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PHYSICS 8867/01

Paper 1 Multiple Choice

1 hour

Additional Materials: Multiple Choice Answer Sheet

READ THESE INSTRUCTIONS FIRST Write in soft pencil. Write your name and tutorial group on this cover page. Write and/or shade your name, NRIC / FIN number and HT group on the Answer Sheet (OMR sheet), unless this has been done for you. Do not use staples, paper clips, highlighters, glue or correction fluid. There are 30 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 (OMR 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. Calculators may be used.

This document consists of 12 printed pages and 0 blank pages.

[Turn over

NAME CLASS 2T

Catholic Junior College JC2 Preliminary Examinations Higher 1

2 DATA: speed of light in free space c = 3.00 x 108 m s-1

elementary charge e = 1.60 x 10-19 C

unified atomic mass constant u = 1.66 x 10-27 kg

The Avogadro constant NA = 6.02 x 1023 mol-1

rest mass of electron me = 9.11 x 10-31 kg

rest mass of proton mp = 1.67 x 10-27 kg

gravitational constant G = 6.67 x 10-11 N m2 kg-2

acceleration of free fall g = 9.81 m s-2

FORMULAE:

uniformly accelerated motion, s = u t + 2

1a t2

v2 = u2 + 2 a s

resistors in series, R = R1 + R2 + ...

resistors in parallel, R

1 = ...++

21

11

RR

3

1 An ammeter connected in series in a circuit gives readings that have low precision but high accuracy. Which row best describes the likely error in readings taken with this ammeter?

random error systematic error

A large large

B large small

C small large

D small small

2 In 1945, an atomic bomb was denoted over the Japanese city of Nagasaki. It was estimated

to have generated an energy of 80 TJ in a time duration of 1.0 s. What was the estimated power produced by the detonation?

A 8 x 104 GW B 8 x 107 GW C 8 x 1010 GW D 8 x 1011 GW

3 A body is moving along a straight line OA. The variation of the displacement of the object,

s, measured from the reference point O, with time t is shown below.

Which of the following statements describes most accurately the motion of the object?

A The displacement decreases at a decreasing rate.

B The displacement decreases at a constant rate.

C The velocity increases at an increasing rate.

D The velocity decreases at an increasing rate.

4 An astronaut is working on the surface of the Moon where the acceleration due to gravity is

1.6 m s-2. He drops a wrench from rest and catches it when it has a speed of 1.7 m s-1 and is 0.60 m above the ground. From which height above the ground did he drop the wrench?

A 0.75 m B 0.90 m C 1.13 m D 1.50 m

[Turn over

t / s 0

s /m

4

5 A stone is thrown with an initial velocity v at an angle to the horizontal and the trajectory of the stone is shown below. Air resistance is not negligible.

Which of the following statements is correct?

A The horizontal component of the velocity of the stone at A is the same as that at B.

B The air resistance acting on the stone increases from A to B.

C The kinetic energy of the stone at point A is smaller than the kinetic energy at point C.

D The time taken for the stone to travel from A to B is shorter than that from B to C.

6 A train consisting of six wagons each of mass 6.0 x 104 kg is pulled at a constant speed by a

locomotive of mass 2.4 x 105 kg along a straight horizontal track. The horizontal force resisting the motion of each wagon is 5000 N.

A link between two of the wagons breaks and the acceleration of the locomotive immediately after the link breaks is 0.021 m s-2. Between which wagons does the link break?

A 2 and 3 B 3 and 4 C 4 and 5 D 5 and 6

7 Object A of mass 1.2 kg and speed 0.025 m s-1 collides head-on and elastically with object B

of mass 0.60 kg moving with a speed 0.20 m s-1 towards it. After the collision, object B moves off with a speed of 0.10 m s-1 opposite to its initial motion.

What is the final speed of object A?

A 0.00 m s-1 B 0.025 m s-1 C 0.125 m s-1 D 0.800 m s-1

v

ground A

B

C

Object A Object B Before collision

After collision

5

8 A resultant force F acts on a body of mass 100 g. The variation with time t of F is shown below.

At time t = 0 s, the body is travelling at 1.0 m s-1 in the same direction as the positive values

of F. What is the speed of the body at time t = 2.5 s?

A 0.25 m s-1 B 0.75 m s-1 C 1.00 m s-1 D 1.75 m s-1

9 A wire is stretched elastically by a force of 200 N, causing an extension of 4.0 mm. The force

is then steadily increased to 250 N. The wire still behaves elastically. How much extra work is done in producing the additional extension?

A 0.225 J B 0.250 J C 225 J D 250 J

10 A uniform plank of weight 60 N is 2000 mm long. It rests on a support that is 600 mm from

end E.

At what distance from E must a 160 N weight be placed in order to balance the plank?

A 150 mm B 225 mm C 375 mm D 450 mm

[Turn over

F / N

0.3

0

-0.3

0.5 1.0 1.5 2.0 2.5 3.0 t / s

600 mm

2000 mm

E

6

11 A steel ball bearing of mass 0.10 kg is projected at a speed v into a vertical loop-a-loop track which has a radius of curvature of 0.09 m as shown below.

What is the minimum value of v if the ball bearing is not to lose contact with the track as it moves round the loop-a-loop track? Assuming the frictional force on the ball by the track is negligible.

A 0.94 m s-1 B 1.3 m s-1 C 1.9 m s-1 D 2.1 m s-1

12 Which of the following statements about the satellite in orbit is incorrect?

A The further the satellite is from the Earth, the longer the period of its orbit.

B The speed and kinetic energy of a satellite is independent of its mass.

C The closer the satellite is from the Earth, the faster its linear speed.

D There is only one geostationary orbit with a fixed distance from the centre of the Earth.

13 A constant force F is applied to a stationary object of mass m on a frictionless surface. A

constant acceleration increases the velocity of the object to some value v in a time t. It covers a distance s during this time. Which value of the kinetic energy is given to the object?

A Fst B Fv C Fs D ms2t

14 An electric pump is used to pump water up into a water tank at a vertical height of 12.4 m at

a rate of 6.2 kg s-1. The efficiency of the electric pump is 61%. What is the electrical power supplied to the pump?

A 0.13 kW B 0.46 kW C 0.75 kW D 1.2 kW

v

7

15 The function of many machines is to change energy from one form to another as efficiently as possible. Which of the following machines is the most efficient?

A A car engine as it converts chemical energy to kinetic energy.

B A tungsten light bulb as it converts electrical energy to light.

C A rice cooker as it converts electrical energy to thermal energy.

D A rocket as it converts chemical energy to gravitational potential energy.

16 The electric current in the electron beam of a cathode-ray oscilloscope is 50 μA. What is the number of electrons arriving at the screen in 1.0 s?

A 1.3 x 1011 B 1.3 x 1014 C 3.1 x 1011 D 3.1 x 1014

17 Two wires X and Y, each of the same length and the same material, are connected in parallel

to a battery. The diameter of X is half that of Y. What fraction of the total current passes through X?

A 0.20 B 0.25 C 0.33 D 0.5

18 A battery of e.m.f. E and internal resistance r delivers a current I through a variable resistance

R. R is set at two different values and the corresponding currents I are measured using an ammeter of negligible resistance. When the resistance of R is 1.0 Ω, the current is 3.0 A and when the resistance of R is 2.0 Ω, the current is 2.0 A.

What is the value of the e.m.f. E?

A 3.0 V B 3.5 V C 4.0 V D 6.0 V

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E r I R

8

19 A generator, with output power P and output voltage V, is connected to a factory by cables of total resistance R as shown in the figure.

What is the power loss?

A PR

V

B

PR

2V

C

2

2

P R

2V

D

2

2

P R

V

20 The diagram shows a network of three resistors. Two of these, marked R, are identical. The

other one has a resistance of 5.0 Ω.

The resistance between Y and Z is found to be 2.5 Ω. What is the resistance between X and Y?

A 0.12 Ω B 0.53 Ω C 1.9 Ω D 4.8 Ω

Y R X 5.0 Ω R Z

generator power = P, output voltage = V

R/2 factory R/2

9

21 Two parallel conductors, carrying equal currents, pass vertically through the two corners X and Y of a square WXYZ. The currents are flowing out of the page.

Which arrow represents the direction of the resultant magnetic field at point O, the centre of the square?

22 A wire of length 3.0 cm is placed at right angles to a region of uniform magnetic field of flux

density 0.040 T as shown below. The wire carries a current of 5.0 A.

What is the magnitude of the magnetic force which the field exerts on the wire?

