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INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION in preparation for General Certificate of Education Advanced Level Higher 1

CANDIDATE NAME CLASS GROUP:

PHYSICS

Paper 1 Multiple Choice Questions Additional Materials: Multiple Choice Answer Sheet

8867/01

14 September 2018

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 index number on the Answer Sheet in the spaces provided unless this has been done for you. 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 Answer Sheet. Read the instructions on the Answer Sheet very carefully. 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. The use of an approved scientific calculator is expected, where appropriate.

This document consists of 16 printed pages.

Innova Junior College [Turn over

2

© IJC 2018 8867/Prelim/18 [Turn over

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 + ½at2

v2 = u2+ 2as

resistors in series, R = R1 + R2 + …

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

3


1 Which SI unit, expressed in base units, is not correct?

A unit of force, kg m s-2

B unit of momentum, kg m s-1

C unit of pressure, kg m-2 s-2

D unit of work, kg m2 s-2

2 What is the best estimate of the kinetic energy of a family car travelling along the

Pan-Island Expressway (PIE)?

A 4.0 × 103 J B 4.0 × 105 J C 4.0 × 107 J D 4.0 × 109 J 3 In still air, a bird can fly at a speed of 10 m s-1. The wind is blowing from the east at

8.0 m s-1.

In which direction must the bird fly in order to travel to a destination that is due north of the bird’s current location? A 37° east of north

B 37° west of north

C 53° east of north

D 53° west of north

4


4 The current in a block of semiconductor is 30.0 mA when there is a potential difference

(p.d.) of 10.0 V across it. The dimensions of the block and the direction of the current in it are as shown.

The electrical meters used are accurate to ± 0.1 mA and ± 0.1 V. The dimensions of the block are accurate to ± 0.2 mm.

What is the resistivity of the semiconductor?

A 10.0 ± 0.2 Ω m

B 10.0 ± 0.3 Ω m

C 10.0 ± 0.5 Ω m

D 10.0 ± 0.8 Ω m

5 A ball is released from rest above a hard, horizontal surface. The graph shows how

the velocity of the bouncing ball varies with time. At which point on the graph does the ball reach maximum height after the first bounce?

5


6 A boy throws a stone with a horizontal velocity of 10 m s-1 from the top of a building.

The height of the building is 8.0 m. The stone travels along a curved path until it hits the ground, as shown in the diagram.

Neglecting air resistance, how long does it take for the stone to reach the ground? A 0.61 s B 0.80 s C 1.3 s D 1.6 s

7 A firework rocket is fired vertically upwards. The fuel burns and produces a constant

upward force on the rocket. After 5 seconds, there is no fuel left. Air resistance is negligible. What is the acceleration before and after 5 seconds?

before 5 seconds after 5 seconds A constant constant B constant zero C increasing constant D increasing zero

8 Two blocks X and Y are falling through a vacuum in a uniform gravitational field.

Block X has weight 2w. Block Y has weight w. The blocks do not move apart.

Which value best represents the force exerted by block X on block Y? A 0 B w C 1.5w D 2w

6


9 Two equal masses X and Y are moving towards each other on a frictionless air track

as shown. The masses make an elastic collision.

Which row gives possible velocities for the two masses after the collision?

velocity of X velocity of Y

A zero 20 cm s-1 to the right

B 10 cm s-1 to the right 10 cm s-1 to the right

C 20 cm s-1 to the left zero

D 30 cm s-1 to the left 50 cm s-1 to the right

10 The diagrams show two ways of hanging the same picture.

In both cases, a string is attached to the same points on the picture and looped symmetrically over a nail in the wall. The forces shown are those that act on the nail. In diagram 1, the string loop is shorter than in diagram 2. Which information about the magnitude of the forces is correct?

A R1 = R2 T1 = T2

B R1 = R2 T1 > T2

C R1 > R2 T1 < T2

D R1 < R2 T1 = T2

7


11 A cross-shaped structure, freely pivoted at O, has arms of lengths 5.0 m, 4.0 m, 3.0 m

and 2.0 m. It is acted on by forces of 2.0 N, 3.0 N, 4.0 N and an unknown force F. The structure is in rotational equilibrium.

What is the magnitude of force F?

A 0.40 N B 2.0 N C 2.6 N D 4.4 N

8


12 A uniform solid block has weight 500 N, width 0.4 m and height 0.6 m. The block rests

on the edge of a step of depth 0.8 m, as shown.

The block is knocked over the edge of the step and rotates through 90˚ before coming to rest with the 0.6 m edge horizontal. What is the change in gravitational potential energy of the block?

A 300 J B 400 J C 450 J D 550 J 13 An object of weight 15.0 N is pulled along a horizontal surface at a constant velocity

of 2.00 m s-1. The force pulling the object is 12.0 N at 30˚ to the horizontal, as shown.

What is the power used to move the object? A 12.0 W B 20.8 W C 24.0 W D 30.0 W

9


14 A turntable has radius R. It is driven by a rubber drive wheel of radius r in contact with

the inside of the rim of the turntable, as shown in the plan view diagram.

The turntable rotates with angular velocity Ω and the linear speed of a point on its rim is V. The drive wheel rotates with angular velocity ω and the linear speed of a point on its rim is v. Which pair of equations show the relationship between the angular velocities and the linear speeds of the turntable and the wheel?

angular velocities linear speeds A Ω = ω V = v B Ω = ω V = (r/R) v C Ω = (r/R) ω V = v D Ω = (R/r) ω V = (r/R) v

15 Which of the following is true about geostationary orbits around Earth?

A There is more than one possible orbital radius.

B The period of the orbit is independent of the satellite’s mass.

C The satellite experiences no acceleration since it is not moving.

D The satellite moves away from a spot above the Earth and returns to it exactly 24 hours later.

10


16 Two lamps are connected in series to a 250 V power supply.

One lamp is rated 240 V, 60 W and the other is rated 10 V, 2.5 W. Which statement most accurately describes what happens?

