ZMSMSASKL
PAPER 2(Section A)
8 Structure Question(Q1 – Q8)
1 hour and 30 minutes
Total Marks for Paper 2 Section A
(60 Marks)
Question 1 (Knowledge, understanding & application)
Diagram 1 shows a micrometer screw gauge when
the jaws are closed.
1
P
P
a)
Name the part label P
b)
Rachet
1 mark
What is the function of the part label
P?
To ensure the pressure exerted does not exceed //
undue pressure is not exerted
1 mark
Question 1 (Knowledge, understanding & application)
c) What is the value of zero error shown by the
micrometer above?
-0.02 mm
1 mark
d) Give one reason why micrometer screw gauge is
more accurate compare to a vernier calliper
The scale is smaller
1 mark
Question 2 (Knowledge, understanding & application)
Diagram 2 shows a boy throwing a ball upwards
at a velocity of 10 m s-1. The ball decelerates to a
maximum height before accelerating downwards.
Diagram 2
Question 2 (Knowledge, understanding & application)
c) Sketch a displacement, s against time,t graph to
describe the motion of the ball.
2 marks
Question 3 (Knowledge, understanding & application)
Diagram 3 shows an arrangement of apparatus used to
determine the atmospheric pressure in a laboratory. The
length of the glass tube is 100 cm and the atmospheric
pressure in the lab is 75 cm Hg.
Question 3 (Knowledge, understanding & application)
(a)
(b)
(c)
Name the apparatus shown in Diagram 3.
Mercury Barometer
1 mark
What is in space X
Vacuum
1 mark
(i) What is the value of H?
75 cm/0.75 m/750 mm
1 mark
(ii) What happens to height, H, when this
apparatus is submerged in water.
Increase
1 mark
Question 3 (Knowledge, understanding & application)
(iii) Give a reason for your answer in c(ii).
The pressure outside glass tube
increases // pressure of water +
atmospheric pressure
1 mark
(d)State one application of atmospheric pressure in
everyday life.
Vacuum cleaner// siphon // straw // syringe
// water pump
1 mark
Question 4 (Knowledge, understanding & application)
Diagram 4 shows a truck pulling a car with a cable. The
cable is at an angle of 600 to the horizontal. The force, F,
of the cable is 1500 N.
Diagram 4
Question 4 (Knowledge, understanding & application)
(a ) What is meant by force?
Anything that can move a stationary object
// stop a moving object //
change direction / shape / speed of an
object.
1 mark
.
Question 4 (Knowledge, understanding & application)
(b)
On Diagram 4, draw the horizontal
component, Fx and the vertical component Fy
for F.
In your drawing show the direction of Fx and
Fy.
2 mark
Fx = 1500 x cos 60
= 750 N
Question 4 (Knowledge, understanding & application)
(c)
Calculate the magnitude of the horizontal
component, Fx.
Fx = 1500 x cos 60°
= 750 N
(d)
1 mark
What is the effect of the component of
forces, Fx and Fy to the towed car?
Fx : to make car move forward //
overcome frictional force
1 mark
(i)
Fy: to lift the car off the ground // to
move the car upwards// to overcome
the weight of the car
1 mark
Question 5 (KUA and Conceptualization Skill)
(a ) Diagram 5.1 and Diagram 5.2 show the hair
shampoo is pressed out with the same force
from a shampoo container.
Question 5 (KUA and Conceptualization Skill)
(a) What is meant by pressure ?
Normal Force / surface area //
Normal force per unit area
1 mark
(b) Based on Diagram 5.1 and Diagram 5.2, compare
(i) the volume of shampoo in the container
Diagram 5.2 is bigger than Diagram
5.1/ greater // vice versa
1 mark
Question 5 (KUA and Conceptualization Skill)
(b) Based on Diagram 5.1 and Diagram 5.2, compare
(ii) the volume of shampoo that spurts out from
the container
Diagram 5.2 is bigger than Diagram
5.1 / greater// vice versa
1 mark
(c ) Based on your answer in (b),
(i) relate the volume of shampoo in the container
with the volume of shampoo that spurts out
from the container.
When the volume of shampoo in the container
is less, the volume of shampoo spurted out
from the container is less // vice versa 1 mark
Question 5 (KUA and Conceptualization Skill)
(c ) Based on your answer in (b),
(ii) relate the volume of shampoo that spurts out
from the container with the pressure
exerted by the shampoo. (1 mark)
When the volume of shampoo spurted out from the
container is big, the pressure exerted towards the
shampoo is big // vice versa
(iii) state the relationship between pressure and the
volume of shampoo in the container.
When the volume of the shampoo in the
container increase the pressure also
increases.
1 mark
Question 5 (KUA and Conceptualization Skill)
(d) State one characteristic of the liquid
Liquid used is non compressible / hard to
compress
1 mark
(e) Name the principle involved that enable the
shampoo to spurt out from the container.
