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Introduction to

1- D 2. B 3. C 4. D 5. C

6. D 7. A B. C 9. C 10. A

11- C 12. C 13. C 14. B 15. B

16. A 17. D lB. B 19. D 20. B

21. A 22. C

*"SIIUIHIDiWiliii!t section A

1. (a) (i)

Electric charge

Energy

Force

Unit

Coulomb, C

Joule, J

Newton, N

(ii) They are all derived quantities

(b) 108 MHz = 1.08 x 10' Hz

(c) Area = 12 cm x 15 cm = 12 x 10-' x 15 x 10-2

= 180 X 10-4 m2

= 1.80 x 10-' m2

2. (a) Vectors: Physical quantities which have magnitude and direction. Scalars: Physical quantities which have magnitude only.

(b)Vec:torSscat1r8 '

Acceleration Speed

Momentum Temperature

3. (a) Sensitivity of an instrument is its ability to detect small changes in the value of the quantity being measured.

(b) Mass (c) Triple beam balance: 0.1 g

Compression balance: 100 g

(d) (i) Triple beam balance (ii) This is because the smallest

division in the scale of the triple beam balance is smaller than the smallest division in the compression balance.

(9) (I) Zero error (ii) It will reduce the accuracy of

the reading.

4. (a) (I) External diameter (ii) Internal diameter

(b) Diagram 3.1: 2.03 cm Diagram 3.2: 1.70 cm

(c) Thickness of the glass = 2.03 - 1.70

= 0.165 cm 2

(d) • The zero error has to be subtracted from both the extemal and internal diameter.

• In determining the thickness of the glass, the intemal diameter is subtracted from the external diameter, thus the zero error will cancel out.

• Hence, it is not necessary to include it in the determination of the thickness of the glass.

Section B

5. (a) Zero error is a systematic error caused by a defect in the measuring instrument. It results in the reading not being exactly zero when the instrument is not measuring anything. [1]

(b) Diagram 4.1: Position of pointer on the scale of balance P = -20 g Diagram 4.1 : Position of pointer on the scale of balance Q = +20g [1] Diagram 4.2 : Mass of X= 240 g Diagram 4.2 : Mass of X = 280 g [1] For balance P, the position of the pointer when no load is placed, that is the zero error, is negative 20 g (-20 g) and the mass of X measured is smaller. [1] For balance Q, the position of the pointer when no load is placed, that is the zero error, is positive 20 g (+20 g) and the mass of X measured is bigger. [1] Hence, when the zero error is negative, the reading becomes smaller than the true value and vice versa. m

..!. (c) The true value of the mass of X =

Reading of balance - zero error. Based on Diagram 4.1, true mass of X =240 - (-20) =260 g [1] Based on Diagram 4.2, true mass of X= 280 - (+20) = 260 g [1) Therefore, the true mass of X is 260 g .!!l.

3

(d) (i) Errors: • Every measurement of

a physical quantity is an estimate.

• It is not possible to obtain the exact value of any physical quantity through measurement due to unavoidable disturbances

or external factors, defects in instruments and also errors made by the observers.

• The uncertainties of measurement caused by these factors are known as errors of measurement. [2]

(ii) causes of random error:

• Random error refer to the inconsistency of the readings obtained when a measurement is repeated several times. The causes of random errors are as follow: - the experimenter being

inconsistent while operating the instrument (personal errors). E.g., a person's reaction time while starting or stopping a stopwatch during a 100m relay.

- the quantity measured is not uniform. E.g., irregular wire will result in irregular measurement.

- disturbances due to inconsistent external factors such as temperature and pressure. [2]

(iii) Causes of systematic errors: • Systematic error refers to

the reading obtained in a measurement being shifted away from the true value.

• It can happen if - there is a defect in the

measuring instrument. E.g., wrong calibration of a voltmeter.

- the procedure used in the measurement is not appropriate. E.g., length of a pendulum should be measured when it is hung on the retort stand as the weight of the pendulum causes extension on the string.