A Slightly less than 0.0060 N

B 0.0060 N

C Slightly less than 0.60 N

D 0.60 N

[Turn over

W X A

B C O

D

Z Y

X X X X X X X X X X X X X X X X X X X X X X X X X

5.0 A

region of uniform magnetic flux

3.0 cm

10

23 The diagram shows two long straight parallel wires, X and Y, carrying currents in the same direction into the paper

X X

What are the directions of the forces on X and Y?

direction of force on X direction of force on Y

A N S

B E E

C E W

D W E

24 Which of the following statements about an electric field is incorrect?

A The resultant electric field strength at a point due to a few charges is the vector sum of the individual electric fields at that point due to the charges.

B The electric field strength at a point is a measure of the force exerted on a unit positive charge at that point.

C A stronger electric field strength is represented by electric field lines which are closer together.

D The direction of the electric field at a point is along the force acting on a negative change placed at that point.

25 In an experiment to investigate the nature of the atom, a very thin gold film was bombarded

with α-particles. Which of the following observations of deflection of the α-particles was made?

A A few α-particles were deflected through angles greater than a right angle.

B All α-particles were deflected from their original path.

C Most α-particles were deflected through angles greater than a right angle.

D No α-particles was deflected through an angle greater than a right angle.

N W E S

X Y

11

26 One reaction which might be used for controlled nuclear fusion reaction is shown

⎯⎯→ +7 2 43 1 2 Li + H 2( He) X

What is particle X?

A an α-particle B a β-particle C a proton D a neutron

27 High energy α-particles can transform Nitrogen-14 to Oxygen-17:

14 4 17 17 2 8 1 N + He O H⎯⎯→ +

The sum of the rest masses of the nitrogen and helium nuclei is 18.006 u. The sum of the rest masses of the oxygen and hydrogen nuclei is 18.007 u. The energy equivalent of 0.001 u is 1 MeV. What do the data show?

A Mass of 0.001 u has been converted into 1 MeV of energy.

B The kinetic energy of the products exceeds the kinetic energy of the reactants by

1 MeV.

C The kinetic energy of the reactants exceeds the kinetic energy of the products by

1 MeV.

D The reactants had only 1 MeV of kinetic energy and all of this was converted into mass.

28 A stationary uranium nucleus of mass 238 units disintegrates by the emission of an α-particle

of mass 4 units.

The ratio kinetic energy of the particle

iskinetic energy of the recoiling daughter nucleus

α -

A 4

234 B

4

238 C

234

4 D

238

4

29 The half-life of a certain radioactive material is 3.0 s.

How long does it take for its activity to reduce by 90 %?

A 0.46 s B 5.4 s C 10 s D 15 s

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12

30 A graph of the natural logarithm (log of base e) of the activity A of a radioactive source plotted against time is given as shown in figure.

What is the half-life of the source in years?

A 0.0044 B 160 C 400 D 860

== END OF PAPER ==

ln (A/s-1)

7.5 -

7.0 -

6.5 -

6.0 -

5.5 -

5.0 -

4.5 -

4.0 -

0 200 400 600 800 time / year

1

PHYSICS 8867/01

Paper 1 Multiple Choice

1 hour

Additional Materials: Multiple Choice Answer Sheet

READ THESE INSTRUCTIONS FIRST Write in soft pencil. Write your name and tutorial group on this cover page. Write and/or shade your name, NRIC / FIN number and HT group on the Answer Sheet (OMR sheet), unless this has been done for you. Do not use staples, paper clips, highlighters, glue or correction fluid. There are 30 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 (OMR 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. Calculators may be used.

This document consists of 18 printed pages and 0 blank pages.

[Turn over

NAME CLASS 2T


2 DATA: speed of light in free space c = 3.00 x 108 m s-1








FORMULAE:


1a t2

v2 = u2 + 2 a s



1 = ...++

21

11

RR

3

1 An ammeter connected in series in a circuit gives readings that have low precision but high accuracy. Which row best describes the likely error in readings taken with this ammeter?

random error systematic error

A large large

B large small

C small large

D small small

Answer: B Precision is related to random error while accuracy is related to systematic error. When precision is low, it indicates that the random error is large while a high accuracy indicates that the systematic error is small.

2 In 1945, an atomic bomb was denoted over the Japanese city of Nagasaki. It was

estimated to have generated an energy of 80 TJ in a time duration of 1.0 s. What was the estimated power produced by the detonation?

A 8 x 104 GW B 8 x 107 GW C 8 x 1010 GW D 8 x 1011 GW

Answer: A = = × = 8 × 10 W = 8 × 10 GW

4

3 A body is moving along a straight line OA. The variation of the displacement of the object, s, measured from the reference point O, with time t is shown below.

Which of the following statements describes most accurately the motion of the object?

A The displacement decreases at a decreasing rate.

B The displacement decreases at a constant rate.

C The velocity increases at an increasing rate.

D The velocity decreases at an increasing rate.

Answer: C The gradient is negative but increasing hence the displacement changes at an increasing rate velocity increases

4 An astronaut is working on the surface of the Moon where the acceleration due to gravity

is 1.6 m s-2. He drops a wrench from rest and catches it when it has a speed of 1.7 m s-1 and is 0.60 m above the ground. From which height above the ground did he drop the wrench?

A 0.75 m B 0.90 m C 1.13 m D 1.50 m

Answer: D v2=u2+2as Displacement

s=v2

2a=

1.72

2×1.6=0.90 m

Height above the ground = 0.90 + 0.60 = 1.50 m

t / s 0

s /m

5

5 A stone is thrown with an initial velocity v at an angle to the horizontal and the trajectory of the stone is shown below. Air resistance is not negligible.

Which of the following statements is correct?

A The horizontal component of the velocity of the stone at A is the same as that at B.

B The air resistance acting on the stone increases from A to B.

C The kinetic energy of the stone at point A is smaller than the kinetic energy at point C.

D The time taken for the stone to travel from A to B is shorter than that from B to C.

Answer: D The horizontal component of the air resistance causes the horizontal component of the stone to decrease. Option A is not true From A to B, the speed of the stone decreases. Since the air resistance is proportional to the speed, the air resistance decreases. Option B is not true. Since there is air resistance, work has to be done to overcome the air resistance. Hence, when back to the same level on the Earth’s surface, the stone will have lesser kinetic energy at C than A. The average vertical component of the velocity from A to B is larger than the average vertical component of the velocity from B to C. The vertical displacement is the same, the time taken to travel from A to B is shorter than that from B to C. Hence, option D is correct.

v

ground A

B

C

6

6 A train consisting of six wagons each of mass 6.0 x 104 kg is pulled at a constant speed by a locomotive of mass 2.4 x 105 kg along a straight horizontal track. The horizontal force resisting the motion of each wagon is 5000 N.

A link between two of the wagons breaks and the acceleration of the locomotive immediately after the link breaks is 0.021 m s-2. Between which wagons does the link break?

A 2 and 3 B 3 and 4 C 4 and 5 D 5 and 6

Answer: C When the train is traveling at constant speed, the forward force on the locomotive is equal in magnitude to the sum of the resistive force on the train, = 6 × 5000 = 30000 N When a link breaks, the forward force by the locomotive at the instance is 30 000 N. Let the number of wagons attached to the locomotive be n Net force = ma 30000 – [n x 5000] = (2.4 x 105 + 6.0 x 104 n) (0.021) n = 4 the link between 4 and 5 breaks.

7 Object A of mass 1.2 kg and speed 0.025 m s-1 collides head-on and elastically with object B

of mass 0.60 kg moving with a speed 0.20 m s-1 towards it. After the collision, object B moves off with a speed of 0.10 m s-1 opposite to its initial motion.

What is the final speed of object A?

A 0.00 m s-1 B 0.025 m s-1 C 0.125 m s-1 D 0.800 m s-1

Answer: C By conservation of linear momentum, + = + 1.2 0.025 + 0.60 −0.20 = 1.2 + 0.60 0.10 = −0.125 m s-1

Object A Object B Before collision

After collision

7

8 A resultant force F acts on a body of mass 100 g. The variation with time t of F is shown below.

At time t = 0 s, the body is travelling at 1.0 m s-1 in the same direction as the positive values

of F. What is the speed of the body at time t = 2.5 s?

A 0.25 m s-1 B 0.75 m s-1 C 1.00 m s-1 D 1.75 m s-1

Answer: D The area under a force time graph gives the change in momentum of an object. From t = 0 s to t = 2.5 s, the change in momentum is, ∆ = × 0.5 × 0.3 = 0.075 kg m s-1

Therefore, the change in velocity is, ∆ = 0.0750.100 = 0.75 ∆ = − = 0.75 + 1.00 = 1.75 m s-1

9 A wire is stretched elastically by a force of 200 N, causing an extension of 4.0 mm. The force

is then steadily increased to 250 N. The wire still behaves elastically. How much extra work is done in producing the additional extension?

