physics 9646 2013 c1 promotional examination suggested...

Physics 9646 2013 C1 Promotional Examination Suggested Mark Schemes

These mark schemes are published to serve as a guideline to teachers and students, to indicate the requirements of the test. They show the basis on which the marks are awarded marks, however it is expected that alternative correct answers and unexpected approaches will sometimes be encountered. Due credit will be given at the discretion of the teachers provided the students have shown understanding to the concepts tested. In the same vein, students who arrived at the correct answer or provided a correct statement but otherwise in their working contradicted themselves or showed that they did not really understand the concept will/may not be credited. Mark schemes must be read in conjunction with the question. HCI will not enter into discussions or correspondence in connection with these mark schemes. Incorrect Physics: No credit is given for correct substitution, or subsequent arithmetic, in a physically incorrect equation. Error-carried-forward ECF: Based on a wrong computation from problem, wrong deduction is made. Answers to later work that are consistent with earlier incorrect physics, no matter how obtained, may

be awarded up to full credit for the later work. Answers not worked out:

Mark for final answer will not be awarded. Particular exceptions – e.g. 2 rad – will be decided by the discretion of the marker.

Significant Figures (s.f.): Answers must be given to the correct no. of s.f. or d.p., else one mark will be deducted from the whole

paper. Answers will wrong s.f. will have the "s.f." written at the side of the part of the question as well as the overall marks to the question. One mark is deducted from the whole paper for this mistake.

Gravitational Acceleration (g): The value of g used should be 9.81 m s-2, unless the question specifically mentions that 10 m s-2 can

be used. One mark is taken off from the whole paper if this mistake is made.

2013 HWA CHONG INSTITUTION (COLLEGE SECTION) C1 H2 PHYSICS Promotional Examination Paper ( 01 October 2013)

2

Paper 1 Suggested Solution

1 2 3 4 5 6 7 8 9 10

D D C A C B B C C A

11 12 13 14 15 16 17 18 19 20

C D B A C D B A C C

Suggested solution

1 D

2

2 24

0.2 100 2(10)2 0.071

25.2 2100 1500

o H

o H

o H

AvL

V f

V fL A

L A V f

2 D

∆𝑣 = √6.02+8.02 = 10 m s−1

𝑡𝑎𝑛𝜃 =6.0

8.0= 36.8°

3 C

Distance travelled by X – Distance travelled by Y = 900 m

½ (5 x 5) x 2 + 5 T = 700 T = 135 s

4 A

The only force acting on the projectile is weight which acts vertically downwards, hence the acceleration will be in the vertical direction and no component of acceleration is horizontal.

5 C

The word "must" in the question is crucial in that though all the other forces mentioned might exist, the presence of the particular force mentioned leads to only one other force to "complete" the Newton pair which in this case is the upward pull exerted on the Earth by the student to form the pair force which is the downward pull of the earth (gravity) on the student.

6 B

The weight of the rice is 10g. The upward normal contact force of the scale is 12g. Hence the net upward force is 2g. Using N2L we get 2g = 10a , hence a = 2.0 m s-2.

7 B

Student applies relative speed and conservation of linear momentum correctly

If velocity of approach of 2kg mass is u, then relative speed of approach is also u, if speed of 2kg mass after collision is u/4 then to maintain a speed of separation of u, the speed of the unknown mass after the collision has to be 5u/4. Then applying COM we get momentum before equals 2u and momentum after is 2u/4 +m5u/4 , yielding a value of m as 1.2kg.

Δv = vf + (-vi) (-vi)

vf = 6.0 m s-1

θ


3

8 C

Let length of pole be L.

In equilibrium,

clockwise moment = anticlockwise moment

1

sin 482

omg L T L

Solving, 21.5T N

9 C

Let Tw be the tension and Uw the upthrust in water

Let To be the tension and Uo the upthrust in oil

V is the volume of the sphere

W is the weight of the sphere

T U W

30.0 10.0 20.0w w wU V g W T

o oo o o w w

w w

T W U W V g W V g W U

0.82

30.0 20.0 13.61.00

oT N

10 A

EP

t

21

2mv

Pt

1

22P

v tm

11 C

The student correctly knows that the centripetal force is the net force acting towards the centre of the circle, regardless of whether the circular motion is uniform or not.

12 D

The tension is smallest at the top and largest at the bottom.

13 B

The gravitational field strength is given by the magnitude of the potential gradient. Since the gradient at B is the largest, it has the largest g.

14 A

2

2 2

3 3

164

GM M GMF

RR

15 C

At equilibrium, the speed of the pendulum is the largest, hence its kinetic energy is maximum.


4

At the lowest point of a circular motion, there is a resultant force acting towards the centre of the circle which is the point of pivot of the swing. Tension of the string = Weight of pendulum + centripetal force.

16 D

The amplitude will be higher but the maximum amplitude will still be at the same driver frequency.

17 B

All other options will have the polarizing light being perpendicular to the next polarizing axis, giving zero intensity.

18 A

Diffraction becomes less apparent when the wavelength is smaller than the slit width.

Increasing the frequency will increase the wavelength and hence results in less diffraction.

19 C

All particles oscillate with SHM except at the displacement nodes.

20 C

Interference:

2

Lx

d

Diffraction grating: 1 sin30od

2 1

3

500

sin 301 10 m

500

o

d d

Lx

A B C D

1

Did not re-arrange before calculating fraction uncertainty

Subtracted fractional uncertainties of quantities in the denominator

square the fractional error for f instead of multiplying by 2.

Correct Option

2 Assumed change in velocity = change in speed

Assumed change in velocity = change in speed

Used the vector addition formula Δv = vf + vi

Correct Option

3 Takes area under the graph and equate to 500 m. 700 = ½ (30 + 35) x 5 x 2 + 35

T T = 10.7 s.

Only takes area under the graph for T portion and equates to 900 m.

700 = 35 T T = 20.0 s

Correct Option

Only considered the portion of T.

5T = 700 T = 140 s.

4 Correct Option

The horizontal velocity is a constant but non-zero.

Refer to b Refer to b

5 Correct Option

6 To achieve a net downward force the weight of the rice would have to more than the contact force of the scale on the rice which it is clearly not.

