[2m]

[1m]

m

kg

kg

kg m S-2

r

F

m

M

By rearranging the equation, F= eM';' r

ii.

(b) i.

ii.

(c) 1.

2.

3.

(c) It is necessary to stir the liquid to ensure that the temperature of the liquid is uniform.

e=~ mM

Derived units of e = (kg m s-2)(m 2 ) =kg- l m' S-2

(kg) (kg) [1m]

One example of scalar quantity: Mass (m or M) or distance (r) [1m] Vector quantity: Force (F). [1m] Consistency is the tendency for the values of measurement to be close to each other and concentrate at a particular value. [1 m] Accuracy is how close the value of a measurement is to the actual value. [1m] The jaws are closed and the zero error is recorded. The zero error is corrected for all the readings taken. [1m] Parallax error is avoided by taking readings at a position perpendicular to the scale. [1m] Repeated measurements at different places on the ping-pong ball are made. The average value of all the readings is calculated to reduce random error.

[1m] (d) i. One example of systematic error is zero error.

[1m] One example of random error is parallax error.

[1m] ii. - The readings taken by Group A are recorded up

to one decimal place while that by Group Bare up to two decimal places. [1m] This shows that both groups are using stopwatches of different sensitivities. [1 m] Analogue stopwatches are probably used by Group A while digital stopwatches are used by Group B. [1m]

iii. - Readings obtained by Group A are less distributed and are close to each other (around 8.3 s).

[1m] Therefore, readings obtained by Group A have a hib' er consistency. [1m]

Section B 6 (a) i. r-----------r----------,

Base unit

113

5 B 10 D 15 C 20 B 25 C 30 A 35 A

4 A 9 A

14 B 19 D 24 A 29 A 34 C

3 C 8 D

13 A 18 C 23 B 28 D 33 D

2 C 7 D

12 C 17 B 22 D 27 D 32 A 37 C

f1l random error is the parallax error. -gauge. aating too much pressure on the object

doe to overtightening of the spindle. +1lD2 mm

Corrected reading =4.86 ~ (+ 0.02) mm =4.84 mm

ii. Thickness of 1 sheet of paper = 4.84 mm 50

= 0.097 mm

(b) i. 12 m sol ii. Time taken = 28 - 12 s = 16 s

~km (c) 12 m sol = 12 m = 1 000 = 12 x 3600 km

Is _1_ h 1000h 3600

= 43.2 km h- ' 2 (a) i. 1.2 cm

ii. Diameter of copper wire P = 1.2 cm = 0.060 cm or 0.60 mm 20

(b) This method can improve the sensitivity of a metre rule. One example of systematic error is the end error or zero £nOL

CHAPTER 1: Introduction to Physics PAPER 1

1 C 6 D

11 A 16 C 21 C 26 B 31 C 36 B

PAPER 2 Section A

1 (a) ~~~~~~l~~~~~~~

r apparatus,

Tangkan satu

rlc/i markah]

rk/i markah]

,ling ofeach the water in

;/10 markah]

rlr kuasa bagi daripada air

variable and 4 (a) i. Ammeter

ii. Electric current. (b) i. P: 0.8A

mboleh ubah Q: 0.84 A ii. Q is more sensitive than P.

(c) To avoid parallax error by making sure that the readings are taken from the correct eye position.

5 (a) i. Mercury ii. Increases in length and volume/Expansion occurs.

(b) i. Thermometer R.

rs/lO markah] ii. Its smallest division is smaller than that of S/It is

able to detect a smaller change in temperature.

II Section C

7 (a) i.

ii.

(b) i.

..11.

(c) i.

ii.

PAPER 3 Section A

1 (a) i. ii.

The density of water is 1 000 kg m-3 means that the mass for every 1 m3 volume of water is 1 000 kg.

[1m] The total volume of the ball bearings is equal to the volume of water displaced by the ball bearings.

[1m] Volume of 50 ball bearings =volume displaced by the ball bearings [1m] =(82 - 75) cm' =7.0 cm3

= 7.0 x 10-6 m3 70 10-6 3 [1m] Volume of 1 ball bearing = . x 50 m

= 1.4 x 10-7 m' [1m]

M f 1 ball b . - 278.2 - 223.6 g - 1 092 ass 0 earmg - 50 -. g

= 1.092 X 10-' kg [1m] Density = Mass = 1.092 x 10-3 kg

Volume 1 s = 7.8 X 103 kg m-3 [1m]

Different types of stones contain varying amount of natural minerals. [1m] These natural minerals have different molecular masses. [1 m] The presence of these minerals give rise to different masses in every 1 m of stone. [1m] Therefore, different types of stones have different values of densities. [1 m]

measuring cylinder

[2m] Each of the stones, P, Q and R are tied to a long string. [1m] A measuring cylinder is filled with some water and then placed on an electronic balance. [1m]

