Suggested Solution to 2002 CE Physics Paper 1

Section A

1.(a)F1:Normal reaction1A

F2:Weight of the cube1A

(b)Since both F1 and F2acts on the same body, the cube,1M

they are NOT an action-reaction pair of forces.1A

(c)It moves with constant velocity to the right.1A

2.(a)Slope = = -1.25ms-21A

Acceleration1A

(b)

(c)Maximum displacement = 1M

= 40 m1A

3.(a)Convex / Converging lens1A

(b)

(d)Move away from the film1A

(a) Pt=mcΔT; P =mcΔT / t1M

2104 = 0.1(4200) (ΔT) / 11M

ΔT = 47.6 ℃

T = 47.6 + 20 =67.6 ℃1A

(b)No heat lost to the surroundings / All heat generated by the engine is absorbed by the water 1A

(c)Regulation of body temperature / Warmer winter and cooler summer in coastal area / Any other reasonable answer 1A

5(a)

(b)To ensure the magnetic field lines are always perpendicular to the coil.1A

(c)Since the air moleculesvibrate at the same frequency of the a.c. (10 Hz)

/ Sound of 10 Hz is produced 1M

which is out of the range that human can hear (20 Hz to 20 kHz) 1A

6. (a) 1A

Input A / Input B / Output
0 / 0 / 0
0 / 1 / 0
1 / 0 / 0
1 / 1 / 1

(b)When the room is bright, the resistance of LDR is small. 1A

The potential difference across R2 is large. (The p.d. on LDR is small.)1A

=> The input B is at high potential.1A

(c)When the room is bright and temperature is high.1A

Section B

7.(a)(i)P.E. increased = mgh

= 500×10×51M

= 25000 J1A

(ii)Average power output = 1M

= 12500 W1A

(b)(i)Energy consumed = 0.05×10250000 1M

= 125000 J1A

(ii)Efficiency = 1M

= 20%1A

(c)

(deduct 1A for each incorrect answer) 3A

(d)The statement is incorrect.1A

This is because without friction, the car cannot go upwards.1A

The friction helps to provide upward force for the car.1A

8.(a)Constructive interference1A

Path difference at S = PS–QS = 01A

(b)The sailor can check if the ship moves along a line which is always at constructive interference (maximum signals). 2A

(c)(i)No.1A

For any frequency used, the path difference along XY is always equal to zero.1A

(ii)No.1A The only difference is that there is always destructive interference along XY(minimum signal). 1A

(iii)No.1A

Increasing the amplitude of the wave would only increase the wave energy.1A

(d)The sailor can determine the wavelength by wave equation λ=v/f. 1A

When the ship moves along PQ, it can detect the number of maximum signals per second by the detector and timer. 1A

Thus, he can determine the distance travelled in one second calculated by number of wavelengths travelled in one second ( no. of wavelength = no. of maximum / 2). 1A

Then the speed can be determined by the speed equation speed = distance / time.1A

9.(a)If the air is wet, the water droplet will vaporize on heating. The total number of gas molecules changes from time to time. Gas laws cannot be applied. 2A

(b)(i)The glass tube should not touch the bottom of the beaker.

Otherwise the gas inside will be directly heated by the flame. The temperature of the air will be higher than that of the water. 1A

(ii)Add more water so that the whole volume of the air is completely immersed in water. 1A

(c)

(d)At absolute zero, the volume of the air shrinks to zero.1A

(e)(i)Leon is wrong.

The amount of heat gained by the air is not measured in the experiment. Only the temperature of the air is significant. 1A + 1A

(ii)Leon is wrong. The air pressure inside the glass tube will change.1A + 1A

10(a)(i) 1M or 1A

(ii) 1M

or 1A

(iii)Power loss on transmission line = 1M

=1A

(iv) 1M+1A

(b)Live (L) - Q1A

Neutral (N) - P1A

(c)Connect to live wire (L)1A

Ensure the kettle still not stand at high potential evenwhen the fuse melts or breaks.

(d)The kettle won’t work properly.1A

Since the voltage across the main power supply is 220 V which is muchgreater than 120 V, 1A

large current flow through the kettle which will melt the wire, cause fire or damage the heater of the kettle. 1A

11.(a)(i)To ensure that the radiations areemitted in one direction.1A

(ii)To prevent the radiations from colliding with gas molecules.1A

(b)

(c)When the foil is inserted, particles are scattered by the foil

 the counter can obtain reading1A+1A

(d)(i) radiation can be absorbed by a 1 mm paper sheet.1A

(ii)The 0.5 mm aluminium sheet is too thin to block the  and  at Q.1A+1A

(iii)He is wrong.1A

The readings are caused by background and -radiation.1A

 The lead block can never completely block-radiation.1A

(e)It can cause cancer cells.1A

because -radiation can ionize and break up the molecules in the cells / DNA.1A

END OF SOLUTION

Solution of spare question

Section A

2.(a)Initial velocity = = 0.47ms-1

Final velocity = = 2.35ms-11M

Acceleration =1M

= 9.4ms-21A

(b)No. It is because air resistance is significant to the rubber.1M

Replace the rubber by a metal ball (or any reasonable method).1M

[Larger rubber is not acceptable.]

4.(a)(i)

(ii)Diffraction1A

(b)(i)AM radio waves are easier to be received than FM in case obstacles are found.1M

It has longer wavelength than FM radio waves.1M

It will bend more when they come across obstacles.1A

5.(a)

(b) The foil is positively charged,  it is attractedby plate B1A

After that  the foil touches plate B, they shares the same charges. 1A

The foil is repelled by plate B.1A

8.(a)By conservation of momentum, we have

mAuA + mBuB = mAvA + mBvB

2(6) + 0 = 0 + 3vB1M

vB = 4 ms-11A

(b)Potential energy of ball B changes gradually into its kinetic energy.1A

(c)Let speed of ball B at R be V. We have, along QR,

K.E. gained = P.E. lost

1/2×3×V2 - 1/2×3×42 = 3×10×11M+1A

(1M is given for the subtraction of the K.E.)

V = 6 ms-1

So, speed of ball B at R is 6 ms-1.1A

(d)(i)Inelastic collision1A

(ii) Yes. (Some mechanical energy is changed into Heat energy / Sound energy / Internal energy of ball B and the plasticine). 1A+1A

(iii)Average force acting on ball B = momentum change / time required

= 3×(0 - 6) / 21M

= - 9 N (9 N in backward direction)1A

(iv)Conservation of momentum CANNOT be applied.

It is because there is an external force (friction between the plasticine and the Earth) acting on the system during collision. 2M+ 1A

Conservation of momentum CAN be applied if we regard the Earth and the plasticine as one body. 2M+1A

P.1/7

Suggested Solution to 2002 CE Physics Paper 1