Correct Option

The student takes the "new" weight of the rice as the net force acting on the rice

The student fails to subtract the weight of the rice in arriving at the net force

7 Student works out answer by simple ratios

Correct Option

Student incorrectly calculates speed of unknown mass after collision as 3v/4

Student takes ¼ of incoming mass

8 Use T = 0.5 mg Use sine instead of cosine in equation

Correct Option

Think that L must be given.

9 T = density of oil/ density of water * tension

T = density of water / density of oil * tension

Correct Option T = 30 – 0.82*T


5

10 Correct Option

v proportional to t2

v is linear to t as power is constant

v proportional at the beginning, turn over to achieve terminal velocity

11 For non-uniform circular motion, the resultant force will not be towards the centre.

Misconception that the centripetal force is another force

Correct Option

Misconception that the centripetal force is another force

12 Correct Option

13 Correct Option

14 Correct Option

15 Student knows that at equilibrium position, the speed of the pendulum is the largest. Since kinetic energy is directly related to the square of the velocity, its kinetic energy is maximum. Student may think that the concept of centripetal force is not applicable here.

Student may mixed up the concept of kinetic energy and potential energy at equilibrium. Student may know the concept of centripetal force well.

Correct Option

Student may not understand the concept of oscillation and circular motion well.

16 Correct Option

17 Correct Option

18 Correct Option

Increasing the speed increases the wavelength and makes diffraction more obvious

Neither the wavelength nor

the slit width is dependent

on the amplitude. No effect.

Students may think that

reducing the length of the

bar affects the wavelength

when in actual fact it does

not.

19 pressure antinode = displacement node

Correct Option

20 Correct Option


6

Paper 2 Suggested Solution with Comments

1 (a)

Comments:

The tutors had originally thought that this was a giveaway question as this question was assessed on H1 student in previous batches and they had done well and hence was surprised that this question became quite discriminatory in the end.

Those who knew their concepts of projectile motion well easily secured 9 of the 11 marks, and those who did not know their fundamentals scored badly. Of the 11 marks only the last two were a little challenging requiring a little bit logical reasoning with only the better candidates able to provide coherent and logical arguments.

Overall there has been an improvement in the presentation of working, with many students putting in to present their train of thoughts, these students should be commended. There are however still a group of students who still persist in unclear working therefore put themselves in a big disadvantage especially when the examiner could not figure out where the numbers came from and hence did not give credit.

It was also found that there were a significant number of candidate who had a bad habits of not putting units for physical quantities – this is actually conceptually incorrect as every physical quantity does have its units unless dimensionless.

(a) (i) Time between each strobe is the same : t

Q1 :

21

2ys gt

Q2 :

2 21 1(2 ) 4( )

2 2ys g t gt

- 4 squares from top

Q3 :

2 21 1(3 ) 9( )

2 2ys g t gt

- 9 squares from top

B1 - Correct position of Q3 with label

B1

Q3

P1

P2

P3

Path of P

Path of R


7

Comments:

Most were able to get this part correct.

(a) (ii) Horizontal velocity of the ball P remains constant – gains two squares every time.

Vertical velocity of the ball P should follow the vertical velocity of the ball Q. Hence the vertical displacements of ball P at each strobe should match that of Q.

B1 - Correct position of P1, P2 and P3 with label.

Allow ecf from above for the vertical displacement of P3.

If penalty for no label has been given in (i), do not apply penalty again.

B1

Comments:

Errors made by student in this part included:

No labels given – throwing away marks.

Not being aware that the horizontal velocity remains unchanged and hence putting the horizontal displacement to 1 sq per flash or fall straight after, or having a varying displacement between the flashes of the strobe.

Not reading question carefully or unable to understand the wants of the question and label P1, P2 and P3 on existing balls.

There were a group of students who thought that the horizontal velocity of ball P would become zero after falling off the table. This clearly showed that they do not understand projectile motion.

o If they had read the whole question parts (b) and (c) later would require them to determine quantities like horizontal velocity, resultant and direction of resultant velocity. These are clear hints that there is a horizontal component.

o As this was considered "wrong physics" and a serious conceptual error, no error carried forward was applied for subsequent workings and these students would tend to get zero for (b) and (c).

Students are encouraged to draw diagrams in pencil, as there were a significant who drew in pen and wanted to erase and scribbled all over and at times making the answer ambiguous as they did not properly indicate which dot they wanted and hence were not given credits.

There were also a handful of students who marked dots there were so small that the examiner could not gauge if it was an ink blot from printing or pen marking and hence credit were also not given.

(b) (i) Time taken of ball Q to fall from rest to position to Q2 = Time for 2 strobes

= 2/8.0 = 0.25 s

A1

Comments:

Most were able to successfully calculate this part, common errors include students who took:

1 1

2(0.8)time . The examiner could not figure out why the students did so. Of course there were

also some other weird combinations of numbers which the examiner could not follow the train of thought of the students.

Student forgetting that when counting the time first dot, the first dot should indicate the time t = 0 and instead counted the first dot as time for 1 flash and hence subsequently multiply T by 3. Students must remember in counting e.g. oscillations when we start the stopwatch, that point is t = 0. Hence 3 dots indicate 2 time intervals. (Tree Counting Problem in Primary School Math?)

1 mark was deducted for students who did premature rounding. i.e. they rounded the period 0.125 to 0.13 and then multiply by 2 to get 0.26. Students are reminded and encouraged not to do premature rounding for intermediate values and should leave the intermediate values to higher number of s.f. so that there will not be compounded rounding errors. (Students should not be the source of error!)


8

(b) (ii) Distance fallen by ball Q from rest to position Q2:

2 2 21 1

( ) (0) ( )(9.81 m/s )(0.25 s) 0.30656 m=0.307 m2 2

y ys u t gt

Length of each square grid, L = 0.30656 m / 4 = 0.0766 m

M1

A1

Comments:

Common Errors:

Count 3 squares between Q1 and Q2 and yet take u to be zero. Q1 actually has an initial velocity.

Significant number of transcription error – e.g. taking time to be 0.025 s, or calculating 0.0768 and yet filling answer blank as 0.768 s.