A 0.225 J B 0.250 J C 225 J D 250 J

Answer: A When force is 200 N, the extension is 4.0 mm. By Hooke’s law, = = = . = 50000 N m-1

When force is 250 N, the extension, = = = 5.0 mm

Therefore, the extra work done is, = 12 − 12 = − = 50000 0.005 − 0.004 = 0.225 J

F / N

0.3

0

-0.3

0.5 1.0 1.5 2.0 2.5 3.0 t / s

8

10 A uniform plank of weight 60 N is 2000 mm long. It rests on a support that is 600 mm from

end E.

At what distance from E must a 160 N weight be placed in order to balance the plank?

A 150 mm B 225 mm C 375 mm D 450 mm

Answer: D

By the principle of moments, taking moments about the triangular pivot, Sum of clockwise moments = sum of anticlockwise moments 60 × 400 = 160 = 150 mm from pivot Hence from E, it is 450 mm.

11 A steel ball bearing of mass 0.10 kg is projected at a speed v into a vertical loop-a-loop track

which has a radius of curvature of 0.09 m as shown below.

What is the minimum value of v if the ball bearing is not to lose contact with the track as it moves round the loop-a-loop track? Assuming the frictional force on the ball by the track is negligible.

A 0.94 m s-1 B 1.3 m s-1 C 1.9 m s-1 D 2.1 m s-1

Answer: D To complete the loop, the ball needs to have sufficient speed at the top of the loop-a-loop track. At the minimum speed, the ball just loses contact with the track at the top,

600 mm

2000 mm

E

600 mm

2000 mm

E

400 mm

60 N 160 N

x

v

9 ∑ = = = = √0.09 × 9.81 = 0.94 m s-1

As the car goes from the bottom of the loop to the top, it loses KE and gains in GPE. Therefore, in order to have the minimum velocity required at the top of the loop, − = ℎ − = ℎ = 2 ℎ + = 2 ℎ + = √2 × 9.81 × 0.18 + 0.94 = 2.1 m s-1

12 Which of the following statements about the satellite in orbit is incorrect?

A The further the satellite is from the Earth, the longer the period of its orbit.

B The speed and kinetic energy of a satellite is independent of its mass

C The closer the satellite is from the Earth, the faster its linear speed.

D There is only one geostationary orbit with a fixed distance from the centre of the Earth.

Answer: B = 2

∝ Therefore, the further the satellite, the longer the period and the slower it travels. Option A and C are true statements. In addition, since geostationary satellite requires a fixed period of 24 h, thus, there is only one geostationary orbit with a fixed r. Even though it is true that the speed of the satellite is independent of the satellite’s mass, the kinetic energy of the satellite depends on the mass of the satellite as KE is ½ mv2.

13 A constant force F is applied to a stationary object of mass m on a frictionless surface. A

constant acceleration increases the velocity of the object to some value v in a time t. It covers a distance s during this time. Which value of the kinetic energy is given to the object?

A Fst B Fv C Fs D ms2t

Answer: C When a constant force F is applied to the stationary object and yet it accelerates constantly means that F is in the direction of the motion. Only option C gives the unit of work done, which has the same units as kinetic energy. By the work energy theorem, the net work done on the object gives rise to the change in kinetic

10

energy of the object. Since the object is initially at rest, the work done expression will give the kinetic energy given to the object. Option A is work done x time; option B is the instantaneous power of the applied force when the object has a speed. In option D, s/t is the speed. It lacks the square to give the usual KE equation, ½ mv2.

14 An electric pump is used to pump water up into a water tank at a vertical height of 12.4 m at

a rate of 6.2 kg s-1. The efficiency of the electric pump is 61%. What is the electrical power supplied to the pump?

A 0.13 kW B 0.46 kW C 0.75 kW D 1.2 kW

Answer: D The rate of gravitational potential energy gain is, = ℎ = 6.2 × 9.81 × 12.4 = 754.2W

Therefore, the input power is, = . = .. = 1236 W

15 The function of many machines is to change energy from one form to another as efficiently as

possible. Which of the following machines is the most efficient?

A A car engine as it converts chemical energy to kinetic energy.

B A tungsten light bulb as it converts electrical energy to light.

C A rice cooker as it converts electrical energy to thermal energy.

D A rocket as it converts chemical energy to gravitational potential energy.

Answer: C Car and rocket both produces large amount of heat and sound during its operations. Hence, the efficiency of its intended use is quite low. In a tungsten light bulb, much of the electrical energy is used to heat up the tungsten which then gives out light when it become white hot. As such, a relatively large amount is used to heat up and maintain the temperature of the light bulb. The rice cooker converts majority of its electrical energy to thermal energy and very minimal to the electrical circuit or buzzer (if any). In addition, there is also no moving parts and therefore, the conversion to other forms of energy is minimal.

11

16 The electric current in the electron beam of a cathode-ray oscilloscope is 50 μA. What is the number of electrons arriving at the screen in 1.0 s?

A 1.3 x 1011 B 1.3 x 1014 C 3.1 x 1011 D 3.1 x 1014

Answer: D I = Q/t = Ne/t 50 x 10-6 = N (1.6 x 10-19)/ (1.0) N = 3.1 x 1014

17 Two wires X and Y, each of the same length and the same material, are connected in parallel

to a battery. The diameter of X is half that of Y. What fraction of the total current passes through X?

A 0.20 B 0.25 C 0.33 D 0.5

Answer: A Diameter of X = ½ diameter of Y Cross sectional area = π d2 / 4 Cross sectional area of X = ¼ cross-sectional area of Y R = ρL/A ρ and L are constants Resistance of X = 4 resistance of Y Since X and Y are in parallel p.d. across X = p.d. across Y Current in X x Resistance of X = Current in Y x Resistance of Y Current in X = ¼ current in Y Current in X = 1/5 total current

18 A battery of e.m.f. E and internal resistance r delivers a current I through a variable resistance

R. R is set at two different values and the corresponding currents I are measured using an ammeter of negligible resistance. When the resistance of R is 1.0 Ω, the current is 3.0 A and when the resistance of R is 2.0 Ω, the current is 2.0 A.

E r I R

12

What is the value of the e.m.f. E?

A 3.0 V B 3.5 V C 4.0 V D 6.0 V

Ans: D E = I ( r + R) E = 3 ( r + 1) = 3r + 3 ……..…(1) E = 2 ( r + 2) = 2r + 4………...(2) Solve (1)& (2) r = 1 Ω & E = 3(1+1) = 6 V

19 A generator, with output power P and output voltage V, is connected to a factory by cables of total resistance R as shown in the figure.

What is the power loss?

A

PR

V

B

PR

2V

C

2

2

P R

2V

D

2

2

P R

V

Ans: D Current through cable I = P/V Power dissipated in cable = I 2(R/2 + R/2) = (P/V)2R

20 The diagram shows a network of three resistors. Two of these, marked R, are identical. The

other one has a resistance of 5.0 Ω.

Y R X 5.0 Ω R Z

generator power = P, output voltage = V

R/2 factory R/2

13

The resistance between Y and Z is found to be 2.5 Ω.

What is the resistance between X and Y?

A 0.12 Ω B 0.53 Ω C 1.9 Ω D 4.8 Ω

Answer: C YZ Total resistance = (2R )(5) /(2R + 5) = 2.5 Ω R = 2.5 Ω XY Total resistance = (2.5) ( 7.5) / (2.5 + 7.5) = 1.875 = 1.9 Ω

21 Two parallel conductors, carrying equal currents, pass vertically through the two corners X and

Y of a square WXYZ. The currents are flowing out of the page.

Which arrow represents the direction of the resultant magnetic field at point O, the centre of the square?

Ans: D Consider the magnetic flux density of X and Y Resultant magnetic field is the vector sum of the magnetic field due to X and that due to Y

W X A

B C O

D

Z Y

W X Z Y

magnetic field due to current in X

magnetic field due to current in Y

resultant magnetic field

14

22 A wire of length 3.0 cm is placed at right angles to a region of uniform magnetic field of flux density 0.040 T as shown below. The wire carries a current of 5.0 A.

What is the magnitude of the magnetic force which the field exerts on the wire?

A Slightly less than 0.0060 N

B 0.0060 N

C Slightly less than 0.60 N

D 0.60 N

Ans: A For a 3.0 cm wire in the field F = BIL sin θ = 0.040 x 5.0 x 0.03 = 0.0060 N But the field over the wire is slightly less than 3.0 cm Hence force is slightly less than 0.0060 N

X X X X X X X X X X X X X X X X X X X X X X X X X

5.0 A

region of uniform magnetic flux

3.0 cm

15

23 The diagram shows two long straight parallel wires, X and Y, carrying currents in the same direction into the paper

X X

What are the directions of the forces on X and Y?

direction of force on X direction of force on Y

A N S

B E E

C E W

D W E

Ans: C The force is attractive. Hence force on Y is to the left (westwards) and force on X is to the right (eastwards).