- The initial volume of water and the electronic balance reading are recorded in the table as VI and m respectively. [1m]

1

- Stone P is held by the string and lowered slowly into water in the measuring cylinder. [1m]

- The final volume of water and the electronic balance reading are recorded again in the table as V, and m, respectively. [1m]

- The experiment is repeated by using stones Q and R. [1m]

- For each stones, the density can be calculated by using the formula,

m, - m, p= [1m]

V, - VI

- Conclusion: Different types of stones are found to have different values of densities. [1 m]

Mass, m of the liquid. Final temperature, 8 of the liquid.

114

iii. Time, t to heat the liquid, or the power of the immersion heater.

(b) Temperature, 8 varies linearly with _1_ . m

(c) By extrapolating the graph to the e - axis, 8

0 = y-intercept = 26.2 °C

(d) i. Gradient, g = (33.0 - 27.3)OC = 11.4 °C kg(0.6 - 0.1) kg 1

ii. c = 4.6 X 10' J = 4.04 X 103 Jkg- 1 °C-l 11.4 °C kg

(e) The liquid should be stirred constantly using a glass rod or stirrer/ Temperature readings are taken only after the thermometer readings are steady.

2 (a) The length of the shadow at the flag pole depends on the angle of depreSSion of sunlight.

(b) As the angle of depreSSion of sunlight increases, the length of the shadow of the flag pole decreases, or The larger the angle of depression of sunlight, the shorter is the length of the shadow of the flag pole.

(c) i. To investigate the relationship between the length of the shadow of the flag pole and the angle of depression of sunlight.

ii. Manipulated variable : Time (Angle of depression of sunlight). Responding variable : Length of the shadow of the flag pole. Constant variable : Length of the stick.

iii. Straight stick (about 2 m in length), metre rule and clock or stopwatch.

iv. mid-noon

),11,afternoon ," " .- morning0I 1"­

(j =angle of depression

1---".;,-- stick

(j(j

length of shadow length of shadow

v. 1. The length of the straight stick is measured using a metre rule.

2. The stick is then planted open ground where there is sunlight.

3. At 8:00 a.m., the length of the shadow of the stick is measured.

4. The measurement of the length of the shadow is repeated at 9:00 a.m., 10:00 a.m., 11:00 a.m., 12:00 noon and 1:00 p.m.

5 The data are recorded in a table. vi. ... .

Time . 8:00 a.m.

9:00 a.m.

10:00 a.m.

11:00 a.m.

12:00 noon

1:00 p.m.

vii. I

'I ()

d 'I

CHAPTER. PAPER 1

1 B 6 B

11 B 16 B 21 C 26 D 31 A 36 B 41 C 46 D 51 D

PAPER 2 Section A

1 (a)

(b)

The' 2J

10';

(c) a =­

2 (a)

(b)

(c)

a =­

The I The certl VeIa. =5C)I

Veq =5. Aver;.

a=­

3 (a) (b) (c) (d)

=5 Elasti Elasl M~ i. E

""

)Ower of the

~I

ng a glass rod

ty after the

t depends on

increases, the reases, or sunlight, the

: flag pole. en the length · the angle of

of depression

shadow of the

k.. Detre rule and

oming

•·low

neasured using

ground where

shadow of the

of the shadow m., 11:00 a.m.,

vii. The length of the shadow against time is plotted as shown below.

Length of shadow/m

8:00 12:00 1:00 Time/h

The above graph shows that the length of the shadow of the stick (flag pole) decreases as the angle of depression of sunlight increases. Therefore, the stated hypothesis is supported.

CHAPTER 2: Forces and Motion PAPER 1

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

11 B 12 C 13 C 14 D 15 B 16 B 17 A 18 D 19 B 20 D 21 C 22 D 23 A 24 A 25 B 26 D 27 A 28 C 29 B 30 D 31 A 32 D 33 C 34 D 35 A 36 B 37 B 38 B 39 C 40 B 41 C 42 C 43 D 44 C 45 C 46 D 47 B 48 A 49 D 50 D 51 D 52 D 53 D

PAPER 2 Section A

(a) The velodty increases uniformly.

(b) 2.0 cm _ 10 -1

10 x 0.02 s - cm s

(c) a = 35 - 10 =25 cm S-2 1.0

2 (a) The speed of the parachutist increases. The acceleration of the parachutist decreases from certain positive value to zero.

