2018 end-of-year exams - the learning space

Candidate Name: _________________________________

This question paper consists of 14 printed pages.

[Turn over

Class Adm No

2018 End-of-Year Exams

Pre-University 2

H1 PHYSICS 8867/01

Paper 1 Multiple Choice 17 September 2018

Additional Materials: OMR Answer Sheet 1 hour

READ THESE INSTRUCTIONS FIRST

Write in soft pencil.

Do not use staples, paper clips, highlighters, glue or correction fluid.

Write your name, class and admission number on the Answer Sheet in the spaces provided.

There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D.

Choose the one you consider correct and record your choice in soft pencil on the separate OMR Answer Sheet.

Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate.

2

Data

speed of light in free space, c = 3.00 × 108 m s–1

elementary charge, e = 1.60 × 10–19 C

unified atomic mass constant, u = 1.66 × 10–27 kg

rest mass of electron, me = 9.11 × 10–31 kg

rest mass of proton, mp = 1.67 × 10–27 kg

the Avogadro constant NA = 6.02 × 1023 mol–1

gravitational constant G = 6.67 × 10–11 N m2 kg–2

acceleration of free fall, g = 9.81 m s–2

Formulae

uniformly accelerated motion, s = ut + 2

1at2

v2 = u2 + 2as

resistors in series R = R1 + R2 + …..

resistors in parallel 1/R = 1/R1 + 1/R2 + …..

3

[Turn over

1 Which of the following is a reasonable estimate for the volume of a taxi in Singapore?

A 0.5 m3

B 5 m3

C 50 m3

D 500 m3

2 The SI unit for force is Newton.

Which of the following is the unit for force expressed in SI base units?

A kg m-1 s-2

B kg m-1 s-1

C kg m s-1

D kg m s-2

3 The figure below shows the movement of a vehicle. It is initially moving at 10 m s-1 with a bearing of 060° and then it moves vertically downwards at 5 m s-1.

What is the direction and magnitude of the change in velocity?

Bearing Direction Magnitude/ m s-1

A 180° 5.0

B 090° 8.7

C 123° 11

D 221° 13

4 A sphere is measured for its diameter using an instrument which gives a percentage uncertainty 3.0%.

What is the percentage uncertainty of the volume of the sphere if the volume of the sphere is 10 cm3?

A 3.0 % B 6.0 % C 9.0 % D 12.0 %

4

5 A special running track has a lap distance of 600 m. Dave completed his 2.40 km run by running at an average speed of 1.20 m s-1 for his first lap, 1.40 m s-1 for his second lap and 1.50 m s-1 for his third lap.

At what speed must he run for the last lap to have an average speed of 1.50 m s-1 for the full 2.40 km run?

A 1.60 m s-1

B 1.80 m s-1

C 2.00 m s-1

D 2.20 m s-1

6 In a tennis match, a ball is hit horizontally with a speed v, as shown in the diagram.

The bottom of the ball is initially 3.0 m above the ground and a horizontal distance of 12 m from the net. The ball just clears the net, which is 1.0 m high.

What is the value of v? (Assume that air resistance is negligible.)

A 16 m s-1 B 19 m s-1 C 30 m s-1 D 38 m s-1

7 Two similar spheres, each of mass m and travelling with speed v, are moving towards each other.

The spheres have a head-on elastic collision. Which statement is correct?

A The spheres stick together on impact.

B The total kinetic energy after impact is mv2

C The total kinetic energy before impact is zero.

D The total momentum before impact is 2mv.

5

[Turn over

8 A balloon is acted upon by three forces, weight, upthrust and sideway force due to the wind, as shown in the diagram below.

What is the resultant force on the balloon?

A 500 N B 1000 N C 1100 N D 1500 N

9 A light cord, placed over a smooth pulley, connects two blocks A and B of masses 2.0 kg and 1.5 kg respectively. Block A is lying on a smooth inclined plane while B is hanging vertically as shown. The masses are released from rest at the positions shown.

How long will it take for block B to hit the floor?

A 0.32 s B 0.55 s C 0.84 s D 0.90 s

10 An astronaut in a spaceship orbiting the Earth experiences “weightlessness”. Which of the following statement is true?

A His reaction force is equal and opposite to his weight.

B He has zero weight.

C His weight is the only force acting on him.

D There is no gravitational force acting on him.

6

11 Which of the following statement about drag force is false?

A Air resistance is a type of drag force.

B Drag force opposes motion.

C Drag force acting on the object is dependent of the density of the object.

