November 2004 GCE O LEVEL PHYSICS (5052) P1 P2 SUGGESTED SOLUTIONS

Nov 2004 O Level Physics (5052) P1 P2 1

PAPER 1 (MCQ):

Q A Remarks

1 B Main scale = 33 mm; vernier scale = 0.5 mm. Thus final reading = (33 + 0.5) mm = 33.5 mm

2 C "Smooth table" Friction is negligible.

Same mass used in all cases. Since F = ma, then a = F/m. Thus the largest resultant force will cause the greatest acceleration.

3 B Distance travelled (1st 3 s) = area under v-t graph = ½(base height) = ½(3)(6) = 9 m

4 D Since the spring is stretched, then force on X due to Y (F1) = force on Y due to X (F2).

Let mass of trolley Y = m. Thus mass of trolley X = 2m.

F1 = F2

(mx)(ax) = (my)(ay)

(2m)(2) = (m)(ay)

Thus, ay = 4 m/s2.

5 D Motion of car is to the left, but the part of the tyre in contact with the ground is moving to the right. Thus, friction (always in opposite direction to motion) is acting to the left.

6 D Mass of a body is not influenced by changes in gravitational field strength.

Gravitational field strength increases as the astronaut moves from the Moon to the Earth. Hence, the weight of the astronaut increases.

7 A In (A), density of block = m / V = 200 / 100 = 2.0 g/cm3. greatest density

In (B), density of block = m / V = 250 / 200 = 1.25 g/cm3.

In (C), density of block = m / V = 300 / 300 = 1.0 g/cm3.

In (D), density of block = m / V = 350 / 400 = 0.875 g/cm3.

8 A Weight of wheelbarrow & load (W) = mg = (60)(10) = 600 N Apply Principle of moments about the pivot, we have: (F)(0.50 m + 0.70 m) = (600)(0.70 m) F = 350 N

9 C Since the toy bird can balance on the finger, it must be in stable equilibrium Centre of gravity of toy must be below the point of suspension (the finger).

Adding the paper clips at the tail is to make the base heavier. Hence, the centre of gravity of the toy is lowered.

10 A Net force acting on car = (3600 2000) N = 1600 N

F = ma a = F / m = 1600 / 800 = 2.0 m/s2.

11 B Work done = (force)(distance moved in the direction of the force) = (weight of car)(vertical height

gained by car) = (500 10)(10) = 5.0 104 J

12 C There is less air at higher altitude than at lower altitude.

Thus, as the boys go higher, there is less air above them (i.e., atmospheric pressure decreases)

height h of mercury column decreases.

13 A Power = work done / time taken = (300 3.0) / 20 = 45 W.

14 D The one that is least likely to sink into soft ground is the one that exerts the least pressure on soft ground (i.e., least force / greatest area).

15 A Pressure acting on bubble (top of lake) = atmospheric pressure (P0) Pressure acting on bubble (bottom of lake) = P0 + pressure due to 20 m of water = P0 + 2P0 = 3P0. Volume of bubble (top of lake) = 6 cm

3.

Assume that the temperature of air in the bubble is constant, then P1V1 = P2V2. (where 1 = bottom of lake; 2 = top of lake)

(3P0)(V1) = (P0)(6) (3P0)(V1) = (P0)(6) V1 = 2 cm3.

So, volume of bubble at bottom of lake = 2 cm3.

16 B Factual recall question: Due to the weak intermolecular forces between gas molecules, gases have no definite shape, and they expand the most when heated.

17 B Oxygen gas is heated. Thus thermal energy is transferred to the oxygen molecules and their average speed increases.

18 B Factual recall question.

19 C Factual recall question.

20 B Equal amounts of heat are supplied to paraffin & water in identical containers.

Temperature of paraffin rises more quickly less heat is needed to cause the temperature of 1kg of paraffin to rise by 1 K than in 1 kg of water (i.e., specific heat capacity of paraffin is lower than that of water).

21 B Cooling at X: the denser, colder water will not rise to the top. Convection current can't be formed.

Cooling at Y: denser cold water sinks to the bottom. Convection current established.

Heating at Y or Z: hotter, less dense water will not sink. Convection current can't be formed.

22 A Concept recall question: P = amplitude; Q = period (Note: horizontal axis = time axis, not distance).

23 B The refractive index of glass is constant for the same piece of glass.

Thus, sin w / sin x = sin y / sin z.

24 C Image is always vertically opposite the object.

Distance between object & plane mirror = distance between image and plane mirror.

