physics 8867/01 - the learning space · 2020. 5. 25. · physics paper 1 multiple choice additional...

This paper has 14 printed pages.

HWA CHONG INSTITUTION

JC2 Preliminary Examinations

Higher 1

CANDIDATE NAME

CT GROUP 17S

TUTOR NAME

PHYSICS Paper 1 Multiple Choice

Additional Materials: Optical Mark Sheet

8867/0120 September 2018

1 hour

INSTRUCTIONS TO CANDIDATES

Write in soft pencil.

Write your name, CT, NRIC or FIN number on the optical mark sheet (OMS). Shade your NRIC or FIN 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 OMS.

Each correct answer will score one mark. A mark will not be deducted for a wrong answer.

Any rough working should be done in this booklet.

2

© Hwa Chong Institution 8867 / 01 / C2 Prelim 2018

Data Formulae

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

uniformly accelerated motion

resistors in series resistors in parallel

s = ut + 2

1at2

v2 = u

2 + 2as

R = R1 + R2 + . . .

1/R = 1/R1 + 1/R2 + . . .

3


1 Which of the following statements is correct?

A Density is mass per cubic metre.

B Potential difference is energy per unit current.

C Speed is distance travelled per second.

D Power is work done per unit time.

2 The coach of a 100 m sprinter wanted to estimate the average speed of his trainee sprinter. He positioned himself at the finishing line of the 100 m track. Upon hearing the sound of the firing pistol, the sprinter started the run and the coach started his stopwatch. The coach stopped the stopwatch when he saw the sprinter crossed the finishing line.

If the measurement in distance is perfect and the coach had an uncertainty of ± 0.2 s each time he pressed the stopwatch. The recorded time for the run is 10.0 s, what is the fastest possible average speed the sprinter had achieved?

A 9.6 m s–1 B 9.8 m s–1 C 10.2 m s–1 D 10.4 m s–1

3 A hot-air balloon is rising vertically with a constant velocity of 1.0 m s–1. A tourist riding the balloon was photographing the surrounding scenery when he accidentally dropped his camera. What is the magnitude of the change in velocity of the camera two second after its release?

A 9.8 m s–1 B 18.6 m s–1 C 19.6 m s–1 D 20.6 m s–1

4 An elevator is moving downwards with an acceleration of 5.8 m s-2. A ball, held 2.0 m above the floor of the elevator and at rest with respect to the elevator, is released.

How long does it take for the ball to reach the floor of the elevator?

A 0.51 s B 0.64 s C 0.83 s D 1.00 s

4


5 A clay pigeon is launched vertically into the air from the ground.

A marksman lies at a horizontal distance of 170 m away from the launching device. When the clay pigeon reaches its maximum height of 60 m, the marksman aims his rifle at the clay pigeon and fires a bullet at it. The bullet leaves the rifle with a speed of 300 m s-1.

At which time after the bullet is fired, does the bullet hit the clay pigeon? Assume air resistance is negligible.

A 0.17 s B 0.57 s C 0.60 s D 1.66 s

6 A projectile of mass m is fired at ground level with velocity u from a point A, as shown below.

Neglecting air resistance, determine the magnitude of the change in momentum of the mass between leaving point A and arriving back at ground level.

A zero B ½ mu C mu D 2mu

7 An empty truck has a mass of 5.0 x 103 kg. Regardless of the mass of load it has, it experiences a fixed retarding force of 70 kN when it decelerates from a speed of 50 m s-1 to 30 m s-1. The duration for the empty truck to decelerate is t1. The duration for it to decelerate when it has a full load of 1.3 x 103 kg is t2.

What is the difference (t2 – t1)?

A 0.37 s B 0.56 s C 0.93 s D 1.43 s

A

θ = 30o

u

5


8 An object of mass M travelling to the right with velocity 2v collides with another object of mass 2M travelling to the left with velocity v. After the collision, the objects stick together.

Which line in the table shows the total momentum and the total kinetic energy of the two objects after the collision?

momentum kinetic energy

A 0 0

B 4Mv 0

C 0 3Mv2

D 4Mv 3Mv2

9 In an α-scattering experiment, an α-particle is moving in a straight line towards a gold nucleus. Due to electrostatic repulsion, the α-particle slows down and comes to a stop at a distance of 10-13 m from the gold nucleus. It then starts moving back in the direction from which it came, reaching a final speed which is the same as its initial speed. As the gold nucleus is so much heavier than the α-particle, it may be considered stationary in this experiment.

Which of the following statements is correct?

A This collision is not elastic, as the kinetic energy is not conserved throughout the interaction.

B This collision is not elastic, because the linear momentum is not conserved.

C This is not a collision, as the particles never touch each other.

D This is an elastic collision, because the kinetic energy before the collision is equal to the kinetic energy after the collision.

M 2M2v v

6


10 A uniform rod has a wooden section and a solid rubber handle, as shown below.

The length of the handle is l and the length of the wooden section is 4.00 l. The rod balances a

distance 2.10 l from the rubber end.

What is the ratio density of rubber

density of wood?

A 1.71 B 2.25 C 2.50 D 3.27

11 Hooke’s law states that

A the extension is proportional to the load when the elastic limit is not exceeded.

B the extension is inversely proportional to the load when the elastic limit is not exceeded.

C the extension is independent of the load when the elastic limit is not exceeded.

D load is dependent on extension.

12 A ladder of weight W rests against a vertical wall.

Friction between the ladder and the ground and also between the ladder and the wall prevents the ladder from slipping.

Which diagram shows the directions of the forces on the ladder?

7


13 When a horizontal force F is applied to a trolley over a smooth horizontal surface of distance x, its kinetic energy changes from 2 J to 6 J.

If a force 2F is applied to the trolley over a distance of 2x, what will be the final kinetic energy of the trolley? Assume the original kinetic energy of the trolley is 2 J.

