VJC H1 PHY Solutions

1 2016 VJC Prelim H1 Paper 1 Suggested Solutions 1 A 11 C 21 C 2 B 12 C 22 B 3 B 13 D 23 C 4 A 14 C 24 C 5 B 15 C 25 D 6 B 16 D 26 A 7 D 17 C 27 C 8 B 18 C 28 B 9 B 19 B 29 C 10 C 20 C 30 B 1 Ans: A Energy produced = Pt = 3.0 x 109 x 2.0 x 10-12 = 6.0 x 10-3 J = 6.0 x 10-15 TJ 2 Ans: B Minimum hypotenuse length is given by sides of 8.8 cm and 6.8 cm. Minimum hypotenuse length = 22 8.68.8  = 11.12 cm Maximum hypotenuse length is given by sides of 9.2 cm and 7.2 cm. Maximum hypotenuse length = 22 2.72.9  = 11.68 cm Average = (11.12 + 11.68)/2 = 11.40 cm Uncertainty = extreme – average = 11.68 – 11.40 = 0.28  0.3 cm

2 3 Ans: B Uncertainty of period = 20 5.05.0  = 0.05 s 4 Ans: A  The ball starts with zero speed. So when s = 0, v = 0.  The ball can move either up or down. So v should change sign.  But the ball is always below the point P. So s should always have the same sign. 5 Ans: B Remember: v = vf - vi The horizontal component of vi = vf since there’s no horizontal acceleration, so v is vertical. v = 12 sin25o = 5.1 m s-1 6 Ans: B Take upwards as positive. Vertically, s = (u sin)t + ½at2 -15.0 = (8.50 x sin40.0o)t – (½ x 9.81)t2 Solving, t = 2.39 s (reject negative answer) 25o vf -vi = 12 m s-1 v  = 40.0o s= -15.0 m u= 8.50 m s-1

3 7 Ans: D Taking moments about pivot: Wrubber x (1.90 L) = Wsteel x (0.10 L) 1.0Lx A x rubber x 1.90Lg = 3.00L x A x steel x 0.10L  𝜌𝑠𝑡𝑒𝑒𝑙 𝜌𝑟𝑢𝑏𝑏𝑒𝑟 = 1.0𝐿𝐴×1.90𝐿 3.00𝐿𝐴×0.10𝐿 = 6.33 8 Ans: B Torque by a couple = F x perpendicular distance between them = F x L sin 9 Ans: B Weight of helicopter is the gravitational force of Earth on the helicopter. So by N3L, the helicopter will exert an equal and opposite gravitational force on the Earth. 10 Ans: C Statement C is false because the collision forces are internal, not external, forces, and so the total momentum of the system should be conserved throughout the whole duration of the collision. 11 Ans: C Area under F-t graph = change in momentum ½ x (30 + 15)x 1500 – ½ x15 x 1500 = m(v – 10) v = 32.5 m s-1 12 Ans: C Since object is at constant speed up the inclined plane: Total w.d by 50 N force = Gain in GPE + w.d against friction 1500 = 50 x 12 + w.d against friction  w.d against friction = 1500 – 600 = 900 J 13 Ans: D When boat is travelling at constant speed v, the driving force is equal to the drag force. Then the power P is given by: P = Fv = Dv = (kv2)v Ie. P = kv3 When both engines are working: 2 x 30 kW = k(10)3 - - - - (1) When only one engine is working: 30 kW = kv3 - - - - (2) (2)/(1): and solve for v: v = 7.9 m s-1

4 14 Ans: C The electrical power is the input power = IV 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 = 𝑃𝑜𝑢𝑡 𝐼𝑉 0.80 = 4.0×106 25×103×𝐼 I = 200 A 15 Ans: C Upon reflection at the wall, both pulses undergo a 180 degree phase change, the earlier pulse will end up positive and the later will end up negative, ruling out B and D. If the reflection of the earlier pulse meets the later pulse before it reflects, their superposition will yield C as the answer. 16 Ans: D After passing through a polarizer, unpolarised light’s intensity is halved (therefore we eliminate options A/B). Amplitude becomes 2/A Since the polarization angle is 75°, using Malus’ Law:  15sin2/75cos2/' AAA  17 Ans: C 2 2 r AkI  2 2 2 2 2 2 2 2 2 W 1125.11 4 3)2(5' ' ''   m A AI r r A A I I 18 Ans: C When a loud sound is heard for the 3rd time: cm 21 25.115 4 5l       19 Ans: B

