Download Year 11 2 Unit Physics Preliminary Exam 2007 General Instructions

Year 11
2 Unit Physics
Preliminary Exam 2007
Total marks (75)
This paper has two Parts, Part A and Part B
Part A
General Instructions




Reading time – 5 minutes
Working time – 2 hours
Write using blue or black pen
Write your student number at the top of
this Answer Booklet and the Multiple
Choice Answer Sheet
Total marks (15)
 Attempt Questions 1 – 15
 Allow about 30 minutes for this part
 Use the Multiple Choice Answer Sheet
and hand in separately
Part B
Total marks (60)
 Attempt Questions 16 – 27
 Allow about 1 hour 30 minutes for this
part
Part A
Total Marks 15
Attempt Questions 1 – 15
Allow about 30 minutes for this Part
Use the multiple choice answer sheet
1
Which of the following best describes a sound wave?
A
B
C
D
2
A sound wave is a transverse, mechanical wave.
A sound wave is a transverse, non-mechanical wave.
A sound wave is a longitudinal, mechanical wave.
A sound wave is a longitudinal, non-mechanical wave.
A microphone was used to convert the pressure variation of two sound waves, sound A
and sound B, into voltage signals. The diagram below shows a cathode ray
oscilloscope (CRO) trace for each of these voltages. Note that the same voltage and
time scale was used to examine each sound wave.
What can we deduce about the original sounds from these CRO traces?
A
B
C
D
Sound A was louder, had a higher pitch and a longer wavelength than sound B.
Sound A was quieter, had a lower pitch and a shorter wavelength than
sound B.
Sound A was louder, had a lower pitch and longer wavelength than sound B.
Sound A was quieter, had a higher pitch and a shorter wavelength than
sound B.
1
3
A light wave is totally internally reflected by the glass/air boundary in an optical fibre
as shown in the diagram below.
Which of the following statements about this light beam is true?
A
B
C
D
4
The angle  is less than the critical angle for the glass/air boundary.
The angle  is greater than the critical angle for the glass/air boundary.
The angle  must be equal to the critical angle for the air/glass boundary.
The frequency of the light in this type of glass must be less than the frequency
of the light in air.
The intensity of two light globes, measured 1 m from each globe, is shown in the table
below.
Light Globe
A
B
Light Intensity 1 m from Globe (Wm-2)
32
12
For an observer who was 4 m away from globe A and 2 m away from globe B how
bright would each globe appear?
A
B
C
D
5
Which of the following scientists believed that ‘animal electricity’ was different from
‘atmospheric electricity’?
A
B
C
D
2
The globes would appear to be equally bright.
Globe A would appear to be brighter than globe B.
Globe B would appear to be brighter than globe A.
It is impossible to determine the relative brightness of the globes from the data
given.
Luigi Galvani
Alessandro Volta
Andre-Marie Ampere
Augustus Coulomb
6
A negatively charged particle with mass m is placed in an electric field of strength E,
directed towards the left as shown below.
What is the acceleration of the charge due to the electric field?
A
B
C
D
7
qEm to the left.
qEm to the right.
qE/m to the left.
qE/m to the right.
Assuming all the globes in the circuits below have the same resistance, in which
circuit would globe x be brightest?
A
B
C
D
3
8
The diagram below shows an air solenoid mounted on a frictionless trolley near a bar
magnet mounted on a trolley. Both the trolleys have the same total mass.
What will happen when the switch is closed?
A
B
C
D
9
Which of the following statements is true for a car traveling at constant velocity?
A
B
C
D
10
The acceleration could be positive or negative.
There is a net force on the car in the direction of the motion.
The rate of change of the displacement is zero.
The net force on the car is zero.
If a car travelling with a velocity of v ms-1 requires a distance X metres to stop, what
would the stopping distance be if it was initially travelling with a velocity of 3v?
(Assume the stopping force on the car is equal in each case.)