[2]

(iv) Effects of the errors on the measurement:

• Random errors: - The readings are

not consistent when repeated. The readings are spread out statistically about a mean value.

© PenerlJitan Pelangi Sdn. 8hd.

Form 4 AnswslB

- This results in an uncertainty in the value measured and reduces the precision of the measurement [lJ

• Systematic errors: - The values obtained will

be either more or less than the true value by a certain amount.

- This will reduce the accuracy of the measurement. [1]

(v) Steps ttIat can be taken to reduce the effect of these errors on the measurements:

• Random errors: - Measurement is

repeated several times and an average value is taken.

- Statistical methods can

be used to estimate the amount of uncertainty and this can be expressed as a percentage error.

• Systematic error:

- Instruments must be checked carefully for defects from time to time and are immediately rectified if it exists.

- The procedure used must also be critically evaluated and improved if a source of systematic error is suspected.

- In certain cases, such as zero error, the amount of the systematic error can be estimated and correction can be made on the reading obtained by subtracting it from the reading obtained. [2J

10

Section C

6. (a) Random errors I systematic errors [1]

(b) The effects of random errors can be reduced by:

- repeating the measurement several times and taking an average, [1 J using statistical methods to determine the amount of uncertainly in the reading. [1]

The effects of systematic errors can be reduced by

- checking for defects in instruments, such as zero error, possibility of parallax error and possibility of wrong calibration, [1]

- rectifying the defects, [1 J - taking into account the zero

error by SUbtracting the zero error from the final reading,

[1]

- improving the measurement procedure. [1]

(Anyone)

Maximum 4

(c) (i) • The features of ammeter are: - zero error. which causes

systematic error, [1] - mirror strip, which helps

to reduce parallax error. [11

- range. which must be within the range of 0.3 A to 0.8 A, [1]

- size of the smallest division. which must be as small as possible. [1]

J1l 5

'f' ~itif~mmetei'

AmlIIeter Zero MInr!r" i ::.~.error ~ .....

W Yes Yes Oto 5 A 0.1 A

X No Yes Oto 1 A 0.01 A -­

y No No oto 1 A 0.D1 A

Z No Yes oto

0.005 A0.5 A

(ii) • W has a zero error and it gives readings to only one decimal place. [1]

• Y has no mirror so it can cause parallax errors. [1]

• Z can give readings to 0.005 A, that is three decimal places. However, its range does not cover the required 0.3 A to 0.8 A. [lJ

• X has no zero error and it has a mirror strip to reduce parallax error. It also has a range that covers the required range of 0.3 A to 0.8 A. The smallest division is 0.01 A. so it gives readings in ampere with two decimal places. [1]

• Therefore, X is the most suitable ammeter. l!l

5

(d) • The micrometer screw gauge can take measurements correct to 0.01 mm. Which is 0.001 cm. [1]

• However. the vernier calipers can only give readings correct to 0.01 cm. [1]

• To obtain a reading for the thickness of the paper correct to 0.001 cm with the vernier

calipers, 100 sheets of paper are stacked together and their total thickness is measured. [1]

• For example. if 100 sheets have a total thickness of 1.82 cm (correct to 0.01 cm), the average thickness of one sheet of paper

1.82 cm is 1"0() = 0.0182 cm [1 J This reading is correct to 0.0001 cm and is comparable to or better than what a micrometer screw gauge can give. l!l

5

~ Section A

(b)

1. (a) (i)

(ii)

(iii)

Mass of load, m

Period, T

Force constant of spring

[lJ [1]

[1 J

I

(c)

1.2

1.3

1.1

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

o 50 100 150 200 250 300

[5J (d) F is directly proportional to the

mass of the load, m. [lJ

Section B

2. (a) The lengths of the two trapezes are equal and both the man with the bigger mass and the boy with the smaller mass complete the swing in the same time. [1]

(b) The period of a pendulum is independent of the mass of the bob. [lJ

[7]

iI'f ~t ·t/s T/s P/ s'

50 4.4 0.44 0.19

100 6.2 0.62 0.38 e--­

150 7.6 0.76 0.58

200 8.8 0.88 0.77

250 9.8 0.98 0.96

300 10.8 1.08 1.17

,:.;;.