Orders of ten errors! There were also significant number of orders of ten errors where student correctly writing the figures in working and ended up with 0.768 and ended with 12.3 m/s.

Appalling to find a handful who still has not figured out what counts as a significant figure rounding answer to 0.08 and a few who not only rounded to 0.08 and still state "(to 2 s.f.)" …..Examiner fainz…..:<

(c) (i) Horizontal velocity of ball P just before it hits the floor

The horizontal distance remains unchanged. Using the point it launches off the table and P2.

1 1

( )4(0.0766 m)

1.226 m s 1.23 m s0.25

y

xxv u

s

t

A1

Comments:

Common Errors:

Used wrong timing 2 squares -> 0.125 s not 0.25 s

Handful still believe that the horizontal velocity before hitting the floor should go to zero.

Projecting the trajectory and using the squares to make estimation…..not the best method and hence credit was not given.

Very weak candidates could not figure out a simple question like this with some mixing and matching the horizontal and vertical quantities in the equation of motion, not appreciating that the equation of motion is a vector quantity.

Interesting to find that many students are not coherent in their thinking, using 2 squares for each time interval but yet in the diagram drawing only one square, vice versa.

Handful of students simply took 2 x (length of one grid) without dividing the value by the appropriate time. Students may want to develop habit of checking whether the equation that they use is homogeneous in the first place.

(c) (ii) Vertical velocity of ball P just before it hits the floor:

2 2

2 2 1

( ) 2

2 (0 ) 2(9.81)(10x 0.0766 m) 3.88 m s

y y y y

y y y y

v u a s

v u a s

Magnitude of the velocity,

2 2 2 2 1(1.23) (3.88) 4.07 m sx yv v v

M1 A1


9

Direction of the velocity:

1 o

tan3.88

tan ( ) 72.41.23

y

x

v

v

The velocity vector is 72.4o below the horizontal. M1 – Correctly determining of vy

A1 - Correctly determining v (with clear working)

A1 - Correctly determining (with clear working) and stating direction clearly.

[-1] Student did not explain what is or there is no accompanying labelled diagram for

[-1] Student states "to the horizontal", " to the left" or "to the right" downward

A1

Comments:

Common Errors:

Forgetting to square root v

Significant number of students had transference error, omitting decimal places or zeros as they transferred the value here.

Most errors were associated with the ability to describe the direction of vector and includes:

o Using "North, South, East, West" to describe direction. Using compass bearings. Note "The sky is not North, the ground is NOT south. If Australia is due North, will you start burrowing into the ground?"

o Students not familiar with convention of describing direction for vectors. Direction is measured from tail, not arrow head of the vector. (See diagram below)

o Using weird phrases like "towards the right downwards" !!!!

Cambridge does not practice ecf within the question. Hence, many would lose all 3 marks due to calculation. Examiner chose to delink the 2 marks for v and 1 mark for direction.

o Hence if vx is incorrectly calculated 2 marks is taken off.

o But 1 mark awarded if the method and description of direction is correct.

o Those who also described direction incorrectly or were vague also lost the last mark

vx = 1.23 m/s

vy= 3.88 m/s

v

30o

This vector is 30o below horizontal, NOT "30o above.


10

o Those with very glaring conceptual errors however did not get credit.

Apparently there is a significant handful who thinks that whenever the vector is sloped at an angle the angle is 45o. This is the first cohort that have made such a fundamental conceptual error in the examiner's 15 years of teaching…..this is quite alarming!

The examiner was also surprised to find (1st time in 15 years of teaching as well) that there was a significant number of students who successfully calculated vx and vy and v but yet did not proceed to find direction ….and wonders why.

(d) Possible.

Although ball R now has a lower horizontal component of velocity. It also has an initial upward component of velocity and hence a greater time of flight and hence there should be an angle of projection that allows it to gain sufficient horizontal displacement for it to land at P.

Correct explanation must be given (M1) before answer mark can be awarded.

Failure to explain why there is a greater time of flight (i.e. stating that there is now an upward component of velocity) but arriving at the correct conclusion – allows 1 mark.

Failure to consider EITHER "lower vx" OR "vy non zero and hence greater time of flight" and arriving at wrong conclusion warrants zero credit.

A1 M1

Comments:

This was a challenging part for most candidates and was meant to be the discrimination for the question to help sieve out the strong candidates.

Many students did not know how to approach the question and were hence unable to explain phenomenon using sound scientific principles

It was pleasing to note that most students learnt their lesson from Block Test 1 understanding that "explain" questions assesses the students conceptual understanding of the situation and their ability to use scientific principles to reason and form a coherent, logical explanation of the phenomenon and hence made a good attempt.

Common Errors however included :

o assume R and P have the same horizontal speed

o assume R and P have the same time of flight

o only considered one physical quantity but not both. (Note, this is a favourite trick of Cambridge examiners, but alas it assesses higher order thinking as now students must be able to make judgement if changing one quantity will affect all other quantities with may affect the final result. So it is a good point to remember and students should slow down their thinking and examine the details more.)

Students should note that if no explanation is given, no credit will definitely be awarded. Hence is it always worth the while to have a try at it. You never know, you may get compassionate marks or partial marks awarded.

There were also a handful of students who tend to beat around the bush, discussing "if" this happens "then" that is possible but refusing to answer the question with respect to the phenomenon on hand. No credit was given to this group of students.

Of course, there would be another handful of students who will change the situation entirely and talk about Ball R projecting at same angle as before and same speed….(hm…then of course they will land at the same spot cause they will have the same trajectory)….of course no credit was awarded as well.

Another group of student tried to use energy to explain, but fail to see that it is an approach that cannot be used cause the energy approach does not differentiate the point of landing.

Students tend to be lose in their use of terms. Mixing terms like "constant" and "equal" interchangeably. Constant means the value do not change with time. Equal compares two values. Hence in this case, for both balls their horizontal velocities are constant, but their horizontal velocities are not equal.

Because of their lose use of terminology in Science, some students also become less precise in their understanding of physics e.g. many thought that since speed is same for both, the horizontal components must also be the same for both balls – clearly demonstrating that they do not understand


11

the properties of vectors.