24 Which of the following statements about an electric field is incorrect?

A The resultant electric field strength at a point due to a few charges is the vector sum of the individual electric fields at that point due to the charges.

B The electric field strength at a point is a measure of the force exerted on a unit positive charge at that point.

C A stronger electric field strength is represented by electric field lines which are closer together.

D The direction of the electric field at a point is along the force acting on a negative change placed at that point.

Ans: D The force is along the field when the charge is positive.

N W E S

X Y

16

25 In an experiment to investigate the nature of the atom, a very thin gold film was bombarded with α-particles. Which of the following observations of deflection of the α-particles was made?

A A few α-particles were deflected through angles greater than a right angle.

B All α-particles were deflected from their original path.

C Most α-particles were deflected through angles greater than a right angle.

D No α-particles was deflected through an angle greater than a right angle.

Ans: A

26 One reaction which might be used for controlled nuclear fusion reaction is shown

⎯⎯→ +7 2 43 1 2 Li + H 2( He) X

What is particle X?

A an α-particle B a β-particle C a proton D a neutron

Ans: D Conservation of mass number 7+2 = 2x4 + A A = 1 Conservation of atomic number 3 + 1 = 2 x 2 + Z Z = 0 X is a neutron.

27 High energy α-particles can transform Nitrogen-14 to Oxygen-17:

14 4 17 17 2 8 1 N + He O H⎯⎯→ +

The sum of the rest masses of the nitrogen and helium nuclei is 18.006 u. The sum of the rest masses of the oxygen and hydrogen nuclei is 18.007 u. The energy equivalent of 0.001 u is 1 MeV. What do the data show?

A Mass of 0.001 u has been converted into 1 MeV of energy.

B The kinetic energy of the products exceeds the kinetic energy of the reactants by 1 MeV

C The kinetic energy of the reactants exceeds the kinetic energy of the products by 1 MeV

D The reactants had only 1 MeV of kinetic energy and all of this was converted into mass

Ans: C Rest mass energy of reactants is 0.001 u less than the rest mass energy of the products. Rest mass energy of reactants + kinetic energy of reactants = rest mass energy of the products + kinetic energy of the products. Hence the kinetic energy of reactants is 0.001 u more than the kinetic energy of the products.

17

28 A stationary uranium nucleus of mass 238 units disintegrates by the emission of an α-particle of mass 4 units.

The ratio kinetic energy of the particle

iskinetic energy of the recoiling daughter nucleus

α -

A 4

234 B

4

238 C

234

4 D

238

4

Ans: C The reaction can be represented by

He ThU 42

23490

23892 +⎯→⎯

By principle of conservation of momentum, (238 – 4) Vth = 4 Vα

where Vth : velocity of the recoiling thorium nucleus and vα : velocity of the

4

234=th

α

V

V

Ratio of the kinetic energies is

4

234

4

234

234

4

234

4

nucleus thorium recoiling the ofenergy kinetic

particle the ofenergy kinetic 2

=

==−

2th

2

V

Vα α

29 The half-life of a certain radioactive material is 3.0 s.

How long does it take for its activity to reduce by 90 %?

A 0.46 s B 5.4 s C 10 s D 15 s

Ans: C (½ )n = (100-90) /100 =0.1 n log 0.5= log 0.1 n = 3.32 t = 3.32 x 3 = 10 s

particle−α

18

30 A graph of the natural logarithm (log of base e) of the activity A of a radioactive source plotted against time is given as shown in figure.

What is the half-life of the source in years?

A 0.0044 B 160 C 400 D 860

Answer B ln (Ao) = 7.5 Ao = 1800 At half-life, A = 1800/2 = 900 In (900) = 6.8 From graph, ln(A) = 6.8 t is approximately 160 years

== END OF PAPER ==

ln (A/s-1)

7.5 -

7.0 -

6.5 -

6.0 -

5.5 -

5.0 -

4.5 -

4.0 -

0 200 400 600 800 time / year

1

CANDIDATE NAME

CLASS 2T

PHYSICS 8867/02 Paper 2 2 hours

Candidates answer on the Question Paper. READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen in the space provided. [PILOT FRIXION ERASABLE PENS ARE NOT ALLOWED]

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 from this section. Section B: Answer one question from this section. The number of marks is given in brackets [ ] at the end of each question or part of the question.

This document consists of 24 printed pages and 0 blank page.

[Turn over

FOR EXAMINER’S USE DIFFICULTY

L1 L2 L3

Q1 /6

Q2 /7

Q3 /5

Q4 /11

Q5 /12

Q6 /19

Q7 /20

Q8 /20

PAPER 2 /80

PAPER 1 / 30

TOTAL / 110

%


2 PHYSICS DATA: speed of light in free space c = 3.00 x 108 m s-1








PHYSICS FORMULAE:


1a t2

v2 = u2 + 2 a s



1 = ...++

21

11

RR

3

Section A

Answer all questions from this section. 1 The drag coefficient CD of a car moving with speed v through air of density ρ is described by

CD=2F

ρv2A

where F is the drag force exerted on the car and A is the frontal cross-sectional area of the car when it is moving forwards.

(a) By using SI base units, show that CD has no unit.

[2]

(b) In an experiment to investigate the drag force acting on a car travelling at a constant speed on a

level road. The following measurements were made: frontal cross-sectional area of the car = (2.60 ± 0.05) m2

drag force acting on the car = (28.3 ± 0.2) N speed of the car = (8.1 ± 0.2) m s-1 density of air = (1.16 ± 0.05) kg m-3

Calculate the drag coefficient of the car. Quote the value of the drag coefficient with its absolute uncertainty.

CD = ……………… ± ………………

[4]

[Total: 6]

[Turn over

4 2 (a) A space station of mass 3.5 x 104 kg orbits the Earth at a height 4.0 x 105 m above the surface

of the Earth. The radius of the Earth is 6.4 x 106 m and the mass of the Earth is 6.0 x 1024 kg.

(i) Calculate the gravitational force on the space station.

gravitational force = …………………… N

[2] (ii) Using the answer to (a)(i), calculate the time for the space station to complete one orbit.

time = …………………… hours

[3] (iii) Explain why the astronaut in a space station experiences ‘weightlessness’.

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

[2] [Total: 7]

5 3 A truck of mass 1.2 x 104 kg collides head-on with a stationary car of mass 2200 kg. The car moves

forwards with a velocity of 5.5 m s-1 as a result of the collision. The time of impact is 0.25 s.

(a) By considering momentum, calculate the average force on the truck.

force on the truck = …………………… N

[3] (b) Whiplash is a common injury that occurs to a person's neck following a sudden rapid forward or

backward movement of the head and neck, most commonly from motor vehicle accidents. A car travelling at a speed of 50 km h-1 is stopped suddenly in a head-on collision with a wall. Explain, using Newton’s law of motion, why the driver who is secured firmly to the seat by the seatbelt is likely to suffer an injury of whiplash.

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

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

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

[2]

[Total: 5]

[Turn over

6 4 This question concerns three different types of field, namely magnetic field, electric field and

gravitational field.

(a) Explain what is meant by the term field in physics.

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

(b) Fig.4.1 shows a beam of electrons entering a region of magnetic field ABCD of magnetic flux

density 8.9 mT perpendicularly. Electrons travelling with a velocity 8.6 x 106 m s-1 enter the field at right angles to the edge BD of the field.

Fig.4.1 (i) Calculate, for the electrons in the magnetic field, the magnitude of the magnetic force on

an electron.

magnitude of magnetic force = ……………………..N

[2]

(ii) On Fig. 4.1, sketch the path travelled by the electrons within and beyond the magnetic

field, assuming that the electrons will leave the magnetic field from the edge AC of the magnetic field.

[2]

electron beam

region of magnetic field vertically down into the paper

A C B D

7

(iii) An uniform electric field of electric field strength 1.7 x 105 N C-1 is now applied in the same region of and in the same direction as the magnetic field. Calculate the magnitude of the resultant force acting on an electron at the instant when it enters the fields. Explain your answers.

magnitude of resultant force = ……………………. N

[4]

(iv) Discuss whether your answer in (b)(iii) be affected by the gravitational field of the Earth.

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

[2]

[Total: 11]

[Turn over

8

5 (a) Uranium nuclei may undergo nuclear fission when bombarded by neutrons. One such reaction is

+ → + +235 1 139 95 192 0 54 38 0U n Xe Sr 2( n) .