A
B
C
D
4
Both trolleys will move apart from one another with the same acceleration.
The air solenoid trolley will move to the left and the magnet trolley will remain
at rest.
The air solenoid trolley will move to the right and the magnet trolley will
remain at rest.
Both trolleys will move towards one another with the same acceleration.
(3/2)X metres
3X metres
9X metres
(1/3)X metres
11
Why do cars sometimes go straight ahead rather than turning when they enter a bend
on a slippery road?
A
B
C
D
12
Why is the sum of the momentum always conserved when cars collide?
A
B
C
D
13
During the collision, the impulse on one car is always equal and opposite to the
impulse on the other car.
In any collision the kinetic energy is always conserved.
The sum of the work done by the cars in a collision is always zero.
Newton’s first law states that a body in motion should remain in motion unless
acted upon by a net unbalanced force.
Hubble showed that the further a galaxy was away from us the faster it was receding
from us. How did Hubble determine the velocity of other galaxies?
A
B
C
D
14
The centripetal force provided by the tyres pushing the road towards the
outside of the bend is reduced and the car continues to move in a straight line
due to Newton’s First Law.
The centripetal force provided when the road pushes the tyres inwards towards
the centre of the bend is reduced and the car continues to move in a straight
line due to Newton’s First Law.
The centripetal force provided by the steering wheel is reduced and the car
goes straight ahead due to Newton’s Second Law.
The reaction force between the steering wheel and the driver is reduced and the
car goes straight ahead due to Newton’s Third Law.
By solving Einstein’s equations of General Relativity.
By plotting stars from other galaxies on a Hertzsprung/Russell diagram.
By comparing the spectra from stars in other galaxies with spectra obtained
from excited gases on the Earth.
By studying the cosmic background radiation.
What type of stars use the fusion of helium into carbon as an energy source?
A
B
C
D
White Dwarf stars.
Main Sequence stars.
Red Giant stars.
Neutron stars.
5
15
The data in the graph below shows information from an experiment to measure the
penetrating power of radioactive particles through 1 mm thick sheets of aluminium.
The radioactive source used for the experiment may have emitted more than one type
of nuclear radiation.
By examining the graph what are the most likely radioactive particle/s emitted by the
source?
A
B
C
D
Beta and gamma radiation.
Gamma radiation only.
Alpha and beta radiation.
Beta radiation only.
End of Part A
6
PART B
60 marks
Attempt Questions 16–27
Allow about 90 minutes for this section
Marks
Question 16 (1 mark)
Identify a feature that all electromagnetic waves have in common.
1
Question 17 (5 marks)
A radio wave is illustrated in the diagram below.
(a)
What is the amplitude of this radio wave?
1
(b)
Calculate the frequency of this radio wave.
1
(c)
Even though radio waves travel in straight lines, some low frequency radio
waves can be sent from one side of the Earth to the other. Explain how this is
possible.
1
Outline how amplitude modulation of radio waves differs from frequency
modulation.
2
(d)
7
Marks
Question 18 (4 marks)
Drawing ray diagrams and wave front diagrams are a useful ways of picturing the path that
waves will take.
(a)
(b)
8
Convex and concave mirrors are shown below. Complete the ray diagram to
show the path of the light rays after they hit each of the reflectors.
2
The diagram below shows a light ray (and three wavefronts) that is approaching
a glass surface from air. Complete the ray diagram by sketching the reflected ray
and the refracted ray. Include three wavefronts on the reflected and refracted
rays. (Exact angles are not required)
2
Marks
Question 19 (6 marks)
In an experiment to measure the refractive index of Perspex, a student carefully measured a
range of angles of incidence (i) and the corresponding angles of refraction (r) for a light beam
entering a block of Perspex from air. The student than calculated the sin of each of the angles
and plotted a graph of sin r against sin i. This graph is shown below.