A 12 J B 16 J C 18 J D 24 J

14 The figure below shows how the force F applied to an object varies with its displacement s.

Which of the following graphs correctly shows the variation of the work done by F with respect to displacement?

A B

C

D

0 2 4 6 8 10

F/N

s / m

W W

W W

s s

s s

8


15 A cyclist exerts an effort of 10 000 J when he is cycling at a constant velocity of 10 m s–1 over a certain level stretch of road. What is the total work done against frictional forces? The combined mass of cyclist and bicycle is 80 kg.

A 4000 J B 6000 J C 10 000 J D 14 000 J

16 Two rough discs of mass m and 2m are placed on a rough, horizontal and level turntable as shown in the diagram. The turntable starts rotating from rest with gradually increasing angular velocity ω. Eventually, both discs will slip off the turntable.

Given that the maximum frictional force acting on mass m is half of that on mass 2m, which of the following is correct?

A Disc of mass m experiences maximum frictional force first.

B Disc of mass 2m experiences maximum frictional force first.

C Both discs experience maximum frictional force at the same time.

D Neither disc will experience maximum frictional force

r

2m

m

ω

2r

turntable

disc

diameter lines

disc

9


17 The diagram shows the magnitude of the gravitational force, acting on a mass of 1 kg, at distances from an isolated point mass Q.

The gravitational force is given at the points on the grid lines at different distances from Q. The grid lines have equal separation.

What would the gravitational force be, when a mass of 1 kg were placed at the two points marked R and S?

force at R / N force at S / N

A 4.0 3.6

B 6.4 1.8

C 6.4 3.6

D 8.0 4.0

18 There are two isolated planets A and B, of mass MA and MB, respectively. Their centres are a distance D apart and they rotate with a uniform angular velocity ω about their common centre of mass, O, as shown in the figure below.

If the distance between planet A and point O is R, which of the following expressions for the centripetal force on planet B is incorrect?

A 2

BA

D

MGM B ( )2RD

MGM BA

−

C MA R ω2 D MB (D - R) ω2

O

D - R

Planet A Planet B

R

16.0 N

32.0 N

10


19 Two different resistance wires, X and Y, are placed in series with one another to form part of a circuit. They both have the same length and cross-sectional area.

Two ideal voltmeters monitor the voltage across them. The reading in each voltmeter is shown in the diagram.

What is the ratio of the resistivity of X to the resistivity of Y?

A 1 : 5 B 1 : 4 C 4 : 1 D 5 : 1

20 Three identical bulbs are connected as shown and all of them light up.

Which of the following statement best suggests their relative brightness?

A All of them have the same brightness.

B All of them have different brightnesses.

C Y and Z have the same brightness, which is greater than the brightness of X.

D Y and Z have the same brightness, which is smaller than the brightness of X.

X Y

10 V

2 V

X Y

Z

11


21

A network is constructed using eight resistors, each of resistance R, and three switches S1, S2 and S3.

Which of the following combinations will give rise to the minimum total resistance between points X and Y?

S1 S2 S3

A closed closed closed

B closed open closed

C open closed closed

D open open open

22 A row of 30 decorative lights, connected in series, is connected to a mains transformer. When the supply is switched on, the lights do not work. The owner uses a voltmeter to test the circuit. When the voltmeter is connected across the fifth bulb in the row, a reading of zero is obtained.

Which of the following scenarios described is not possible?

A Only the filament of the fifth bulb has broken.

B The fuse in the mains transformer has blown.

C The filament of at least one of the other bulbs has broken.

D There is a break in the wire from the supply to the transformer.

23 A neutral sub-atomic particle is at rest in a magnetic field of flux density B. It splits into two particles, each of mass m, one of which has a negative charge –q. The particles then move with velocities perpendicular to the magnetic field.

After what time will the particles collide?

A B C

D 2

X Y

R S1

S2S3

R R

R

R

R

R

R

12


24 The diagram below illustrates one of the earliest designs of a galvanometer. A coil of wire is wound around a circular iron core which is placed between two magnets with circular surfaces such that the magnetic field (indicated by the arrows) on the surface of the iron core is directed perpendicularly onto the surface and of the same magnitude across the surface of the iron core as shown in the diagram.

When a constant current flows in the coil, the needle will be deflected to an angle from the vertical direction. Which of the following graphs show the variation of the torque on the soft iron core τ, due to magnetic forces acting on the coil, with the angular displacement θ of the needle as the needle rotates from 0 to ?

A

B

C

D

0

Graduated Scale of Galvanometer

Needle connected to soft iron core

Magnet

Soft iron core

Coil of wire linked to external circuit

1-1

θ

τ

θ

τ

θ

τ

θ

τ

13


25 A charged particle, initially travelling in a vacuum in a straight line, enters a uniform field. This causes the particle to travel in a curved path that is not the arc of a circle.

Which type of field and which initial direction of the particle with reference to the field causes this to happen?

field type initial direction of the particle compared to the field

A electric parallel

B electric perpendicular

C magnetic parallel

D magnetic perpendicular

26 Two long straight and parallel wires, carrying currents in the same direction, separate the surrounding space into three regions 1, 2 and 3.

In which region(s) can there be a neutral point (i.e. a point of zero magnetic field)?

A Region 2 only.

B Both regions 1 and 3.

C Either region 1 or region 3, but not both.

D None of the regions.

27 Which statement about radioactive nuclides is correct?

A The half-life is proportional to the probability of decay per unit time of each nucleus.

B The higher the temperature, the larger the probability of decay per unit time of each nucleus.

C The probability of decay per unit time of each nucleus decreases with time.

D The smaller the probability of decay per unit time, the longer the half-life.