Data: Binding energy per nucleon of 235

92U = 7.60 MeV,

Binding energy per nucleon of 13954 Xe = 8.39 MeV

Binding energy per nucleon of 9538Sr = 8.74 MeV.

(i) State what is meant by the binding energy per nucleon of a nucleus.

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

[1] (ii) Calculate the energy released per fission event.

energy released per fission = ………………………………….MeV

[2] (iii) In a proposed nuclear fission reactor using this reaction to produce electrical energy, the

efficiency of converting the available energy into electrical energy is 8.0 %. The electrical energy is used to supply a city with an average demand of 1.5 x 109 W.

1. Calculate the amount of energy that can be obtained from 1 kg of the fissionable isotope 235

92U .

amount of energy = ……………………………..MeV

[2]

9

2. Calculate the duration of time this would supply the electrical energy to the city.

time duration = ………………………..hours

[4] (b) A radioactive isotope of thallium 207

81Tl emits a β-particle and is thought to emit a gamma ray

photon.

The radiation from a radioactive source of 20781Tl is allowed to pass through a vertical uniform

magnetic field perpendicularly in a cloud chamber as shown in Fig. 5.1. The photograph of the traces produced by the β-particles and gamma ray photons is obtained under certain conditions. vertical uniform magnetic field

Fig. 5.1

Explain as many features of the traces as you can.

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

[3] [Total: 12]

[Turn over

tracks of beta particles with different radii

faint tracks of gamma ray

radioactive source

top view of cloud chamber

10 6 The bow is a powerful two-arm string machine used for archery. Figure. 6.1 shows the three types of

bows, namely the Longbow, the Recurved and the Compound bow.

Fig. 6.1

Each bow consists of a limb, a handle and a string. The distance between the string and the bow handle at rest is known as the bracing distance. When a bow is drawn by the fingers of the archer, the string is not stretched but the shape of the bow is changed and bent and the string is displaced by a drawing distance. The shaft of the arrow is rested on the handle and the tail of the arrow is rested on the middle of the string. Fig.6.2.

Fig. 6.2

In a Longbow, if we suppose that the bow is initially unstressed and the string is almost slack, then the archer starts to draw his arrow with a pulling force which is nearly zero and the pulling force increases with the drawing distance. The energy stored in the bow is equal to the work done in drawing back the string. In practice, a typical archer can draw an arrow back about 0.6 m and with a maximum force of about 350 N as shown in Fig. 6.3.

Longbow Recurved bow Compound bow

drawing distance 0.6 m

bracing distance0.15 m

handle

limb

string

Pulling force 350 N

11

Fig. 6.3

The efficiency of the bow, η, can be defined as η = kinetic energy of the arrow _ elastic potential energy stored in the bow

When an arrow is shot, the force due to the tension in the string accelerates the arrow. A larger part of

the energy stored in the bow is transferred to the arrow. Since this transferred energy is converted into the kinetic energy of the arrow ½ mv2 (where v is the speed of the arrow as it leaves the bow), the increase in the length and therefore the mass of the arrow has two opposing effects: an increase in efficiency but a decrease in speed. When a bow string is released, the string exerts a forward force on the arrow and causes it to accelerate forwards. At the same time, there is a sudden compressive force along the length of the arrow which causes it to buckle. Hence the arrow will undergo lateral vibrations as it accelerates forwards. Fig. 6.4 shows the view from above the arrow leaving bow (not to scale).

[Turn over

0.0 0.1 0.2 0.3 0.4 0.5 0.6 drawing distance x/m

Force F / N 500

400

300

200

100

0

12

top view of the arrow

Fig. 6.4

Both the frequency and the amplitude of these vibrations must be matched to the bow if the arrow is to avoid hitting the side of the handle during its discharge. Ideally the arrow will make 1¼ vibrations from the moment of release until it finally clears the handle of the bow. A Recurved bow is one in which the end of each limb curves away from the archer such that the limbs are braced with a bracing distance. The archer must start his pull with a non-zero force which is about 1/3 of the maximum force. When using a Recurved bow, the average force will be higher as compared to a Longbow without bracing. In a Compound bow, which utilizes levers, the drawing force increases and decreases with the drawing distance as shown in Fig. 6.5.

Fig. 6.5

Force / N 350 0

0.0 0.6 drawing distance x / m

handle

arrow shaft

string

moment of release

moment of clearing the handle of bow

½ cycle of vibration of the arrow shaft

1 cycle of vibration of the arrow shaft

13 (a) Explain what is meant by the term lateral vibrations.

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

[1] (b) Use the graph in Fig. 6.3 to calculate the energy stored in the Longbow when the maximum pulling

force of 350 N is exerted on the string.

energy stored = …………………………………J

[2] (c) On Fig 6.6, sketch a graph to show how the work done on the Longbow changes with the drawing

distance of the string. Label the work done axis clearly.

[2]

Fig. 6.6

(d) It is thought that the efficiency of the bow obeys a relation of the form

m

m kη =

+

where m is the mass of the arrow and k is a constant depending on the mass of the bow. A student performed an experiment to investigate how η, the efficiency of the bow varies with m, the mass of the arrow. He obtained data which allows him to plot the graph of 1/η against 1/m as shown in Fig. 6.7.

[Turn over

Work done on the bow / J

0.0 0.1 0.2 0.3 0.4 0.5 0.6 drawing distance x / m | | | | | |

14

Fig. 6.7

(i) Draw the line of best fit for the points. [1] (ii) Explain whether the relation in (d) is valid using the line drawn in Fig. 6.7.

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

[3] (iii) Use the line drawn in Fig. 6.7 to determine the magnitude of the constant k in the expression

in (d).

k = …………………………. [2]

0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 (1/m) / g-1

1/η 2.0

1.6 1.2

0.8

0.4

0.0

15 (iv) An arrow of mass 70 g is being shot from an initially unstressed Longbow drawn back as

shown in Fig. 6.2. Use the graph in Fig. 6.7 and the definition of efficiency to determine the speed of the arrow leaving the bow.

speed of the arrow = …………………..m s-1

[3] (e) Calculate the frequency of vibration for an arrow leaving the string at 50 m s-1, from a bow of

bracing distance of 0.15 m, and shot by an archer with a draw distance 0.6 m, as shown in Fig. 6.2.

frequency of vibration = …………………………..s-1

[2] (f) (i) On Fig. 6.3, sketch the force-drawing distance graph for a Recurved bow which is already

braced to 150 N. The maximum drawing force of 350 N is exerted on the string at the maximum drawing distance of 0.6 m.

[1]

(ii) State and explain, in terms of the energy stored, why Recurved bow is better than of that a

Longbow.

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

[1] (g) Refer to the force-drawing distance graph of a Compound bow as shown in Fig. 6.5. Suggest an

advantage of a Compound bow as compared to the Longbow.

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

[1] [Total: 19]

[Turn over

16

Section B

Answer one question from this section.

7 A steel nut of mass 200 g on the top of a hemispherical dome becomes loose and moves down the dome from A to B as shown in Fig 7.1 with a negligible initial speed. The surface of the dome is smooth. At point B, the nut loses contact with the surface of the dome. From B to C, it moves in air under the gravitational force only. It hits the ground at point C at a horizontal distance x from B. The radius of the dome is 30 m

Fig 7.1

(a) Consider the motion from A to B,

(i) using energy considerations, explain why the speed of the steel nut increases.

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

[2]

(ii) explain why the centripetal acceleration of the nut increases.

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

[2]

D

` `

30 m

10 m

17 (b) The nut loses contact with the surface of the dome at point B, which is 10 m vertically below

the top of the dome.

(i) State the resultant force acting on the nut at B.

resultant force = ……….……………N [1] (ii) Show that the velocity is 14 m s-1 directed at an angle of 48.2o below the horizontal.

[3]

(c) For motion from B to C, assuming air resistance is negligible,

(i) explain why the motion of the steel nut is parabolic.

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

(ii) calculate the time taken for the steel nut to fall from B to C,

time taken = …………….……..s [2]

[Turn over

18 (iii) calculate x, the horizontal distance between B and C.

x = …………….…..m [2] (d) Fig. 7.2 shows the variation of the vertical component acceleration of the steel nut ay with time

from point A to point B.

Fig. 7.2

(i) Sketch the variation of vertical component acceleration ay of the plate with time from point B to C.

[1] (ii) Describe qualitatively the variation of the vertical component of the velocity with time

from A to C.

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

[2]

ay / m s-2

t / s time at B

time at A

time at C

19 (iii) On Fig.7.3, sketch the variation of the vertical component of the velocity vy of the nut

with time from point A to point C.

Fig. 7.3

[2]

(iv) Explain how the radius of the dome may be determined from the graph you sketched in (d)(iii).