D The drag force acting on the object is zero when the object is stationary.

12 A hinged trapdoor is held in the horizontal position by a cable. Three forces act on the trapdoor: the weight W of the trapdoor, the tension T in the cable and the force H at the hinge.

Which of the following option gives the three forces in increasing order of magnitude?

A H,T, W B T, H, W C W, H, T D W, T, H

7

[Turn over

13 In which situation could the pair of forces applied to the rigid object produce a couple?

14 A solid rubber ball has a diameter of 8.0 cm. It is released from rest with the top of the ball 80.cm above a horizontal surface. It falls vertically and then bounces back up so that the maximum height reached by the top of the ball is 45 cm, as shown.

If the kinetic energy of the ball is 0.75 J just before it strikes the surface, what is its kinetic energy just after it leaves the surface?

A 0.36 J B 0.39 J C 0.40 J D 0.42 J

8

15 An object of mass 400 kg is lifted through a vertical height of 1200 m in 2.0 minutes by an electric motor. What is the amount of electrical power needed if the overall efficiency of the system is 80%?

A 3.1 kW B 4.9 kW C 49 kW D 2900 kW

16 A car mass m moving at a constant speed v passes over a humpback bridge of radius of curvature r. (A humpback bridge is curves in a semicircle above a river.)

Given that the car remains in contact with the road, what is the net force R exerted by the car on the road when it is at the top of the bridge?

A mg + mv2

r

B mv2

r

C mg D mg -

mv2

r

9

[Turn over

17 The figure below shows a bowl of food on a circular turntable, which is turning at constant speed.

Which of the following best shows the forces acting on the bowl of food when in the position shown?

A

B

C

D

18 A disc in an engine is rotating in a circle of radius 8.0 cm at 3000 revolutions per minute. What is its centripetal acceleration?

A 25 m s-2 B 7900 m s-2 C 3.1 x 104 m s-2 D 7.2 x 107 m s-2

19 Which of the following statement is true about an orbiting geostationary satellite?

A The satellite orbits with a period of 365 days.

B The only force acting on the satellite is its weight.

C Only one geostationary satellite can be in orbit at any one time.

D Geostationary satellite orbits from the North Pole to the South Pole.

10

20 Electric lamp uses tungsten filaments. The resistance of these tungsten filaments increases as their temperature increases.

Which graph shows how the current I in the filament varies with the potential difference V across it?

21 In between a pair of charged plates is a gas at low pressure. When a radioactive particle passes through the space between the plates, the gas molecules are ionised to become pairs of positive ions and electrons.

What is the force acting on each positive ion if the electric field strength between the plates is 5.0 x 10-5 N C-1?

A 8.0 × 1024 B 1.6 × 1025 C 1.9 × 1025 D 3.0 × 1025

22 A cylindrical piece of electrically conducting copper rod has resistance R. A new rod is manufactured such that its length is increased by three times by connecting 3 similar cylinder in series.

What is the resistance of the new rod?

A

2

R

B R C 3 R D 9 R

11

[Turn over

23 What physical quantity would be the result of multiplying a potential difference with an electric charge?

A Electric Power

B Electric Current

C Electric Potential Energy

D Electromotive force

24 A potential divider circuit is formed using a light-dependent resistor (LDR) and a thermistor as shown in the following diagram.

Under which set of conditions will the potential difference across the thermistor have the smallest value?

illumination temperature

A Low Low

B High Low

C Low High

D High High

12

25 The figure below shows a circuit with four voltmeter readings V, V1, V2 and V3.

Which equation relating the voltmeter readings is true?

A V = V1 + V2 + V3

B V + V3 = V1 + V2

C V – V3 = V1

D V3 = 3(V2)

26 The diagram shows a 4.0 cm straight conductor carrying a current of 5.0 A, and placed in a region of uniform magnetic field of flux density 0.04 T.

What is the magnitude of the force acting on the conductor?

A 4.0 × 10-3 N

B 6.0 × 10-3 N

C 7.0 × 10-3 N

D 8.0 × 10-3 N

13

[Turn over

27 The diagram below shows a compass placed inside a solenoid with current flowing in the direction as shown.

What will happen to the compass needle if the current in the coil is increased?

A The needle will turn more to the right.

B The needle will turn more to the left.

C The needle will remain in the same position.

D The needle will oscillate about the original position.