(a)
Why did the student ignore the point marked x when drawing the line of
best fit?
1
Explain why the student measured several angles of incidence and refraction and
plotted the results rather than just measuring one angle of incidence and
refraction to find the refractive index.
2
(c)
Find the gradient of the line of best fit and the refractive index for Perspex.
2
(d)
Use the experimental data to find the speed of light in Perspex.
1
(b)
9
Marks
Question 20 (5 marks)
A magician decides to use an electric field to levitate (float) a light plastic ball in the air. The
ball has a mass of 15 grams and can be given a charge of + 0.5 coulombs.
+0.5 C
(a)
Sketch the electric field around the positively charged ball. Assume the ball is in
‘free space’ with no other external electric fields.
1
(b)
Calculate the weight of the ball.
2
(c)
Determine the strength and direction of the external electric field that the
magician would require to levitate the ball.
2
10
Marks
Question 21 (6 marks)
Consider the circuit shown below.
(a)
On the circuit diagram shown above show how a voltmeter and ammeter would
be connected to measure the voltage across and the current passing through the
5  resistor.
2
(b)
Calculate the total resistance of the circuit.
2
(c)
Determine the total power used by the circuit.
2
11
Marks
Question 22 (5 marks)
High voltage AC can be extremely dangerous. Describe how these dangers are minimised in
household electric circuits.
5
12
Marks
Question 23 (5 marks)
The displacement of a car over a short period is plotted in the graph below.
(a)
How far has the car travelled after 10 minutes?
1
(b)
What is the car’s instantaneous velocity (in ms-1) 3 minutes into the journey?
2
(c)
What is the car’s average velocity during the first 10 minutes of the journey?
2
13
Marks
Question 24 (5 marks)
A car of mass 1600 kg is initially at rest. The engine then provides a constant force of
1200 newtons which pushed the car to the right.
(a)
Assuming there is no friction, find the acceleration of the car.
1
(b)
If this force remained constant, what would the car’s velocity be after
40 seconds?
1
(c)
How much work is done by the engine for the car to reach this velocity?
1
(d)
If a frictional force of 240 newtons opposed the driving force, what would the
velocity of the car be after 40 seconds?
2
14
Marks
Question 25 (6 marks)
A truck (A) collides with a car (B) travelling in the opposite direction as shown in the diagram
below.
(a)
Find the initial momentum of the truck and the car.
1
(b)
After the collision, the cars become entangled and move as one. Find the
velocity of the wreckage after the collision.
2
Explain with reference to Newton’s laws why seat belts reduce the risk of injury
in collisions like this.
3
(c)
15
Marks
Question 26 (7 marks)
The Hertzsprung/Russell diagram has helped astronomers unlock the secrets of stars.
(a)
(b)
16
Sketch a Hertzsprung/Russell diagram on the axes below. Label the axes and
illustrate the positions of the main star groups reflecting the evolutionary
sequence of stars.
4
Explain how astronomers can determine the temperature of a star by examining
the light emitted by the star.
3
Marks
Question 27 (5 marks)
Describe the key elements of the Big Bang theory and the evidence that supports the
theory.
5
End of paper
17
YEAR 11 PHYSICS
PRELIMINARY EXAM 2007
MULTIPLE CHOICE ANSWER SHEET
Student number ____________________
Select the alternative A, B, C or D that best answers the question. Fill in the response oval completely.
Sample: 2 + 4 =
(A) 2
(B) 6
(C) 8
(D) 9
A
B
C
D
If you think you have made a mistake, put a cross through the incorrect answer and fill in the new answer.