25 A Image is virtual & enlarged object distance (u) < focal length (f).

Only option (A) is less than the focal length of the lens (10 cm).

26 D Factual recall question: all electromagnetic waves travel at the same speed through vacuum.

27 D A magnet consists of countless "tiny magnets" that are neatly arranged. Even when the original magnet is broken to three smaller pieces, the arrangement of the "tiny magnets" is still maintained.

November 2004 GCE O LEVEL PHYSICS (5052) P1 P2 SUGGESTED SOLUTIONS

Nov 2004 O Level Physics (5052) P1 P2 2

Fig. 2.2 Fig. 2.3

Q A Remarks

28 D By definition, potential difference (V) = W / Q W = QV = (100)(12) = 1200 J

29 C In the original circuit, we can determine that the emf of the cell = IR = (3.0)(2.0) = 6.0 V

In the 2nd

circuit, the 2.0 m metre wire has a resistance of 2(2.0) = 4.0

Current through the 2nd

circuit (I) = V / R = 6.0 / 4.0 = 1.5 A

30 B Points A & D are on the main circuit. Thus, the current cannot be the smallest there.

Current B + Current C = current D. Thus the answer is either B or C.

Current B is the smallest because the resistance in B is bigger than in C.

31 A When lamp Z is added in parallel to lamp X, the effective resistance of X and Z decreases. Thus, current through lamp Y should increase and lamp Y should be brighter than normal.

32 D Some facts to remember:

The fuse must always be on the live wire.

The earth wire must always be connected to the metal frame of the appliance.

33 A Use Fleming's Left Hand Rule: 1st finger (magnetic field) points to the right, 2

nd finger (conventional

current) points to the top of the page. The thumb (force) will point into the page.

34 C When magnet enters the coil from the top, Lenz's Law will dictate that the top of the coil be induced north (i.e., the induced current flows anticlockwise there.

When the magnet leaves the coil from the bottom. Lenz's Law will dictate that the bottom of the coil be induced north.

There is thus a reversal in direction of the induced current. Thus, the ammeter needle should show a change in direction of deflection (i.e., quickly to the left) before returning to zero.

35 C Speed of rotation is doubled number of waves on the screen is doubled, & the amplitude of waves on the screen is also doubled.

36 B Options (C) and (D) are wrong because a thermistor (instead of an LDR) is shown in the circuit.

Option (B) is correct because of the following:

Less light shines on the LDR resistance of LDR increases p.d. across LDR increases p.d. across the lamp increases & the lamp lights up.

37 B Y = gamma radiation, because it is not deflected by any magnetic field and electric field.

Z = beta radiation, because its path of travel is the sharpest (beta particles are lighter than alpha particles.)

38 A Concept recall question.

39 C YHeeeX 230

90

4

2

0

1

0

1

234

90 (where Y = daughter nucleus)

Proton number of daughter nuclide = 90 1 1 + 2 = 90

Nucleon number of daughter nuclide = 234 4 = 230

Thus, daughter nuclide has 90 protons & (230 90) = 140 neutrons.

40 C Isotopes must have the same number of protons but different number of neutrons.

PAPER 2: Section A 1(a) The distance-time graph of train P has a steeper gradient than that of train Q. 1(b) 90 minutes = 1.5 h Speed of train Q = gradient of distance-time graph = 120 km / 1.5 h = 80 km/h. 1(c) The graph should be parallel to the distance-time graph of train P, starting at t = 30 minutes. 2(a) Pressure inside gas pipe in Fig 2.1 is larger than atmospheric pressure. 2(b) Referring to Fig. 2.1, the difference between the two water levels = 30 mm

Pressure difference = pressure due to this column of water of height 30 mm = hg = (30/1000)(1000)(10) = 300 Pa. (Note: remember to convert height h to metres!)

2(c)(i) Pressure (P) due to a liquid column = hg. There is no change in the height of the liquid column, no change in the density of the liquid and no change in the value of g. Hence, a change in diameter will not bring about any change to the water levels. We should still see a height difference of 30 mm between the two water levels. (See Fig. 2.2)

2(c)(ii) P = hg. Given that pressure difference (P) is constant, gravitational field strength (g) is constant, and that density is halved, the height (h) of the water column should be doubled. Thus, we should expect a height difference of

60 mm (60 = 2 30) between the two water levels. 3(a) From the graph, when the speed of the pole-vaulter is 8.4 m/s, the resistive force acting on him is 320 N. This is equal and opposite to his maximum forward force. Since his forward force cannot be more than 320 N, his resistive force cannot exceed that value. Hence his maximum speed stays at 8.4 m/s. 3(b)(i) Max KE = ½mv

2 = ½ (60)(8.4)

2 = 2100 J (to 2 s.f.)