28 A detector of ionising radiation gives a background count rate of 28 counts per minute. Samples of two radioactive nuclides, X and Y, are individually measured by the detector and each sample gives the same reading of 508 counts per minute. X has a half-life of 4 months and Y a half-life of 3 months. The samples are mixed together. Assuming no change in background radiation levels, what will be the reading of the mixture after one year?

A 88

B 90

C 95

D 118

29 A nucleus of polonium Poxy decays to thallium Tl210

81 by a sequence of particle emissions.

polonium Poxy lead + α

bismuth + β

thallium Tl21081 + α

How many neutrons are there in a nucleus of Poxy ?

A 84

B 86

C 132

D 134

14

30 A radioactive source emits radiation which enters an evacuated region and follows a curved path in the plane of the paper as shown. The region has a uniform magnetic field perpendicular to the plane of the paper and is divided into two by a sheet of aluminium 1 mm thick.

Which of the following correctly describes the type of radiation and its point of entry?

Type of radiation Point of entry

A Alpha X

B Alpha Y

C Beta X

D Beta Y

Millennia Institute 8867 PU2 H1 Physics Paper 1 Solution

1 B Estimation 2 D Using F = Ma

[F] = kg m s-2 3 D

By cosine rule, (ΔV)2 = 102 + 52 – 2 (10)(5) cos (120°) ΔV = √175 = 13.2 m s-1

By sine rule, √175sin 120°

10sin

θ = 41° Hence the bearing is 180 + 41 = 221°

4 C Express area in terms of diameter.

∆ 3 ∆ 3 3.0% 9.0%

5 Total Distance / Ave Speed = Time To achieve an ave speed of 1.50 ms-1, he needs to complete 2.4 km in 1600 s. Time taken for lap 1 = 500 s Time taken for lap 2 = 428.57 s Time taken for lap 3 = 400 s Time taken for lap 4 = 1600 – (500 + 428.57 + 400)s = 271.43 s Hence his ave speed for lap 4 = 600 / 271.43 = 2.2 m s-1

6 B Horizontal distance traveled = 12 m Vertical distance traveled = 2.0 m (taking downwards as positive) Time taken, t = (12 / v) Using s = ut + ½ a t2, 2.0 = (0) + ½ (9.81) (12 / v)2, v = 19 m s−1

7 B For elastic collision, the total momentum of the system (2 masses m) and kinetic

energy of the system is conserved.

-10 m s-1

5 m s-1

120°

θ ∆V


Total KE before collision = ½ mv2 + ½ mv2 = mv2.

8 C Net force vertically = 10000 – 9000 = 1000 N (upwards) Net force horizontally = 500 N (right wards)

Resultant force = 2 21000 500+ = 1100 N

9 C Consider the 2 block as a whole system, Fnet = 1.5 x 9.81 – 2.0 x 9.81 x sin 30˚ = 4.905 N Fnet = ma 4.905 = (2 + 1.5) a a =1.401 m s-2 (downwards) s = u t + ½ a (t2) 0.50 = 0 + ½ (1.401) t2

t = 0.84s 10 C No normal force is acting on him, hence the only force he experience is his weight11 C Dependent on surface area of the object and speed of the object but not the

density of the object.

12 B Sketch a possible force diagram with the 3 forces forming a closed triangle.

13 C Net force is 0. Produces a net turning effect on the body.14 D KE before striking the surface = (GPE before releasing the ball) which is proportional to a

height of 80 cm. = 0.75 J KE just after leaving surface = (max GPE after bounce) which is proportional to a height of 45 cm. Hence, by proportion of height, KE just after leaving surface = (max GPE after bounce)/(GPE before releasing) x 0.75 J = (45/80) x 0.75 = 0.42 J

15 C Power required to lift the object = mgh/time = 400 x 9.81 x 1200 / [2 x 60] = 39240 W - 80% 100 % = [39240 / 80] x 100 = 49000W

16 D Fnet = Fc = mv2

r = mg – R

Hence R = mg - mv2

r

17 A Centripetal force is a net force. It cannot co exisit with other free body diagram. 18 B Radius = 0.08 m

ω =3000 rev per min = 3000 x 2 π/ 60 = 314.16 rad s-1

a = r ω 2 = 0.08 x 314.162 = 7895 m s-2 19 B The only force acting on the satellite is its weight.20 D The resistance increases as V and I increases 21 A F = qE


= 1.6 x 10-19 x 5 x 10-5 = 8.0 x 10-24 N 22 C Resistance

A

LR

ρ=

If L is increased 3 times, R is increased by 3 times. 23 C By definition of potential difference, multiplying p.d. with charge will yield an

energy quantity

24 C Potential across the thermistor will be the lowest when its resistance is low and the resistance across the LDR is high. Hence, its temperature must be high and the illumination on the LDR must be low.