I1

I2

Region 1

Region 2

Region 3

14


28 Which of the following combinations of radioactive decay results in the formation of an isotope of the original nucleus?

A one α and four βdecays

B one α and two βdecays

C two α and two βdecays

D four α and one βdecays

29 A radioactive source consists of a mixture of two isotopes P and Q.

P has a half-life of 60 minutes and Q has a half-life of 30 minutes. The initial activity recorded by a suitable counter is 800 min-1. After 120 minutes, the counter registers an activity of 80 min-1.

What is the initial contribution of P to the count rate?

A 160 min-1 B 240 min-1 C 270 min-1 D 480 min-1

30 Alpha, beta and gamma radiations are absorbed to different extends in solids, and behave differently

in an electric and magnetic field.

The diagrams illustrate these behaviours.

Which three labels on these diagrams refer to the same kind of radiation?

A L, P, X B L, P, Z C M, P, Z D N, Q, X

End of Paper

1

Hwa Chong Institution C2 2018 Preliminary Examinations H1 Physics Paper 1 Solutions and Explanations:

Qn Ans Explanation

1 D Density is mass per unit volume.

Potential difference is energy per unit charge.

Speed is distance travelled per unit time.

Power is defined as the rate at which work is done, which is equivalent to the amount of work done per unit time.

2 D The distance is fixed, so the fastest speed is achieved when the time taken is the shortest.

Shortest time after taking into account the uncertainty = 10.0 – 0.2 – 0.2 = 9.6 s.

Fastest speed = 100/9.6 = 10.4 m s–1.

3 C Once the camera is released, it is subjected to a constant acceleration due to gravity of 9.81 m s–2.

Δv = a(Δt) = (9.81)(2) = 19.6 m s–1. 4 D

Distance travelled by ball = + 210 (9.8)

2bs t ,

Distance travelled by elevator = + 210 (5.8)

2es t ,

= −2 21 12.0 (9.8) (5.8)

2 2t t = 1.00 st

5 C θ −= 1 6 0

1 7 0ta n

Consider horizontal motion:

1700 60 s

300x

x

st .

u cosθ= = =

6 C From the vector addition diagram:

∆u = 2u sin 30o = u

∆p = m ∆u = mu

7 A When the racing car is almost out of fuel, its deceleration a = F/m = 7000 / 500 = 14.0 m s-2. The time for it to slow down t1 = (v – u) / a = (30 – 50) / (-14.0) = 1.43 s.

When the racing car has a full load of fuel, its deceleration a = F/m = 7000 / 630 = 11.1 m s-2. The time for it to slow down t2 = (v – u) / a = (30 – 50) / (-11.1) = 1.80 s.

The difference t2 – t1 = 1.80 – 1.43 = 0.37 s.

8 A Since the momenta of the two objects are initially equal in magnitude and opposite in direction, the total momentum is zero. Hence, the objects are not moving after the collision, which implies that the kinetic energy is zero.

9 D An interaction in which the initial kinetic energy is equal to the final kinetic energy is called an elastic collision.

30o

30o

u

u

Δu

2

10 B By the principle of moments, taking moments about the balancing point,

the sum of clockwise moments = the sum of anticlockwise moments,

(Wrubber handle) x (distance of CG to pivot) = (Wwooden section) x (distance of CG to pivot)

(ρrubber handle A l) x (1.60 l) = (ρwooden section A 4l) x (0.90 l)

1.60 (density of rubber) = 3.60 (density of wood)

density of rubber

density of wood= 3.60

1.60= 2.25.

11 A Hooke’s Law states that the magnitude of force F exerted by a spring on a body attached to the spring is proportional to the extension x of the spring from its natural length provided the proportional limit of the spring is not exceeded.

Increasing the load causes the extension to increase; changing the extension does not cause the load to increase.

12 C The ladder tends to slide down, so the frictional force by the wall on the ladder is pointing upwards. Since there is also a normal contact force by the wall on the ladder, the total contact force by the wall on the ladder is to the right and upwards.

The ladder tends to slide to the right, so the frictional force by the floor on the ladder is pointing towards the left. Since there is also a normal contact force by the floor on the ladder, the total contact force by the floor on the ladder is to the left and upwards.

13 C By conservation of energy, work done by the force W = gain in kinetic energy ΔK

Δ = = = Δ =Δ

= + =

' ' (2 )(2 )4 ' (4)(6J-2J)=16 J

New kinetic energy ' 16 2 18 J

K W F xK

K W FxK

14 B Area under the F – x graph = work done by force. The work done keeps increasing until the force becomes negative.

15 C Energy supplied goes into work done against frictional forces, since the kinetic energy remains constant.

16 B The frictional force provides the centripetal force.

For m: required centripetal force = mrω2

For 2m: required centripetal force = (2m)(2r)ω2 = 4 mrω2

As long as the masses are not slipping the frictional force on 2m is four times than on m.

17 C Let the length of each square be 1 unit.

Then S is 3 units away Q. Hence, the magnitude of the gravitational force is 2

32 3.553

= N.

R is 5 units away from Q. Hence, the magnitude of the force is

( )232 6.45

= N.

18 B Gravitational force on A and on B is 2D

MGM BA

For A : 22

ωRMD

MGMA

BA =

For B : 22

)( ωRDMD

MGMB

BA −=

B A D - R R

3

19 B = = since and are the same. = since = and is the same for both (in series) = = =

20 D The currents through Y and Z are the same. The current through X is larger than the currents through Y or Z. Brightness is related to power and hence related to I2R. Since R is the same for all of them, the brightness is related I2.

21 A The larger the number of parallel connections for the resistors, the lower the total resistance. Short-circuiting any resistance will also reduce the net resistance.

22 A If only the filament of the fifth bulb has broken, the voltmeter will register a non-zero voltage across the transformer.