3(b)(ii) When the pole-vaulter rises, KE during running = GPE gained during rising 2100 = mgh = (60)(10)(h) h = 3.5 m

November 2004 GCE O LEVEL PHYSICS (5052) P1 P2 SUGGESTED SOLUTIONS

Nov 2004 O Level Physics (5052) P1 P2 3

7(a)(ii) +

+ + +

+

+ + + +

+

-

-

- -

-

+ +

+

+ +

7(c) +

+ +

+ +

+ 7(d) Glass

Polythene (or plastic) (Note: wood is not a good insulator).

4(a) The smoke particle moves this way because it was hit by the (invisible) air molecules at different angles, at different times and unevenly. 4(b) The air molecules around the smoke particles are moving at high speeds and randomly. 5(a), (b) Note:

Ray 3 is perpendicular to the surface of the bubble (i.e., angle of incidence = 0). Thus ray 3 will not be refracted. The ray will maintain a straight line as it enters (and leaves) the air bubble. This ray of light will also pass through the centre of the bubble. Using a ruler, the centre of the bubble is located.

The normal axes for rays 1 and 2 are then drawn (dotted lines). These two dotted lines must pass through the centre centre of the bubble (property of a circle).

Ray 1 moves from dense medium (water) to less dense medium (air). Furthermore, its angle of incidence is greater than the critical angle. Thus ray 1 will not enter the air bubble. Instead, it undergoes total internal reflection. The reflected ray is reproduced such that angle of reflection = angle of incidence.

Angle of incidence of ray 2 is less than the critical angle. Thus ray 2 will enter the air bubble and be refracted away from the normal (because it is travelling from dense to less dense media).

5(c)(i) Refractive index of water (n) is the ratio of the speed of light through vacuum (c) to the speed of light through water (v). That is: n = c / v. OR: For a light ray travelling from air to water, the refractive index (n) is defined as the sine of the angle of incidence (i) divided by the sine of the angle of refraction (r). That is: n = sin i / sin r.

5(c)(ii) n = c / v = 3.0 108 / 2.2 10

8 = 1.4

6(a)(i) The hand should be drawn to move vertically.

6(a)(ii) f = 4.0 Hz T = 1/f = 0.25 s The hand must move vertically such that it completes an up-and-down motion every 0.25 s.

6(b) v =

= v / = 0.80 / 4.0 = 0.20 m 6(c) To double the wavelength, the frequency of the wave must be halved. 7(a)(i) From the Law of electrostatics, we understand that like

charges repel. When the negatively charged rod is brought near the uncharged sphere, electrons on the left hand side sphere are repelled by the electrons on the rod. This causes the electrons on the left hand side of the sphere to be repelled towards the right hand side.

7(b) 8(a) Transformer B is used to step down voltage. 8(b)(i) When high voltages are used, the electric current flowing in the cables will be low. Hence the power loss (I

2R) would

then be minimised. 8(b)(ii) From the graph, we see that thermal energy loss is lower for thicker cables. This is because thicker cables have

large cross-sectional area A. Since resistance R is inversely proportional to A, the resistance of the cable is less if the cables were thicker. Since power loss (P) = I

2R, less resistance would mean less power loss.

8(c)(i)

P

S

P

S

V

V

N

N

where: NS = number of turns in the secondary coil, NP = number of turns in the primary coil, VS = voltage across the secondary coil, VP = voltage across the primary coil. 8(c)(ii)

20000

27500048000

PP

S

P

S

NV

V

N

N

Thus, NP = 3490. (to 3 s.f.) PAPER 2: Section B 9(a)(i) When thermal energy reaches the copper molecules, they will vibrate about their fixed positions with a larger

amplitude and pass the vibrations to the adjacent layers of copper molecules. This process is repeated until thermal energy reaches the alcohol.

November 2004 GCE O LEVEL PHYSICS (5052) P1 P2 SUGGESTED SOLUTIONS

Nov 2004 O Level Physics (5052) P1 P2 4

Moment / Nm

Time / s T 0.50 T 0.75 T 0.25 T 0

9(a)(ii) When alcohol vaporises under the Sun, latent heat of vaporization is absorbed by liquid alcohol. When the vapours reach the tank of cold water, the vapours condense and give out latent heat of vaporization to the water. It is through this repeated sequence that thermal energy is transferred.