A
B
C
D
If you have changed your mind and have crossed out what you consider to be the correct answer, then indicate
this by writing the word correct and drawing an arrow as follows:
correct
A
B
C
D
ATTEMPT ALL QUESTIONS
Question
18
1
A
B
C
D
2
A
B
C
D
3
A
B
C
D
4
A
B
C
D
5
A
B
C
D
6
A
B
C
D
7
A
B
C
D
8
A
B
C
D
9
A
B
C
D
10
A
B
C
D
11
A
B
C
D
12
A
B
C
D
13
A
B
C
D
14
A
B
C
D
15
A
B
C
D
Formula Sheet
v  f
I
1
d2
v1 sin i

v 2 sin r
E
F
q
R
V
I
Energy = VIt
a av 
m1m2
r
F  mg
v x  ux
2
2
v  u  at
v y  u y  2a y y
2
2
x  u x t
P  VI
v av 
E p  G
r
t
v
vu
therefore a av 
t
t
1
y  u y t  a y t 2
2
r 3 GM

T 2 4 2
F
Gm1m2
d2
F  ma
mv 2
F
r
Ek 
1 2
mv
2
W  Fs
p  mv
Impulse = Ft
19
Physics Preliminary 2007
Physics Data Sheet
Numerical values of several constants
Charge on an electron, qe
-1.602  10-19 C
Mass of electron, me
9.109  10-31 kg
Mass of neutron, mn
1.675  10-27 kg
Mass of proton, mp
1.673  10-27 kg
Speed of sound in air
340 ms-1
Earth’s gravitational acceleration, g
9.8 ms-2
Speed of light, c
3.00  108 ms-1
Magnetic force constant, k
2.0  10-7 NA-2
Universal gravitational constant, G
6.67  10-11 Nm2kg-2
Mass of Earth
6.0  1024 kg
Planck’s constant, h
6.626  10-34 Js
Rydberg’s constant, RH
1.097  107 m-1
Atomic mass unit, u
1.661  10-27 kg
931.5 MeV/c2
1 eV
1.602  10-19 J
Density of water, r
1.00  103 kgm-3
Specific heat capacity of water
4.18  103 Jkg-1K-1
20
Product code: 736101
Mapping grid
Question Mark
1
1
Content
8.2.1 col 2 dot 3, 8.2.2 col 2 dot 2
Outcome
Band
P8
2–3
2
1
8.2.2 col 2 dot 3, 3 col 3 dot 1
P8 and P14
4–5
3
1
8.2.4 col 2 dot 7, 8
P8 and P14
4–5
4
1
8.2.3 col 2 dot 4, col 3 dot 1
P7 and P8
5–6
5
1
8.3.1 col 3 dot 1
P1 and P8
2–3
6
1
8.3.2 col 2 dot 3, col 3 dot 2
P6 and P9
3–5
7
1
8.3.3 col 2 dot 1, 2
P7
4–6
8
1
P7 and P14
4–6
9
1
8.3.5 col 2 dot 5, col 3 dot 2, 8.4.2 col 2
dot 9
8.4.2 col 2 dot 3, 4 col 3 dot 6
P6 and P14
3–5
10
1
P6 and P7
4–6
11
1
8.4.2 col 2 dot 6, 10 or 8.4.3 col 3
dot 1
8.4.2 col 2 dot 8, col 3 dot 1, 7
P6
3–5
12
1
8.4.4 col 2 dot 3
P6 and P7
3–5
13
1
8.5.2 col 2 dot 1
P1 and P2
2–4
14
1
8.5.3 col 2 dot 4
P7
2–4
15
1
8.5.4 col 2 dot 1, 2, col 3 dot 1
P10 and P14
3–5
16
1
8.5.4 col 2 dot 6
P9
2–4
17
5
9.2.1 col 2 dot 4, 6 col 3 dot 6, 8.2.3
col 2 dot 5, 8.2.4 col 2 dot 2
P7 and P8
2–5
18
4
8.2.4 col 3 dot 1, 2, 8.2.4 col 2 dot 1, 3
P8 and P13
2–5
19
6
8.2.4 col 2 dot 5, col 3 dot 4 & 5
P8, P11, and
P12
3–5
20
5
8.3.2 col 3 dot 1, 8.4.2 col 2 dot 7, 8.3.2
col 2 dot 3, col 3 dot 2
P6 and P9
2–5
21
6
8.3.2 col 2 dot 1, 2, 3, 4 col 3 dot 1, 8.3.4
col 3 dot 2
P7, P9 and P11
3–5
22
5
8.3.6 col 2 dot 1, 2
P3, P4 and P13
2–6
23
5
8.4.1 col 2 dot 4, col 3 dot 3
P6 and P14
2–4
24
5
8.4.2 col 2 dot 4, 5, col 3 dot 6
8.4.3 col 2 dot 1, col 3 dot 1
P6, P7 and P14
3–6
25
6
8.4.4 col 2 dot 1, 3, col 3 dot 1, 3
P3, P4, P6 and
P14
3–5
26
7
8.5.3 col 2 dot 1, 2, 3 col 3 dot 1, 3
P1, P2, P7 and
P8
2–5
27
5