Pascal’s principle
1 mark
Question 6 (KUA and Conceptualization Skill)
Diagram 6.1 and 6.2 show a virtual image produced
by a plane mirror and a convex mirror
respectively.
Diagram 6.1
Diagram 6.2
Question 6 (KUA and Conceptualization Skill)
(a) What is meant by virtual image?
Image that cannot be formed on
screen
1 mark
(b) Based on Diagram 6.1 and Diagram 6.2,
(i) compare the size of the image
Image in Diagram 6.1 is bigger than
Diagram 6.2// vice versa
1 mark
Question 6 (KUA and Conceptualization Skill)
(b)(ii) state one other similarity of the image
formed besides virtual.
Upright
1 mark
Question 6 (KUA and Conceptualization Skill)
(c ) In Diagram 6.3, draw a ray diagram to show
how the image in Diagram 6.2 is formed.
DIAGRAM 6.3
Question 6 (KUA and Conceptualization Skill)
(e) Name the light phenomenon that occurs.
Reflection
1 mark
(f) State one advantage of using convex mirror as the
side mirror of a car.
Wide angle of reflection // wider the
vision
1 mark
Question 7 (KUA and Problem Solving)
Diagram 7 shows a hydrometer used to
determine the density of a liquid.
Diagram 7
Question 7 (KUA and Problem Solving)
(a) (i)Name the physics principle involved .
Archimedes’s principle // Forces in
equilibrium // Principle of flotation
1 mark
(ii) Explain why the hydrometer floats on the
surface of the liquid.
The buoyant force = weight of the hydrometer
// density of liquid > density of hydrometer //
the volume of the liquid displaced by the
hydrometer is large
Question 7 (KUA and Problem Solving)
(a)(iii) The volume of the hydrometer under the
surface of the liquid is 25 cm3. The density of liquid
•
measured is 0.8 g cm-3. Calculate the buoyant force exerted to the hydrometer.
F =Vg
= (8 x 102) x ( 25 x 10- 6 ) x 10 //
= 0.2 N
[ 2 marks ]
(b)(i) hydrometer in Diagram7 is unsuitable to
measure the density of an acid solution which has
smaller density.
[ 2 marks ]
the plastic wall is replaced by a glass wall
he
increase the volume of the air filled bulb //
reduce the mass of lead shot
Question 7 (KUA and Problem Solving)
(b)(ii) Give the reasons for your answer in b(i)
to avoid corrosion
1 mark
to increase the buoyant force // to float
the hydrometer
1 mark
(iii) Suggest one method to increase the stability of
the hydrometer
add more lead shot
1 mark
(c ) What happens to the hydrometer in Diagram 7
when it is placed in a higher density liquid?
Less submerged //hydrometer floats higher
1 mark
Question 8 (KUA and Decision Making)
Diagram 8.1 and Diagram 8.2 show a block of iron
and a block of aluminium, each of mass 250 g, are
heated by an immersion heater. The power of the
immersion heater is 50 W.
Diagram 8.1
Diagram 8.2
Question 8 (KUA and Decision Making)
The temperature-time graph for the two experiments
are shown in diagram 8.3
.
Diagram 8.3
Question 8 (KUA and Decision Making)
(a) What is meant by heat?
Energy transfer from higher temperature body to
lower temperature // type of energy that flow
1 mark
(b)
Based on Diagram 8.3, calculate the change in
temperature per minute for;
(i) Iron
80
2
= 40 °C per minute
1 mark
1 mark
Question 8 (KUA and Decision Making)
(b)
Based on Diagram 8.3, calculate the change in
temperature per minute for;
(ii) Aluminium
100 -20
7
= 80
7
= 11.43 °C per minute
1 mark
1 mark
Question 8 (KUA and Decision Making)
(c) Based on your calculation, which metal gets hot
faster? Explain your answer.
Iron
1 mark
the rate of change of temperature is
higher
1 mark
(d) Determine the specific heat capacity for iron and
aluminium.
C iron = 50 x 2 x 60
0.25 x 80
= 300 J kg-1 °C -1
2 mark
C Al = 50 x 8 x 60
0.25 x 80
= 1200 J kg-1 °C -1
1 mark
Question 8 (KUA and Decision Making)
(e) Table 8 shows the specific heat capacity of
materials which could be used to make a frying
pan.
Specific heat
Material
•
Material
Bahan
A
B
C
Material
Bahan
capacity/ J kg-1 oC-1
Muatan haba tentu
780
1528
1415
Specific heat
capacity/ J kg-1 oC-1
Muatan haba tentu
Based on your answer in (c) and (d) , which material
would be suitable to make a frying pan?
Explain your answer.
Question 8 (KUA and Decision Making)
Based on your answer in (c) and (d) , which
material would be suitable to make a frying
pan? Explain your answer.
A
1 mark
it has lowest specific heat capacity // easily to
get hot
1 mark