A Both lamps light up at less than their normal brightness.

B Both lamps light up normally.

C Only the 60 W lamp lights up.

D The 10 V lamp blows.

17 A simple circuit is formed by connecting a resistor of resistance R between the

terminals of a battery of electromagnetic force (e.m.f.) 9.0 V and constant internal resistance r.

A charge of 6.0 C flows through the resistor in a time of 2.0 minutes causing it to dissipate 48 J of thermal energy.

What is the internal resistance r of the battery?

A 0.17 Ω B 0.33 Ω C 20 Ω D 160 Ω 18 The resistance of a thermistor depends on its temperature, and the resistance of a

light-dependent resistor (LDR) depends on the incident illumination. Under what conditions will the resistance of both a thermistor and an LDR be highest? thermistor LDR

A B C D

highest temperature

highest temperature

lowest temperature

lowest temperature

highest illumination

lowest illumination

highest illumination

lowest illumination

11


19 In the circuit below, P is a potentiometer of total resistance 10 Ω and Q is a fixed

resistor of resistance 10 Ω. The battery has an electromotive force (e.m.f.) of 4.0 V and negligible internal resistance. The voltmeter has a very high resistance.

The slider on the potentiometer is moved from X to Y and a graph of voltmeter reading V is plotted against the slider position. Which graph would be obtained?

12


20 A 2 Ω resistor and a 4 Ω resistor are connected to a cell as shown below.

Which graph shows how the potential V varies with distance between X and Y?

13


21 A long, horizontal, straight wire is placed close and parallel to the plane of a coil as

shown in the diagram.

Both the coil and the wire carry currents I1 and I2 respectively in the directions indicated.

The force at point P on the wire due to these currents is

A in the direction of X.

B in the direction of Y.

C out of the plane of the paper.

D into the plane of the paper. 22 A 20-turns square coil of side 8.0 mm is pivoted at its centre and placed in a magnetic

field of flux density 0.010 T. The two sides of the coil are parallel to the field and two sides of the coil are perpendicular to the field as shown below.

A current of 5.0 mA is passed through the coil.

What is the magnitude of the torque acting on the square coil?

A 1.6 × 10-9 N m

B 3.2 × 10-8 N m

C 6.4 × 10-8 N m

D 3.2 × 10-5 N m

14


23 A small square coil of N turns has sides of length L and is mounted so that it can pivot freely about a horizontal axis PQ, parallel to one pair of sides of the coil, through its centre, as shown below. The coil is situated between the poles of a magnet which produces a uniform magnetic field of flux density B. The coil is maintained in a vertical plane by moving a rider of weight W along a horizontal beam attached to the coil. When a current I flows through the coil, equilibrium is restored by placing the rider a distance x along the beam from the coil. Which of the following gives the correct expression for the distance x?

A 2NLBWI

B NLB

WI

CWNLBI D

WNLB 2I

24 Three particles 1, 2 and 3 travel through a region of space where the magnetic field is

out of the page, as shown in the figure below.

Which statement below about their charges is correct?

A 1 is negative, 2 is neutral, and 3 is positive.

B 1 is neutral, 2 is negative, and 3 is positive.

C 1 is positive, 2 is negative, and 3 is neutral.

D 1 is positive, 2 is neutral, and 3 is negative.

1 2

3

magnetic field is pointing out of page

15


25 The diagram shows three long parallel straight wires P, Q and R normal to the plane of the paper. They are placed at three corners of a square.

All three currents have the same magnitude. Wire R carries current directed out of the plane of the paper while wires P and Q carry currents directed into the plane of the paper.

Which arrow best shows the direction of the resultant force on wire P?

26 In an experiment on α-particle scattering, α-particles are directed normally onto a gold foil and detectors are placed at positions P, Q and R as shown in the diagram below.

gold foil P Q α R What is the distribution of α-particles as recorded at P, Q and R respectively?

P Q R A none none all

B most some none

C most some few

D few some most

27 23892

U decays through a series of transformations to a final stable nuclide.

The particles emitted in the successive transformations are

α → β → β → α → α

Which nuclide is not produced during this series of transformations?

A 22888

Ra B 23090

Th C 23491

Pa D 23492

U

P Q

R

A

B C

D

16


28 The equation

235 1 121 113 192 0 45 47 0U + n Rh + Ag + 2 n→

shows the fission of a uranium−235 nuclide by a slow-moving neutron into a rhodium-121 nuclide, a silver-113 nuclide and two neutrons.

binding energy per nucleon of 23592U = 7.59 MeV

binding energy per nucleon of 12145Rh = 8.26 MeV binding energy per nucleon of 11347 Ag = 8.52 MeV

What is the energy released during this fission process?

A 9.19 MeV B 24.4 MeV C 73.9 MeV D 179 MeV

29 Radioactive atoms of X disintegrate with a half-life of T to give atoms of Y which are

stable. The initial ratio of number of X atoms to number of Y atoms is 1:2.

What is the ratio after an interval of 2T?

A 1:7 B 1:8 C 1:11 D 1:12 30 The rest masses of three particles, tritium H1

3 , proton H11 , and neutron n0

1 , are 3.01605u, 1.00728u and 1.00866u respectively.

A tritium may disintegrate into the respective proton(s) and neutron(s) if it

A captures a photon of energy 6 MeV. B captures a photon of energy 8 MeV. C emits a photon of energy of 6 MeV. D emits a photon of energy of 8 MeV.