There were also a significant number who were not specific in use of terms, and simply state that "horizontal velocity / time of flight / velocity is different". Generally, we would like to know "how different" is different? Tell the examiners whether the terms have become bigger or smaller. In this case, it is possible for the increase in the time of flight to compensate the effects brought about by the decrease in horizontal component.

Other common errors :

o Serious Misconceptions: Students thinking extra energy is required to overcome horizontal motion and hence smaller range.

o Using the horizontal range equation, without understanding that it is not valid in this case. By the way, students need to know that the equation cannot be used without proof. It is stated in the SEAB syllabus that students can only use fundamental equations, all other equations needs to be proven before use.

o Students who think, that trajectory must be same / horizontal speed must be the same / vertical speed must be the same / time of flight / same angle of velocity at the point of landing / projection must be the same before they can land on same spot

o Students who actually think that launching at an angle have a higher energy? Or greater potential energy? or greater kinetic energy.

2 (a) Net external force acting on the body is zero.

Net torque (or moment) acting on the body about any point is zero.

B1

B1

Comments:

1. Common mistake of not including “about any point” for rotational equilibrium. 2. Some students rephrase the rotational equilibrium as “Net torque about any point on the body is

zero”. This is incorrect because the pivot can be outside the body. 3. It is a mistake to state “no net force” or “no net torque”. Correct version is “zero net force” or “zero net

torque”. Force and torque have magnitudes. Zero is a magnitude. “No” is not a magnitude.

(b) (i)

Draw and label the 4 tensions correctly.

Draw and label the weight of man and plank correctly.

Remarks:

It is a mistake to draw normal contact forces between man and plank as these are internal forces for this system that comprises man and plank. Only external forces are indicated on free body diagrams.

A1

A1

T T

T T

Wman

Wplank

T: tension in rope

Wplank: weight of plank

Wman: weight of man


12

(ii) Consider the free body diagram of the system consisting of the man and plank.

4 70 10 9.81T

196T N Tension = 196 N

M1

A1

(iii) Let R be the normal reaction the man exerts on the plank.

Based on the free body diagram of the plank:

W is the weight of the plank.

2R m g T

2 196 (10.0)9.81 294R N downwards

ALTERNATIVE

Let R be the normal reaction the plank exerts on the man.

Based on the free body diagram of the man

W is the weight of the man.

2R T m g

(70.0)9.81 2 196 294.7R N upwards

Based on Newton’s 3rd Law, the normal reaction the

man exerts on the plank (R’) is equal and opposite to R.

Therefore, ' 294.7R N downwards.

M1

A1

(iv) The tension of the right rope will decrease.

Using the point where the left rope is attached to the plank as the pivot, the clockwise moment provided normal reaction force exerted by the man on the plank decreases because the perpendicular distance (pivot length) decreases.

The tension must decrease to decrease the anti-clockwise moment.

M1

A1

Comments:

1. This question should be approached from the concept of rotational equilibrium. 2. A pivot defines the axis of rotation. Some students “take moments about the left

(or right) rope”. The rope is not the axis of rotation, hence this is incorrect. In fact, the axis of rotation is perpendicular to the plane of the ropes.

3. The position of the pivot must be defined when discussing torque or moments. A clockwise moment about one pivot can become an anti-clockwise moment about another pivot. For example, the weight of the man produces a clockwise moment about the left end of the plank but an anti-clockwise moment about the right end of the plank.

T T R

W

T T

W

R


13

3 (a) Forces acting: weight downwards and friction acting inwards, with proper labels.

Length of arrow for normal force must be equal to length of arrow for weight and length of arrow for friction must be half the length of the arrow for the normal force

B1

B1

Comments:

No marks were given if either the arrow for the friction or the arrow of the weight was missing, if one or both of them were pointing in the wrong direction, or if there were any extra arrows (e.g., to indicate the centripetal force). Centripetal force is a net force and should not be drawn in a free-body-diagram.

The first mark was given if the arrows had the correct point of action, correct direction and proper labels. The second mark was given if both arrows had the correct length.

The arrow indicating the weight has to be the same length as the arrow indicating the normal force, as the net force in the vertical direction is zero. Some students drew an extra arrow for the normal force, which shows that they did not read properly: the normal force was indicated by the arrow labelled R.

The arrow indicating the friction has to be half the length of the arrow indicating the normal force. This was also indicated in the question: the magnitude of the friction is 0.50 R, where R is the normal force.

(b) The centripetal force is provided by the friction between the tires and the road.

f= Fc

0.50R = 75 v2 / r

0.50W = m v2 /

0.50 m g = m v2 /

r = v2 (0.50 g) = (28 / 3.6)2 / (0.5 x 9.81) = 12 m (2 s.f.)

M1

A1

Comments:

Generally, this question was well done. Some students forgot to convert km h-1 to m s-1, while others forgot to convert mass to weight. This is simply carelessness.

(c) No he can’t. The road will be wet and the maximum friction will be lower.

The frictional force will be less than the centripetal force required.

Alternative for the second mark: as the friction is lowered, there is a net clockwise moment and David will topple.

B1

B1

Comments:

The first mark was practically free and was given if the students stated that David cannot round the corner at the same speed, because the friction is reduced as a result of the rain.

The second mark was not free. Many students used, implicitly or explicitly, that the friction is equal to the

W

R

N

f

W: weight of man and

bicycle

f: friction


14

centripetal force, or f = m v2 / r. However, if David cannot round the corner, he may actually skid and go straight (Newton’s first law), in which case there is no circular motion and hence also no centripetal force.

The correct argumentation is that, since the friction is lower, it may not be large enough to provide the centripetal force required to round the same corner at the same speed.

Another error was to say that 0.5R becomes lower. Rather, the frictional force becomes less than 0.5R, while R remains the same (in technical terms, the coefficient of friction becomes less, so that the friction < 0.5R).

4 (a) The gravitational potential at a point in a gravitational field is the work done per unit mass, by an external force, in bringing the mass from infinity to the point without any change to the kinetic energy.