(b) Velocity before parachute opens = 50 m sol Velocity after parachute opens = 5 m S-l

(c) Average acceleration, a = v - u

t a = (38 - 18) m sol

v = 5.0 m S-2

3 (a) Elastic potential energy (b) Elastic potential energy ~ kinetic energy (c) More distance, (less resistance) (d) i. Elastic potential energy

=..!.Px 2

=+x 8 x 0.2 =0.8 J

ii. Kinetic energy =+mY'

= Elastic potential energy

115

1

:. "2 m~ : ~::4 m S-I

Elastic collision 4 m S-l

Maximum height = mgh 2m x 10 x h ="21

mv

h=..!.N 2 10

= 0.8 m No change/maximum height is the same, because the maximum height does not depend on the mass.

5 (a) The principle of the conservation of momentum. (b) Momentum of the boy (towards the direction of the

jetty) is the same as the momentum of the boat (away from the jetty)

(c) m,v, =m2v

2

48 x 2 = 125v 2

V = 48x2 2 125

= 0.77 m sol (d) i. Thrust, drag, lifting force, weight.

ii. Thrust> drag; Because: There is a change in velocity horizontally. Weight = lifting force Because: Velocity remains zero vertically.

6 (a) Horizontal component =5 cos 50· =3.2 N (b) Frictional force

=horizontal component of the force = 3.2 N

(c) i. Balance of the force = 10 N - 3.2 N =6.8 N Acceleration of the block, a

=~ = 34 m S-22 .

ii. Reduce the frictional force by placing grease between the wooden block and the surface or reduce the mass.

(d) Resultant force =J52 + 122 =13 N 7 (a) He had to increase his speed so that he can obtain

maximum kinetic energy before his jump. (b) The pole was bent so that it stored elastic potential

energy that can be converted to gravitational potential energy of Chong Wei.

(c) Maximum height = 5.2 + 0.3 m = 5.5 m

Maximum potential energy = mgh = (58 kg)(lO N kg-')(5.5 m) = 3 190 J or 3.19 kJ

(d) Acceleration due to gravity, g= 10 m S-2

(e) The thick rubber mattress prolonged the time of impact of Chong Wei. Impulsive force is inversely proportional to the time of impact. Therefore the impulsive force was reduced.

8 (a) i. 1. Weight 2. Buoyant force

ii. 1. Weight 2. Tension T, and T

2 in the strings

(b) i. Buoyant force = weight ii. Resultant tension in the strings =weight

4 (a) i. ii.

(b) i.

ii.

i

(c) 1 cm represent 4 N

\ \

\ \

\ T, \

\ 18 N \

\ \

\ 60'

T2

From scale drawing, T ~ 20.8 N

I

T, ~ lOA N 9 (a) Kinetic energy

Kinetic energy ~ (2) (4)2 ~ 16 J+(b) Speed decreases (c) Change of kinetic energy

=1- (2) W- 1') = 15 J2

(d) Potential energy due to gravity ~ mgh =2(10) (OA) =8 J

(e) Energy loss ~ 15 J ~ 8 J~ 7 J 10 (a) i. Newton

ii. x is directly proportional to W. iii. Hooke's Law

(b) i. 1.5 cm ii. 0.067 kg

(c) x/em

r / th!eker spring

LW/N 13

Section B 11 (a) Both persons bend their knees when landing [1 m]

Both of them stop. [1m] - To increase the time of contact when landing [1m]

Impulsive force acts [1m] - Increasing the time of contact will reduce the

impulsive force. [1 m] (b) i. The safety barriers made of metal are stronger[ 1m]

ii. To increase the strength [1 m] iii. Kinetic energy ~ Heat energy/sound [1m]

Work is done to bend the safety barriers. [1 m] (c) Use a wider safety belt

- To reduce the pressure on the driver [2m] Use a safety belt that can be extended further - To extend the time of collision and reduce the

impulsive force [2m] The front and back of the car can be made more collapsible ~ To extend the time of collision and reduce the

impulsive force [2m] Use of air bag for the driver - To reduce the impulsive force and limit the damage

to the victims [2m] The frame of the car is made of steel - To produce a strong body frame for the car. [2m]

Section C 12 (a) Acceleration is the rate of change of velocity. [1m]

(b) - Crash helmet is streamline [1m]

116

- Tight fitting clothes for trapping heat. [1 m] - The gait as of a jockey when riding a horse is to

produce a streamline shape. [1m] - Aerodynamic shape of the bicycle. [1 m] - Clothes made of smooth material and can withstand

cold. [1m] (c) Density of substance use is low

- So that the bicycle is light. [2m] Tyre with is small - So that the pressure on the road is big to give a good

grip. [2m] Low fragility - The tyre do not disintegrate easily at low

temperatures. [2m] Low rate of expansion - So that the size of the tyre remains constant and do

not change easily. [2m] The most suitable tyre is R because it has low density, small tyre width, low rate of desintegration and low rate of expansion. [2m]