End of paper

2018 JC2 Prelims Exam : H1 Physics Paper 1 Solution

2018 H1 Physics Paper 1

Ques Ans Ques Ans Ques Ans 1 2 3 4 5

C B C C C

11 12 13 14 15

B C B C B

21 22 23 24 25

A C D A D

6 7 8 9 10

C C A D B

16 17 18 19 20

B C D A D

26 27 28 29 30

D A D C B

1 Ans: C P = F/A

Units of pressure = kg m s-2/m2 = kg m-1 s-2

2 Ans: B Estimated mass of car = 1200 kg Estimated speed of car along PIE = 90 km h-1 = 25 m s-1 Kinetic energy = ½ mv2 = 3.8 × 105 J 3 Ans: C cosθ = 8/10 θ = 37° Thus bird is flying 53° east of north

4 Ans: C Resistivity ρ = RA/L = (Vxy)/(IL)

ρ = 3 3

3 3

(10.0)(15.0 10 )(30.0 10 )(30.0 10 )(15.0 10 )

− −

− −

× ×× ×

= 10.0 Ω m

0.1 0.2 0.2 0.2 0.1

10.0 15.0 30.0 15.0 30.0ρ

ρΔ = + + + + = 0.047

Δρ = 0.5 Ω m (to 1 s.f.)

ρ = 10.0 ± 0.5 Ω m

5 Ans: C A: the turning point when the ball hit the surface abruptly (instantaneously at rest) B: the point where the ball just leaves the surface and is moving upwards C: the maximum height where its velocity is instantaneously zero (gradually transiting in

direction) D: the point where the ball just hits the surface for the 2nd time 6 Ans: C sy = uyt + ½ ayt2 8.0 = 0 + ½ (9.81)t2


t = 1.3 s 7 Ans: C Fnet = ma T – W = ma (T=rocket thrust , W = weight) The rock thrust is assumed to be constant and so there is a constant upward force due to the propulsion of fuel. However, the mass of the fuel is depleting as it ascends. Therefore, the more probable answer is the net acceleration is increasing. After 5 s, the thrust is zero and the (new) mass remains constant. As the resultant force is due to its weight alone, the acceleration is constant (free fall). 8 Ans: A Since both blocks are free-falling (ie acceleration = 9.81 m s-2), the resultant force acting on each block is their own respective weights. So there could not be any force exerting on block Y by block X (or vice versa). 9 Ans: D collision: m1u1 + m2u2 = m1v1 + m2v2 where m1 = m2 u1 + u2 = v1 + v2 (+ 50) + (- 30) = v1 + v2 v1 + v2 = 20 elastic u1 – u2 = v2 – v1 (+ 50) – (- 30) = v2 – v1 v2 – v1 = 80 Solving the above equations, v1 = -30 m s-1 & v2 = +50 m s-1 Or the relative speed of approach = 50-(-30) = 80 the relative speed of separation must be 80 (Option D) Or by checking the options

v1 + v2 v2 – v1 A 0 + 20 = 20 20 – 0 = 20 B 10 + 10 = 20 10 – 10 = 0 C (-20) + 0 = - 20 0 – (-20) = 20 D (-30) + (50) = 20 50 – (-30) = 80


10 Ans: B Since the tension R1 and R2 in the vertical strings are used to support the same weight of the picture, R1 = R2.

Since the string loop in 1 is shorter than that in 2, the angle θ1 of the tension T1 to the vertical is larger than angle θ2 of T2. For vertical equilibrium, ΣFy = 0 For picture1: R1 = T1 cos θ1 + T1 cos θ1 For picture2: R2 = T2 cos θ2 + T2 cos θ2 Thus, 2 T1 cos θ1 = 2 T2 cos θ2

T1T2

= cosθ2cosθ1

Since θ1 > θ2, cos θ1 < cos θ2 T1 > T2 11 Ans: B Rotational equilibrium: ΣM = 0 Since 4.0 N force passes through the pivot, it does not create any rotational moment. Clockwise moment about O = Anticlockwise moment about O F × 5.0 = 2.0 sin 30˚ × 4.0 + 3.0 × 2.0 F = 2.0 N 12 Ans: C ∆GPE = mg∆h where ∆h = difference in height of CG = 500 × (0.2 – 1.1) = -450 J The GPE drops by 450 J. 13 Ans: B P = Fv = 12.0 cos 30˚ × 2.00 = 20.8 W 14 Ans: C Since the drive wheel and the turntable are in contact, they should be moving with the same linear speed, i.e. V = v Therefore, RΩ = rω Ω = (r/R)ω 15 Ans: B

θ1 θ2


A: Kepler’s 3rd law specifies T2∝R3 for circular orbits. As the period is fixed, the orbital radius is also fixed.

B: Period of orbit is independent of mass of satellite. = C: Geostationary does not mean that the satellite is literally stationary. It only appears to be ‘stationary’

to the observer as it is positioned directly above the observer on the equator such that it moves with the Earth in the same direction and same angular speed. Because it is undergoing circular motion, it must have centripetal acceleration.

D: If the satellite moves away from the spot on Earth at any time, it is no longer geostationary.

16 Ans: B Since they are connected in series to a 250 V source, based on the principle of potential divider, the p.d. across the 60 W lamp is 240 V and the p.d. across the 2.5 W lamp is 10 V. Thus, both lamps light up normally. 17 Ans: C Using E = V + Ir

Finding current = = 6.02 × 60 = 0.050 Finding V = 482 × 60 = (0.050) = 8.0 Thus , 9.0 = 8.0 + (0.050)r

r = 20 Ω 18 Ans: D Thermistor and LDR are semiconductors. More charge carriers are liberated for current conduction when energy (thermal and light respectively) is absorbed. Thus, resistance is highest at lowest temperature and lowest illumination. 19 Ans: A The voltmeter is connected parallel to the combined resistors (P and Q). The voltmeter can be considered to be connected directly to the battery as its resistance is very large compared to P and Q (10Ω each). Hence, the p.d. as measured by the voltmeter will always register the value of 4 V irrespective of the position of the slider. 20 Ans: D Because X is attached to the positive terminal it has a higher electric potential than Y. This rules out Option A and B. Because V = IR (current is constant as X and Y are in series). Thus, the potential gradient is steeper across the 4Ω. The potential drop across the 4Ω is twice as much as that for 2Ω.