25 C Since resistor R and 3R are parallel, V2 = V3. V = V1 + V3 Hence, V – V3 = V1

26 D F = BIL sin θ θ is the angle between the magnetic field and the current. In this case, θ = 90° Hence, F = BIL F = (0.04)(5.0)(0.04) = 8.0 x 10-3 N

27 A In the absence of the current, the compass needle would point northwards. With the current turned on, the magnetic field due to the solenoid points to the right. Combining the two components, the compass needle points toward the right. With a further increased in the current in the same direction, the magnetic field set up in the solenoid is even larger and hence, the needle points even more to the right.

28 D 118: correct value, after adding in background count rate

118283060)background plus year,1(

30)480()2

1()lives-half 4 year,1(

60)480()2

1()lives-half 3 year,1(

48028508

4

3

=++=

==

==

=−==

mixture

Y

X

YX

C

C

C

CC

29 D 134: correct neutron number

30 D Beta particle because it can penetrate the aluminium. Initial radius must be larger because velocity is higher because kinetic energy is loss after travelling. Hence Y must be the point of entry.

Class Adm No Candidate Name:

This document consists of 21 printed pages and one blank page. [Turn over

2018 End-of-Year Exams Pre-University 2

H1 PHYSICS 8867/02

Paper 2 Structured Questions 12 September 2018

Candidates answer on the Question Paper. 2 hours

No Additional Materials are required.

READ THESE INSTRUCTIONS FIRST Do not turn over this page until you are told to do so. Write your full name, class and Adm number in the spaces at the top of this page and on any separate answer paper used. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Section A Answer all questions.

Section B Answer one question only.

You are advised to spend one and half hours on Section A and half an hour on Section B.

At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.

For Examiner’s Use

Sect A

1 /8

2 /10

3 /8

4 /14

5 /7

6 /13

Sect B (one question only)

7 /20

8 /20

Presentation

Total

/80

3

[Turn over

Section A Answer all the questions in the spaces provided.

1 The velocity-time graph in Fig. 1.1 shows the first 1.6 s of the motion of a ball which is thrown downward at 6.0 m s-1.

Fig. 1.1

(a) Define displacement and velocity.

……………………………………………………………………………………………………...........

....................................................................................................................................................

....................................................................................................................................................

................................................................................................................................................[2]

(b) Determine the distance travelled by the ball before hitting the ground.

distance = …………………… m [2]

(c) Determine the maximum height attained by the ball after it hits the ground.

maximum height = …………………… m [1]

4

(d) Calculate the acceleration of the ball when it is in the air.

acceleration = ………………… m s-2 [2] (e) Determine the time the ball next reach the ground.

time = ……………………. s [1]

5

[Turn over

2 (a) A simple two-stage rocket system consists of two components A and B as shown in Fig. 2.1. They are stacked one on top of another and fired in stages.

Fig. 2.1

When the two components separate in space, the variation of the force that component A exerts on component B is shown in Fig. 2.2 below.

Fig. 2.2

(i) On Fig. 2.2, sketch the graph of the force that component B exerts on component A. [1]

(ii) Explain how your answer to (i) is consistent with the Principle of Conservation of Linear

Momentum. ……………………………………………………………………………………………………..

...........................................................................................................................................

...........................................................................................................................................

.......................................................................................................................................[2]

(iii) The two-stage rocket consists of component A of mass 3m and component B of mass 2m. The rocket is travelling in space at a constant speed Vo when an internal explosion causes component A to move backwards with a speed of ⅓Vo.

Show that the speed of component B after the internal explosion is 3Vo. [2]

6

(b) Fig. 2.3 shows speed-time graphs for three spherical objects made of the same material but different radii, a1, a2, a3, falling through a column of liquid after they are released from rest.

Fig.2.3

(i) Explain qualitatively why the speeds tend to a constant value.

……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

…………………………………………………………………………………………………[3]

(ii) If the constant speed of a falling sphere is proportional to the square of its radius, calculate a2 given that a1 = 2.0 mm.

radius a2 = ……………………… mm [2]

7

[Turn over

3 (a) State the principle of moment. ……………………………………………………………………………………………………………

……………………..…………………………………………………………………………………..[1]

(b) To increase the extension of a stiff spring for a given load, a student set up the system shown in Fig. 3.1. The weight of the metal bar was 5.0 N and the tension the student achieved in the spring is 37 N when the bar is horizontal. The spring constant k of the spring is 550 N m-1.