~h',:,',"'. ,';;,

,...•~ .J ::

© Penaroitan Palangi Sdn. Bhd.

--

--

s of paper If and tl'leir .-.ad. [1]

l sheets have 1.82 cm

I. the average -' of paper

12 em [1]

mel to 0.0001 .. toar miCrometer -;we. 1!l

~

" [1] [1]

of spring [1l

ti­ 'PI" [).44 0.19

0.62 0.38

0.76 0.58

0.88 0.77

0.98 0.96

1.08 1.17

[7]

I 250 300

[5]

rtional to the ,m. [1]

e two trapezes are lie man with the I lie boy with the lft1IIele the swing

[1 1 lJlIfIlIuIum is tie mass of the

[1 ]

(c) (i) Aim: To investigate the variation of the period of a pendulum with the mass of the bob [1]

(ii) Varlebles: Manipulated variable: Mass of bob

Responding variable: Period of bob

Constant variable: Length of pendulum [1]

(iii) List of apparstull:

2 pendulum bobs of different mass, stopwatch, two small pieces of plywood, retort stand and clamp, half metre ruler.

List of material:

String [1] (iv) Arrangement of apparatus:

Plywood

Bob

[1 ]

[1 ]

(vii) Analysis of data: From the data, it is seen that the period for the pendulum is the same for both pendulum bobs. Therefore,

the period is independent of the mass of the bob. .1!l

10

....iiIIff!t11f•• Objective Questions

1. C 2. A 3. D 4. C 5. D 6. D 7. 8 8. D 9. 8

Written Practical

Section A

1. (a) (i) Length of the pendulum [1l (ii) Period of oscillation [1 ] (iii) Mass of the bob [1]

(b) (i) Diagram 1.2 45.50

Diagram 1.3 52.70 Diagram 1.4 58.90 Diagram 1.5 64.60

Diagram 1.6 69.80 [1 ]

(ii) Diagram 1.2 4.55

Diagram 1.3 5.27 Diagram 1.4 5.89 Diagram 1.5 6.46 Diagram 1.6 6.98 [1]

(iii) Diagram 1.2 20.70 Diagram 1.3 27.77 Diagram 1.4 34.69 Diagram 1.5 41.73 Diagram 1.6 48.72 [1]

(c)

30.0 4.55 20.70

40.0 5.27 27.77

50.0 5.89 34.69

60.0 6.46 41.73

70.0 6.98 48.72 [2]

(d)

[5]

(e) T' is directly proportional to I. [1]

..E.~r~

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.. Ill.

~l '. HHE

~ Forces and Motionf~"

1. A 2. C 3. 8 4. C 5. D

6. 8 7. C 8. A 9.8 10. 8

11. D 12. D 13. A 14. D 15. A

16. B 17. A 18. D 19. C 20. C 21. D 22. A 23. D 24. 8 25. 8 26. C 27. A 28. C 29. D

Section A

1. (a)

20 +-4 --1---+--+ --t'·_·· .-i- .

K 18+---+---''k''''-+-!i-+--+----i----i 16 +--+-+--'i<c--+----+---:---~

14+---+-+-+-i"'\~1T-+_-+----i 12 +-+---ji~.+-+,i\l-+l ---\---1

! \ 10+--+-+--+-Hr+-:--~

8 jI, \1 6 ~. I~+---t\--\'j------1T' ! 1\ 4 +"--+-r,--+---i-~ ! \!Ii

2+---+----:--+--+---+-1-\~~

+--+--i--+-I-+---1i---l~Time I s 0.5 1.0 1.5 2.0 2.5 3.0

(b) Because the driver took 0.5 second before he stepped on the brakes.

(c) (i) Refer to graph in 1(a).