23 B Since one of the particles has a negative charge –q, by the principle of conservation of charges, the other particle will have to have a positive charge +q. By the principle of conservation of linear momentum, the particles will move in opposite directions. Hence, both particles will feel a magnetic force in the same direction.

As the particles move with velocities perpendicular to the magnetic field, they undergo circular motion, with the magnetic force providing the centripetal force:

FB = Fc Bqv = mv2/r r = mv/Bq

Since m, v, B and q are the same for both particles, the radius of their circular motion is also the same for both. The particles move in opposite directions and collide after moving through half a circle each, i.e. after 0.5T = (πr)/v = (πmv)/(Bqv) = (πm)/(Bq).

24 A Since the force on the coil is given by = , the force is constant even as the coil rotates. Hence, the torque is also a constant.

25 B A charged particle moving in a magnetic field would either undergo either circular motion or linear motion. Hence, the field must be an electric field. If the particle were travelling parallel to the field, it would either speed up or slow down, but it would continue its motion in a straight line. Therefore, the particle must be travelling perpendicular to an electric field.

26 A Current I1 sets up a magnetic field out of the page in region 1 and into the page in regions 2 and 3. Current I2 sets up a magnetic field out of the page in regions 1 and 2 and into the page in region 3. For there to be a neutral point, the magnetic fields due to both currents must point in opposite directions. This is only the case in region 2.

27 D If for each nucleus the probability of decay per unit time is larger, the half-life is shorter. Radioactive decay is a spontaneous process, independent of the temperature. And for any given nucleus, the probability of decay per unit time does not change.

28 B Isotopes have the same number of protons but different number of neutrons. By emitting one α- and two β-particles, the number of protons of the element will be conserved.

29 A Initially: P + Q = 800

After 120 min, ¼ P + 1/16 Q = 80

Therefore P = 160

30 C M, P, Z are all beta particles.

This paper has 22 printed pages.

HWA CHONG INSTITUTION

JC2 Preliminary Examination

Higher 1

CANDIDATE NAME

CT GROUP 17S

CENTRE NUMBER

INDEX NUMBER

PHYSICS Paper 2 Structured Questions

Candidates answer on the Question Paper.

No Additional Materials are required.

8867/02 14 September 2018

2 hours

INSTRUCTIONS TO CANDIDATES

Write your Centre number, index number, name and CT class clearly on all work you hand in.

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, paperclips, highlighters, glue or correction fluid.

Section A

Answer all questions.

Section B

Answer one question only. Circle the question number of the cover page.

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

The number of marks is given in brackets [ ] at the end of each question or part question.

You are reminded of the need for good English and clear presentation in your answers.

For Examiner’s Use

SECTION A

1 8

2 9

3 9

4 11

5 8

6 15

SECTION B

7 20

8 20

Deductions

Total 80

2

© Hwa Chong Institution 8867 / 02 / C2 Preliminary Examinations 2018

Data Formulae

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

uniformly accelerated motion

resistors in series

resistors in parallel

s = ut + 2

1 at2

v2 = u2 + 2as

R = R1 + R2 + . . .

1/R = 1/R1 + 1/R2 + . . .

3


Section A

Answer all questions in the spaces provided.

1 (a) A spring, which has an unstretched length of 0.650 m, is attached to a fixed point. A mass of 0.400 kg is attached to the spring and gently lowered until equilibrium is reached. The spring has then stretched elastically by a distance of 0.200 m.

Calculate, for the stretching of the spring,

(i) the loss in gravitational potential energy of the mass,

loss = ………………………… J

[1]

(ii) the elastic potential energy gained by the spring.

gain = …………………………. J

[2]

(b) Explain why the two answers to (a) are different.

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

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

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

4


(c) Assume the lowering of the mass is done very slowly such that there is no change in the kinetic energy of the mass. On the axes provided, sketch how the gravitational potential energy (label as GPE) and elastic potential energy (label as EPE) vary from the unstretched position of the spring to the extension of 0.200 m. The point X indicates the value of the GPE when the spring had extended by 0.200 m.

[3]

0 0.200

X

Energy

Extension of spring/m

5


2 (a) State the principle of conservation of linear momentum.

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

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

(b) A light and long string (string 1), runs over two smooth and light pulleys (pulleys A and B). One end of the string is fixed to the ceiling and the other end is attached to mass M1. Pulley A is fixed to the ceiling while pulley B is movable and is attached to mass M2 via another light string (string 2).

Fig. 2.1

(i) Given that M1 = 4.0 kg and M2 = 8.0 kg and they are both at rest, determine the tension in string 1 and string 2.

tension in string 1 = …………………………. N

tension in string 2 = …………………………. N

[2]

M2

M1

Pulley A

Pulley B

String 1

String 2

6


(ii) A little disturbance is made to the system and M1 starts to move upward with speed v1 while M2 starts to move downwards with v2. By considering linear momentum, express v2 in terms of v1.

v2 = ……………………… [2]

(iii) Hence, determine the energy introduced to the system due to the disturbance in terms of v1.

energy = ……………………………. [2]

(iv) Given that the masses move with constant speed, show that total energy of the system is conserved. Explain your working clearly. You may assume air resistance is negligible.

[2]

7


3 Fig. 3.1 is a diagram of a human arm lifting an object.

Fig. 3.1

The lower arm is horizontal and its centre of gravity is 0.150 m from the elbow joint. The weight of the lower arm is 18 N. The bicep muscle exerts a force F at an angle of θ to the vertical.

The horizontal distance between the elbow joint and the point of attachment of the muscle to the lower arm bone is 0.040 m. The weight of the object held in the hand is 30 N and its centre of gravity is 0.460 m from the elbow joint. The arm is in equilibrium.

(a) Define centre of gravity.

[1]

(b) Determine the value of F when θ = 15o.