B2

Comments:

Students were penalised one mark each (to a maximum of two) for each of the following phrases that were missing: “at a point”, “work done per unit mass”, “external force” and “from infinity to the point”.

Many students included the phrase “to the point” or “to that point” or even “to that particular point”, without mentioning which point they were talking about. This shows that they memorised the definition (incompletely), possibly without understanding it, and did not read back what they had written.

(b) The centripetal force is provided by the gravitational force.

G M m / r2 = m v2 / r = m (2π r / T)2 / r

(2π / T)2 = G M / r3

M = (2π / T)2 r3 / G = 4π2 (9378 x 103)3 / [(7.6 x 3600)2 x (6.67 x 10-11)] = 6.5 x 1023 kg.

M1

A1

Comments:

The first step was required. Just a bunch of equations, with possibly numbers plugged in, will not get you the mark.

Students were also penalised if the jumped from the final equation straight into the final value, without showing which values they plugged in. They have to show that they use the correct values.

Some students misinterpreted the term “orbital distance” and took this to be the distance from the respective moon to the surface of Mars, or twice the radius between Mars and the moon, or the diameter of the moon’s orbit. For r they then took, e.g., the orbital distance plus the radius of Mars.

Many students bluffed the last step, plugged in some (wrong) numbers, gave the final answer and wrote that they had proven what had to be proven. The smarter ones, who figured out that the last step was wrong, went back to check which numbers would give them the correct final answer, thus scoring that one mark.

(c) The escape velocity is the minimum velocity required at the surface of Mars to reach infinity with no kinetic energy left.

KEat surface + PEat surface = KEat infinity + PEat infinity

KEat surface + PEat surface = 0

KEat surface = -PEat surface

½ m vesc2 = G M m / r

vesc = (2 G M / r)1/2 = 5.1 x 103 m s-1 (2 s.f.)

M1

C1

A1

Comments:

Some students started by stating that “loss in kinetic energy equals gain in potential energy” or even (and much worse) that “kinetic energy equals gravitational potential energy”. Others started with the equation ½ m v2 = G M m / r, or even with v = (2 G M / r)1/2. This was not accepted. The students have to indicate that the total energy is conserved and that, therefore, the velocity is equal to the escape velocity if and only if the total energy at infinity is zero.


15

Occasionally, students used m g h for the gravitational potential energy. However, that formula can only be used in a uniform gravitational field, not when moving from Mars to infinity.

5 Overall comments:

Most students can do very well for this question.

The main problems for those who did not do very well for this question are:

(i) never learn the proper definition of SHM (ii) careless while keying in the numbers into the calculator and hence did not get the correct answers (iii) not clear about the difference between angular frequency and frequency (iv) did not realize that the displacement is a cosine graph and hence the potential energy-time graph

is a cosine2 graph. Some did not sketch the graph with decreasing amplitude.

(a) Simple harmonic motion is a periodic motion in which the acceleration of the oscillator

is directly proportional to the displacement from its equilibrium point, and

is always in opposite direction to the displacement.

B1

B1

(b) (i) For the system on the horizontal plane, the total energy is given by

E = ½mv2

E = ½m(A)2

0.060 = ½(0.30)()2 (0.050)2

=[ 2(0.060)

(0.3)(0.05)2 ]1

2= 12.6 rad s-1

M1

(ii) 1 At x = 3.0 cm,

𝑣 = √𝐴2 − 𝑥2

v=12.6√0.0502-0.0302

=0.504 m s-1

2 a = ()2x

=(12.6)2(0.030)

=4.76 m s-2

3 Potential energy, Ep = ½m(x)2

Ep = ½(0.30)(12.6x0.03)2 = 0.0214 J

A1

A1

A1


16

(c)

Shape of sketch correct (cos2)

Amplitude decay

Correct period

B1

B1

B1

6 (a) (ii)

A1

Comments:

About 40% of the students could not identify the locations of maximum/minimum displacements. Students who marked only 1 point correctly were not awarded with any mark.

(i) Sound waves cannot be polarized because they are longitudinal waves.

Longitudinal waves propagate parallel to the direction of oscillation.

A1

A1

Comments:

In general, this question was done well. Only few thought that it was a transverse wave hence it should be polarized. Students lost a mark for not being able to explain the property of longitudinal wave which is the direction of oscillation is parallel to the direction of wave propagation.

(b) (i) Pressure Amplitude = 0.50 Pa. A1

Comments: This question was done pretty well. Almost all students got it correct.

(ii) 0.25 = 0.50 × sin(𝜔𝑡 + 𝜑), for t = 0, Phase angle: 𝜑 =𝜋

6 rad = 0.524 rad A1

Comments:

This question posed a great challenge to the students. Only about 45% of students managed to get the phase angle. Students who are not well versed in manipulating trigonometric functions could not find the phase angle.

X X X

P/Pa


17

(iii) Angular Frequency: 𝜔 =2𝜋

𝑇=

2𝜋

(1190−350)×10−6𝑠= 7.5 × 103 rad s-1. A1

Comments:

Although more students managed to calculate the angular frequency correctly, students who read off period wrongly from the graph could not get the right answer. Another common mistake was to think that micro (μ) refers to 10-9. Some students alternatively use the sine function to get the angular frequency. However, majority of these students did not use sine function correctly and hence could not find the angular velocity.

(iv) Wavelength: 𝜆 = 𝑣𝑇 = 340 × (1190 − 350) × 10−6 = 0.29 m. A1

Comments:

Although this question was regarded as easy, about 45% of students could not find the wavelength. The

common mistake is using the answer in b(iii) “angular frequency” as frequency in the relation v = fλ.

(c) A point source propagates power uniformly in all directions.

Intensity is power per unit surface area.

At distance r, power is spread over a spherical surface of area 4πr2.