(ii) At point T, the gradient of the graph becomes steeper. This shows that the deceleration became higher because the driver stepped on the brakes harder.

(d) Distance travelled by the car during the 3.0 seconds shown is equal to the area under the graph.

Area under the graph

20 + 14 = (20 x 0.5) + [(-2-) x (2.0 ­

0.5)] + [(i- x 14 x (3.0 - 2.0)]

= 10 + 25.5 + 7 = 42.5 m As he moves a distance of 42.5 m in 3.0 second, he managed to stop at a distance of 2.5 m before reaching the junction.

2. (a) (i) Unifonn motion

(ii) This is because the hand straps are hanging vertically.

(b) (i) Acceleration in the forward direction

(ii) This is because the hand straps are inclined backwards.

© Penerbitan Pelangi Sdn. Bhd.

~

-

-

_

~ Pby8iCII Form 4 Answers

(c) Inertia

(d) Inertial seat belt

3. (a) Momentum of an object is the product of its mass aM its velocity, p ~ mv.

(b) The velocity of the tennis ball is 50 times more than the velocity of the ship, but the mass of the tennis ball is many times smaller than the mass of the ship. Hence, momentum of the ship is bigger.

(c) Momentum ~ Mass x velocity ~ 50 x 10" kg x 50 m S-1

~ 2.5 kg m So,

(d) (i) Impulse Change in momentum

~ Final momentum - Initial momentum

~ 2.5 - 0 ~ 2.5 kg m s-'

(ii) Average force

_ Change in momentum - Time

~ ~ ~ 125 N0.2 .

(e) Newton's second law of motion states that the rate of change of momentum is directly proportional to the force acting on a body and takes place in the direction of the net force.

4. (a) 10cm

(b) Original length of the spring

(c) Hooke's law states that the extension of spring is directly proportional to the force applied to It if the elastic limit is not exceeded.

(d) (i) It obeys Hooke's law.

(ii) Because the graph is a straight line where the extension is directly proportional to the force

(e) (i) When force ~ 20 N, X ~ 20 cm. Extension, x ~ X - original length ~ 20 - 10 ; 10 cm ~ 10 x 10-2 m

(ii) Force constant, k F x

20 = 10 x 10"

o

~ 200 N m'

Section B

5. (a) (i) If the tape chart has a steeper positive gradient, the acceleration is higher. [1]

(ii) • The force, that is the weight of the load of mass 100 g hanging vertically, is the same for both diagrams. [1]

• The total mass is bigger in Diagram 5.2 than in Diagram 5.1 [1]

© Panero/tan Pa/angi Sdn. Bhd.

• However, the acceleration is higher in Diagram 5.1 than in Diagram 5.2. [1]

• For the same amount of force, the bigger the mass, the smaller the acceleration. [2]

5

(iii) Newton's second law of motion [1]

(b) • The bigger the mass, the larger the weight. [1]

• Based on Newton's second law of motion, F ~ rna, where F is the force, m is the mass and a is the acceleration, the weight of an object is the gravitational force acting on it. Hence, weight, W ~ mg. [1]

• Since g is the same for all objects, weight is proportional to mass. l!l.

.L (c) (i) Engine capacity:

• The engine capacity must be large so that a large force can be produced to result in a large acceleration of the car. [1]

• A large acceleration is necessary for the racing car to overtake other cars at increasing velocity during a race. [1]

(ii) Mass of car: • The mass of car must

be small so that a large acceleration can be obtained based on Newton's second law, F; rna. [1 ]

(iii) Shape of car:

·

• The shape of car must be aerodynamically designed so that air resistance can be minimised. [1 ] If air resistance is too large, the net force acting on the car will become small and thus it is not possible to obtain high acceleration. [1]

• The final cruising speed of the car will not be very high if air resistance balances out the force of the engine.

[1] • The car must also be

designed to have a low centre of gravity so that the car is stable while travelling at high velocity.