8.5.2 col 2 dot 2, 3, 4, col 3 dot 1
P1, P2 and P7
3–6
Marking guidelines
Section I
1
2
3
4
5
6
C
D
B
C
A
D
7
C
8
A
9
10
D
C
11
B
12
A
13
C
14
15
C
A
16
D
Amplitude related to loudness and period to pitch and frequency.
 >C is required for total internal reflection.
Intensity is proportional to 1/r2.
F = qE and F = ma. Hence ma = qE or a = qE/m in the opposite
direction of the field, as the particle is negatively charged.
The voltage across the globe (and hence the power produced by the
globe) is greatest in the parallel circuit.
Current produces magnetic field that opposes that of the bar magnet
and accelerations are equal, as the masses are equal.
Application of Newton’s First Law of motion.
Work done to stop the car = Fs = initial KE = mv2/2.
As the stopping force (F) is constant the stopping distance (s) must be
proportional to the square of the initial velocity (sv2).
Centripetal force is a reaction force directed towards the centre of the
circle and is provided by the road to the tyres. The tyres push the road
outwards and the road pushes the tyres and the car inwards.
Impulse = p = Ft. As the impulse on each body is equal and opposite,
the total change of momentum of the system is zero.
This is how Hubble discovered that the light from distant galaxies was
red shifted.
About half the radiation (beta) is stopped by a cm of aluminium, but a
significant fraction is not affected by several cm of aluminium
(gamma).
Section II
Question 17(a)
Answer: Amplitude = 20 Vm-1.
Marking guidelines
Criteria
 Correct numerical answer.
Marks
1
Question 17 (b)
Answer: v = f and hence f = v/ = 3 × 108/200 = 1.5 × 106 Hz = 1.5 MHz.
Marking guidelines
Criteria
 Correct numerical answer.
Marks
1
Question 17(c)
A good answer would include:
AM frequency radio waves are reflected by the ionosphere and hence can be sent over the
horizon. (A diagram illustrating this process would also suffice, provided the ionosphere and
radio waves are labelled.)
Marking guidelines
Criteria
 Correctly explains reflection from ionosphere.
Marks
1
Question 17 (d)
A good answer would include:
 In order for a radio wave to carry information it must be modulated in some way.
 In AM, the amplitude of the carrier wave is continually modulated (changed) to transmit
information, while in FM, information is encoded by continually modulating (changing)
the frequency of the carrier wave slightly.
 Diagrams could also be used to show how the carrier wave is altered by amplitude and
frequency modulation.
Marking guidelines
Criteria
Marks
 Correctly explains the difference between AM AND FM.
2
 Correctly explains how AM OR FM carries signals or a vague explanation
of the difference between AM and FM that makes a correct statement about
AM or FM.
1
Question 18(a)
Answer:
Marking guidelines
Criteria
Marks
 Correctly completes both ray diagrams.
2
 Correctly completes one of the diagrams.