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

[2]

[Total:20]

[Turn over

vy / m s-2

t / s time at B

time at A

time at C

20 8 (a) (i) Define electrical resistance of a component.

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

(ii) Explain why it is difficult to quote a value for the resistance of a filament lamp.

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

(b) The circuit of Fig. 8.1 consists of a 12 V battery and a potential divider is used to produce

the I-V characteristic of the lamp. A voltmeter and ammeter are connected to measure the current in the lamp and the potential difference across the lamp.

Fig. 8.1

The characteristic is shown in Fig.8.2.

Fig. 8.2

0 2 4 6 8 10 12 V / V

I/A 3

2

1

0

AA

AV

21 (i) For the lamp operating under normal conditions with a potential difference of 10.0 V

across it, calculate the resistance of the lamp

resistance = ………………. Ω

[2]

(ii) Calculate the resistance of the lamp when the potential difference across it is only

2.0 V.


[1]

(iii) Explain in microscopic terms why the resistance in b(ii) is less than the resistance in (b)(i).

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

[3]

(iv) With the connection of the meters, explain whether the resistance of the lamp obtained

from Fig. 8.2 is an under-estimation or over-estimation of the expected resistance of the lamp if the voltmeter and the ammeter are not ideal.

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

[2]

[Turn over

22

(c) (i) Show, by considering energy conversion, that V the terminal potential difference of a cell of e.m.f. E and internal resistance r is given by V = E – Ir where I is the current flowing in the cell.

[2] (ii) Fig. 8.3 shows a battery with e.m.f. E and an internal resistance r connected to a uniform

nichrome resistance wire MN. J is a movable contact J which can slide along wire MN. The positive terminal of the battery is earthed. The voltmeter and the ammeter are taken to be ideal in this question.

Fig. 8.3

23

The voltmeter readings V and ammeter readings I obtained for different lengths of JN are used to plot the graph in Fig. 8.4.

Fig. 8.4

1. Deduce from Fig. 8.4 the e.m.f. E and the internal resistance r of the cell.

E = ……..………….. V

r = ………..……….. Ω [4]

[Turn over

24

2. J is placed at the position such that maximum power is delivered from the cell to the wire JN. Determine the potential at J and N.

potential at J = …..………V

potential at N = …..………V

[3] 3. On Fig.8.5 sketch the graph to show how electric potential varies with position.

Label the vertical axis clearly.

Fig. 8.5 [1]

[Total: 20]

== END OF PAPER ==

J N distance from J

electric potential / V

position along JN

CANDIDATE NAME

MARK SCHEME

CLASS 2T

PHYSICS 8867/02 Paper 2 2 hours

Candidates answer on the Question Paper. READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen in the space provided. [PILOT FRIXION ERASABLE PENS ARE NOT ALLOWED]

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 from this section. Section B: Answer one question from this section. The number of marks is given in brackets [ ] at the end of each question or part of the question.

This document consists of 31 printed pages and 1 blank page.

[Turn over

FOR EXAMINER’S USE

Q1 /6

Q2 /7

Q3 /5

Q4 /9

Q5 /12

Q6 /19 Q7 /20

Q8 /20

PAPER 2 /80

PAPER 1 / 30

TOTAL / 110

%


2 PHYSICS DATA: speed of light in free space c = 3.00 x 108 m s-1








PHYSICS FORMULAE:


1a t2

v2 = u2 + 2 a s



1 = ...++

21

11

RR

3

Section A Answer all questions from this section.

1 The drag coefficient CD of a car moving with speed v through air of density ρ is described by

CD=2F

ρv2A

where F is the drag force exerted on the car and A is the frontal cross-sectional area of the car when it is moving forwards.

(a) By using SI base units, show that CD has no unit.

[2]

Solution: = 2

= ( ) =

Hence CD has no units.

(b) In an experiment to investigate the drag force acting on a car travelling at a constant speed

on a level road. The following measurements were made: frontal cross-sectional area of the car = (2.60 ± 0.05) m2

drag force acting on the car = (28.3 ± 0.2) N speed of the car = (8.1 ± 0.2) m s-1 density of air = (1.16 ± 0.05) kg m-3

Calculate the drag coefficient of the car. Quote the value of the drag coefficient with its absolute uncertainty.

4

CD = ……………… ± ………………

[4] Solution = 2

= 2(28.3)(1.16)(8.1) (2.6) = 0.286

∆ = ∆ + ∆ + 2∆ + ∆ ∆ = 0.228.3 + 0.051.16 + 2 0.28.1 + 0.052.60 = 0.1188 ∆ = 0.1188 × 0.286 = 0.034 = 0.03

= (0.29 ± 0.03)

[Turn over

2 (a) A space station of mass 3.5 x 104 kg orbits the Earth at a height 4.0 x 105 m above the surface of the Earth. The radius of the Earth is 6.4 x 106 m and the mass of the Earth is 6.0 x 1024 kg.

(i) Calculate the gravitational force on the space station.

gravitational force = …………………… N

[2] Solutions: = = (6.67 × 10 )(6.0 × 10 )(3.5 × 10 )((0.4 + 6.4) × 10 ) = 3.03 × 10 N

(ii) Using the answer to (a)(i), calculate the time for the space station to complete one orbit.

5

time = …………………… hours

[3] Solutions:

Gravitational force provides for the centripetal force for the satellite to go in orbit. = 2

3.03 × 10 = 3.5 × 10 × (0.4 + 6.4) × 10 ) 2 = 5569 = 1.55 h

(iii) Explain why the astronaut in a space station experiences ‘weightlessness’.

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

[2] Solutions:

The gravitational force solely provides for the centripetal force on the person. Or The acceleration due to gravity is equal to the centripetal acceleration. Hence, the person is no longer in contact with the space station and experiences no normal contact force. Thus, the person experiences weightlessness.

6 3 A truck of mass 1.2 x 104 kg collides head-on with a stationary car of mass 2200 kg. The car moves

forwards with a velocity of 5.5 m s-1 as a result of the collision. The time of impact is 0.25 s.

(a) By considering momentum, calculate the average force on the truck.

force on the truck = …………………… N

[3]

Change in momentum of the car, ∆ = ∆ ∆ = 2200(5.5 − 0) ∆ = 1.21 × 10 kg m s-1

By Newton’s 2nd law, ⟨ ⟩ = ∆∆ ⟨ ⟩ = . ×. = 4.84 × 10 N

By Newton’s third law of motion, Force on truck is equal and opposite to the force on car = 4.84 × 10 N

(b) Whiplash is a common injury that occurs to a person's neck following a sudden rapid forward

or backward movement of the head and neck, most commonly from motor vehicle accidents. A car travelling at a speed of 50 km h-1 is stopped suddenly in a head-on collision with a wall. Explain, using Newton’s law of motion, why the driver who is secured firmly to the seat by the seatbelt is likely to suffer an injury of whiplash.

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

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

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

[2] Solution:

The car is stopped suddenly by the braking force on the car. For the head to stop, the neck exerts a large backwards force on the head, by the muscles in the neck. If the muscles at the back of the neck is not strong enough to provide this large backward force, they will be torn apart and the driver will suffer whiplash injury.

7 4 This question concerns three different types of field, namely magnetic field, electric field and

gravitational field.

(a) Explain what is meant by the term field in physics.

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

Region of space in which an object experiences a force.

(b) Fig.4.1 shows a beam of electrons entering a region of magnetic field ABCD of magnetic flux density 8.9 mT perpendicularly. Electrons travelling with a velocity 8.6 x 106 m s-1 enter the field at right angles to the edge BD of the field.

Fig.4.1 (i) Calculate, for the electrons in the magnetic field, the magnitude of the magnetic force on

an electron.

magnitude of magnetic force = ……………………..N

[2]

Magnetic force = Bqv = 0.0089 x 1.6 x 10-19 x 8.6 x 106

= 1.22 x 10-14 N

(ii) On Fig. 4.1, sketch the path travelled by the electrons within and beyond the magnetic

field, assuming that the electrons will leave the magnetic field from the edge AC of the magnetic field.

[2]

electron beam


A C B D

8

Circular arc to the right inside field M1 Straight line outside field tangent to path just before leaving edge AC M1

(iii) An uniform electric field of electric field strength 1.7 x 105 N C-1 is now applied in the same

region of and in the same direction as the magnetic field. Calculate the magnitude of the resultant force acting on an electron at the instant when it enters the fields. Explain your answers.

magnitude of resultant force = ……………………. N

[4]

Electric force = qE = (1.6 x 10-19 x 1.7 x 105) = 2.72 x 10 -14 N Magnetic force and electric force are perpendicular Resultant force = √ (2.72 x 10 -14)2 + (1.22 x 10-14)2

= 2.98 x 10-14

= 3.0 x 10-14 N

(iv) Discuss whether your answer in (b)(iii) be affected by the gravitational field of the Earth.