[1]

·(iv) Braking system:

The tyres of the car must have wide base area so as to obtain good grip on the ground and to avoid from skidding. [1 ]

• Tyres must have appropriate th read if the car is driven on

(b4)

wet road as water can be drained away between the tyre and the ground. [1 J

• An anti-locking braking system is also necessary to avoid from skidding during braking. This can also help in controlling the car while taking comers. l!l

10

Section C

6. (a) (i) Elasticity is a property of an object that enables the object to return to its original size and shape after the force acting on the object is removed. [1]

(ii) • When a force is applied to a spring, the spring will extend. [1]

• The spring obeys Hooke's Law. [1]

• According to this law, the extension of the spring is proportional to the force applied provided the proportional and elastic limits of the spring are not exceeded. m

4

(b) Initial length:

At full load, the total length of spring must not exceed 20.0 cm.

Total length ~ Initial length ­extension.

To check for this condition, initial length of spring must be taken into account. [2]

Force constant:

Force constant is needed for calCUlation of extension of spring when the spring is loaded to 10.0 N, which is the maximum load. It can be used to determine whether the total length of spring has exceeded 20 cm when fully loaded. [1J

Force constant is also needed to determine the weight that will cause the extension to increase by the smallest division of the rufer, which is 1 mm. Hence, it can be used to determine whether the spring satisfies the condition where the smallest division of the weight scale must be 0.1 N or less. [1]

Maximum extension before elastic limit is exceeded: • The spring must obey Hooke's

law within the range of weight to be measured, that is 0 to 10 N.

[1] • The elastic limit must be known

so that this limit will not be exceeded. [1]

(c) (iJ

k. : =;;;:. ­

;":;~" mr ' 0

!~:. -­~ ­

s water can be rry between the I ground. [1]

ICking braking Iso n8CElssary to Skilllling during

is can also help III the car while M>;. J!l

10

property of .enables the ~ to its original Ie after the n the object is

[1]

a! is applied , the spring will

[11 Obeys Hooke's

[1]

10 this law, the of the spring is :II to the force provided the BI and elastic e spring are not

.0 4

lDIal length of exceed

itiaI length ­

s rondition, spring must be no [2]

is needed • extension Ihe spring is ". which is the " can be used IIllher the total has exceeded

IJ Io8.ded. [1]

is also needed II weight that will 1!iion to increase I division of the

1 mm. Hence, d to determine mg satisfies the Ie the smallest IUight scale musl s. [1]

ion before 1Ceeded: It Obey Hooke's Mlge of weight to ~t isO to 10 N.

[1 J I must be known I will not be

[1)

Determination of the best choice:

• Spring X: - When fUlly loaded (with 10.0

N), the extension,

=!.-= ~ =20cm k 0.5

- Hence, the total length of spring when fUlly loaded is 10 + 20 = 30 em, which exceeds the maximum length allowed.

- Spring X is not suitable. [1]

• Spring Y: - The smallest division of the

ruler is 1 mm or 0.1 em. The weight that can cause this extension is

F = lex = 4 xO.l

=OAN - Hence, the smallest division

on the weight scale is 0.4 N, which exceeds 0.1 N.

- Spring Y is also not suitable. [1]

Spring Z: - The maximum extension

before elastic limit is . F

exceeded IS k = 10 em

Therefore, the maximum length of spring is 10 + 10 = 20 em, which fulfils the condition that the balance must not be longer than 20.0 em.

The force that can cause an extension of 1 mm is

F = lex = 1 x 0.1

= 0.1 N

which also fulfils the condition that the smallest division on the weight scale required is 0.1 N.

At full load, the extension is 10 em, which is less than

the elastic limit of 12 em. Therefore, the elastic limit is not exceeded.

Spring Z is the most suitable. .0

10

(c) (i) F= lex 0.5 = k x (10.0 - 8.0) x 10"

k= 0.5 (10.0 - RO) x 10"

k = 25 Nm-1 [2]

(ii) Elastic potential energy, 1

P.E. = 72FX

= 721 x 0.5 x (2.0 x 10-')

= 5 x 10-3 J [3)

PbylIlcs Form 4 ~ II

The intercept of the v1 -axiS

..IS 2

m-1 s.