B1

B1

B1

Comments: This question was challenging for most of the students. Since the question is 3 marks, students’ answers should have been explicit. The common mistakes

not being able to mention about the assumption that power does not diminish as the sound wave travels radially

not being able to realize that power is uniformly distributed in all directions

not being able to relate the spread of power to spherical surface area

using the relationship between intensity and amplitude to explain the inverse square law

7 Overall Comments: This question is very poorly done, with the mode at 2 marks out of 9. There were also a

significant number who gave up on it, leaving whole or most of the question blank. For a question of this

standard, an average student is expected to obtain 5 marks for the first 2 parts. The remaining 2 parts are the

more difficult parts in terms of understanding data provided in context as well as details in answering qualitative

question. If you did not obtain 5 marks, you basically need to restudy this chapter during the holidays so that

you can do better next year. FYI, Superposition is a big chapter in the GCE A levels, you can’t afford not to

know.

(a) The principle of superposition states that when two or more waves of the same kind simultaneously exist in the same region of space, the resultant displacement at any point is given by the vector sum of displacements of all waves at that point.

B2

Comments:

Many students did not memorise the statement which is totally unacceptable. There aren’t many definitions or statements that you need to memorise for this chapter plus this is the most popular one in the syllabus. So you can only blame yourself for not doing the basic work.

The question is marked by overall concept then by deduction since there were student who wrote nonsense so it is unfair to mark by deduction and award them one mark just because they have some keyword in their statements.


18

1 Mark is deducted for missing the following terms:

two or more: the principle is not limited to just two waves but can be applied to infinite number of waves

of the same kind: stating they are similar is not accepted since the exact property is not stated. Neither is same amplitude, wavelengths, frequencies necessary for the principle to be applied. It is really a very general principle.

At that point/at that instant

1 Mark is deducted when these words are used:

Amplitudes instead of displacements.

Waves interfere…: Concept is interference is defined based on principle of superposition so you cannot use interference to explain principle of superposition.

Particles: The principle can be applied to EM waves such as light and microwaves which do not require a medium and has no “particles” in them.

(b)(i) The sound wave reflects off the bench.

The incident and reflected waves superpose/interfere with each other.

Maximum amplitude is detected at when the waves meet in phase and interfere constructively. Minimum amplitude is detected when the waves meet in antiphase and interfere destructively.

OR

The incident and reflected waves superpose/interfere to form a stationary wave.

The maximas correspond to positions of anitnodes while the minimas correspond to the positions of nodes.

B1

B1

B1

Comments:

the incident and reflected waves do not have the same amplitude, which explains why the minimas of resultant wave do not have zero amplitude. Some students who explained the formation of stationary waves stated that the 2 waves have the same amplitude. Though it was initially penalised, the deduction was removed and the standard of marking relaxed as this question was poorly attempted.

Marks are not awarded when the two waves superposing are not explicitly identified.

Some students explained constructive and destructive interference using path difference approach. The concept we want is phase difference as reflection of waves can sometimes cause a phase change in the wave, and this affects the conditions for path difference that will result in constructive or destructive interference. This marking point was also removed due to poor attempts by students. Providing the condition leading to constructive interference either using path or phase difference was accepted in the end. However, just stating that the maximas are constructive and minimas destructive warrants no marks.

Many students are confused between path difference and phase difference. Path difference is the difference in the distance travelled by the two waves. Its units would be

metres, or in terms of . Phase difference is an angle quantity, units would be degree or

radians or in terms of . They cannot be used interchangeably.

Many students did not even realise this is a superposition question and instead commented on sound being a longitudinal wave and the maximas and mininas are regions of compression and rarefaction. This shows poor exam skills; the fact that you are asked to state the principle of superposition at the start of the question already hints at the application you are going to be tested on. Besides it is not possible to obtain the wave profile graph for a single sound wave using the set-up given. Please refer to Chapter 9 Waves Tutorial D10 part (c) for the experimental set-ups.

There are also a significant number of students who thought the loud speaker produces many sound waves and these sound waves are interfering amongst themselves. If this was true, the detector (microphone) would probably be moving along either an antinodal line and hence would most likely be constantly high and not alternating between high and low.


19

Unacceptable misconceptions include students who believed that the microphone emits sound waves and is the “other” source in the question. The microphone is just a detector. One way to analyse the situation is to actually remove the detector and review what is happening. Also it is not very reasonable to state that the microphone absorbs the energy of the sound wave so the amplitude of the wave becomes smaller after passing through the microphone. This effect is likely to be negligible.

(b) (ii) Dips in the waveform Position of nodes, and

Distance between 2 consecutive nodes = 2

Therefore, 3( ) 18.0 2.6 cm2

= 10.3 cm

Hence speed of the sound wave 3 2 1(3.20 10 Hz)(10.3 10 m) = 330 m sv f

Deduct one mark from this part if

student does not use three consecutive nodes/antinodes to determine wavelength

student read off incorrectly

M1

A1

Comments:

This part is extremely poorly attempted. Many students use the distance between two consecutive maximas or minimas as the wavelength. Students who explained using stationary waves were generally able to deduce that the distance between adjacent maximas or minimas is only half a wavelength. However, because the distance between adjacent nodes/antonides actually do not remain the same, they are expected to use the method of averages to find the wavelength. It is also a shame that many weren’t able to read the values off the graph accurately or correctly as they did not notice the scale that was used.

For students who did not approach the question using stationary waves but path difference, it is important to note that when we move from one maxima to another maxima, there is an increase of path difference of one wavelength. However, in context of the question, it is represented by 2 times of x, as the reflected wave travels an addition 2x when the detector moves by x alone. The conclusion is the same as using stationary wave, that the distance between 2 maxima/minimas is only half a wavelength.

(b)

(iii)

As x increases, the reflected wave is weaker (because the reflected wave spreads out /disperses and hence the intensity is reduced) and the incident wave stronger.

There is more complete cancellation of the incident and reflected waves near the

B1

Wave from loudspeaker reaches the detector.

Wave from loudspeaker reaches the bench and

reflects. Reflected wave travels upwards and

reaches the detector. Additional distance

travelled by reflected wave is 2x.


20

bench-top and less complete cancellation as x increases.

OR

As x increases, the difference in amplitudes becomes more significant and the amplitude is the reflected is less able to alter the amplitude of the incident wave, hence the contrast between constructive and destructive interference decreases.

B1

Comments:

The biggest problem with students’ answers is the lack of a subject in their statements. Eg “distance travelled increases so the amplitude decreases.” There are 2 waves in this application so which wave are you talking about? We cannot emphasise enough the need to be explicit, so you have to craft your answer properly and with details.

