [1] v = 44.7 m/s [1]

3 (ii) Power generated = 3600 x 10 x 100 x 60% [1] = 3600000 x 60% = 2160000 W [1] 4 (a) The dust are kept in suspension as they are continuously hit by the air molecules/particles [1] which are at constant random motion and moving at high speeds. [1] (b) (i) As the gas is heated, the gas molecules gain K.E/speed up and hit the piston with a greater force and more often [1], hence pushing the piston outwards. (ii) As the piston moves outwards the volume of the gas increases and the rate of collision decreases, hence pressure decreases. [1] The piston stops moving when the pressure inside equals the pressure outside. [1] 5 (a) ]1[23r ]1[rsin 35sin47.1 rsin isinn °= = = (b) Both have the same refractive indices [1] therefore the speed of light does not change when it reaches the interf ace and no bending will be observed.[1] (c) (d) No. [1] The incident ray from air can only enter the corn oil at max. 90 °, this will give a refracted angle = critical angle of corn oil. [1] But for total internal reflection to occur, the incident angle at P needs to be greater than the critical angle of corn oil. [1] 6 (a) positive charges on the inside and negative charges on the outside of the can. (equal number must be drawn.) [1] 35° [1]

4 (b) The can will be positively charged. [1] The excess electrons will move from the can to the earth. [1] (c) No. [1] As metal rod and the hand are conductors of electricity, the electrons will flow from the rod to the hand and to earth and the rod will be neutral. [1] 7 (a) ]1[4R 20 5 5 1 20 1 R 1 // // Ω= = += R T = 2 + 4 = 6 Ω [1] (b) ]1[A0.2 6 12 R VI T T 1 = = = 4 I 2 = I3 I 2 + I3 = 5 I2 = 2.0 A I 2 = 0.4 A [1] I 3 = 1.6 A [1] 8 (a) W 2 experiences a force toward W1. Or W 2 experiences a force to the left. [1] (b) W 2 experiences a force of attraction to W1. W2 also experiences a force of attraction toward W3. [1] However the force between W2 and W3 is smaller as the distance between W2 and W3 is larger than between W1 and W2 and the current in W3 is smaller. [1] The net force on W2 is one of attraction toward W1 [1] but the magnitude is reduced due to the presence of W3, as compared to the force experienced in Fig. 8.1. [1]

5 9 (a) - Since power = voltage x current, a high voltage means a small current flows in the cable, so less power lost in the cable as thermal energy. - Alternating voltages will produce a changing magnetic field which is needed for a transformer to work. (b) Step-down transformer. [1] (c) (i) 11 1 kV275 kV25 V VratioTurns S P = = = Turns ratio is 1:11 [1] (ii) Power output = power input 275000 x I S = 400 IS = 1.45 x 10 -3 A [1] Assumption: transformer is 100 % efficient [1] (d) Heating effect of the coils or Heating effect in the soft-iron core due to eddy currents [1] (e) Combined resistance = 2 x (5000 x 0.0012) = 12 Ω [1] Power loss = I 2R = 402 x 12 [1] = 19200 W [1] 10 (a) Slip rings [1] and carbon brushes [1] correctly drawn and labelled. Current is clockwise in coil. [1] (b) Point Z. [1] Point X indicates the max output when the coil is at a horizontal position with maximum cutting of the magnetic field lines. Point Y indicates no output as the coil is in a vertical position whereby there is no cutting of magnetic field lines. [1] When the coil is inclined as shown, there will be some cutting of magnetic field lines which is responsible for an output value between the max and zero. [1] (c) The output graph shows a continuous curve – this is due to the continuous turning of the coil. [1] The graph has positive and negative values at alternate phase – this is due the current changing direction every half cycle as the coil rotates in the magnetic field – effect of electromagnetic induction. [1m] (d) The peak output value is halved [1] and the period is twice. [1]

6 Either 11 (a) A [1], the component of weight along the slope is greatest at A and friction has no effects on the skateboarder yet. [1] (b) GPE A = KE B mgh = ½ mv 2 v 2 = 2 gh = 2 x 10 x 1.8 [1] v = 6.0 m/s [1] (c) (i) Energy lost = mg (1.8 – 1.6) = 54 x 10 x 0.2 = 108 J [1] ]1[N5.13 ]1[0.8 108 .Dist lostEnergyForceFrictional = = = (ii) Energy needed at D = GPE A + WDf GPE D + KED = GPE A + WDf ½ mv 2 = GPE A – GPED + WDf 0.5 x 54 x v 2 = 108 + 108 [2] v = 2.8 m/s [1] OR 11 (a) Thermal energy needed to increase the temperature of 1 kg of the substance by 1°C (1K) (b) So that the liquid can be heated up by convection. [1] Hot liquid will expand and become less dense, moves up. Cool liquid on top is denser, sinks, to be heated up [1] (c) As temperature of liquid rises, more energy is lost to the surroundings. [1] Temperature will stop rising when the rate of thermal energy absorbed by liquid equals to rate of thermal energy given out to surroundings by the hot liquid. [1] (d) Energy received by surroundings equals to energy given out to surroundings [1] (e) P t = m c θ 240 (16 x 60) = (6) c (10) [1] c = 3840 J/(kg°C) [1] (f) In latent heat of vaporization, more energy needed to break liquid bonds and move them much further apart. [1] Energy also needed to lift molecules into atmosphere (work done against gravity and atmospheric pressure) [1]