Fig. 3.1

(i) Calculate the mass of the load that the student used.

mass of load = ………………… kg [2]

(ii) Calculate the magnitude of the force exerted on the metal bar at the pivot.

force = ………………………. N [2]

(iii) Draw on Fig. 3.1 an arrow labeled F, to show the direction of the force calculated in part (ii).

[1]

(iv) Calculate the energy stored in the spring.

energy = ……………………. J [2]

8

4 (a) Explain whether an object moving at constant speed in a circular motion is experiencing acceleration. ………………………………………………………………………………………………………….…

……………………..…………………………………………………………..………………………[1]

(b) A bucket of water of total mass 10.4 kg is attached to an old piece of rope and whirled in a vertical circle. The effective radius of the rotation is 1.65 m.

(i) State the position in the circular motion at which the rope is most likely to break. ……………………………………………………………………………………………………

…………………..………………………………………………………………………………[1]

(ii) The maximum tension that the old rope can withstand is 300 N. Determine the maximum speed the bucket of water can have without the rope breaking when whirled in a vertical circle.

maximum speed = ………………….. m s-1 [3]

(iii) Determine the minimum speed the bucket can have at the top of its motion without the water spilling from the bucket.

minimum speed = …………………… m s-1 [3]

9

[Turn over

(c) State Newton’s Law of Gravitation.

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

……………………………….………………………………………………………………………..[2]

(d) (i) Define angular velocity for an object travelling in a circle. ………………………………………………………………………………………………………

……………………………………………………………………………………………………[1]

(ii) Calculate the angular velocity of the Earth in its orbit around the sun. Assume that the orbit is circular and give your answer in terms of SI unit for angular velocity.

angular velocity = ………….……………..

SI Unit = ………………………………..[3]

10

5 (a) Define the tesla. ……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

……………………..…………………………………………………………………………………..[2]

(b) A large horseshoe magnet produces a uniform magnetic field of flux density B between its poles. Outside the region of the poles, the flux density is zero. The magnet is placed on a top-pan balance and the wire XY is situated between its poles, as shown in Fig. 5.1.

Fig. 5.1

The wire XY is horizontal and normal to the magnetic field. The length of wire between the poles is 4.4 cm. A direct current of magnitude 2.6 A is passed through the wire in the direction from X to Y. The reading on the top-pan balance increases by 2.3 g.

(i) State and explain the polarity of the pole P of the magnet. ……………………………………………………………………………………….…………

……………………………………………………………………………………………………

………………………………………………………………………..………………..…………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

…………………………………………………………………..……………………...…..… [3]

(ii) Calculate the magnetic flux density between the poles.

magnetic flux density = ……………………….T [2]

11

[Turn over

6 A series of data on the performance of one particular modern car are extracted from the manufacturer’s handbook. The mass of the car under test is 1400 kg. Study the following information in Table 6.1, Fig. 6.1 and Fig. 6.2, and answer the questions that follow.

Table 6.1 Time to reach the speed from rest

Speed, v / m s−1 13.0 18.0 22.0 27.0 31.0 35.0 36.5 Time to reach the speed from rest, t / s

3.5 5.0 7.0 10.0 13.5 19.5 28.0

Fig. 6.1 Graphs of available force at the wheels (for different gears) and total resistive forces plotted

against speed

12

Fig. 6.2 Climbing resistance of the car on a particular slope. Note: A 10% gradient means the slope rises 10 m vertically for every 100 m of horizontal distance.

(a) (i) On Fig. 6.3, plot a graph of speed v against time t for the car as it accelerates through

the gears. [2]

Fig. 6.3

13

[Turn over

(ii) From the above plot, determine the acceleration when the car is travelling at 25 ms-1.

acceleration = ……………………. m s-2 [2]

(b) (i) Determine the optimum gear for maximum acceleration when the car is travelling at a speed of 25 m s-1. Justify your choice. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. …………………………………………………………………..…………………….……… [2]

(ii) Calculate the maximum theoretical acceleration at 25 m s-1.

maximum acceleration = ……………………………. m s-2 [3]

(iii) Hence, comment on whether the information provided by the manufacturer is consistent. ………………..……………………………………………………………………………………

……………………….………………………………………………………………………….[1]

14

(c) The total resistive force FT to the car’s motion on a slope is given by

FT = R + F

S

where FS is a constant climbing resistance on a particular slope.