F = N [2]

bicep muscle

lower arm

8


(c) For the lower arm to be in equilibrium, the elbow joint also needs to exert a force R on the lower arm bone.

(i) Draw a labelled arrow on Fig. 3.1 to represent the force R that the elbow exerts on the lower arm. [1]

(ii) Explain how you determined the direction of this force R.

[2]

(d) As the lower arm is slowly moved away from the body, the angle θ increases.

(i) Sketch in Fig. 3.2 the graph of how F varies with θ. (You may assume that the lower arm remains horizontal and is in equilibrium at all times.)

Fig. 3.2 [1]

(ii) Explain the shape of your graph in Fig. 3.2.

[2]

0

F

θ 90o

9


4 (a) The variation with temperature of the resistance RT of a thermistor is shown in Fig. 4.1.

Fig. 4.1

The thermistor is connected in series with a resistor R as shown in the circuit in Fig. 4.2.

Fig. 4.2

The battery has e.m.f. 9.00 V and negligible internal resistance. The voltmeter has infinite resistance.

(i) For the thermistor at 22.5 °C, determine the resistance of the thermistor.

RT = ……...……………………. Ω [1]

(ii) Given that the voltmeter reading is 2.70 V, determine the resistance of resistor R.

resistance of R = ……...……………………. Ω

[2]

9.00 V

V

R

10


(b) A student had a lamp rated at 3.0 V, 0.20 A. He had only a battery of 12 V with negligible internal resistance. In order to obtain a voltage of about 3.0 V, he connected a circuit as shown in Fig. 4.3. Before connecting the lamp, the student used a voltmeter, with a resistance of 11.0 kΩ, to check the voltage between terminals A and B. The resistance wire CD has a total resistance of 1000 Ω.

Fig. 4.3

When the sliding contact, J, was at point C in Fig. 4.3, the voltmeter reading was 12 V. The contact J was then moved to point D.

(i) Determine the reading of the voltmeter when the contact J was at point D.

voltmeter reading = ….…………… V

[2]

The student then modified the circuit to that shown in Fig. 4.4. He is now able to adjust the sliding contact J to obtain a voltmeter reading of 3.0 V.

Fig. 4.4

J 12 V

J 12 V

11


(ii) 1. Calculate the current flowing through the voltmeter when it gives a reading of 3.0 V.

current = ……...……………………. A

[1]

2. Assuming this current is negligible compared with the current flowing through the resistance wire, determine the distance from C that the sliding contact J would be at. Express your answer as a fraction of the length CD.

……...……………………. length of CD

[1]

3. Calculate the power dissipated in the length you calculated in (b)(ii) part 2.

power = ……...……………………. W

[2]

(iii) The student then replaced the voltmeter with the lamp across AB. However, the lamp did not light up. If the lamp is not defective, how do you explain this?

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

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

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

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

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

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

[2]

12


5 (a) Fig. 5.1 shows the variation of binding energy per nucleon with the number of nucleons in nucleus.

Fig. 5.1

(i) Define binding energy of a nucleus.

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

[1]

(ii) Using Fig. 5.1, estimate the binding energy of the nucleus Iridium-170, 17077Ir .

binding energy of 17077Ir =……………..…… MeV

[2]

(iii) Hence, calculate the mass defect of 17077Ir .

mass defect of 17077Ir =………..…………… u [2]

13


(b) Stellar nucleosynthesis is a collective term for nuclear reactions taking place in stars. These reactions create nuclei of elements heavier than hydrogen.

The “triple” alpha process is a nuclear fusion reaction that occurs in stars where three alpha particles combine to form carbon-12, 12

6C .

(i) Determine the energy released in this reaction.

Given:

mass of alpha particle = 4.002603 u

mass of 126C = 12.000000 u

energy released =………….………… J

[2]

(ii) “Silicon” burning is the final stage of fusion in massive stars. This process involves silicon-28, 28

14 Si capturing multiple alpha particles, until the sequence terminates

at 5628Ni . At this point the star can no longer release energy via nuclear fusion. This

eventually results in a catastrophic collapse of the star.

Suggest a reason why the star can no longer release energy via nuclear fusion.

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

[1]

14


6 A typical mobile device has an accelerometer that can detect acceleration on two or three axes, allowing it to sense motion and orientation. A man enters an elevator of his residential building which is already loaded with a few of his neighbours. He switched on the accelerometer in his mobile device at time t = 0 and placed it on the floor of the elevator.

The elevator started moving up after a while. The output from the mobile device’s acceleration sensor is shown as a form of graph in Fig. 6.1. It is a graph of apparent acceleration A against time t.

Fig. 6.1

(a) State the value of A when the elevator is not moving.

acceleration = ………….………… m s-2 [1]

(b) Hence, suggest a reason why the vertical axis of the graph is labelled as apparent acceleration instead of acceleration.

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

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

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

(c) State the time at which the elevator first starts to move.

time = ………….………… s [1]

(d) How much time does it take for the elevator to pick up speed?

time = ………….………… s [1]

15


(e) What is the average actual acceleration of the elevator over the time duration quoted in part (d)? Briefly explain how you arrive at your answer.

average acceleration = ………….………… m s-2 [2]

(f) During the movement of the elevator, there is a part where it is travelling at constant speed. Determine the value of this constant speed.

speed = ………….………… m s-1

[1]

16


(g) Sketch the speed-time graph for this elevator from t = 0 to t = 16 s.

Label axes with all the values you have quoted in part (c), (d) and (f).

[3]

(h) Hence, estimate the distance moved by the elevator before it first comes to a rest.

distance = ………….………… m [2]

(i) For a typical residential building, suggest how many storeys the elevator may have passed.

number of storeys = ………….…………[1]

(j) Suggest one reason why the graph is not smooth.