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

[2]

Gravitational force = mg = 9.11 x 10-31 x 9.81 = 8.9 x 10-30 N This is 8.9 x 10-30 / 3.0 x 10-14 = 3 x 10-16 times smaller than the resultant force hence it will not affect the answer.

electron beam


A C B D

9

5 (a) Uranium nuclei may undergo nuclear fission when bombarded by neutrons. One such reaction is

+ → + +235 1 139 95 192 0 54 38 0U n Xe Sr 2( n) .

Data: Binding energy per nucleon of 235

92U = 7.60 MeV,

Binding energy per nucleon of 13954 Xe = 8.39 MeV

Binding energy per nucleon of 9538Sr = 8.74 MeV.

(i) State what is meant by the binding energy per nucleon of a nucleus.

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

[1] Average energy per nucleon to split a nucleus into separate nucleons. (ii) Calculate the energy released per fission event.

energy released per fission = ………………………………….MeV

[2]

Energy released per nuclear fission = B.E of products – B.E of reactants = [8.39(139) + 8.74(95)] – 7.60(235) = 211 MeV

(iii) In a proposed nuclear fission reactor using this reaction to produce electrical energy, the efficiency of converting the available energy into electrical energy is 8.0 %. The electrical energy is used to supply a city with an average demand of 1.5 x 109 W.

1. Calculate the amount of energy that can be obtained from 1 kg of the fissionable isotope 235

92U .

amount of energy = ……………………………..MeV

[2] number of 235

92U nuclei in 1 kg

= 1/ 235 u = 1/ (235 x 1.66 x 10-27 ) = × 242.563 10 nuclei Energy available = 2.563 x 1024 x 211 MeV = 5.408 x 1026 MeV = 5.4 x 1026 MeV

2. Calculate the duration of time this would supply the electrical energy to the city.

time duration = ………………………..hours

[4]

10

Energy available = 5.408 x 1026 x 106 x 1.6 x 10-19 J = 8.653 x 1013 J Efficiency = useful output energy/ input energy 8/100 = useful output energy / 8.653 x 1013

useful output energy = 8.653 x 1013 x 0.08 = 6.922 x 1012 J

time taken = electrical energy / power = 6.922 x 1012 / 1.5 x 109

= 4615 s = 1.3 h

(b) A radioactive isotope of thallium 207

81Tl emits a β-particle and is thought to emit a gamma photon.

The radiation from a radioactive source of 20781Tl is allowed to pass through a vertical uniform

magnetic field perpendicularly in a cloud chamber as shown in Fig. 5.1. The photograph of the traces produced by the β-particles and gamma ray photons is obtained under certain conditions. vertical uniform magnetic field

Fig. 5.1

Explain as many features of the traces as you can.

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

[3] Circular paths of small radii show that beta particles are present.

tracks of beta particles with different radii

faint tracks of gamma ray

radioactive source

top view of cloud chamber

11

Circular paths of different radii show that the beta particles have different speeds and hence different kinetic energy. The faint straight line path shows the present of gamma ray photons which have no charge and not deflected by the magnetic field and the path is faint shows that the gamma ray photon has much less ionization ability as compared to the beta particles. (any 3 points)

12 6 The bow is a powerful two-arm string machine used for archery. Figure. 6.1 shows the three types of

bows, namely the Longbow, the Recurved and the Compound bow.

Fig. 6.1

Each bow consists of a limb, a handle and a string. The distance between the string and the bow handle at rest is known as the bracing distance. When a bow is drawn by the fingers of the archer, the string is not stretched but the shape of the bow is changed and bent and the string is displaced by a drawing distance. The shaft of the arrow is rested on the handle and the tail of the arrow is rested on the middle of the string. Fig.6.2.

Fig. 6.2 In a Longbow, if we suppose that the bow is initially unstressed and the string is almost slack, then

the archer starts to draw his arrow with a pulling force which is nearly zero and the pulling force increases with the drawing distance. The energy stored in the bow is equal to the work done in drawing back the string. In practice, a typical archer can draw an arrow back about 0.6 m and with a maximum force of about 350 N as shown in Fig. 6.3.

Longbow Recurved bow Compound bow

drawing distance 0.6 m

bracing distance0.15 m

handle

limb

string

Pulling force 350 N

13

Fig. 6.3

The efficiency of the bow, η, can be defined as η = kinetic energy of the arrow _ elastic potential energy stored in the bow

When an arrow is shot, the force due to the tension in the string accelerates the arrow. A larger part

of the energy stored in the bow is transferred to the arrow. Since this transferred energy is converted into the kinetic energy of the arrow ½ mv2 (where v is the speed of the arrow as it leaves the bow), the increase in the length and therefore the mass of the arrow has two opposing effects: an increase in efficiency but a decrease in speed. When a bow string is released, the string exerts a forward force on the arrow and causes it to accelerate forwards. At the same time, there is a sudden compressive force along the length of the arrow which causes it to buckle. Hence the arrow will undergo lateral vibrations as it accelerates forwards. Fig. 6.4 shows the view from above the arrow leaving bow (not to scale).

[Turn over

0.0 0.1 0.2 0.3 0.4 0.5 0.6 drawing distance x/m

Force F / N 500

400

300

200

100

0

Answer to f(i)

14

top view of the arrow

Fig. 6.4

Both the frequency and the amplitude of these vibrations must be matched to the bow if the arrow is to avoid hitting the side of the handle during its discharge. Ideally the arrow will make 1¼ vibrations from the moment of release until it finally clears the handle of the bow. A Recurved bow is one in which the end of each limb curves away from the archer such that the limbs are braced with a bracing distance. The archer must start his pull with a non-zero force which is about 1/3 of the maximum force. When using a Recurved bow, the average force will be higher as compared to a Longbow without bracing. In a Compound bow, which utilizes levers and as such the drawing force increases and decreases with the drawing distance as shown in Fig. 6.5.

Fig. 6.5

Force / N 350 0

0.0 0.6 drawing distance x / m

handle

arrow shaft

string

moment of release

moment of clearing the handle of bow

½ cycle of vibration of the arrow shaft

1 cycle of vibration of the arrow shaft

15

(a) Explain what is meant by the term lateral vibrations.

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

[1] Solution:

Periodic to and fro motion of a segment of the arrow from its position before release in a direction perpendicular to the limb of the bow/ left-right.

(b) Use the graph in Fig. 6.3 to calculate the energy stored in the Longbow when the maximum pulling

force of 350 N is exerted on the string.

energy stored = …………………………………J

[2] Energy stored

= area under the force-displacement graph = ½ (350)(0.6) = 105 J

(c) On Fig 6.6, sketch a graph to show how the work done on the Longbow changes with the drawing

distance of the string. Label the work done axis clearly.

[2]

Fig. 6.6

L2

Work done on the bow / J


16

Curve line from 0 J to 105 J when X changes from 0 to 0.6 m Zero gradient at x = 0 and increasing gradient

(d) It is thought that the efficiency of the bow obeys a relation of the form

η = m / (m + k) where m is the mass of the arrow and k is a constant depending on the mass of the bow. A student performed an experiment to investigate how η, the efficiency of the bow varies with m, the mass of the arrow. He obtained data which allows him to plot the graph of 1/η against 1/m as shown in Fig. 6.7.

Work done on the bow / J 105


17

Fig. 6.7

(i) Draw the line of best fit for the points.

[1]

0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 (1/m) / g-1

1/η 2.0

1.6 1.2

0.8

0.4

0.0

18

Fig. 6.7 (ii) Explain whether the relation in (d) is valid using the line drawn in Fig. 6.7.

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

[3] From η = m / (m + k)

1/η = (m + k)/ m = 1 + k/m If the graph of 1/η against 1/m is a straight line with y-intercept equal to 1 Since the graph is a straight line with a y-int equal to 1, the relation is valid.

(iii) Use the line drawn in Fig. 6.7 to determine the magnitude of the constant k in the expression

in (d).

k = ………………………….