By referring to Fig. 6.1 and Fig. 6.2, determine the maximum possible acceleration of the car on a 5 % slope at 15 m s−1.

maximum acceleration = …………………………… m s-2 [3]

15

[Turn over

Section B Answer one question in this section in the spaces provided.

7 A thermistor is often used in an electrical circuit to detect temperature changes.

(a) In the space below, draw the I-V characteristic graph of a thermistor and explain how the resistance of the thermistor changes with temperature. ……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

………………………………………………………………………………………………………..[3]

(b) The setup shown in Fig. 7.1 was used to detect the variation in the temperature of the environment through observing the brightness of a filament lamp in the setup.

Fig. 7.1

Describe and explain the variation of the potential difference across the lamp over time when temperature of the environment and also the lamp increases. ……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

………………………………………………………………………………………………….…………

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

…………………………………………….…………………………………………………………..[4]

16

(c) The electrons within an electrical circuit are moving at a speed (also known as drift velocity) of 10-4 m s-1.

(i) Estimate the time taken for one electron to travel from the negative terminal of the cell E to the positive terminal of the cell E via the two long wires, AB and CD, each 1 m in length, at drift velocity.

Fig. 7.2

time = …….……………. s [1]

(ii) Explain how it is possible for a light bulb connected by metre long wires as shown in Fig. 7.2 to be lighted up almost immediately when the switch is closed despite the low drift velocity.

……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

…………………………………………………………….…………………………………..[3]

17

[Turn over

(d) Fig. 7.3a and Fig. 7.3b shows two similar fixed resistors of resistance R in two different connection setup, supplied by identical e.m.f. source E.

Fig. 7.3a Fig.7.3b

(i) State and explain whether the cell in Fig. 7.3a or Fig. 7.3b will last longer before it stops supplying power to the resistor. ……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

…………….………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

…………………………………………………………………………………………….…..[4]

(ii) The internal resistance of the e.m.f. source in Fig. 7.3a and Fig. 7.3b is 2R.

Determine the power supplied to the external circuit for the setups in Fig. 7.3a and Fig. 7.3b in terms of R and E.

power supplied in Fig. 7.3a = ………………………..…….

power supplied in Fig. 7.3b = …………………………. [3]

18

(e) Distinguish between electromotive force (e.m.f) and potential difference using energy consideration. ……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

……………………………………………………………………………………………………….. [2]

19

[Turn over

8 (a) α-particles of the same velocity, directed normally in a parallel beam at a thin metal foil in a vacuum, are scattered at various angles.

(i) State and explain two deductions made from the scattering distribution about the structure of the atoms in the foil. ……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

…………….………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

…………………………………………………………………………………………….…..[4]

(ii) Suggest two reasons for using a very thin metal foil. ……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………….. [2]

(iii) On Fig. 8.1, complete the paths taken by the parallel α-particles of the same energy

values which all come close enough to the nucleus to be deflected. [3]

Fig. 8.1

20

(b)

When uranium-235 ( 23592U ) absorbs a slow-moving neutron, one possible nuclear reaction is

+ → +235 1 141 92 1

92 0 56 36 0U n Ba Kr + 3 n + energy

(i) State the name of this type of nuclear reaction.

…..………………………………………………………………………………………………[1]

(ii) The masses of some particles and nuclei are given in Fig. 8.2.

mass / u

uranium-235 23592U 235.044

neutron 10n 1.009

barium-141 14156Ba 140.914

krypton-92 9236Kr 91.926

Fig. 8.2

For the reaction in (b), calculate

1. the change in the rest mass before and after the nuclear reaction. Leave your answer in atomic mass unit, u.

change in mass = ……………………….. u [2]

2. the energy released, in MeV, to three significant figures.

energy = ……………………….. MeV [3]

21

[Turn over

(iii) Fig. 8.3 shows the binding energy per nucleon – nucleon number graph.

Fig. 8.3

Explain with reference to Fig. 8.3 why energy is released for the reaction. ……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………….. [2]

(c) An isotope of Potassium-40 decays to form Argon-40. This isotope has a half-life of 1.4 x 109

years. This process is random and spontaneous.

(i) Define half-life of a nuclear decay. ……………………………………………………………………………………………………

……………………………………………………………………………………………….. [1]

(ii) Explain what is meant by the term random and spontaneous when referring to the nature of nuclear decay. ……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………….. [2]

22

Blank Page


1(a) Displacement – A vector whose length is the shortest distance from the point of reference. Velocity – Rate of change of displacement with respect to time.