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

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

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

[1]

0

Speed / m s-1

Time / s

17


Section B

Answer one question from this Section in the space provided.

7 (a) State Newton’s law of gravitation.

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

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

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

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

(b) Earth rotates about its axis with a period of 24 hours. Assume Earth has a uniform density.

Radius of Earth = 6.37 x 106 m

Mass of Earth = 6.0 x 1024 kg

(i) Calculate the centripetal force of an 80 kg man standing at Earth’s equator.

centripetal force = ……………………. N

[2]

(ii) Hence, calculate the acceleration of free fall at the equator of Earth, and explain why this value may be different from that at the poles.

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

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

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

[4]

18


(c) The International Space Station (ISS) revolves the Earth in a circular orbit at a height of just 408 km above Earth’s surface.

(i) For a circular orbit, the radius, r, and period, T, are related by the relationship 2 3T r∝

This result is known as Kepler’s Third Law. Derive Kepler’s Third Law.

[2]

(ii) Hence, show that the period of the ISS is 1.5 hours.

[1]

(iii) A student noted that a point on Earth’s equator rotates with a period of 24 hours but the ISS orbits with a period of just 1.5 hour.

Explain the apparent discrepancy between the student’s observation and Kepler’s Third Law.

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

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

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

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

(iv) An astronaut on the ISS deduced that he must be weightless since he was floating. Comment on his deduction.

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

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

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

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

[2]

19


(d) X and Y are two stars of the same mass. The points P and Q are equidistant from X and Y.

Fig. 7.1

(i) Sketch the variation in magnitude of the gravitational force on mass m due to the stars along PQ.

[2]

(ii) Describe the subsequent motion of a mass released at point P.

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

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

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

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

F

P Q

20


8 (a) State and define the S.I. unit for magnetic flux density.

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

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

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

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

[3]

(b) Joseph John Thomson was the first who demonstrated that atoms are actually composed of aggregates of charged particles. Prior to his work, it was believed that atoms were the fundamental building blocks of matter. The first evidence contrary to this notion came when people began studying the properties of atoms in large electric fields.

If a gas sample is introduced into the region between two charged plates, a current can be observed between them. The source of these charged particles is a cathode that causes the atoms of the gas sample to ionize. In 1897, Thomson set out to prove that the cathode rays produced from the cathode were actually a stream of negatively charged particles. A schematic setup is shown in Fig. 8.1.

Fig. 8.1

(i) Assume that electrons are released from rest at the cathode. As the cathode is negatively charged and the anode is positively charged, the electrons feel a force towards the anode.

1. Given that the electric field between the cathode and the anode has a magnitude of 6.0 x 106 N C-1, determine electric force acting on an electron released at the cathode.

electric force = ……………………. N

[2]

21


2. Hence, given that the distance between the cathode and the anode is 25 μm, calculate the speed of the electron as it reaches the anode.

speed = ……………………. m s-1

[3]

3. With what fraction of the speed of light does the electron reach the anode?

The electron reaches the anode at ……………………. of the speed of light.

[1]

(ii) In general, electrons may be released from the cathode with a non-zero velocity. The electrons then enter the region between the horseshoe magnet with an unknown velocity v in the horizontal direction. In order to determine this velocity, electric and magnetic fields are both applied, each giving rise to a force on the electrons. These forces are in the vertical direction. Assume both the electric and the magnetic field are uniform in this region.

1. State the electric force FE on an electron, given that E is the magnitude of the electric field between the two horizontal plates at the horseshoe magnet.

magnitude of the electric force is ………………………….

direction of the electric force is ………………………….

[2]

2. State the magnetic force FB on an electron, given that B is the magnitude of the magnetic flux density between the horseshoe magnet.

magnitude of the magnetic force is ………………………….

direction of the magnetic force is ………………………….

[2]

3. Hence, derive an equation of v in terms of E and B for particles that travel through the region between the horseshoe magnet undeflected.

[3]

22


4. The setup of horseshoe magnet and horizontal metal plates is called a velocity selector, as only charged particles with a velocity equal to v will pass through it undeflected. These particles will hit the zinc sulfide coating at location B.

Calculate the speed of an electron that hits the sulfide coating at location B, given that E = 2.0 x 105 N C-1 and B = 25 mT.

speed = ……………………. m s-1

[2]

5. Explain whether an electron with a speed greater than the value calculated in (b)(ii) part 4 may end up at location A or location C on the zinc sulfide coating.

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

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

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

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

[2]

END OF PAPER

Hwa Chong Institution

2018 Suggested Solutions to H1 Paper 2

Question Answer Marks

1(a)(i) Loss in gravitational potential energy

∆GPE = mg∆h = (0.400) (9.81) (0.200) = 0.785 J (3 or 4 s.f.)

A1

(a)(ii) Method 1:

By Newton’s second law, taking upwards as positive,

Fnet = m a

Fspring – W = 0

k x = m g

k = m g = (0.400 x 9.81) / (0.200) = 19.62 N m-1

The elastic potential energy stored

∆EPE = ½ k x2 = ½ (19.62) (0.200)2 = 0.392 J

Method 2:

In equilibrium, the spring force is equal to the weight of the object,

Fspring = W

k x = m g

k = m g = (0.400 x 9.81) / (0.200) = 19.62 N m-1

The elastic potential energy stored

∆EPE = ½ k x2 = ½ (19.62) (0.200)2 = 0.392 J

C1

A1

C1

A1

(b) The difference is work done against the external force needed to support the mass while lowering it gently. This force is the difference between the mass’s weight and the tension in the spring.

OR

A variable external upward force is required when the spring mass is stretched gently downwards towards its equilibrium point. Thus some gravitational potential energy is lost due to the negative work done by the external force, while the remainder is converted to elastic potential energy..

