1. Give best answer by estimating: 20km x 15 km = 3.0 x 108 m2

b) acc= 20ms-1 in 5s= 4ms-2

c) Power = Fxv

Estimate car mass =1000kg

acc = 4ms-2

v - 80kmh-1

2. T= 3.5 ms

b) 0.87m

c) v = 0.87/3.5x10-3 =250 ms-1

4b)The two forces are not action and reaction forces as they act on the book . They are also not of the same nature. One is gravitational and the other is contact force.

c) The total force is zero. Hence Sum of the forces is zero .The change in momentum of of A + the change in momentum of B = 0.There fore the total momentum is constant . This is principal of conservation of momentum.

5i) when two waves meet the resultant displacement at a point is the sum of the displacement of the individual displacements of the two waves.

ii) The diffraction is the bending of waves as it passes an aperture or obstacle.

iii) Coherence is the condition when two waves has a constant phase difference.

b)i) The wavefront AB reaches the slit A and also at slit B .

ii) Mark out the crosses on the screen YZ . You have 3 marked C including at O . In between is marked D . You can find 2 marked D.

Label C-C and marked it x

vi) wavelength = ax/D

v) The equation is only approx as the length L is not large enough .

7. The time taken using formula s= 0.5 at2 = 2.89 s

aii) The actual time is 2.9 s . The air resistance is negligible.

bi) The extension of the rope = 73 - 41 = 32 m

ii) The elastic potential energy = 0.5 at2 or ( area under the graph = area of traingle)=5.4 x 104 J

ci) complete the table :

GPE : 5.4 x 104 // 2.4 x 104J // 0J

EPE : 0 //0 // 5.4 x 104 J

KE : 0 // 3.0 x 104 J // 0

ii) Use graph to find extenstion when Tension = weight of man , extension = 7m

distance fallen when he has Ke max = 41 + 7 m = 48 m

iii)KE max = total energy - epe - gpe

KE max = 5.4 x 104 - 2600J - 1.84 x 104 J=3.3 x 104 J

To draw graph , plot the points with values calculated

GPE: Draw a straight line sloping downwards to 0 at 73 m

KE: Draw a straight line with postive gradient until 41 m . From 41m to 73 m , the graph curve up to peak at 48 m and 0 at 73 m

EPE: The graph starts from 0 at 41 m and U shape increase to 73 m.

8. b) Parallel to each other hence , i) 400 ohm ii) 100W iii) 23 ohm

c) lamp: 200// 0.50// 400// 100.

tele: 200// 1.2// 167// 240.

cooker: 200// 6.0 // 28.6 // 1400

d) Resistance on ammeter and internal resistance is taken to be negligible.

e) The lamp must have smaller cross sectional area than the cooker as the resistance is much higher. The lamp resistance is 14 X higher than the cooker , so the cross sectional area must be 14 times smaller . Using a longer length of wire may not fit the lamp . So the length should not be increased but change the wire to a thinner one. Use R=pL/A.

9. E=hc/~ = 1.63 x 10-18 J ,

ii) The current will increase when the intensity is increased.

iii) The faster photoelectrons have higher kinetic energy can overcome the electric force to reach the anode .

ii) eV = 0.5 mv2

v = 1.16 x 106 ms-1

iii) work function = hf - KE= 1.63 x 10-18 -6.08 x 10-19 J= 1.0 x 10-18 J.

c) The reduction in intensity does not change the photon energy and therefore the ke of the electron remain the same. The rate of photon will decrease and hence will decrease the photocurrent and not the KE max.

d) The transition from level 2 to level 1 will give the same energy as the 122 nm uv radiation.

The data analysis:

6. I choose W = 3N , and looking vertically upwards I recorded the values of d for 10 different L values in the table. The taking the log , I plotted the graph suggested , the value of n is found to be 3.00