21 Ans: A Magnetic field set up by flat coil points into the plane of the coil. Using FLHR, force acts upwards, in the direction of X. 22 Ans: C

B

-3 -3 2

-8

=== ×= 20×0.01×5×10 ×(8×10 )= 6.4×10 N m

FdF LNBIL L

τ

23 Ans: D Clockwise moment abt pivot = Anticlockwise moment abt pivot Wx = BILN(L) x = BINL2/W 24 Ans: A Particle 2 is neutral since its path is straight and not affected by the magnetic field. Using Fleming’s Left Hand Rule, the curved paths of the particles 1 and 3 indicates that particle 1 is negatively charged whereas particle 3 is positively charged. 25 Ans: D P is repelled by R (due to anti-parallel current) and so has a force acting on P vertically upwards. At the same time, P is attracted to Q (current in same direction) and has a force acting on P to the right.So the net force on P is in the direction D. 26 Ans: D Majority of the α -particles passed through the gold foil without being deflected since the atom consists of mostly empty space. 27 Ans: A

Pb22688αT23090

αS23492βR23491

βQ23490αU23892 ⎯→⎯⎯→⎯⎯→⎯⎯→⎯⎯→⎯

aR22888

is not produced during this series of transformations.

28 Ans: D 235 1 121 113 1

92 0 45 47 0U + n Rh + Ag + 2 n→

Energy eqv inherent in parent nuclide 23592U

= (7.59)(235) = 1783.65 MeV

Energy eqv inherent in daughter nuclides 12145Rh and 113

47 Ag

= (8.26)(121) + (8.52)(113) = 1962.22 MeV Energy released = 1962.22-1783.65 = 178.57 = 179 MeV (3sf) 29 Ans: C


Initial ratio X:Y = 1:2 = 13 : 23

t=0 t=T t=2T

= 13 = 13 × 12 = 16 = 16 × 12 = 112 = 23 = 23 + 16 = 56 = 56 + 112 = 1112

1:2 1:5 1:11

Alternatively, just focus on the decay of X. After 2 half-lives,

Final fraction of X = (Initial fraction of X) = × = Final fraction of Y = 1 − = Therefore, X:Y = 1:11 30 Ans: B

H13 → H1

1 + 2 n01 Rest mass of reactant = 3.01605u Rest mass of product = 1.00728u + 2×1.00866u = 3.02460u Since combined rest mass of product is greater, photon must be captured to maintain the mass-energy equivalence for the fission process. ∆ = ∆ ∆ = (0.00855 × 1.66 × 10 )(3.0 × 10 ) = 1.28×10-12 J = 7.98 MeV (A photon of 8 MeV is more than the min. energy (7.98 MeV) to cause fission. 6 MeV is insufficient.)

This document consists of 24 printed pages. Innova Junior College [Turn over

INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION in preparation for General Certificate of Education Advanced Level Higher 1

CANDIDATE NAME CLASS INDEX NUMBER PHYSICS Paper 2 Structured Questions Candidates answer on the Question Paper

No Additional Materials are required

8867/0229 August 2018

2 hours

READ THESE INSTRUCTIONS FIRST Write your name, class and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Section A Answer all questions. Section B Answer one question in this section. At the end of the examination, fasten all your work securely together. The number of marks is given in the brackets [ ] at the end of each question or part question.

For Examiner’s Use

Section A

1 10

2 10

3 12

4 10

5 18

Section B

6 20

7 20

Penalty

P1 Total 30

P2 Total 80Total

Percentage

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 constall 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 + ½at2

v2 = u2+ 2as

resistors in series, R = R1 + R2 + … resistors in parallel, 1 / R = 1 / R1 + 1 / R2 + …

3


For Examiner’s

Use

Section A

Answer all questions in this section.

1 A paraglider with his parachute P of mass 95 kg is pulled by a wire attached to a boat as shown in Fig 1.1. The wire makes an angle of 25° with the horizontal water surface. P moves in a straight line parallel to the surface of the water.

Fig. 1.1

(a) State Newton’s second law of motion. …………………………………………….…………………………………………............. …………………………………………………………………………….………………….. .…………………………………………………………………………………………… [2]

(b) While P is moving at a speed of 4.6 m s-1, P starts to accelerate at a constant rate for a time of 3.0 s. P then moves at a constant speed of 8.8 m s-1. Calculate the acceleration of P.

acceleration = ………………………. m s-2 [2]

4


For Examiner’s

Use

(c) The tension in the wire during the acceleration is 280 N. Calculate, for the horizontal motion,

(i) the vertical lift force F supporting P,

F = ……………………. N [2] (ii) the force R due to air resistance acting on P in the horizontal direction.

R = ……………………. N [2] (d) A second paraglider of smaller mass moves with the same motion. Explain why the

angle of the wire to the horizontal is larger. …………………………………………….…………………………………………............. …………………………………………………………………………….………………….. .…………………………………………………………………………………………… [2]

5


For Examiner’s

Use

2 (a) State the two conditions for an object to be in equilibrium. 1. …………………………………………………………………………………………….. ……………………………………………………………………………………………….. 2. …………………………………………………………………………………………….. …………………………………………………………………………………………… [2]

(b) A uniform beam AC is attached to a vertical wall at end A. The beam is held horizontal

by a rigid bar BD, as shown in Fig. 2.1.

Fig. 2.1 (not to scale)

The beam is of length 0.40 m and weight W. An empty bucket of weight 12 N is

suspended by a light metal wire from end C. The bar exerts a force on the beam of 33 N at 52˚ to the horizontal. The beam is in equilibrium.

(i) Calculate the weight W of the beam.