(d) Weight component that is parallel to the road ~ mgsin 30° [1m] ~ 600 (0.5) = 300 N Frictional force = 100 N Force opposing the movement = mg sin 30° - frictional force [1m] Assuming the force necessary is x Resultant force = rna [1m] x - mg sin 30° - frictional force = rna [1m] x = rna + mg sin 30° + 100

= 60 x 2 + 300 + 100 = 520 N [1m]

(a) i. The length by which a shortens from its original length when a force acts on top of it. [1 m]

ii. The distance compressed will become half of the distance compressed whep using one spring. [1 m] This is because the force that is on each spring is half the original force. [1 m]

(b) - The spring is made from material of high density. [1m]

This produces a more stable position. [1m] - The diameter of the spring coil is big. [1 m]

This will make the spring softer and more comfortable to sit on. [1m]

- The spring is made from metal that can withstand rusting. [1m]

- This will ensure that the sofa will last. [1 m] - The length of the spring is big. [1 m] - This will make the spring softer [1m] - The choice is spring Z because it has a high density,

big coil diameter, big length and a low rate of rusting. [1m]

(c) i. Gradient of the graph 500

=-5- gcm-1 [1m]

= 100 g cm- I [1m] ii. Compression of the spring

=1- x 12 = 4 cm [1m]3

From the graph, the mass required = 400 g [1m] (d) i. 85000 g is divided equally among 250 springs [1m]

One spring carries a load of 340 g [1m]

ii

PAPER 3 Section A

1 (a) i. ii iii

(b)

v. 1. II •

2. AI 3. ..

5i 4. 11

tI 5. P

II

6. HI 3

[1m] se is to

[1m] [1m]

ithstand [1m]

[2m]

e a good [2m]

at low [2m]

It and do [2m]

, density, I low rate

[2m]

[1m]

frictional [1m]

[1m] [lm]

[1m] ts original

[lm] half of the ,ring. [1m] h spring is

[1m] density.

[1m] [1m] [1m]

Jmfortable [1m]

withstand [1m] [1m] [1m] [1m]

~ density, w rate of

[1m]

[1m]

[1m]

[1m]

g [1m] ings [1m]

[1m]

From the graph each spring is wm"n:sscd = 3.4 em Um]

ii. The constant of the spring is IIBIIl mating the sofa soft and comfortable. [1m]

PAPER 3 Section A

1 (a)

(b)

30.0 34.5 39.5 44.0 49.0

6.0 6.9 7.9 8.8 9.8

20 30 40 50 60

i.

(make measurements of I in the diagram and calculate the values v based on the measurements)

(c)

(d) The velocity increases linearly with the height. (e) Avoid errors due to parallax by placing the eye directly

opposite the scale.

Section B 2 (a) The mass of a load affects the acceleration of the

lorry. (b) The bigger the mass of a load, the lower the acceleration

of the lorry. (c) i. To study the relationship between the mass and the

acceleration of an object under a constant force ii. Manipulated: Mass

Responding: Acceleration Constant: Force acting on the body

iii. Three trolleys, ticker timer, ticker tape, power supply, pulley, thick string, weight (500 g)

iv. ticker timer trolley

ticker ,~~ """' ­tape

power pulley supply

weight 0/ SOO g

v. 1. Elevate one end of the wooden plane until it is a friction - compensated runway.

2. Attach one end of a ticker tape to the trolley. 3. A thick string connects the trolley to a weight of

500 g. 4. Switch on the power supply and released the

trolley. 5. Find the acceleration from the ticker tape and

record in the table. 6. Repeat above steps of the experiment with 2 and

3 trolleys.

(a) Weight (b) i. Have the same weight

ii. The tyre sinks more into the muddy surface in Diagram 1.2 is than in Diagram 1.1

iii. Surface area of tyre in Diagram 1.1 is bigger than surface area of tyre in Diagram 1.2.

iv. Pressure (c) The pressure increases when the surface area decreases. (d) Pressure acted by the shoes on the field is high, giving

a firm grip 2 (a) i.

1Z-~I 2 3

A straight line graph is obtained. Hence, the stated hypothesis is accepted.

(b)

ii. Liquid pressure increases when the depth of liquid increases.

iii. A

CHAPTER 3: Forces and Pressure PAPER 1

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

11 B 12 D 13 D 14 B 15 D 16 C 17 A 18 D 19 D 20 B 21 C 22 C 23 B 24 D 25 D 26 C 27 D 28 D 29 C 30 C 31 B 32 A 33 B 34 C 35 B 36 C 37 A 38 C

PAPER 2 Section A

117