B1 B1

B1 B1

(c)

B1: linear, decreasing graph for GPE

B3

0 0.2

X

Energy

Extension of spring/m

GPE

EPE

B1: parabolic, increasing graph for EPE

B1: EPE starts from zero; the change in EPE is half the change in GPE

Max Marks 10


2(a) The total linear momentum of a system is conserved if no net external force acts on the system.

B1

2(b)(i)

T1 = M1g = (4.0)(9.81) = 39.2 N (2 or 3 s.f.)

T2 = M2g = (8.0)(9.81) = 78.5 N (2 or 3 s.f.)

A1

A1

(b)(ii) Since the disturbance is small and the net force on the system can be taken to be zero, the total momentum remains at zero throughout.

The momentum of M1 is equal in magnitude to the momentum of M2.

Since the mass of M1 is half the mass of M2,

v2 = ½ v1

C1

A1

(b)(iii) The disturbance results in the increase of total kinetic energy of the system.

Energy introduced = ½(4)(v12) + ½(8)(v2

2) = 3v12

M1

A1

(b)(iv) There is no change in kinetic energy when the masses are moving at constant speed.

Loss in gravitational potential energy per unit time of M1 = 4gv1

Gain in gravitational potential energy per unit time of M2 = 8gv2 = 4gv1

Since the kinetic energy does not change and there is no net change in the gravitational potential energy, the total energy of the system remains the same.

M1

M1

Max Marks 9

M2

M1

Pulley B M1g

T2 M2g

T1 T1

T1

T2

Qn 3 Answer Marks

(a) It is the point where the weight appears to act. B1

(b) Taking moments about the elbow joint, o( cos15.0 )(0.040) (18)(0.150) (30)(0.460)F = +

F = 427 N

M1

A1

(c)(i)

Correct direction of R, the length of R is not marked.

A1

(c)(ii) ‘The rightward horizontal component of R is to balance the leftward horizontal component of F.

Taking moments about the point at which the muscle is attached to the bone, the vertical component of R needs to act downwards to provide an anticlockwise moment to counter the clockwise moment provided by the 18 N and 30 N forces.’

OR

‘The rightward horizontal component of R is to balance the leftward horizontal component of F.

The net vertical force due to the weights and tension is upwards, hence the vertical component of R must act downwards so that there is no net vertical force.’

OR

‘A vector diagram is drawn to scale.

Using drawn vector diagram, we can deduce the direction of the force R should point rightward and downward.’

OR

‘The forces of 18 N and 30 N can be combined to form one downwards force which should be drawn between the two mentioned forces.

Since the forces of R, F and combined force due to 18 N and 30 N should intersect at a point, we can deduce the direction of the force R should point leftward and downward.’

B1

B1

B1

B1

B1

B1

B1

B1

bicep muscle

R

(d)(i)

Curve sloping upwards, asymptote at 90o

B1

(d)(ii) Consider moments about the elbow joint.

The perpendicular distance between the elbow and the line of action of F decreases.

OR

The clockwise moment due to the18 N and 30 N forces remains the same; hence F cos θ (or the vertical component of F) remains constant.

Hence, as θ increase (cos θ decreases), F increases.

OR

F is inversely proportional to cos θ.

When θ approaches 90°, F tends to infinity.

A1

A1

Max Marks 9


4(a)(i) Reading off graph, at 22.5 °C, RT = 1600 Ω

A1

(a)(ii) Since p.d. across thermistor, Vthermistor = 2.70 V RT / (RT + R) = 2.70 / 9.00 R = 3730 Ω

M1

A1

(b)(i)

By the potential-divider principle = = 12 1100011000 1000

= 11 V

M1

A1

(b)(ii)1.

= . = 2.7 ×10 A A1

(b)(ii)2. Since 9 V must exist across C and the contact point, by the potential-divider principle, the contact should be 0.75 x the length of CD.

A1

(b)(ii)3. = = 9.00.75 × 1000 M1

0

F

θ

90o

= 0.108 W A1

(b)(iii) Unlike the voltmeter, the lamp does not have a very large resistance. The effective resistance of the lamp and the part of CD it is in parallel with is less than 250 Ω.

Hence, the potential difference across the lamp will not be 3.0 V and may be too low to provide a noticeable brightness.

In fact, taking the operational values, the lamp’s resistance is 3.0/0.20 = 15 Ω when it is operating normally. (It would be even lower if the lamp were cool.)

A lamp of 15 Ω in parallel with a resistance of 250 Ω gives an effective resistance of 14

Ω. The potential difference across the lamp is then only × 12 = 0.22 V, which is

much less than the rating of 3.0 V.

B1

B1

Max Marks 11


5(a)(i) The binding energy of a nucleus is the energy required to separate the nucleus into its constituent neutrons and protons.

B1

5(a)(ii) At A=170, BE/A ~ 8.10 MeV

(Accept all values from 8.00 to 8.25 MeV, accurate to half the smallest square.)

BE of = (8.00 to 8.25) x 170 = 1360 MeV to 1400 MeV (2 or 3 s.f.)

M1

A1

5(a)(iii)

( )( )−

−

× × ×Δ = = =× ×

6 19

22 27 8

1377 10 1 6 101 47 u

1 66 10 3 10

BE .m .

c .

1360 MeV to 1400 MeV 1.46 u to 1.50 u

M1

A1

5(b)(i) Mass defect = (3 x 4.002603 – 12.000000) u

Energy released

( )−

−

= Δ

= × × × ×

= ×

2

227 8

12

0 007809 1 66 10 3 00 10

1 167 10 J (3 or 4 s.f.)

E mc

. . .

.

M1

A1

5(b)(ii)

After this point, after nickel, the BE of the reactants exceeds that of the products so no net energy is released after the reaction.