1
Question 18 (b)
Answer:
Marking guidelines
Criteria
Marks
 Correctly completes both refracted and reflected rays and includes both sets
of wavefronts with correct spacing. (All four required for 2 marks.)
2
 Two of the four factors required above (both rays and both wavefronts)
drawn correctly.
1
Question 19(a)
Answer: Because it goes against the trend of all the other points, the student would have
decided the data were probably incorrectly recorded or measured and hence could be
reasonably excluded from the analysis of the experimental results.
Marking guidelines
Criteria
 Correct explanation for excluding the outlying data point.
Marks
1
Question 19(b)
A good answer would include:
 By using multiple points, plotting them and drawing the line of best fit to find the
gradient, the random measurement errors that occur in all measurements are minimised. In
addition, out-riding points (serious errors) such as the one shown in part (a) can be easily
identified.
 Hence, by taking multiple readings over a range of values, we ensure the results obtained
are more accurate and hence this technique is more valid than taking single measurements.
Marking guidelines
Criteria
Marks
 A clear answer that explains why a more accurate and more valid result will
be obtained by using multiple points.
2
 A vague answer about the use of multiple points OR an answer that only
mentions the possibility of spotting out-riding points (as per part (a)).
1
Question 19(c)
Answer: Gradient = refractive index = 1.4.
Marking guidelines
Criteria
Marks
 Correct numerical answers.
2
 Correct numerical answer for the gradient or refractive index OR the same
(incorrect) value for both the gradient and the refractive index.
1
Question 19(d)
Answer: v = c/n = 3 × 108 /1.4 = 2.1 × 108 ms-1.
Marking guidelines
Criteria
 Correct numerical answer.
Marks
1
Question 20(a)
Answer:
Marking guidelines
Criteria
 Field drawn as straight lines radiating away from the ball and direction of
the field indicated on the lines.
Marks
1
Question 20(b)
Answer: Weight = mg = (0.015)(9.8) = 0.147 newtons.
Marking guidelines
Criteria
Marks
 Correct numerical answer AND units.
2
 Correct numerical answer but incorrect units OR correct working and units
but failure to convert to kilograms.
1
Question 20(c)
Answer: To levitate the ball the forces must be balanced, i.e. qE = mg or E = mg/q =
0.147/0.005 = 29.4 NC-1 or 29.4 Vm-1 upwards.
Marking guidelines
Criteria
Marks
 Correct numerical answer and direction of the field.
2
 Correct numerical answer and incorrect direction OR incorrect numerical
answer but correct field direction.
1
Question 21(a)
Answer:
Marking guidelines
Criteria
Marks
 Correctly shows ammeter wired in series with the 5 resistor and the
voltmeter wired in parallel with the 5 resistor.
2
 One of the above wired correctly.
1
Question 21(b)
Answer: For parallel resistors, R = 1/(1/30 + 1/60) = 20; hence, the total resistance
= 20 + 5 = 25.
Marking guidelines
Criteria
Marks
 Correct numerical answer.
2
 Incorrect final answer but correctly calculates the sum of the parallel
resistors.
1
Question 21(c)
Answer: P = V2/R = 1002/25 = 400 Watts OR I = V/R = 100/25 = 4 Amps and
P = IV = 4 ×100 = 400 Watts.
Marking guidelines
Criteria
Marks
 Correct numerical answer.
2
 Incorrect answer but correctly calculates the total current or uses the correct
equation for power.
1
Question 22
Students could cover a range of safety features here. A good answer could include:
 To minimise the dangers of electrocution and fires, household circuits incorporate the
following safety features:
– Fuses are wired in series with all household circuits. These are designed to burn out and
break the circuit if too much current flows. This reduces the risk of fires.
– The external metal casing of appliances are earthed. If a fault occurs in the appliance
and a live wire comes in contact with the metal casing, the earth wire carries the current
safely to earth.