Many students were able to identify that the amplitude of the reflected wave is smaller using the concept of spreading out of the wave, however they did not apply the same concept to the incident/direct wave from the loudspeaker. Hence no marks was awarded.

Many also attributed the drop in the amplitude of the reflected wave to loss of energy when the reflection, which is a reasonable assumption. Unfortunately, it would not be able to explain the observation here. If the amplitude of the incident and reflected wave are not the same, we would just not be able to get complete cancellation for destructive interference, but the magnitude of the destructive interference will be a constant. The same argument can be applied to constructive interference.

There was also a number who attributed the loss in amplitude of the reflected wave to air resistance/friction and drag. The particles are oscillating about their equilibrium position with SHM. If there was damping of any sort, all particles would experience the same effect and suffer the same loss in amplitude. It would not be decreasing with increasing distance from x. Besides, if there is damping then both the incident and reflected wave will be affected, not just the reflected wave.

There were a huge number of students who obviously did not read past tests’ markers’ reports. It was already emphasised previously that you cannot use or state equations and manipulated them and use them for an “explain” question. All arguments must be presented in full statements, eg intensity is inversely proportional to the square of the distance. Also, the equations were totally undefined with symbols used everywhere, so it is really just a waste of time, How would the examiner know what “I” or “r” or “A” stand for? We hope this will be the last time we have to comment on this.

It was actually heartening to know that many students are able to quote “intensity is inversely proportional to the square if the distance” relationship. However, this relationship strictly cannot be applied here. This relationship is only true with you have a point source of constant power. In this context, the loudspeaker is unlikely to be a point source giving out spherical waves, spreading its energy over a spherical shell. However the idea that the sound wave is likely to spread out is true so with increasing distance the intensity and amplitude should drop.

8 (a) The total momentum of a system is unchanged/constant

if no net external force acts on the system.

B1

B1

Comment:

This part is well answered. Some students lost marks because they omitted the word “net”. PCOM is applicable even when there are external forces, as long as the external forces cancel themselves out.

Only a very small minority still state PCOM in the narrow context of collisions.

Students are advised not to repeat the word “conserve” since it does not add more clarity to the answer. Words like “remain unchanged”, “same”, “constant” are preferred.


21

(b)(i) By principle pf conservation of momentum:

(6.0)(5.0) + (8.0)(-3.0) = (6.0+8.0)v

v = 0.43 m s-2

M1

Comment: This part is well answered.

(b)(ii) WD = KEf-KEi = ½ (6.0)(0.43)2 – ½ (6.0(5.0)2

= -74.4 J

Must be negative. Only 1 mark if positive.

M1

A1

Comment:

Many students thought that the negative sign in their answer was a mistake. They put up the modulus signs, or they did “initial – final” instead, and ended up losing the mark. Since the 6.0 kg lost KE, the work done on it must be negative.

Is (22-12) equal to (2-1)2? Too many students are making the algebraic mistake of writing change in KE as ½ m (v-u)2

Some students calculated the loss in KE of the 8.0 kg instead. Perhaps they thought that the gain in KE of the 6.0 kg is the loss in KE of the 8.0 kg. Unfortunately, this is not so.

Some students used the WD=Fs approach. While most of them successfully calculated the average force (using <F>=Δp/Δt), most of them calculated s as either vfΔt or viΔt. The average velocity should have been used. (<v>Δt).

(b)(iii) Impulse = Δp = 8.0(0.43 – (-3.0))

= 27.4 N s

M1

A1

Comment:

Many students did not know impulse can be equated to change in momentum. Many obtained their answer in a long roundabout way. Like impulse = m (v-u)/Δt x Δt.

Many students did not do the double negate and ended up with the difference in speed, rather than the difference in velocity.

O (c)(i) Since trolleys exert equal (but opposite) impulses on each other, the (magnitude of) change of momentum are the same for both trolleys.

OR

Since trolleys exert equal (but opposite) forces on each other, the (magnitude of) rate of change of momentum are the same for both trolleys.

The 8.0 kg trolley comes to rest first because it had a smaller initial momentum.

B1

B1

Comment:

Many students were comparing either only inertia or only velocity of the trolleys.

Of those who were correctly comparing momentum, many did not point out that Newton’s 3rd Law is at play here.

Some students formed their conclusion based on the initial KE of the trolleys. Unfortunately, there is no law that says that both trolleys will be losing KE at the same rate. So this approach is flawed.

Many students still had not internalised Newton’s 3rd Law, and thought that one trolley exerted a larger force on the other.

The spring was often the focal point of students’ analysis. Some students argued that the spring being attached to the 8-kg trolley will push the 6-kg more. Others argued the exact opposite because


22

the spring helps to cushion the blow. Many students thought the EPE stored in the spring would be passed onto the 8-kg trolley only. Yet others thought the spring should pass on the energy onto the 6-kg only.

(c) (ii) 1. By PCOM: Total initial p = Total final momentum at this instant

(6.0)(5.0) + (8.0)(-3.0) = (6.0)(-2.0)+(8.0)v

v = 2.25 m s-1

2. By PCOE: Initial (KE + EPE) = Final (KE + EPE) at this instant

½ (6.0)(5.0)2 + ½ (8.0)(3.0)2

= ½ (6.0)(2.0)2 + ½ (8.0)(2.25)2 + EPE

EPE = 78.75 J

½ k e2 =78.75

e = 0.407 m

M1

M1

C1

A1

Comments:

Part 1 is generally well answered. Part 2 is not.

Students who used the dynamics approach had no chance. Of those who used the energy approach, some equated EPE to KE, instead of Gain in EPE to Loss in KE. Of those who tried to calculate loss in KE, some did not include the KE of the 6 kg trolley. But the 6 kg trolley is part of the system.

Too many students could not recall the formula for EPE. Popular mistakes were ½ kx, kx2, and kx.

(d)(i)

Correct final velocity for trolley B. (Worked out using Relative Speed of Approach = Relative Speed of Seperation)

B1

Comments:

Since it’s an elastic collision, the two trolleys approach and separate at the same relative speed.