B1 B1

(b) Area under graph = ½ (6 +12)(0.6) = 5.4 m

M1 A1

(c) Area under graph = ½ x 1 x 10 = 5 m A1 (d) a = . = 10 m s-2

M1 A1

(e) From graph, time taken to reach max height on rebound = time taken to next reach ground = 1.0 s

A1

2 (a) (i) B1

2 (a) (ii)

When a variable force acts on an object, the change in momentum of the object is given by the area under the force-time graph.

The answer to (i) implies that the change in momentum of A and B are equal in magnitude and opposite in directions.

Consider A and B as one system, their changes in momentum cancel out vectorially thus the total momentum of the system remains unchanged.

B1 B1

2 (a) (iii)

By Conservation of Linear Momentum,

MA(UA) + MB(UB) = MAVA + MBVB

(2m + 3m)V0 = (3m)(- 1

3Vo) + 2mVB

VB = 3V0

M1 A1

2 (b) (i) The object experiences three forces acting on it namely weight W, and a viscous force R when it moves in the liquid which opposes its motion. Net force on object, W – U - R = ma As the speed of object increases, viscous force increases, acceleration of the object decreases until zero which implies that the object is moving at constant velocity. Hence the speed increases to a constant terminal value.

B1 B1 B1

2 (b) (ii)

Given v ∝ a2 0.5 = k(2.0)2 ----- (1) 0.25 = k(a2)2 -------(2)

C1


(1)(2)

2

22

0.5 2.00.25 r

=

r2 = 1.4 mm

A1

3 (a) Sum of clockwise moments about the point equals to the sum of anti-

clockwise moment about the same point (pivot).M1

3 (b) (i)

Taking moment about pivot, 37(0.15) = 5 (0.45) + L (0.85) L = 3.88 N 3.88 / 9.81 = 0.3957 kg Hence mass = 0.396 kg (3 SF)

M1 A1

3 (b) (ii)

Sum of all forces Y direction = 0 37 + T = 5 + 3.88 T = 28.1 N

C1 A1

3 (b) (iii)

A1

3 (b) (iv)

Energy stored = ½ k x2

= ½ (550) (37/550)2

= 1.24 J

C1 A1

4(a) An object moving in a circular motion with constant speed is

accelerating because the velocity of the object is changing as the direction of the velocity is changing.

B1 B1

4(b)(i) Bottom of the circular motion A1 4(b) (ii) Consider the bucket at the bottom of the circular motion,

T – mg =

300 – (10.4)(9.81) = ..

v = 5.60 ms-1

C1 C1 A1

4(b) (iii)

Consider the water in the bucket at the top of the circular motion,

N + mg = If water just stays in bucket, N = 0

Hence mg = Solving v, v = 4.02 m s-1

C1 C1 A1

4(c) Gravitational Force between two objects, F = G M m / r2 B1


Where M and m are the masses of the 2 object and r is the distance between their centre of masses

B1

4(d)(i) It is the rate of change of angular displacement with respect to time. B1 4(d)(ii) T = (365)(24)(60)(60) = 3.156 x 107 s

ω = 2π/T = 1.99 x 10-7 rad s-1

C1 A1 A1

5(a) One tesla is the unit for magnetic flux density which causes a

force per unit length of one newton per metre on a straight wire carrying a current of one ampere and is at right angles to the direction of the magnetic field.

B1 B1

5(b) (i) As seen from the increased balance reading, there is a downward force on magnet due to wire carrying current. By Newton’s third law, there is an upward force on wire by magnet. By Fleming’s left hand rule, pole P is a north pole

M1 M1 A1

5(b) (ii) By Newton’s 2nd law, W – BIL =0 W= BIL 2.3 × 10–3 × 9.81 = B × 2.6 × 4.4 × 10–2 B = 0.20 T

M1 A1


6 (a)(i) B1 – All points plotted correctly B1 – Relatively smooth Graph

6(a)(ii)

Acceleration = 0.00.145.110.33

−−

= 1.536 = 1.54 ms-2

for drawing tangent at v = 25 ms-1

for correct calculation. (acceptable range: 1.39 – 1.69)

M1 A1

6(b)(i) Available force for forward motion, F, is the greatest at this gear 2. Gear 2

M1 A1

6(b)(ii) From Fig. 6.2, F = 2500 N and R = 375 N ( 300 ≤ R ≤450) Newton’s 2nd law,

2

2500 375 1400

1.52 m s

F R ma

a

a −

− =− =

=

C1 C1 A1

6(b)(iii) It is almost equal to the acceleration found in (a)(ii), and hence the information provided by the manufacturer had been consistent.