– Many homes now have safety switches that monitor any current leakage and switch the
electricity off faster than a heart beat if any leakage of current is detected.
– Many appliances are double insulated; that is, their external case is made from an
insulating material such as plastic.
Marking guidelines
Criteria
Marks
 Clear discussion of at least FOUR factors that make electrical circuits safer.
5
 Clear discussion that mentions TWO or THREE safety measures or a vague
answer that mentions four factors.
3–4
 A vague answer that mentions TWO or THREE factors or a good
explanation of one factor.
Question 23(a)
1–2
Answer: Displacement = 6 km.
Marking guidelines
Criteria
Marks
 Correct numerical answer.
1
Question 23 (b)
Answer: Gradient = rise/run = 5000/(5 × 60) = 16.7 ms-1 (accept 16 to 17 ms-1).
Marking guidelines
Criteria
Marks
 Use of gradient to find correct numerical answer.
2
 Finds the correct gradient, but fails to obtain correct numerical answer
in ms-1.
1
Question 23(c)
Answer: Average velocity = total displacement/total time = 6000/(10 × 60) = 10 ms-1.
Marking guidelines
Criteria
Marks
 Correct numerical answer.
2
 Uses correct equation for average velocity, but does not get the correct
answer.
1
Question 24(a)
Answer: F = ma and, hence, a = F/m = 1200/1600 = 0.75 ms-2 to the right.
Marking guidelines
Criteria
 Correct numerical answer.
Marks
1
Question 24(b)
Answer: v = at + u and, as u = 0, v = at = 0.75 x 40 = 30 ms-1.
Marking guidelines
Criteria
 Correct numerical answer.
Marks
1
Question 24(c)
Answer: Work done = kinetic energy gained = mv2/2 = (1600)(302)/2 = 7.2 × 105 J.
Making guidelines
Criteria
 Correct numerical answer.
Marks
1
Question 24(d)
Answer: FNET = FDRIVING – FFRICTION = 1200 – 240 = 960 N to the right. Hence, the
acceleration of the car would be a = F/m = 960/1600 = 0.6 ms-2 and the velocity after 40 s
would be v = at = 0.6 × 40 = 24 ms-1.
Marking guidelines
Criteria
Marks
 Correct numerical answer.
2
 Incorrect answer for the final velocity, but correct calculation of net force
(960 N) or acceleration (0.6 ms-2).
1
Question 25(a)
Answer: p = mv and hence pA = 80000 kgms-1 to the right and pB = 60000 kgms-1 to
the left.
Marking guidelines
Criteria
 Correct numerical answer and direction for each car.
Marks
1
Question 25(b)
Answer: Call motion to the right positive. The initial momentum of the system will be given
by, pinitial = 80000 – 60000 = + 20000 Ns = pfinal = mv = (4000 + 1500) v.
Hence, v = 20000/5500 = 3.64 ms-1 to the right (as the answer is positive).
Marking guidelines
Criteria
Marks
 Correct numerical answer.
2
 Correctly applies conservation of momentum but fails to obtain the correct
answer (e.g. fails to take directions into account).
1
Question 25(c)
A good answer would include:
Using Newton’s laws we can see why passengers without safety belts are more likely to
experience very large forces that cause injuries.
Newton’s First Law: When a car is involved in a collision it comes to rest quickly and
unrestrained passengers keep moving forward until they hit something that will stop them.
Newton’s Second Law: When a passenger hits the dashboard, they decelerate very quickly
and hence the force exerted on them will also be very large.
Seat belts reduce injuries by stretching to decrease the deceleration in a collision (and hence
reducing the force on the passenger by Newton’s Second Law) and by bringing the passengers
to rest with the car preventing them from continuing to move forward into the dashboard etc.
Marking guidelines
Criteria
Marks
 A well-written coherent answer that uses TWO OR MORE of Newton’s
laws to explain why injuries occur without seat belts and why they are
reduced by using seat belts. The key physical principles involved must be
expressed correctly for full marks.