[2]

0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 (1/m) / g-1

1/η 2.0

1.6 1.2

0.8

0.4

0.0

19

k = gradient of the line = 1.86 – 1.0 / 0.030 – 0 = 29 g Calculate gradient correctly, read coordinate to ½ smallest grid

(iv) An arrow of mass 70 g is being shot from an initially unstressed Longbow drawn back as

shown in Fig. 6.2. Use the graph in Fig. 6.7 and the definition of efficiency to determine the speed of the arrow leaving the bow.

speed of the arrow = …………………..m s-1

[3] Let m = 70 g,

1/m = 1/70 = 0.0143 From the graph 1/ η= 1.42 η = 0.704 η = KE/Energy stored = ½ m v2 / E 0.704 = ½ (0.070) v2 / 105 v = 46 m s-1

(e) Calculate the frequency of vibration for an arrow leaving the string at 50 m s-1, from a bow of

bracing distance of 0.15 m, and shot by an archer with a draw distance 0.6 m, as shown in Fig. 6.2.

frequency of vibration = …………………………..s-1

[2]

time to clear handle = (0.15 +0.60)/ 50 = 0.015 s 1 ¼ T = 0.0150 s T = 0.012 s f = 1/T = 1/0.0120 = 83 Hz

20

(f) (i) On Fig. 6.3, sketch the force-drawing distance graph for a Recurved bow which is already braced to 150 N. The maximum drawing force of 350 N is exerted on the string at the maximum drawing distance of 0.6 m.

[1]

Straight line from F = 150 N to F = 350 N (ii) State and explain, in terms of the energy stored, why Recurved bow is better than of that a

Longbow.

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

[1] Energy stored = ½ (350 + 150) x 0.6 = 150 N

This is greater than that of the Longbow (105 J).

(g) Refer to the force-drawing distance graph of a Compound bow as shown in Fig. 6.5.

Suggest an advantage of a compound bow as compared to the Longbow.

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

[1] As compared to the fully drawn Longbow at same max drawing force (350 N) at the drawing

distance of 0.6 m. 1. It can store more energy for a fully drawn bow. 2. It allows the archer to hold and aim a fully drawn bow without as much strain or fatique

since the drawing force at the fully drawn distance is smaller

[Turn over

21

Section B

Answer one question from this section.

7 A steel nut of mass 200 g on the top of a hemispherical dome becomes loose and moves down the dome from A to B as shown in Fig 7.1 with a negligible initial speed. The surface of the dome is smooth. At point B, the nut loses contact with the surface of the dome. From B to C, it moves in air under the gravitational force only. It hits the ground at point C at a horizontal distance x from B. The radius of the dome is 30 m

Fig 7.1

(a) Consider the motion from A to B,

(i) using energy considerations, explain why the speed of the steel nut increases.

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

[2]

Loss in gravitational potential energy becomes gain in kinetic energy, kinetic energy = ½ x mass x square of speed therefore its speed increases

(ii) explain why the centripetal acceleration of the nut increases.

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

[2] Centripetal acceleration is directly proportional to the square of the velocity and inversely

proportional to the radius of the circular path, which in this case is the radius of the dome which remains constant. Therefore, the centripetal acceleration of the wrench increases as it slides from A to B.

D

` `

30 m

10 m

22

(b) The nut loses contact with the surface of the dome at point B, which is 10 m vertically below the top of the dome.

(i) State the resultant force acting on the nut at B.

resultant force = …….……………N [1] Resultant force = weight = 0.200 x 9.81 = 1.962 = 2.0 N (ii) Show that the velocity is 14 m s-1 directed at an angle of 48.2o below the horizontal.

[3]

Gain in kinetic energy = loss in gravitational potential energy

= ℎ = (10.0)(9.81).2 = 14.0 cos = = = 48.2°

(c) For motion from B to C, assuming air resistance is negligible,

(i)

explain why the motion of the steel nut is parabolic.

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

There is a constant vertical force equal to the weight The horizontal force is zero

(ii) calculate the time taken for the steel nut to fall from B to C,

time taken = …………….……..s [2] = + ( . ) . = . ( . )) + ( . ) . = . + .

= .

23

(iii) calculate x, the horizontal distance between B and C.

x = …………….…..m [2] = + = ( ) = . ( . )

= .

(d) Fig. 7.2 shows the variation of the vertical component acceleration of the steel nut ay with time

from point A to point B.

Fig. 7.2

(i) Sketch the variation of vertical component acceleration ay of the plate with time from point B to C.

ay / m s-2

t / s time at B

time at A

time at C

24 L1

Fig. 7.2

[1] (ii) Describe qualitatively the variation of the vertical component of the velocity with time

from A to C.

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

[2]

A to B: it increases at an increasing rate B to C: it increases at a constant rate

(iii) On Fig.7.3, sketch the variation of the vertical component of the velocity vy of the nut with time from point A to point C.

Fig. 7.3

[2]

ay / m s-2

t / s time at B

time at A

time at C

vy / m s-2

t / s time at B

time at A

time at C

25

Fig. 7.3

A to B: curve up showing increasing velocity with increasing gradient from zero to maximum B to C: straight line showing increasing velocity with constant gradient

(iv) Explain how the radius of the dome may be determined from the graph you sketched in (d)(iii).

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

[2]

Area under graph from A to C = vertical distance travelled = radius of the dome

vy / m s-2

t / s time at B

time at A

time at C

26 8 (a) (i) Define electrical resistance of a component.

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

[1]

Ratio of potential difference to resistance (ii) Explain why it is difficult to quote a value for the resistance of a filament lamp.

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

[1]

The resistance changes with the temperature which depends on the current flowing in the lamp.

(b) The circuit of Fig. 8.1 consists of a 12 V battery and a potential divider is used to produce the I-V characteristic of the lamp. A voltmeter and ammeter are connected to measure the current in the lamp and the potential difference across the lamp.

Fig. 8.1

The characteristic is shown in Fig.8.2.

[Turn over

AA

AV

27

Fig. 8.2

(i) For the lamp operating under normal conditions with a potential difference of 10.0 V

across it, calculate the resistance of the lamp


[2]

read coordinates to ½ of the smallest grid R = 10//2.95 = 3.4 Ω

(ii) Calculate the resistance of the lamp when the potential difference across it is only 2.0

V.


[1]

R = 2.0 / 1.4 = 1.43 = 1.4 Ω

0 2 4 6 8 10 12 V / V

I/A 3

2

1

0

28 (iii) Explain in microscopic terms why the resistance in b(ii) is less than the resistance in

(b)(i).

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

[3]

As V increases, power dissipated in the lamp is increased and the temperature of the lamp gets higher. Higher collision frequency of the drifting electron with the atoms which vibrate with greater amplitude. Hence the drift velocity is lower and the current is smaller. Hence the resistance increases.

(iv) With the connection of the meters, explain whether the resistance of the lamp obtained

from Fig. 8.2 is an under-estimation or over-estimation of the expected resistance of the lamp if the voltmeter and the ammeter are not ideal.

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

[2]

The ammeter measures a current higher than the current through the lamp since some of the current in the ammeter flow though the voltmeter. resistance = potential difference/ current, the calculated value is lower than the actual one an under-estimation.

(c) (i) Show, by considering energy conversion, that V the terminal potential difference of a

cell of e.m.f. E and internal resistance r is given by V = E – I r where I is the current flowing in the cell.

[2] Energy supplied by the cell = energy dissipated in the external and internal

resistance Energy supplied by the cell per unit charge = energy dissipated in the external and internal e.m.f. of cell per unit charge = p.d across the external circuit + p.d across the internal resistance cell E = V + Vr E = V + Ir V = E – Ir

29

(ii) Fig. 8.3 shows a battery with e.m.f. E and an internal resistance r connected to a uniform

nichrome resistance wire MN. J is a movable contact J which can slide along wire MN. The positive terminal of the battery is earthed. The voltmeter and the ammeter are taken to be ideal in this question.

Fig. 8.3

The voltmeter readings V and ammeter readings I obtained for different lengths of JN are used to plot the graph in Fig. 8.4.

Fig. 8.4

1. Deduce from Fig. 8.4 the e.m.f. E and the internal resistance r of the cell.

E = ……………….. V

r = ……………….. Ω [4]

30

V = E – Ir Graph of V against I From graph, gradient = 0.0 - 1.6/ 0.400 – 0.00 = -4 Gradient = - r; -r = -4 r = 4 Ω y-intercept = E y-intercept = 1.6 E = 1.6 V

2. J is placed at the position such that maximum power is delivered from the cell to the

wire JN. Determine the potential at J and N.

potential at J = …..………Vpotential at N = …..………V

[3] Solution:

VJ = 0 V The emf = 1.6 V For maximum power to be delivered to NJ, resistance of JN = internal resistance. p.d.across JN = p.d. across internal resistance = ½ e.m.f = 1.6 / 2 = 0.8 V since J is at 0 V and N is 0.8 V lower than that J VN = -0.8 V

31

3. On Fig.8.5 sketch the graph to show how electric potential varies with position. Label the vertical axis clearly.

Fig. 8.3 [1]

== END OF PAPER ==

J N distance from J


position along JN

J N

-0.8

distance from J


position along JN

32

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