Therefore,

v= ------'--1--. ­Intercept on V -axIs

V = '21 =0.5 m s' [2]

(b) 2. increases linearly with m [1]v

6-2 (c) (i) Gradient, k = 1.0 _ 0 = 4 [3]

1 1 (ii) M = kV = 4 x 0.5 = 0.5 kg

[2]

(d) If the final common velocity, v is 0.2 m s·"

2. = ..2.. = 5 m-'s. v 0.2

From the graph, when v1 = 5,

m = 0.75 kg [3]

(e) Ensure that there is very little friction on the wheels of the trolleys. Use smooth bearings on the wheels of the trolleys. [1]

Section B

3. (a) When the slope is steeper, is needed to inclined the plane. [1]

(b) When an object is placed on an inclined plane, the component of its weight along the slope is bigger when the angle of inclination is bigger. [1]

(c) (i) Aim: To investigate the

relationship between the

componenlof the weight of an object along an inclined plane and the angle of inclination of the plane when the object is placed on the inclined plane.

[1]

(ii) Variables:

Manipulated variable: Angle of inclination of the plane

Responding variable: Component of the weight down the plane

Constant variable: Mass of the trolley [1]

(iii) List of apparatus: Inclined plane, blocks of wood, trolley, smooth pulley with clamp, slotted weights, protractor.

List of materials:

String [1]

(iv) Arrangement of apparatus: [1]

Trolley

Hook

Slotted weightsileJ:r:=- Jl:i:I:::::::= Supporting .Ii bfocks 01 wood

© Penerbitan Pelangi Sdn. Bhd.

: -

-'

t' is directly proportional to the distance, d, travelled by the trolley. [1]

m'~g

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-~ ~

~,

Section A

1. (a) (i) (ii)

(iii)

(b)

(c)

Distance, d [1] Time, t [1] Angle of inclination of the inclined plane [1]

.i clint tfs t" f Sl':

0.20 lA 2.0

OAO 2.0 4.0

0.60 2A 5.8

0.80 2.8 7.8

1.00 3.2 10.2

1.20 3.5 12.3

[7]

0.2 0.4 0.6

[5]

(d)

2. (a)

l.'m-1 s V

--

~1m""'" Form 4 AnSWElIS

(v) Procedure: 1. The apparatus is set-up

as shown in above diagram.

2. The angle of inclination, 8, is adjusted to 1(y> by shifting the supporting blocks of wood. The angle, 8 is measured with a protractor and is recorded.

3. Slotted weights are added to the hoOk until the trolley remains stationary when the weights are released.

4. The total weight of the slotted weights, W" is recorded.

5. Step 3 is repeated and the total weight is recorded as W,. The average of W and W, is taken. '

6. Steps 2 to 5 are repeated for 8", 2(Y>, 30", 40", 5(y>

and 60°. [4]

(vi) Tabulation of data:

10

20

30

40

50

60

[1]

(vii) Analysl8 of data: The component of the weight of the trolley along the slope is equal to the total weight oIlhe slotted weights when equilibrium is reached. A graph 01 Wagainst 8 is plotted.

WIN

o 10 20 30 40 50 60 IJ

It can be seen that as the angle, 8 increases, the component of the weight of the trolley along the plane increases. 1!l

1!!.

C PenerbitBn Pelangi Sdn. Bhd.

Objective Que8110n8

1. A 2. C 3. B 4. B 5. D 6. A 7. B 8. A 9. D 10. D

11. D 12. C 13. C 14. A 15. A 16. C 17. B 18. D 19. C 20. C 21. B

Subjective Queslion~

Section A

1. (a) Work done by anOllject is the product of the applied force, F on the object and its displacement, S in the directon of the net force.