3
 A reasonable answer that only links ONE of Newton’s laws to the
reduction of injury OR a vague answer that mentions more than one of
Newton’s laws but fails to explain the physics involved correctly.
2
 A poor answer that fails to link Newton’s laws to the use of seat belts to
reduce injury, but makes one or more correct statements about Newton’s
laws and/or the use of seat belts.
1
Question 26(a)
Answer:
Marking guidelines
Criteria
Marks
 Diagram has both axes correctly labelled and shows the correct position of
Main Sequence, Red Giant and White Dwarf stars (i.e. five factors: two
axes and three star types).
4
 As above, but with one axis incorrectly labelled or with one star type
shown in the wrong position (i.e. any four factors correct).
3
 Any THREE of the above correct.
2
 Any TWO of the above correct.
1
Question 26(b)
A good answer would include:
 The spectral distribution of the light emitted from stars is similar to that emitted from
black bodies (objects that absorb all the light that falls on them) on Earth.
 By examining the spectra from a star and comparing it to the spectra from black bodies at
various temperatures on the Earth, the temperature of the star can be determined.
OR
 By measuring the wavelength at which most electromagnetic radiation is emitted from a
star and comparing it to values of the wavelength of maximum emission from black
bodies on Earth at various temperatures, the temperature of the star can be determined.
OR
 By measuring the wavelength at which most electromagnetic radiation is emitted from a
star and applying Wein’s Law (T = k/max) the temperature of the star can be determined.
Marking guidelines
Criteria
Marks
 Clear answer explaining how astronomers determine the surface
temperature of stars.
3
 A clear answer that states that the colour of a star enables astronomers to
estimate its temperature OR an answer that relates surface temperature to
the emission spectra but fails to talk about the wavelength of maximum
emission (or black body radiation or Wein’s Law).
2
 A vague answer that fails to answer the question, but includes a correct
statement about black body radiation, Wein’s Law or stellar spectra.
1
Question 27
A good answer would include:
 The key elements of the Big Bang theory include:
– The Big Bang theory is a cosmological theory of the creation and evolution of the
universe.
– The theory states that the universe began with all the mass/energy and space of the
current universe concentrated in an infinitely small, dense, hot point that exploded
approximately 13.5 billion years ago and has been expanding ever since.
– As the Universe expanded after the big bang it cooled and mass condensed out from the
energy.
– Initially, sub atomic particles formed and then atoms. Approximately 80% of the matter
formed consisted of hydrogen atoms and about 20% was helium.
– As the universe expanded further, the particles cooled and lost more kinetic energy.
Eventually, stars were formed by the gravitational attraction between particles in vast
clouds of gas and dust and galaxies and galactic clusters formed from the gravitational
attraction between stars and galaxies.
 The Big Bang theory is supported by three main pieces of evidence:
– Hubble’s Law: The light from distant galaxies is red shifted and the further the galaxy
is away from the Earth the more its light is red shifted.
– Microwave background radiation: The universe is bathed in microwave radiation that
the Big Bang theory predicted as the remnant of the gamma ray radiation that once
filled the universe. The observed spectral distribution of this radiation was also
predicted by the Big Bang theory before it was observed.
– Nuclear synthesis: The Big Bang theory correctly predicts the observed distribution of
elements in the universe (80% hydrogen and 20% helium).
Marking guidelines
Criteria
Marks
 Clear and extensive answer that discusses at least three key elements of the
Big Bang theory and gives THREE pieces of evidence to support it.
5
 A good answer that looks at key features of the theory and evidence that
support it, but includes less than three key elements of the theory or less
than three pieces of the evidence that supports it OR an extensive answer
that is confused or vague.
3–4
 Vague answer that includes one or two correct statements about the Big
Bang theory and/or one or two pieces of evidence that supports the theory.
1–2