5 20 Ans: C   0.5x m m ,difference path    8.0400 320 4.06.12.16.1 22 f v x The 2 waves arrive out of phase at all times, resulting in destructive interference. 21 Ans: C Recall “P = IV ”, so V = P/I The electrical potential difference between two points in a wire carrying a current may be defined as the ratio of the power supplied to the current between the points. 22 Ans: B dd d dd dd d L d L d LR d L d LR A lR S S S S SS S S S SAL AL 71.0 2 1 2 1 12 2 2 )2( ),2()1( )2( 2 )1( 2 )2( 2 22 22 22 2 22                                     

6 23 Ans: C Effective resistance,              00.250.050.000.1 00.1 1 00.1 1 00.1 1 00.1 100.1 11 effR 24. Ans: C Effective resistance of the 3 parallel resistors,        588.00.5 1 0.2 1 0.1 1 1 effR Potential difference across each resistor, V = IReff = (5.0)(0.588) = 2.94 V Current through 2.0  resistor, I2 = V/R = 2.94/2.0 = 1.47  1.5 A 25. Ans: D - The direction of the flux density due to 1 and 2 must be the same - The resultant direction of the flux density due to 1 and 2 must upwards Therefore using Right hand grip rule, we can conclude that the current in 1 and 2 must be anticlockwise and - The direction of the flux density due to 2 and 3 must be opposite to each other. And also that the current in 3 is opposite to 2 and hence must be clockwise 26 Ans: A Since the currents in QR and XY are out of phase, the currents in PS and XY are in phase as PSRQ are in a loop. Like currents attract, therefore the attractive force between them is always positive. Also the currents vary from zero to a maximum, so the attractive force also varies from zero to a maximum. 27 Ans: C m N 8 233 104.6 )108(10501.020 22        LNBIL lF Fd B 

7 28 Ans: B de Broglie wavelength   m106.2 )105.1(1067.1 1063.6 14 727 34       mv h p h 29 Ans: C Increasing the frequency of the radiation will increase the maximum kinetic energy of the electrons. Therefore the stopping potential of the electrons will increase. 30 Ans: B    119 7 834 sx10 2.0 104.4 )100.3)(1063.6( 6015.0 tλ Nhc        hc P t N t EP

8 2016 VJC Prelim H1 Paper 2 Suggested Solutions 1(a) Note: 150 km h-1 = 41.7 m s-1 (b) Motorcycle overtakes car: both have travelled the same distance. Areas under the 2 graphs are equal. Using v = u + at, maximum speed of motorcycle = a (t1 – 3.0) ½ a(t1 – 3.0)2 = 41.7 t1 ½ (12)(t1 – 3.0)2 = 41.7 t1 Solving, t1 = 12 s (reject the answer which is less than 3.0 s) 2(a) (b) As R, W and T intersect at a point, taking moment about that point would mean that there is no net moment. This would allow the person to be in rotational equilibrium. (c)(i) R, W and T must form a closed triangle so that there is no net force acting on the body. t / s v / m s-1 0 t13.0 41.7 Pelvis Hip joint pivot W = 400 N Tension T of extensor muscle Head Spine 70.0o 8.0o R R must intersect with W and T

9 (ii) Using sine rule: 𝑊 𝑠𝑖𝑛8𝑜 = 𝑅 𝑠𝑖𝑛102𝑜 = 𝑇 𝑠𝑖𝑛70𝑜 400 𝑠𝑖𝑛8𝑜 = 𝑅 𝑠𝑖𝑛102𝑜 = 𝑇 𝑠𝑖𝑛70𝑜 Solving for R and T we get: R = 2.8  103 N T = 2.7  103 N 3(a) Magnetic Flux Density is the force per unit current, per unit length experienced by a wire at right angles to the magnetic field. (b) (i) (ii) NBIyF  KN Taking moments about the mid-point of LM,  coscos KN NBIyxxF  (c) (i) At rotational equilibrium, the sum of torques must be zero. springcoil   I Ik NBA kNBIA       (ii) From (b)(ii), in a uniform field, the torque applied on the side of the coil depends on the cosine of the angle to the field as it rotates and hence does not vary linearly. However, in a radial field, the magnetic forces on the coil are always perpendicular to the plane of the coil and the torque will be independent of the position of the coil. So, a linear relationship between the current and the magnetic flux dens