W = ……………………. N [2]

6


For Examiner’s

Use

(ii) Hence or otherwise, calculate the magnitude and direction of the force exerted on the beam by the wall.

force = ……..…………………. N

direction = ………………………. [4]

(c) The empty bucket is then filled with paint. State and explain if there is a change to the force exerted on the beam by the bar BD. …………………………………………….…………………………………………............. …………………………………………………………………………….………………….. .…………………………………………………………………………………………… [2]

7


For Examiner’s

Use

3 A car of mass 800 kg starts from 7.0 m s-1 and travels upwards along a straight road of distance l inclined at 5.0° to the horizontal as shown in Fig. 3.1. The speed of the car increases at a constant rate to reach 28 m s-1 at the top of the slope and it takes 32.1 s. Resistive force is not negligible.

Fig. 3.1 (not to scale)

(a) (i) Determine l.

l = ……………………. m [2] (ii) The actual uncertainty in each measurement of the speed of the mass is

± 0.1 m s-1. The percentage uncertainty in the time for the car to travel up the slope is ± 0.6%. Calculate the percentage uncertainty of l.

percentage uncertainty = ……………………. % [2] (iii) Use your answers in (a)(i) and (a)(ii) to determine the value of l, with its actual

uncertainty, to an appropriate number of significant figures.

l = ……………. ± ……………. m [1]

8


For Examiner’s

Use

(b) (i) Explain what is meant by work done. …………………………………………….…………………………………………... …………………………………………………………………………….…………… .…………………………………………………………………………………….. [1]

(ii) For the car moving up the slope, 1. calculate the change in kinetic energy.

change in kinetic energy = ……………………. J [2] 2. calculate the change in potential energy.

change in potential energy = ……………………. J [2] 3. hence determine the average resistive force that acts on the car when it

travels up the slope of distance l, given that the power of the car’s engine is 320 kW.

resistive force = ……………………. N [2]

9


For Examiner’s

Use

4 Some of the planets in the Solar System have several moons (satellites) that have circular orbits about the planet. The planet and each of its moons may be considered to be point masses.

(a) Considering the force acting on each moon by the planet, show that the radius x of the moon’s orbit is related to the period T of the orbit by the expression

π=2 3

2

4 xGMT

where G is the gravitational constant and M is the mass of the planet. Explain your working.

[3]

(b) In July 2015, the New Horizons space probe made its closest approach of Pluto. Charon is one of the moons of Pluto. Data, including the radius r and the period of rotation about the axis T, are given for Pluto and Charon in Table 4.1.

Table 4.1

r / km mass / kg T / days Pluto 1.20 x 103 1.31 x 1022 6.36

Charon 0.600 x 103 1.52 x 1021 6.36 When viewed from above, Pluto and Charon rotate in the same direction about their axes.

(i) The orbital speed of Charon is 0.200 km s-1 and its orbital radius is 1.75 x 104 km. Show that the period of the orbit of Charon around Pluto is 6.36 days.

[2]

10


For Examiner’s

Use

(ii) Hence, determine the mass of Pluto.

mass = …………..…….. kg [1]

(iii) A space probe on the surface of Pluto is able to observe Charon over a time of several days. Suggest what the space probe observes as a result of

1. the period of rotation of Pluto about its axis equalling the orbital period of Charon, ………………………………………………………………………………….. …………………………………………………………………………………..

2. equal periods of rotation about their axes for both Pluto and Charon. ………………………………………………………………………………….. ……………………………………………………………………….……….…

[2]

(c) Another moon of Pluto, Keberos was discovered in June 2011. It has a mass of 1.65 x 1016 kg and was found to be orbiting at 5.8 x 104 km about the center of Pluto. State and explain if it could be a geostationary moon orbiting around Pluto. ………………………………………………………………………………….…………….. ……………………………………………………………………………………..……… [2]

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5 The large amount of energy released in a nuclear fission reaction, together with the emission of more than one neutron, has made it possible for neutron-induced fission to be used as a source of useful energy. When a neutron is captured by a Uranium-235 nucleus, it causes the nucleus to fission. On average, 2.5 neutrons are emitted in these fission reactions. This is illustrated in Fig. 5.1.

Fig. 5.1

When the conditions are suitable, a chain reaction can occur. If this chain reaction is not controlled, an explosion is likely to occur. However, if the chain reaction is controlled, as in a nuclear reactor, a source of continuous power may be created.

(a) Suggest why, in an uncontrolled chain reaction where all neutrons are captured by

Uranium-235 nuclei, the majority of the energy is released during the final stages of the fission of a sample of the uranium. ………………………………………………………………………………………………... ………………………………………………………………………………………………... …..………………………………………………………………………………………… [2]

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(b) The graph in Fig. 5.2 shows the variation of the energy release E during each stage of the chain reaction with its stage number n. (Stage n = 1 represents the stage where the energy is released when the first neutron interacts with first Uranium nucleus and releases 2 or 3 neutrons. Stage n = 2,3… represents the stage where the energy is released when the neutrons from previous stage interacts with other Uranium nuclei and releases 2 or 3 more neutrons for each reaction)

2

2.5

3

3.5

4

4.5

5

5.5

0 2 4 6 8 10

lg(E/MeV)

Fig. 5.2 n

5.5

2.0

3.0

2.5

3.5

4.0

5.0

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(i) The total energy released per chain E would thus increase with the number n of stage that the fission has proceed till and is given by

E 1na b−=

where a is the average number of neutrons emitted per fission reaction and b is a constant. Use Fig. 5.2 to determine the value of b. Show your working clearly.

b = ……………………… MeV [3]

(ii) Suggest what is represented by the value of b. …………………………………………………………………………………………. ……………………………………………………………………………………… [1]

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(c) The induced fission reaction of Uranium-235 may be represented by a nuclear equation of the form

+ → + +235 1 a c92 0 b dU n P Q (2 or 3) +10n 198MeV

Usually when any two nuclei are fissioned, the fission products may not be the same. If a large sample of Uranium-235 is fissioned, many different fission products will be produced. The percentage amount of each fission product in the fissioned material is referred to as percentage yield. The variation with nucleon number of the percentage yield of different products is illustrated in Fig. 5.3.

