Answers and Comments for Post Prelim Mock
Exam (Structured Questions)
1(a)
Normal contact force
Normal contact force
Friction
Friction
Weight
(400 N)
Comments:
1. Weight must be drawn from the CG.
2. Use ‘normal contact force’ instead of ‘contact force’
3. Qn ask for labeling so must name the force or give the quantity of the force
4. Many forgot to label friction. Without friction, the hands will tend to move forward and the feet backwards.
Since friction act to oppose the direction of motion or tendency of motion, it will act backwards at the hand and
forward for the legs.
(b)
Taking moments about his feet,
Total clockwise moment = total anticlockwise moment
400 x 0.8 = F x 1.3
F = 246 N
Comments:
1. Distance must be from pivot to force. The distance from the force acting on his hand to the pivot (leg) should be
0.8 + 0.5 m (not 0.5 m).
2(a)(i) KE at Q = GPE at P (since the slope is smooth)
= mgh = 0.6 x 10 x 2.7 = 16.2 J
(ii) KE = ½ mv2
16.2 = ½ x 0.6 x v2
v = 7.35 m/s
(b)(i) a = (v – u)/t = (4.8 – 7.35)/2.3 = - 1.11 m/s2
F = (1.11 x 0.6) = 0.666N
(c)
The box will reach a height less than 2.7 m because some of its energy is lost as heat when it overcomes friction
along QR. With less total energy, its maximum height reached will be lower.
Comments:
1. For part (c), it is not enough to mention that there is friction at QR. Must include ‘energy is lost when
overcoming friction hence total energy is less’.
3(a)
1/RE = 1/6 + 1/6
RE = 3 Ω
(b)
(c)
Current that pass through cell = total V / total R = 1.5 / 3 = 0.5 A
Current that pass through Z = 0.5 ÷ 2 = 0.25 A
Or:
I = V/R = 1.5/6 = 0.25 A (the pd across WZY is 1.5 V and the total resistance in the branch is 6 Ω)
(d)
pd across WXY = 1.5 V hence every bulb has a voltage of 0.5 V.
Pd across XY (i.e. pd across 2 bulbs) is 0.5 + 0.5 = 1V
Comments:
1.
The circuit is basically a parallel circuit with 2 branches. Each branch has 3 resistors in series. Each branch has a
pd of 1.5 V (pd is the same for branches that are in parallel) and the total resistance for each branch is 6 Ω.
2.
When drawing current direction, must draw on all the branches and not just the wire linking the cell.
i
4(a)
(b)
Ray 1 will undergo total internal reflection when it meet the air bubble because its angle of incidence is larger
than the critical angle.
Ray 2 will enter the air bubble because its angle of incidence is less than the critical angle. It will bend when it
enters air and it is refracted away from the normal as it is now entering an optically less dense medium.
Ray 3 will continue its path and not bend because when its angle of incidence is zero hence its angle of
refraction is also zero.
(c)
sin c = 1/n
n = 1 / sin c = 1 / sin 48.6 = 1.33
Comments:
1.
Many students mistook the bubble to be water and the medium in which the light initially travelled to be air.
Students must recognize that in this case, the light is travelling in the optically denser medium of water into the
optically less dense medium of water.
2.
Students should use ‘optically denser or optically less dense’ rather than ‘denser and less dense’.
3.
For ray 3, it is insufficient to say that light will not bend. Support your answer by mentioning that when angle of
incidence is zero, angle of refraction will also be zero.
5 (a) (b)
N
5 (c)
S
N
S
When the switch is closed, both soft iron will be magnetized.
The ends of the iron bar facing each other have opposite polarity hence they will attract each other as opposite
poles attract.
(d)(i) The two steel bars will still attract each other but it takes a little longer for them to attract compared to iron bars
because it is harder to magnetise steel than iron.
(ii)
There will be no change, the two iron bars will still attract each other as the ends facing each other will still have
opposite polarity.
Comments:
1.
For part (a), the compass needle must be drawn with a ruler. It cannot be curved or crooked. The arrow
direction must be shown clearly. Note the arrow direction for the centre compass as many students made
mistake labeling that compass.
It is insufficient to label only one end of each iron bar. Qn requires both ends to be labeled.
2.
For part (c), the iron bar is magnetized when it is placed in the solenoid. DO NOT USE ‘iron bar is induced with
magnetism’. An iron bar will have magnetism induced in it only if it is placed near a magnet. The magnet will
induced an opposite pole on the iron. Do not use induced magnetism for iron bar that is placed in a solenoid.
3.
For part (d), must compare the change in effect with what is mentioned in part (c).
6.
Pulling force is 8000 N and it acts exactly between the 2 pulling force of 8000 N.
(b)
There is no resultant force acting on the barge.
Resultant force = ma, since the barge is moving at a constant speed, its acceleration is zero and hence the
resultant force will be zero.
Comments:
1.
When drawing vector diagram, it is very important to ensure that it has the same orientation as the diagram
given i.e. the 2 diagrams must look alike and you cannot rotate the forces as you wish.
Question requires you to give the direction of the resultant force which many students forgot to mention. Since
it is confusing to give the direction with respect to the pulling force since they have the same magnitude, you
can say that it acts in the middle of both forces in this case.
2.
For part (b), you must support your explanation by stating formula F = ma.
7(a)
The coolant is melting and temperature does not change during change of state as energy absorbed is used to
break the intermolecular bonds instead of increasing its kinetic energy.
(b)
The temperature of the food falls because heat is transferred from the food to the coolant.
(c)
A white Styrofoam should be used. White is a poor absorber of heat so it will reduce heat gained by radiation.
Styrofoam is a good insulator hence it prevents heat gain by conduction.
Comments:
1.
The temperature of food and coolant in the box is lower than the surrounding. Since heat travel from higher to
lower temperature, heat will move into the box and the box will act as an absorber (and not a radiator/emitter)
2.
Since coolant is colder than food (see graph), heat is transferred from food to coolant. Note that heat cannot be
transferred to the surrounding because the surrounding is hotter than the food.
8(a)
(b)
‘x’ can be at any corner of the card.
X
Comments:
1.
Compass needle must be drawn straight and a ruler should be used. Arrows must be drawn clearly.
2.
Magnetic field decreases with distance from the wire hence X must be at a point on the card that is furthest
from the centre which are the corners of the card.
9 (a)
(b)
It will roll to the right.
The rod experience a force because a current carrying rod will experience a force when it is placed in a magnetic
field. The direction of the force can be determined by the Fleming’s left hand rule.
When P is moved downwards, the resistance of the variable resistor will decrease and hence more current flows
in the circuit. The movement of the rod will be faster.
Reverse the direction of the cell.
(c)
(d)
Comments:
1.
For part (b), it is not enough to just state ‘Fleming’s left hand rule’. Must explain that when the current flows in
the direction of the centre finger, the magnetic field is in the direction of the forefinger, the thumb will point in
the direction of the force. Or, can say that ‘a force acts on a current carrying wire that is placed in a magnetic
field’
2.
For part (d), it is not enough to state ‘change the direction of the current’. Students must suggest a way in which
the current can be reversed i.e. reverse the polarity of the cell.