(b) (i) Work clone

'" Force x displacement ",540x3

'" 1620 J

(ii) Work done

------"'Weight x height ",mgh ",50x10x3",15OOJ

(c) (i) More work is clone by the man compared to the work clone to lift the load of bricks by (1620 - 1500 "') 120 J

(ii) The man has to overcome friction in the pulley. Therefore, he has. to do more work compared to the work done to lift up the load.

2. (a) (i) Magnitude: F, '" F, and F, '" F.

Direction: F, is opposite to F, and F, is opposite to F.

(ii) Zero (iii) Forces in equilibrium

(b) (i) The car will experience an acceleration.

(ii) A situation of unbalanced forces which will produce a net force, f=; ma

3. (a) Impulsive force (b) The glass is not strong enough

to withstand the impulsive force. A short impact time acts on the glass and hence, a high impulsive force cracks the glass.

(c) F", m(v- u) t

_ 0.15(0-10) - 0.02

'" 75 N

(d) Place a thick sponge on the concrete floor.

secllon B

4. (a) (i) Elasticity is a property of matter that enables an object to retum to its original shape and size when the forces that

are acting on it are removed. [1]

(ii) Diameter of coils of the spring X < Diameter of coils of spring Y [1] Maximum height in Diagram 5.1 (a) > Maximum height in Diagram 5.2 (b) [1] The maximum height of the ball decreases as the diameter of coils of the springs increases. [1] Diameter of coils of the springs is inversely proportional to the spring constant, k. [1]

The elastic potential energy of the spring increases as the diameter of coils of the springs decreases. ill

5

(b) (i) F is bigger than Fy • [1]x

Spring X has a smaller diameter of coils of the spring than spring Y. 1!l

2

(ii) Elastic potential energy ~ Kinetic energy ~

Gravitational potential energy [2]

(c) (i) Tight and light attire - Less air friction to enable him to jump further [2]

(ii) Run up and take off. Run as fast as possible and jump high in the air. - To increase the momentum [3]

(iii) Rough surface - To allow the long jumper to have a good grip for taking off [2]

(iv) Use sandy landing pit - increase the time of stopping and to decrease the impulsive force during landing

[3]

!! Section C

5. (a) Speed is the rate at which an object is travelling. It is measured by distance travelled per unit time.

[1]

(b) • A body with mass has inertia. Hence, a force is required to change the velocity of a body which has inertia. If the mass is bigger, a bigger force will be required to cause certain acceleration. [1]

• The engine of a lorry has a certain maximum capacity to produce force. If the mass is too big, the engine may not be able to move the lorry efficiently. Therefore, for each lorry, based on its engine capacity, the mass must be limited to a certain value. [1]

:::1" .,. --~-~--

=1= ==- ~

-= __ H

l:H

(c) •

N

---

-

~

...,... Form 4 ~ III Ie removed.

[1] ;of the IIIIIr of coils

[1]

in Diagram m height in

[1]

_of IS as the of the l. [1) .01 MI8Iy .. spring

[1]

IIiII energy _ses as oils Qf the IS. l!l

5

I Fy • [1)

mailer J of the II Y. l!l

2

iIrlergy,..... Inlial energy

[2]

lire - Less bIe him to

[2) off. Run as IJId jump To increase

[3) To allow the 1l!ve a good

[2]

9 pit IIll of Iacrease the uring landing

.0 10

~an

• measured .. unit time.

[1]

has inertia. required to

, 01 a body If the mass II' force will IUS9 certain

[1]

yhas a capacity to he mass is may not be

ry efficienUy. lorry, based

11, the mass D a certain

[1)

• Momentum = Mass x velocity. If the velocity is high, it means that the momentum will also be large. [1]

• To bring a lorry with large momentum to a stop within a certain distance, a very large force will be needed. However, the braking force has a maximum limit which might not be able to stop the lorry in a reasonable time. Therefore, to avoid such situation, a lorry must have a limit on its speed.

l!l 4

(e) Number of tanks: • It is safer to have a full tank

than a half filled tank.