Fig 5.3

(i) State the proton number of the other fission product for the fission of one uranium if one of the products is 8936Kr .

proton number = ………………………… [1]

(ii) Suggest why the percentage yield is shown on a logarithmic scale. …………………………………………………………………………………………. ……………………………………………………………………………………… [1]

Mo-99

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(iii) Show that the percentage yield Mo-99 are about 400 times more than those having masses equal to each other.

[2]

(d) (i) The energy released during one fission reaction of a uranium nuclei occurs partly as kinetic energy of the fission products (167 MeV) and of the neutrons. Suggest one other mechanism by which energy is released in the fission reaction. ……………………………………………...……………………………..……….. [1]

(ii) In 2016, nuclear plants have managed to supply 2477 TWh of electricity

worldwide. Nuclear energy generated via nuclear reactions generate heat to produce steam, which is used to generate electricity. Assuming that a typical power plants can achieve an efficiency of 35%, determine the amount of heat required to generate the electrical energy worldwide in 2016.

amount of heat = …………………….. J [2]

(iii) Determine the average rate at which uranium is consumed to generate the amount of heat required in 2016.

amount required = ……………………….. kg s-1 [3]

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(e) The fission products are usually radioactive and give rise to a series of radioactive decay products. Each decay product has its own half-life. Two such fission products with their decay products and half-lives are shown below.

9042Mo →

9943Tc →

9044Ru (stable solid)

14054 Xe →

14055Cs →

14056Ba →

14057La →

14058Ce (stable solid)

Consider equal amounts of these two products. Suggest why there are very different problems for the storage of this nuclear waste. ………………………………………………………………………………………………... ………………………………………………………………………………………………... ……………………………………………………………………………………………….. …………………………………………………………………………………………….. [2]

67 hours 2 x 105 years

16 seconds 1.1 minutes 13 days 40 hours

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Section B

Answer one question in this section.

6 Fig. 6.1 shows a mass spectrometer used for measuring the masses of isotopes. It consists of an ion generator and accelerator, a velocity selector and an ion separator, all in a vacuum.

Fig. 6.1

In one experiment, tin ions, each of which carries a charge +1.6 × 10-19 C, are produced in the ion generator and are then accelerated by a p.d. of 20 000 V. Tin has a number of isotopes, two of which are tin-118 (118Sn) and tin-120 (120Sn).

(a) (i) State one similarity and one difference between the isotopes of tin.

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

(ii) Assuming that an ion of tin-120 is at rest before being accelerated, show that the final speed after acceleration is 179 km s-1.

[2]

Magnetic field into paper

Ion separator

Velocity separator

Ion accelerator

Photographic plate P

+200 V 200 V

Ions

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(iii) Determine the final speed of an ion of tin-118.

speed = ………………………… m s-1 [1]

(b) In practice, all ions produced by the ion generator have a range of speeds. A velocity selector is used to isolate the ions with a single speed. In the velocity selector the force produced by the electric field is balanced by that due to the magnetic field which is perpendicular to the plane of the paper.

(i) The plates producing the electric field have a separation of 2.0 cm. The potentials of the plates are marked in Fig. 6.1. Determine the magnitude of the force on an ion due to this electric field in the velocity selector.

force = ………………………… N [2]

(ii) Write down the equation which must be satisfied if the ions are to emerge from the exit hole of the velocity selector. Define the terms in the equation. ………………………………………………………………………………………… ……………………………………………………………………………………… [2]

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(iii) Calculate the magnetic flux density required if ions travelling with a speed of 179 km s-1 are to be selected.

flux density = ………………………… T [1]

(c) After selection, the ions are separated using a magnetic field on its own, as shown

in Fig. 6.1.

(i) Explain why the ions move in circular paths in this region. ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ……………………………………………………………………………………… [2]

(ii) Show that the radius of the path is directly proportional to the mass of the ion.

[3] (iii) The ions are detected using the photographic plate P. Determine the distance

between the points of impact on the photographic plate of the two isotopes of tin when a magnetic flux density of 0.75 T is used in the ion separator.

distance = ………………………… m [3]

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(d) Explain whether the distance between the points of impact of the ions would be the same, greater or smaller for two isotopes of uranium, one with a nucleon number of 236 and the other 238, assuming that they have the same velocity and same charge as the tin ions. ……………………………………………………………………...………………………… ……………………………………………………………………..………………………… ……………………………………………………………..………………………………… ……………………………………………………………….…………………………… [2]

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7 The unlabelled lines on the graph in Fig. 7.1 show how the current through an ohmic resistor, a semiconductor and a filament lamp vary with the applied potential difference.

Fig. 7.1

(a) Use the line for ohmic resistor to find, when the applied potential difference is 0.5 V,

(i) its resistance,

resistance = ……………………………. Ω [2]

(ii) the power supplied to it.

power = ……………………………….. W [2]

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(b) The filament lamp has a filament wire made from tungsten.

(i) Use the line for the filament lamp on the graph in Fig. 7.1 to determine its resistance when the applied voltage is 0.50 V.

resistance = ……………………………. Ω [2]

(ii) Explain in terms of the movement of the charged particles, the shape of the graph for the filament lamp. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. ……………………………………………………………………………………… [3]

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(c) If a potential difference of 12 V is applied across the semiconductor diode, it will be destroyed. A protective resistor is placed in series with it to limit the current to 18 mA. This is shown in Fig. 7.2.

Fig. 7.2

Calculate the value of the required series resistor.

resistance = ……………………………. Ω [2]

(d) A negative temperature coefficient (NTC) thermistor circuit is used to measure room temperature. A needle pointing voltmeter is to be calibrated to function as a ‘thermometer’ such that as the temperature rises, the voltmeter deflection increases. Draw a suitable labelled circuit diagram for such a system.

[2]

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(e) The electrical circuit shown in Fig. 7.3 is unusual because the presence of resistor R makes it impossible to apply the usual rules for either series or parallel connections.