Almost all students made keen attempts to score this mark. A popular bet was that uB=uA, which would have been the correct outcome for a completely inelastic collision.

Velocity

Time

uA

uB


23

(d)(ii)

Force is zero before and after collision, and peaks at the middle of collision.

Forces are equal but opposite and labeled correctly. (FA must be the negative pulse)

B1

B1

Comments:

The shape of FA can be obtained directly from the gradient of the v-t graph in Fig 8.4a. FB would just be the opposite of FA since they are action-reaction pair.

Most common answers were blanks, followed by

Violation of Newton’s 3rd Law

Does not tally with the v-t graph, which shows that acceleration, and thus net force is zero at the beginning and end of the collision.

Peaks at different time, which contradicts Newton’s 3rd Law.

Velocity

Time

uA

Force

FB

FA


24

Totally wrong. Students could be sketching the momentum graphs.

(d)(iii)

KE is equal before and after collision, dips only during the collision.

EPE and KE are mirror images of each other during the collision.

B1

B1

Comments:

Since collision is elastic, total KE before and after collision should be the same. KE is temporarily stored as EPE in the spring during the collision, but since total energy must be conserved, KE and EPE should be mirror images of each other.

This is the instant in time when there is maximum loss of KE, and thus maximum gain of EPE. These two insights were however not required for answering the question.

The v-t graph in Fig 8.4a, if drawn with a little more thought, allows one to make the following two conclusions: 1) that trolley A comes to rest before trolley B, and 2) both trolleys travel at the same speed at the mid-point of the collision. Point 1 tells us that total KE should not reach zero. Point 2 tells us that maximum KE loss (and thus maximum EPE gain) occurs at the mid-point of the collision. Students however were not penalised for neglecting the last 2 points.

Many blanks. The more common flawed answers are:

Totally wrong. Students probably had SHM in mind.

Velocity

Time

uA

Energy

KE

EPE


25

The dip in KE is more than the rise in EPE, violating PCOE.

Many students did not take care to draw EPE and KE to be mirror images of each other.

This reveals students’ misconception that in an elastic collision, KE is conserved throughout the collision (correct understanding is KE is conserved after the collision, not during). This is impossible as during some parts of the collision; both trolleys slow down at the same time.

Some student seemed to have in mind an inelastic collision.


26

Paper 3 Suggested Solution and Mark Scheme with comments

Basic Procedure

[1]

A description of a monochromatic light source shining through the glass slide, and the transmitted light is

incident onto a screen located some distance away, with attempts to determine d using dsinθ = nλ.

Diagram

[1]

Clear, workable labelled diagram showing

Correct arrangement of laser -> glass slide -> screen

EITHER: A laser pointer/ laser source

OR: Monochromatic sodium light with a pin hole or single slit to create point source beam

OR: White light source with monochromatic filters

Method of illuminating the slide: Laser beam can be directly pointed at the glass slide since it is already

a coherent beam, but other non-coherent light sources must go through a single slit or a pinhole to

create a point source.

Methods

[3]

Characteristics of the light source:

Monochromatic and coherent across the illuminated glass slide so as to produce observable interference

fringes.

Determination of n:

Description of the interference fringes with a monochromatic light source: a central bright fringe (zeroth

order), followed by alternating bright and dark fringes symmetrically about the zeroth order (First bright

fringe after the zeroth order is identified correctly as the first order maximum, and so on).

Note: They need not mention the fringes are circular or arranged in circular patterns. They just need to

recognize zeroth order or central maximum is the undiffracted beam, and the next bright fringe beside it

is the first order and so on.

A method of determining angle θ of the bright fringe

for the corresponding order n of the bright fringe. If trigonometric ratios are used, distances must be

measured with a ruler (up to 1 m) or measuring tape (>1m). Direct angle measurements must be

conducted with a protractor or spectrometer

Analysis of data &

results.[1]

Calculate d from the grating formula: d = nλ / sinθ

or from the gradient of a suitably linearized form of the equation.

Control of other

Variables [1]

Any one of this explicitly mentioned:

Ensuring the glass slide and screen is perpendicular to the laser beam using a set square.

Ensuring distance between glass slide and the screen is fixed through measurement with a measuring tape or by marking down their positions.

Other Details

(to improve

reliability of data &

results. Explicit

mention or

explanation

expected for full

credit)

[max 4]

Any good further design details, some of these might be:

Conducting the experiment for six monochromatic light sources with different λ each time, plot the graph

of nλ against sin θ, and calculate d as the gradient of the graph.

NO CREDIT for changing distance between the glass slide and screen to generate additional readings.

Using higher order maxima to reduce percentage error in d.

Using large distance between glass slide and the screen to reduce percentage error in d (conduct

preliminary experiment to determine a suitable distance that will produce a significant separation in the

interference fringes on the screen)

Measuring the angle θ in ≥ 2 different directions for a given n and calculate average θ.

Measuring the position of the peak of the bright fringe more accurately using a light intensity meter with

small sensor, instead of judging the peak with a human eye.

Conduct the experiment in a dark room to improve the visibility of the fringe pattern

Calculate average d from different orders of n for the same λ (or from different λ for the same n)

Pasting a graph paper on the screen to aid in measuring the fringe position/separation

OTHERS: Discussion about benefits or using red light (longer wavelength -> larger θ for the same order

-> lower fractional uncertainty in θ)

Safety Precautions

[1]

Any relevant safety precaution, e.g:

Wear laser safety glasses (or laser safety goggles) to protect the eyes from laser beam or bright

monochromatic light source (At least need to mention goggles used for this purpose)

The glass slide is fragile, clamp it gently or use a proper holder.

OTHERS:


27

Common mistakes: Many students

Did not explain the required question about the characteristic of light to produce observable interference fringes.

Did not use laser when we have already shown in Lab 12 and in lecture that the laser is used to show the diffraction grating interference fringes.

Assumed that the fringes are equidistant on the screen. This is not true.

Did not state the proper instrument to measure distances (metre ruler, measuring tape)

Did not explain how to identify the order of the bright fringes.

physics 9646 2013 c1 promotional examination suggested...

Documents