• This is because in a full tank, there is less space for the liquid to move about. [1) '.

If a large tank is half filled when the lorry start to move or comes to a stop, the liquid will move about violenUy inside the tank due to its inertia. This will make the lorry unstable and unsafe.[1)

If many tanks are used and the volume of liquid is transported is small, a few tanks can be filled until full so that the inertia of the liquid within the tanks will be smaller in separate tanks. This will lead to a safer way of transporting the cargo. [1]

Number of ty,...: • If fewer tyres are used, the

tyres will have to be pumped to a higher pressure in order to support the weight of the lorry and cargo.

• The probability of tyres exploding is higher if the air pressure of tyre is higher.

• However, if more tyres are used, the tyres do not have to be pumped to very high pressure. [1)

• Furthermore, having more tyres means that there will be a larger area of contact with the ground. This will enable betler grip on the road. [1)

Type of brake.: • Since the lorry will be heavy,

it will have a large momentum while moving. A large force will be needed to bring it to a stop.

[1)

• If manually operated hydraulic brakes are used, the driver will have to apply a very large force on the brake pedals to bring the lorry to a stop. [1)

• However, if air brakes are used, the driver just needs to trigger the air compressor with the brake pedals. The pressurised

air will flow into the brake tubes and press the brake pads with great force. Braking will be more effective. [1)

• Based on the above considerations, lorry C is the most suitable because it has 5 tanks, 8 tyres and an air brake system. m

..!!!. (d) (il Acceleration, a

v- u =-/­

0-36 =~

=-3ms-2

The deceleration is 3 m S-2.

[3)

(ii) Force, F =ma = 20 000 x (-3) = -60 oooN The braking force is

60 000 N. [2]

Written Practical

section A

1. (a) (i) a is direcUy proportional to h. [1]

(ii) When h = 0.5 m, a=1.5ms.... [3]

(iii) Gradient = 1.5 - 0 0.5-0

= 3 s'"

al m s-2

2.0

1.5

1.0

0.5

0.1 0.2 0.3 0.4 0.5 0.6 0.7

[3)

(b) m = {, g = m x s

g = 3 x 3.3 = 9.9 m s'" [2)

(c) a = {x h

= 9.9 x 0.15 5

= 0.30 m s'" [2)

(d) Make sure that the tape runs freely through the ticker timer. l!l

.!!

2. (a) (i) When n increases, x decreases. m

(ii) x/em

4.0

3.5

3.0

2.5

2.0

1.5

1.0

0.5

o {(0.0) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.0 1.0

When n = 4, n1 = '4

1 = 0.25

From the graph, when

.!- = 0.25, x = 1.0 em [3] n

. 4.0-0 (bl (i) Gradient = 1.0 _ 0

= 4.0 em [4]

(ii) k= 200 = 200 = 50 N nr' h .4 [1)

1 1 1 (c) k' = k + k

1 1 1 2 -=-+-=­k' 50 50 50

k' = 50 =25 N nr' [3)2

$13 4 L.:::.J1 Forces and Pressure

1. C 2. A 3. B 4. A 5. C 6. D 7. C 8. B 9. A 10. C

11. C 12. D 13. A 14. B 15. C 16. A 17. A 18. D 19. C 20. D 21. D 22. C 23. B 24. C 25. A

26. C Xl. B

SectIon A

1. (a) (i) 1 x 10S Pa

(ii) 1 x 10S Pa

(b) (i) Atmospheric pressure

(ii) Density of water and density of liquid X

(c) Pressure at point U = Atmospheric pressure - hpg = (1 x 1OS) - (12 x 10"")(1000)(10)

= 98 800 Pa

(d) Let density of liquid X be p. Atmospheric pressure - ~p

= 98 800 Pa (1 x 1OS) - (15 X 10-2) p. (10)

= 98 899

1200 p. = '1.5 = 800 kg m-3

C Penerbitan Pelangi Sdn. Bhd.

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