Fig. 7.3

The battery has no internal resistance. Complete the following table for each of the components shown in the Fig. 7.3. Show your working clearly in the spaces provided.

component current / A potential difference / V resistance / Ω

battery

from battery = 3.0 emf = 12 zero

resistor P

1.2 6.0

resistor Q

resistor R

6.0

resistor S

3.0

resistor T

[7]

END OF PAPER

2018 H1 Physics Prelims Solution (8867 Paper 2)

1 (a)(i)Newton’s second law of motion state that the rate of change of momentum of an object is directly proportional to the resultant force acting on it [B1] and the change in momentum occurs in the direction of the resultant force [B1]. 1 (b)v = u +at 8.8 = 4.6 + (a × 3) [C1] a = 1.4 m s-2 [A1] 1 (c)(i)Since the mass is moving horizontally, there is no net vertical force acting on it. L = W + Ty [C1] = (95 × 9.81) + (280 sin 25˚) = 1050 N [A1] (No mark if L = W + T) 1 (c)(ii) Fn = ma Tx – R = ma 280 cos 25˚ - R = 95 × 1.4 [C1] R = 121 N [A1] (No mark if T – R – ma) 1 (d)To maintain the same motion, the angle made by the wire with the horizontal water surface is larger to create a larger downward vertical tension to compensate a smaller downwards weight [B1]. At the same time, the smaller horizontal tension is required to accelerate a smaller mass [B1]. 2(a) The resultant force in any direction is zero. [B1] The resultant moment/torque about any point is zero. [B1] 2(b)(i)Taking point A as the pivot, Clockwise moment abt A = Anticlockwise moment abt A W × 0.20 + 12 × 0.40 = 33 sin 52˚ × 0.30 [C1] W = 15 N [A1] 2(b)(ii)ΣFx = 0 F cos θ = DB cos 52˚ F cos θ = 33 cos 52˚ [C1] F cos θ = 20.32 --- (1) ΣFy = 0 F sin θ + DB sin 52˚ = W + T F sin θ + 33 sin 52˚ = 15 + 12 [C1] F sin θ = 1.0 --- (2) (2)/(1): tan θ = 1.0 /20.32 θ = 2.82˚ clockwise from negative x-axis


Mark can be awarded as long as the direction is correctly phrased.[A1] F cos 2.82˚ = 20.32 F = 20.3 N [A1] 2 (c) Filling paint in the empty bucket increases the tension in the wire, increasing the clockwise momentum about point A [B1]. To maintain equilibrium, the anticlockwise moment about point A has to increase as well. Hence the force exerted by the bar BD increases [B1]. 3(a)(i)Using s = 1/2(u+v)t, l =1/2(7.0+28)32.1 = 562 m 3 (a)(ii) Maximum value of l = 1/2(7.1+28.1)32.3 = 568 Minimum value of l = ½(6.9+27.9)31.9 = 555 Uncertainty of l = 7 Percentage uncertainty = 7/562 = 1.2% 3 (a)(iii) 562±7 3 (b)(i)Work done is the product of force and distance moved/displacement in the direction of the force. [B1] 3 (b)(ii)1. Change in KE = ½(800)(282 – 72) = 294000 J = 2.94 x 105 3 (b)(ii)2. Change in GPE = (800)(9.81)(562sin(5.0°)) = 384000 J = 3.84 x 105 3 (b)(ii)3. Energy supplied by car engine = 320000 x 32.1 =1.03 x 107 [M1] Energy lost due to resistive force = (1.03 x 107) – (2.94 x 105 + 3.84 x 105) = 9.59 x 106 J Average resistive force = 9.59 x 106 / 562 = 1.71 x 104 N [A1] 4(a) The gravitational force acting on the moon provides the centripetal force. [B1] Fc = mac

ω=2

2GMm mx

x [M1]

Since πω = 2

T

π =

2

22GMm mx

x T [M1]


π=

2 3

24 xGM

T [A0]

4 (b)(i)T = 2πr/w = 2π(17500)/0.2 [C1] = 5.498 x 105 s [M1] = 6.36 days 4 (b)(ii) GM = 4π2x3/T2 (6.67 x 10-11)M = 4π2(1.75 x 107)3 /(5.498 x105)2 M = 1.05 x 1022 kg 4 (b)(iii)1. Charon will be above the same plan on the surface of Pluto at all times/ Charon will appear to be geostationary 4 (b)(iii)2. The same face of Charon will be seen from Pluto all the time 4 (b)When device is operating in normal condition, Resistance of device = 0.53Ω [B1] Effective resistance of circuit needed = 4.4/1.5 = 2.93 [B1] Additional resistance required = 2.93 – 0.53 – 1.1 = 1.3 Ω [Draw an additional 1.3 Ω of resistor in series with the emf and device] [A1] 4(c) No. Base on GM = 4π2x3/T2, given Keberos has a difference orbitting radius compared to Charon, it would have an different orbitting period compared to Charon. Hence its orbitting period will not be the same as Pluto’s own period of rotation. 5 (a)The number of neutrons emitted will be very large as the stages proceed on [1] and hence leading to a very large amount of energy [1] 5 (b)(i)E 1na b−= -> lgE=(n-1)lg(a) + lg(b) Thus the intercept of the graph is lg(b) – lg(a) [M1] Gradient of graph , lg (a)= (6.6-3.1)/(10.8-2.0) = 0.398 6.6 = (0.398)(10.8)+c c = 2.30 [M1] therefore lg(b) = 2.30 b=202MeV [A1] 5 (b)(ii) Energy release due to one nuclear fission reaction of uranium 5 (c)(i)92 – 36 = 56