9(a)(iii) The pressure difference within the tube causes the alcohol vapours / liquid alcohol to move quickly back and forth. This allows thermal energy to be transferred at a fast rate.

9(b)(i) Amount of heat needed to change unit mass (1 kg or 1 g) of a substance from its liquid state to its gaseous state (or vice versa) without a change in temperature.

9(b)(ii) Assume that heat losses to the surroundings is negligible, then: Heat lost by alcohol = Heat gained by water

malv = mwc

(25)(840) = (500)(4.2)() = 10 C

10(a)(i) NOT gate input NOT gate output Buzzer

1 0 OFF

0 1 ON

10(a)(ii) thermistor. 10(b)(i) As the temperature rises, the resistance of P decreases. 10(b)(ii) As the temperature rises, the resistance of P decreases. Hence more current flows through P. 10(b)(iv) Since the resistance of P decreases in (b)(i), the potential difference across P must also be reduced (potential

divider principle). 10(b)(iii) The total voltage across P and Q is 6 V. If the potential difference across P is reduced, then the potential difference

across Q must increase. 10(c) By changing the resistance of component Q (variable resistor), it allows us to change the potential difference across

Q. This in turn allows us to influence the temperature at which the buzzer is triggered. 10(d) If the circuit were wrongly wired as described in the question, then the potential difference across P will be high

when the temperature is low. This implies that the voltage across Q will be low, and the input of the NOT gate would be state 0. The output of the NOT gate would then be state 1, switching on the buzzer.

11(EITHER): (a)(i) If the carbon brushes touch both halves of the split-ring commutator at the same time, it would lead to a short circuit

as the current will not flow through the rectangular coil. (a)(ii) The carbon brushes may not be in contact with either half of the split-ring commutator (i.e., open circuit exists even

when the switch is closed). Thus, no current can flow in the coil and the motor cannot start. (b)(i) Moment of a force about a point (also known as the turning effect of a force about a point) is defined as the product

between the force and the perpendicular distance between the pivot and the line of action of the force. (b)(ii) Distance between pivot and side AB (or CD) of coil = ½(0.065 m)

Moment due to force acting on AB (or CD) of coil about the pivot = (3.0)½(0.065) Hence, maximum moment = 2[(3.0)½(0.065)] = 0.195 Nm

(b)(iii) Referring to Fig 11.1. When the coil is horizontal, the perpendicular distance between the pivot and the vertical forces acting on the sides AB and CD is at its greatest. Hence, the moment is maximum there.

(b)(iv) Note to students:

T = time taken for one complete revolution.

When the coil starts to rotate, and half a rotation later (at time = 0 and 0.50 T respectively) the perpendicular distance between the pivot and the line of action of the force is at its maximum. Hence, the moment is maximum at these 2 instances.

At ¼ of a rotation and ¾ of a rotation later (at time = 0.25 T and 0.75 T respectively), the perpendicular distance between the pivot and the line of action of the force is zero. Thus, there is zero moment at these 2 instances.

In between, as the coil rotates, the perpendicular distance between the pivot and the line of action of the force can be found by trigonometry. Hence, the graph is actually something like a trigonometry curve (the cosine curve).

11(OR): (a) Beta particles are high energy electrons while gamma rays are high frequency electromagnetic radiation. Beta particles are negatively charged, gamma rays have no charge. Beta particles are less penetrating than gamma rays. Beta particles can be stopped by a few millimetres (e.g., 5 mm)

of aluminium but gamma rays can only be stopped by a few centimetres (e.g., 2 cm) of lead. etc, etc.. (See textbook for details)

(Note to students: Since this part contains 3 marks, I assume that 3 differences are needed). (b)(i) The difference in count rate is due to (1) the random nature of radioactive decay, and (2) it takes time for the

radioactive source to travel through the blood stream. (b)(ii) Average count rate = ¼[40 + 38 + 36 + 40] = 38.5 per second. (b)(iii) If 2.0 cm

3 of blood yield a count rate of 38.5 per second, then a count rate of 144000 per second would have come

from 0.25.38

144000 = 7480 cm

3 (3 s. f.)

(b)(iv) Count rate after 2 half lives = 22

40 = 10 per second. (16 days = 2 half lives)

(c) The doctor should wear a lead-lined suit and lead-lined gloves when handling this radioactive source.

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