Cmock Exam IV Paper 2

2016-17 F4 Physics Second Term Exam Marking Scheme

1. (a) (i) Energy absorbed when the ice cubes are heated up to 0 °C = mcΔT = (5 × 0.05)(2100)[0 − (−10)] = 5250 J 1M Energy absorbed when the ice cubes are melted 5 = mℓf = (5 × 0.05) (3.34 × 10 ) = 83 500 J 1M Total energy absorbed by the ice cubes = 88750 J 1A (ii) energy lost by the milk tea = energy gained by the ice cubes (before and after they melt) (0.25)(4200)(96 − T) = (0.05)(4200)(T + 0) + 88750 1M 2100T = 12050 T = 5.74 °C 1A (b) The ice bath cools the nearby air and form a layer of cold air around the cup of the milk tea. 1A Since air is a poor conductor, this reduces heat gain from the surroundings by conduction./ 1A Before heat is conducted into the milk tea, the surrounding air has to be heat up first. (No mark for heat loss to the surroundings)

2. (a) n = PV/RT (1M for formula) = 1.03 x 105 x 1.2 x 103 x 10-6 / [8.31 x (15+273)] = 0.051 mol (1A) 23 22 N = nNA = 0.051 x 6.02 x 10 = 3.11 x 10 (1A)

(b) By volume-temperature relation (Charles’ law), final volume of the gas (1M+1A)

(c) Average K.E. of the gas molecules (1M+1A) 1 (d) m  c2  average KE (1M for formula) 2 = 6.90 x 10-21 x 2 / 2.326 x 10-26 = 593293

crms = 593293 =770 m/s (1A)

3. (a) (i) The substance exists in both solid and liquid states. (1A) (ii) The substance exists in liquid state. (1A) (b) Applying E = mcT, the specific heat capacity of substance X in solid state is given by 2000 x 16.9 = 0.5 x c x (53-25) (1M) c = 2414 J kg-1 oC-1 (1A)

(c) Applying E = mlf, the specific latent heat of fusion of substance X is given by

(2000)(62.9 16.9)  (0.5)(lf ) 1 (1M+1A) lf  184 000 J kg (d) If the power is halved, the time spent in each section of the curve will be doubled. (1A for correct melting point + 1A for the same initial temperature + 1A for less steep slopes for both parts) old temperature / C new 63

25 time / s 0 16.9 33.8 62.9 66.9 125.8 133.8

P.1 4. (a) From 0.1 to 0.3 s. 1A 1 (b) Area under graph = x 6 x 0.2 = 0.6 Ns 1M+1A 2 Physical meaning is impulse or change in momentum. 1A (c) Change in momentum = 0.6 = 0.5 x ( v - -1) 1M v = 0.2 m s-1 1A

Constant speed before collision (1A), change from positive to negative (1A) (Accept constant speed after collision or speed decreases to zero)

5.(a) Take the downward direction as positive. In the vertical direction, 1 by s = ut + at 2 , 1M 2 1 1M 2 = 0 +  a 12 2 a = 4 m s2 1A The acceleration due to gravity near the surface of the planet is equal to the acceleration due to gravity in magnitude, i.e. 4 N kg1. s (b) (i) Initial speed of the ball = = 8/1 = 8 m/s 1M+1A t 1 (ii) Initial kinetic energy of the ball = mv 2 1M 2 1 1A =  3  82 = 96 J 2 GM (c) (і) By g = , 1M r 2 GM 6.671011  2.431023 1A radius of planet = = = 2.013  106 m g 4 (ii) By a = r2, a  = r 2 r 2.013106 Period of the satellite = = 2 = 2  a 4 1M = 4457s 1A

P.2 6. (a) 1A for weight and normal reaction 1A for friction (Deduct 1A for any wrong additional force drawn) (b) F = mr2 = 1 x 0.3 x 3.14162 (1M) = 2.96 N (1A) (c) KE = (1/2) mv2 = (1/2) m (r)2 (1M for v = r ) = (1/2) x 1 x (0.3 x 3.1416)2 = 0.444 J (1A) (d) No. of revolutions made in 1s = 3.1416 x (1 rev / 2) = 0.5 rev/s (1A) (e)

7. (a) A larger bulb improves the sensitivity of the thermometer (1A) OR A larger bulb minimizes the effect on the temperature reading due to other parts of the thermometer stem that are exposed to different temperatures. (b) (i) E = mcT = 0.015 x (2.9 x 103) x (20-15) (1M) = 217.5 J (1A) (ii) Time taken to reach air temperature = 217.5 / 0.5 (1M) = 435s (1A) (iii) The thermometer would be in direct contact with the cooler air and would cool down quickly. The temperature reading would be less than the actual soil temperature.

P.3 8. (a) From the graph, his reaction time is 0.4 s 1A (b) Separation between the car and the bus = area under the v–t graph between t = 0 and t = 4.0 s 1M+1A (c) Slope of v–t graph equals acceleration. Acceleration 1M+1A (Accept deceleration = 2.22 m s−2 ) The decelerating force is F = ma = (2400)(2.22) = 5330 N. 1A (d) Impact force acting on the car 1M+1A (e) Difference in total kinetic energy of the car and the bus before and after the collision 1 1 = [(15 000)(1.6)2 + 0] – [(2400)(10)2 + 0] = 19200 – 120000 = −100 800 J ≠ 0 1M 2 2 It is an inelastic collision. 1A

1 9. (a) Total KE = (1400) (27.8)2 = 5.41 x 105 J 1A 2 Average output power = E / t = 5.41 x 105 / 9.3 1M = 58.2 kW 1A (b) (i) As the car is accelerating up an inclined road, the potential energy of the car is increasing. 1M+1A (OR both the KE and PE are increasing) As the power of the car remains unchanged, it takes longer time for the car to accelerate up an inclined road. Alternatively When the car is travelling up an inclined road, a component of weight (mg sin ) acts along the road. The net accelerating force is reduced. So it takes a longer time for the car to accelerate up an 1M+1A inclined road. (ii) Let h be the increase in height of the car. Increase in PE + increase in KE = power x time 1M 1 mgh + mv2 = Pt 2 1400 x 9.81 x h + 5.41 x 105 = 58.2 x 103 x 16.2 1M h = 29.2 m 1A Alternatively, Extra energy output = increase in PE 1M 58.2 x 103 x (16.2-9.3) = 1400 x 9.81 x h 1M h = 29.2 m 1A (c) (i) KE of the car is converted to internal energy (or 1A+1A to do work against friction) (ii) KE of the car = friction x skid mark length 1M 1 x 1400 (v2) = 11200 x 30.5 2 v = 22.09 m s-1 1A = 22.09 x 0.001 / (1/3600) = 79.5 km h-1 > 70 km h-1 So Peter was not telling the truth. 1A MC 1-5 C A A C A 6-10 A C D A B 11-15 C D D D A 16-20 B D D A C 21-25 A A D C A 26-30 C B C B C 31-33 C D B P.4