5 (c)(ii) The range of percentage yield is very large and hence in order to show the variation of the values in a single graph, a logarithmic scale is used. 5 (c)(iii) Percentage yield of having 139 and 99 = 6 [M1] Percentage yield of masses equal = 1.5 x 10-2 [M1] Both values to get 1 mark each Therefore, the ratio = 6/(1.5 x 10-2) = 400 5 (d)(i)Gamma photon. 5 (d)(ii) 2477 TWh = 8.917 x 1018 J [C1] Since the efficiency is 35%, The amount of heat required = 100/35 x 8.917 x 1018 = 2.55 x 1019 J [A1] 5 (d)(iii) Power required = (2.55 x 1019)/(336x24x60x60) = 8.78 x 1011 [B1] Number of reactions needed in one second = (8.78 x 1011)/(167 x1.6 x 10-13) = 3.28 x 1022 [B1] Amount required = 3.28 x 1022 x 235u = 5.45 x 10-5 kg [A1] 5 (e) For the decay chain of Mo, it has a high half life and hence it would be required to be stored for a long period of time For the decay chain of Xe, as its half life is short, a large amount of radioactive particle will be released in a short period of time, hence it needs to be stored very quickly 6 (a)(i)Tin has the same number of protons [B1] but different number of neutrons. [B1] 6 (a)(ii) Mass of tin-120 = 120u = 1.992× 10-25 kg [C1] By conservation of energy, Kinetic energy gained = electric potential energy lost ½ mv2 =q(20000) v = = 179 km s-1 [M1] 6 (a)(iii) Mass of tin-118 = 118u = 1.9588 × 10-25 kg [C1] By conservation of energy, Kinetic energy gained = electric potential energy lost

½ mv2 = q(20000)


v = −

−

××

19

27

2(1.6 10 )(20000)118(1.66 10 )

= 1.81 × 105 m s-1 [A1]

6 (b)(i)Electric field strength = ΔV/d = (200 – (-200)/0.020 = 20000 V [C1] Electric force = qE = (1.6 × 10-19)(20000) = 3.2 × 10-15 N [A1] 6 (b)(ii) For the ions to emerge from the exit, net force on ions must be zero.

Thus electric force is balanced by the magnetic force, qE = qvB v = E/B where B is the magnetic flux density, E is the electric field strength and v is the speed of the ions. 6 (b)(iii) B = E/v = 20000/ 179 × 103 = 0.112T [A1] 6 (c)(i)The ions enter the magnetic field with a velocity at right angle to the magnetic field. Thus, the ions experience a magnetic force that is always perpendicular to the velocity. [B1] The magnetic force acting on the ion changes the direction of the velocity but not its magnitude. [B1] Thus the ions move in circular paths in this region. 6 (c)(ii) Magnetic force provides the centripetal force. [B1] Bqv = mv2/r [M1] r = mv/Bq Since m, v and B are constant values, therefore r is directly propotional to the mass of ion. [A1] 6 (c)(iii)

For tin-120, r1 = −

−

× ××

27 5

19

120(1.66 10 )(1.79 10 )(1.6 10 )(0.75)

= 0.29714 m [C1]

For tin-120, r2 = −

−

× ××

27 5

19

118(1.66 10 )(1.79 10 )(1.6 10 )(0.75)

=0.29219 m [C1]

Distance = 2r2 – 2r1 = 9.90 × 10-3 m 6 (d)


The radius is directly proportional to the mass. [B1] Mass of uranium is larger than the mass of tin, thus the radius will be larger. [B1] Hence, the difference in the distance will also be larger. [A1] 7 (a)(i)R = 0.50 / 0.0060 [C1] = 83 Ω [A1] 7 (a)(ii) power = VI = 0.50×0.0060 [C1] power = 3.0 × 10-3 W [A1] 7 (b)(i)When V = 0.50 V, current = 13.5 mA [C1]

R = 0.50 / 13.5×10-3 [A1] 7 (b)(ii) When the applied voltage is increased, the conduction electrons acquires more kinetic energy and upon colliding with the metal lattice ions impart more energy to them causing them to vibrate more. As a result, the temperature of the metal rises with increasing voltage. [B1] As the rate of collisions between the electrons and lattice ions increases, the electrons are scattered more which impede the progression (drift velocity) of the electrons. [B1] This makes it more difficult for the current to flow and so the V/I ratio increases (resistance increases) as shown in the graph. [A1] 7 (c) Then graph shows that when current in diode is 18 mA, its p.d.=0.60 V Thus, the p.d. across resistor is 12.0-0.60 = 11.4 V [C1] resistance of resistor = 11.4 / 0.018 = 633 Ω [A1] 7 (d)

• The circuit has a thermistor connected in series to a fixed resistor. Because the resistance of the NTC thermistor decreases with temperature, the p.d. across it will fall.

• In order for the voltmeter to have a reading which increases with temperature, it must be connected parallel to the fixed resistor.


7 (e)

component current / A potential difference / V

resistance / Ω

battery

from battery = 3.0

emf = 12 zero

resistor P

1.2 1.2×6.0 = 7.2 (step#1)

6.0

resistor Q

1.2+0.3 = 1.5 (step#6)

12-7.2 = 4.8 (step#7)

4.8/1.5= 3.2 (step#8)

resistor R

1.8/6.0 = 0.3 (upwards) (step#5)

7.2-5.4 = 1.8 ] (step#4)

6.0

resistor S

3.0-1.2 = 1.8 (step#2)

1.8×3.0 = 5.4 (same mtd as step#1) (step#3)

3.0

resistor T

1.8-0.3=1.5 (step#9)

12-5.4 = 6.6 (step#10)

6.6/1.5 = 4.4 (step#11)

The marking scheme for the table is as follows: Correct values in boxes Scoring0 0 1 1 2 or 3 2 4 or 5 3 6 or 7 4 8 or 9 5 10 6 11 7

If the working is not shown, the total scores is divided by 2. Example: If 7 boxes has correct numerical answers but no working is shown. Then, effective values = 7/2 = 3.5 which scores 2 marks. If working were shown, then the score is 4 marks.

voltmeter

thermistor

Fixed resistor V

SA2 Innova Junior College

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