Cmock Exam IV Paper 2 s1

F4 Physics Second Term Examination Marking Scheme (1 3 -1 4 ) 1. (a) s = (1/2) (u+v)t 1.5 = (1/2) (u + 5) (0.43) (1M) u = 1.98 m/s (1A) (b) (i) Required force = mg sin  = 75 x 9.81 x sin 25o = 311 N (1M+1A) (ii) Work done against resistive force = Fs = 311 x 2 = 622 J (1M+1A) (c) PE is lost because work is done against friction (or resistive force) which finally converted to internal energy (heat) in the surroundings. (1A+1A+1A)

2. (a) E = mc T 8500 = 0.12 x 4200 T (1M) T = 16.9oC (1A) (b) E = Pt = mc T 8500 t = 0.41 x 4200 x (100-26) (1M) t = 15.0 s (1A) (c) The daily and annual temperature differences in coastal areas are smaller than inland. It is because the specific heat capacity of water is much greater than that of land. (1M+1M)

3. (a) (i) n = PV/RT = 3.2 x 105 x 1.9 x 10-3 / (8.31 x 285) = 0.257 mol (1M+1A)

5 5 (ii) P2 = (T2 / T1) P1 = (295/285) x 3.20 x 10 = 3.31 x 10 Pa (1M for formula, 1A)

3RT 3(8.31)(295) 21 (iii) Average KE = = 23  6.1110 J (1M for formula +1A) 2N A 2(6.0210 ) (b) When the temperature of a fixed volume of gas increases, the molecules move faster and hit the container walls more frequently and more violently. This increases the gas pressure. (1M+1M+1M) (c) (i) similar : random motion / range of speeds (1M) (ii) different: mean KE / rms speed / frequency of collisions (1M) 4. (a) Velocity vector tangential to path in correct direction and drawn from the ball (1A) Acceleration vector vertically downwards (1A) 1 (b) (i) s = ut + at 2 2 24 = (1/2) (9.81) t2 (1M) t = 2.21 s (1A) (ii) Horizontal distance travelled = vt = 12.2 x 2.21 (1M for formula) = 27.0 m s-1 (1A)

2 2 2 2 v  vx  vy  12.2  (9.81 2.21) 2 2 (iii) (1M for v  vx  vy ,1M for 9.81 x 2.21, 1A)  24.9m / s 5. (a) 24 hours GMm (b)  m 2r (1M) r 2

2 GMm  2   m  r (1M for 2π/T) r 2  T  GMT 2 6.67 1011  5.981024  (24 60 60)2 r 3    7.541022 4 2 4 2 r = 42000 km (1A) (c) (i) v  r (1M for formula) 2 2  7370103 v  r   7350m / s (1A) T 105 60 (ii) Centripetal force = mv2 / r = 650 x 73502 / 7370 x 103 = 4764 N (1M+1A)

2 2 6. (a) (i)     16.89s 1 (1A) T 1.86 / 5 (ii) v  r  16.891.25  21.1m / s (1M+1A) (b) (i) (1M for pointing towards the center)

(ii) Centripetal force = mv2 / r = 7.30 x 21.12 / 1.25 = 2600 N (1M+1A) (iii) Weight of hammer acts downwards, and has to be balanced by the component of tension in the chain. (1M+1M) (c) 1M for tangent to circle, 1M for tangent just passes edge of cage

7. (a) Vacuum reduces heat loss by conduction and convection. (1M+1M) (b) Plastic/cork is a poor conductor (1A), this can help reduce heat loss by conduction.(1A) (c) The vacuum flask can slow down the heat transfer from outside the flask to the tea (1A). Therefore, it can help to keep the tea cold (1A).

8. (a) 5 min (300s) (1A) (b) When the heater is switched off, its temperature is higher than the metal. (1A) Or Heat continues to conduct to the metal until the same temperature is attained. Or It takes time for them to attain the same temperature. (c) (i) Pt = mc T (1M) 48 (5 x 60) = 0.80 c (45=20) c = 720 J kg-1 oC-1 (1A) (ii) Experimental result is higher than the actual value. Not all the energy supplied by the heater goes to the metal. Or Some energy is absorbed by the heater itself and or by the thermometer. Or Some heat is lost to the surroundings. (d) Glass is not a good conductor of heat. Or Prolonged time is required for the whole block to attain uniform temperature. 9. (a) Work done by the force  Fs  50 0.3 (1M+1A)  15 J (b) By conservation of energy, the work done by the force is equal to the gain in kinetic energy of the football. 1 1 15  mv2   0.45v2 2 2 (1M+1A) v  8.16497 m s1 The football leaves the student’s hands at a speed of 8.16 m/s. (1A)

(c) By conservation of momentum, 0  0.458.16497  (50  2)v (1M for conservation of momentum) v  0.07066 m s1

The recoil speed of the trolley is 0.0707 m s1 . (1A) 10. (a) (i)

11. (a) Area under graph = 3 x 10-3 x 160 / 2 (1M) = 0.24 Ns (1A) Area represents impulse or change in momentum (1A) (b) Impulse = change in momentum (1M for knowing impulse = change in momentum) m (20 – 0) = 0.24 m = 12 g (0.012 kg) (1A)

MC 1-5 B B C D A 6-10 C B A C D 11-15 B B A A C 16-20 D B D D A 21-25 D A B B B 26-30 C B A D A 31-33 C B C Explanations to selected MC 3. E = mcT E/m = slope = c T => (27-5) x 103 / 0.5 = c (10) => c = 4400 J kg-1 oC-1 23. By the law of conservation of momentum:

mXuX + mYuY = mXvX + mYvY

2  5 + 5  0 = 2  (–5) + 5  vY v = 4 m s–1 (forward) 24. By the law of conservation of momentum:

mAuA + mBuB = mAvA + mBvB

0 = mA  1 + mB  (– 2)

mA : mB = 2 : 1 28. Vertically, L sin 60o = mg L/m = g / sin 60o = 11.3 N /kg

1 33. s = ut + at2 2 On the Earth, the travelling time is: 1 s = gT2 2 2s T = g On the Mars, the travelling time is: 1 s = 0 + (0.376g)T’2 2 2s 1 2s T’ = = =1.63T 0.376g 0.376 g