A1

6(c) From Fig. 6.2, F = 4300 N and R = 250 N (4350 ≤ F ≤ 4450) From Fig. 6.3, FS = 500 N Newton’s 2nd law,

M1 M1 A1

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

0.0 5.0 10.0 15.0 20.0 25.0 30.0

t/s

v/m

s-1


( )2

4300 250 500 1400

2.54 m s

TF F ma

a

a −

− =− + =

=

7(a)

As the current of the thermistor increases, the temperature of the thermistor increases. The resistance of the thermistor decreases. This is because the number of free electrons and holes increased due to the increase in temperature, hence the resistance decreases.

B1 B1 B1

7(b) As temperature increases, resistance of thermistor decreases This causes the effective resistance across AB to decrease. As the filament lamp heats up, the resistance of the lamp increases as well. Hence, effective resistance across CD increases. By potential divider principle, (may use mathematical relationship) the potential difference across the bulb increases.

B1 B1 B1 (Accept mathematical relationship) B1

7(c)(i) Time = Distance / Speed = 2 / 1 x 10-4 = 2 x 104 s

A1

7(c)(ii) When the switch is closed, an electric field is applied across the circuit due to the cell. All electrons start moving simultaneously at all points within the circuit almost immediately due to the electric field applied across the circuit due to the cell. The electrons in the light bulb moves as well and this will cause the current necessary to light up the light bulb.

B1 (any 3 of the 4 marks) B1 B1


It is not necessary for the electrons from the cell to reach the light bulb to light it up.

B1

7(d)(i) In Fig. 7.3a, as the total resistance is larger, the power supplied is lower than that in Fig. 7.3b as the power supplied to the entire setup in Fig. 7.3a is E/(2R)2 while that in Fig. 7.3b is E/(R/2)2 Therefore, with a larger power supply and an equal amount of total energy it can supply, the cell in Fig. 7.3b will stop supplying power earlier.

B1 B1 M1 A1

7(d)(ii) 22

2

2

2( )4 =

2 8

2( )2 22 =

252

a

b

R E ERPR R

RERR EP R R

×=

+=

A1 C1 A1

7(e) For EMF, the total amount of chemical energy in the battery is transferred to electrical energy by each coulomb of charge. For PD, the total amount of electrical energy is transferred to other forms of energy by each coulomb of charge.

B1 B1

8(a)(i) 1. Most of the α particles are undeflected from its original path.

Deduction: Hence the deduction is that an atom is mainly made up of empty space or the radius of nucleus is much smaller than that of the atom.

2. Some of the α particles are deflected at large angles or even almost reflected backwards Deduction: Hence the deduction is that the positive charges is concentrated in the nucleus of the atom or most of the mass is contained in the nucleus

B1 B1 B1 B1

8(a)(ii) The metal foil should be very thin to:-1. prevent too many α particles from being absorbed 2. allow the α particles to be scattered only once (as few times as possible)

B1 B1

8(a)(iii)

Marking point: 1. The α-particles should be deflected downwards. 2. The angle of deflection should be the greatest for the top α-particle and

B1 B1


smallest for the bottom α-particle.3. The path is symmetrical between the incoming, outgoing α-particle and the

position of the nucleus.

B1

8(b)(i) Nuclear Fission reaction B1 8(b)(ii) 1.

Change in mass ∆m = Total mass of reactants – Total mass of products = (235.044 + 1.009) – (140.914 + 91.926 + 1.009 + 1.009 + 1.009) = 0.186 u

M1 A1

8(b)(ii) 2.

Energy = ∆m × c2

= 0.186 × 1.66 × 10-27 × (3.00 × 108)2 = 2.778 × 10-11 J = 1.74 MeV

C1 C1 A1

8(b)(iii) From the graph U-235 has a lower BE per nucleon than its product Ba-141 and Kr-92. This implies that nuclear energy is released.

M1 A1

8(c)(i) The amount of time it takes for half of the atoms in a sample of a radioactive substance to decay.

B1

8(c)(ii) Random: No way to tell which nucleus will decay and cannot predict when it is going to decay Spontaneous: Process cannot be affected by any external factors such as temperature or pressure.

B1 B1

2018 end-of-year exams - the learning space

Documents