B1

Max Marks 7

17077Ir

2.B E mc= Δ


6(a) 9.8 m s-2

(Accept all values from 9.78 to 9.82 m s-2, accurate to half the smallest square.)

A1

6(b) When the elevator remains at rest, there should be no acceleration experienced.

However, a value related to acceleration is registered by the accelerometer which is not in accordance to reality. Hence, this recorded value is known as an apparent acceleration.

M1

A1

6(c) 2.4 s

(Accept all values from 2.2 to 2.6 s, accurate to half the smallest square.)

A1

6(d) t = 2.4 s to t = 7.2 s. Hence, 4.8 s.

(Accept 4.4 to 5.2 s.)

A1

6(e) 0.2 to 0.3 m s-2

Estimate the area under the graph from t = 2.4 s to t = 7.2 s.

Propose a rectangle of same area over the same time interval. The height of the rectangle is the average value of the apparent acceleration. Subtract 9.8 to obtain the actual acceleration.

A1

M1

6(f) v = at = (0.25)(4.8) = 1.2 m s-1 [error carried forward from parts (d) and (e)] A1

6(g) Deduct 1 mark for incorrect shape

Deduct 1 mark for each missing/wrong value

A3

6(h) Estimate the area under v-t graph.

Around 10 m

A1

A1

6(i) 3 to 4 storeys

[Error carried forward from part (h), taking 2.5 to 3.5 m per storey.]

A1

6(j) Movement of passengers in the elevator A1

Max Marks 16

v / m s-1

t/s 7.2 2.4 10.4 15.2

0

1.2


7(a) Every particle attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

A2

(b)(i)

( )

2

26

2

280 6.37 10

24 60

2.70 N

c c

c

c

F ma mr

F

F

ω

π

= =

= × × =

C1 A1

(b)(ii)

( )( )( )

11 24

22 6

6 67 10 6 0 10 80789 04 N

6 37 10

. .GMmF .

r .

−× ×= = =

×

2

789.04 2.70

80

9.83 m s

cfreefall

freefall

freefall

F Fa

m

a

a −

−=

−=

=

A small fraction of the gravitational force acted by the earth on any body near the equator is to provide for the centripetal acceleration;

at the pole, the centripetal acceleration is zero and the acceleration of free fall is entirely due to the pull of earth.

B1 A1 A1 A1

(c)(i) 2

2

2 32

Gravitational force provides for centripetal force:

2

4

GMmmr

r T

rT

GM

π

π

=

=

Since π 24

GM is a constant, ∝ .

B1

B1

Comments:

This is a “show” question, so students have to explain what they are doing; an answer that only consists of equations will not warrant any marks.

(c)(ii) ( )( )( )

32 6 3

2

11 24

4 6.37 10 408 10

6.67 10 6.0 10

5542 s

55421.5 hr (shown)

3600

T

T

T

π−

× + ×=

× ×

=

= =

(Note: use of correct radius)

M1

(c)(iii) Kepler’s Law holds for scenarios where the gravitational force provides entirely for the centripetal force. Hence, it holds for the satellite in orbit.

However, for a student on Earth’s surface, there are other forces acting on him (e.g. normal contact force) and the resultant (centripetal) force is not the gravitational force

A1

A1

(c)(iv) Both the astronaut and the space station are undergoing circular motion while in orbit, with the weight providing the centripetal force.

There is no contact force acting on him by his surrounding and hence he perceived himself to be weightless.

The astronaut is not truly weightless, but only apparently weightless.

B1 B1

(d)(i)

Marking points:

Correct shape

Zero at midpoint

Gravitational force is a vector quantity. From P to midpoint of PQ, the net gravitational force acts to the right with decreasing magnitude. At midpoint of PQ, the net gravitational force is zero. From midpoint of PQ to Q, the net gravitational force acts to the left with increasing magnitude.

A1

A1

(d)(ii) Mass accelerates in a straight line towards the midpoint of X and Y.

Mass decelerates from midpoint to Q.

Motion repeats itself, oscillate between P and Q.

B1

B1

B1

Max Marks 20


8(a) The S.I. unit of magnetic flux density is the tesla (T).

The magnetic flux density of a magnetic field is said to be 1 tesla, if the force per unit length per unit current which acts on a straight current-carrying conductor placed perpendicular to the magnetic field is 1 newton per metre per ampere.

A1

A2

(b)(i)1. FE = qE = (1.6 x 10-19) (6.0 x 106)

FE = 9.6 x 10-13 N

M1

A1

(b)(i)2.

The work done on the electron is the electric force times the distance travelled

WD = FE d = (9.6 x 10-13) (25 x 10-6)

The final kinetic energy is (9.6 x 10-13) (25 x 10-6) = 2.4 x 10-17 J

Since KE = ½mv2, the speed at the anode is 7.26 x 106 m s-1

M1

M1

A1

(b)(i)3. The fraction of the speed of light is (7.26 x 106) / (3.00 x 108) = 0.024. A1

(b)(ii)1. The magnitude of the electric force FE = eE

The direction of the electric force is downwards

A1

A1

(b)(ii)2. The magnitude of the magnetic force FB = Bev

The direction of the magnetic force is upwards

A1

A1

(b)(ii)3. By Newton’s second law, taking upwards as positive

Fnet = ma

Bev – eE = 0

Bev = eE =

M1 M1 A1

F

(b)(ii)4. = = 2.0 ×1025 ×10

v = 8.00 x 106 m s-1

M1 A1

(b)(ii)5. If the speed is greater, the magnitude of the magnetic force will be greater.

However, the magnitude of the electric force will be unaffected.

Hence, the upwards force will be larger in magnitude than the downwards force and the electron may end up at location A.

B1

B1

A0

Max Marks 20

physics 8867/01 - the learning space · 2020. 5. 25. · physics paper 1 multiple choice additional...

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