18ibtopic5 20200922 [119 marks]

18IBTopic5_20200922 [119 marks]

1. A negatively charged particle in a uniform gravitational field is positionedmid-way between two charged conducting plates.

The potential difference between the plates is adjusted until the particle is held atrest relative to the plates.What change will cause the particle to accelerate downwards relative to theplates? A. Decreasing the charge on the particleB. Decreasing the separation of the platesC. Increasing the length of the platesD. Increasing the potential difference between the plates

MarkschemeA

Examiners report[N/A]

2. A thin copper wire and a thick copper wire are connected in series to anelectric cell. Which quantity will be greater in the thin wire?A. CurrentB. Number of free charge carriers per unit volumeC. Net number of charge carriers crossing a section of a wire every secondD. Drift speed of the charge carriers

[1 mark]

[1 mark]

MarkschemeD


3. The diagram shows a resistor network. The potential difference betweenX and Y is 8.0 V.

What is the current in the 5Ω resistor?A. 1.0AB. 1.6AC. 2.0AD. 3.0A

MarkschemeA


[1 mark]

4. When a wire with an electric current I is placed in a magnetic field ofstrength B it experiences a magnetic force F. What is the direction of F?A. In a direction determined by I onlyB. In a direction determined by B onlyC. In the plane containing I and BD. At 90° to the plane containing I and B

MarkschemeD


5. Two power supplies, one of constant emf 24 V and the other of variableemf P, are connected to two resistors as shown. Both power supplies havenegligible internal resistances.

What is the magnitude of P for the reading on the ammeter to be zero?A. ZeroB. 6 VC. 8 VD. 18 V

MarkschemeB

[1 mark]

[1 mark]


6. The force acting between two point charges is when the separation ofthe charges is . What is the force between the charges when theseparation is increased to ?A.

B.

C.

D.

MarkschemeC


Fx

3xF3

F

3x2

F9

F

9x2

7. A capacitor of capacitance 1.0 μF stores a charge of 15 μC. The capacitoris discharged through a 25 Ω resistor. What is the maximum current in theresistor?A. 0.60 mAB. 1.7 mAC. 0.60 AD. 1.7 A

MarkschemeC

[1 mark]

[1 mark]


8. Two cells each of emf 9.0 V and internal resistance 3.0 Ω are connectedin series. A 12.0 Ω resistor is connected in series to the cells. What is thecurrent in the resistor?A. 0.50 AB. 0.75 AC. 1.0 AD. 1.5 A

MarkschemeC

Examiners reportThis question was well answered by candidates and had a higherdiscrimination index.

[1 mark]

9. Charge flows through a liquid. The charge flow is made up of positive andnegative ions. In one second 0.10 C of negative ions flow in one directionand 0.10 C of positive ions flow in the opposite direction.What is the magnitude of the electric current flowing through the liquid?A. 0 AB. 0.05 AC. 0.10 AD. 0.20 A

MarkschemeD


[1 mark]

10. A beam of negative ions flows in the plane of the page through themagnetic field due to two bar magnets.

What is the direction in which the negative ions will be deflected?A. Out of the page B. Into the page XC. Up the page ↑D. Down the page ↓

MarkschemeA

Examiners reportThis question was well answered by candidates.

⊙

[1 mark]

11. A particle with a charge ne is accelerated through a potential differenceV.What is the magnitude of the work done on the particle?A. B. C.

D.

MarkschemeB


eV

neV

nVe

eVn

[1 mark]

12. The resistance of component X decreases when the intensity of lightincident on it increases. X is connected in series with a cell of negligibleinternal resistance and a resistor of fixed resistance. The ammeter and voltmeterare ideal.

What is the change in the reading on the ammeter and the change in the readingon the voltmeter when the light incident on X is increased?

MarkschemeA


[1 mark]

13. What is the unit of electrical potential difference expressed infundamental SI units?A. kg m s CB. kg m s CC. kg m s AD. kg m s A

MarkschemeC

Examiners reportThe most popular answer was B giving a low discrimination index for thisquestion. It should be a relatively straightforward question provided thecandidate can remember which of ‘C’ or ‘A’ is the fundamental unit.

-1 -1

2 -2 -1

2 -3 -1

2 -1

[1 mark]

14. Three resistors of resistance 1.0 Ω, 6.0 Ω and 6.0 Ω are connected asshown. The voltmeter is ideal and the cell has an emf of 12 V withnegligible internal resistance.

What is the reading on the voltmeter?A. 3.0 VB. 4.0 VC. 8.0 VD. 9.0 V

MarkschemeD

Examiners reportMost candidates at both levels gave option A as the correct response insteadof D. This would indicate that they have misread the diagram thinking thevoltmeter was across the 1.0Ω resistor not the parallel combination.

[1 mark]

15. A horizontal wire PQ lies perpendicular to a uniform horizontal magneticfield.

A length of 0.25 m of the wire is subject to a magnetic field strength of 40 mT. Adownward magnetic force of 60 mN acts on the wire.What is the magnitude and direction of the current in the wire?

MarkschemeB


[1 mark]

16. Two parallel plates are a distance apart with a potential differencebetween them. A point charge moves from the negatively charged plateto the positively charged plate. The charge gains kinetic energy W. The distancebetween the plates is doubled and the potential difference between them ishalved. What is the kinetic energy gained by an identical charge moving betweenthese plates?A. B. WC. 2WD. 4W

MarkschemeA

Examiners reportThe correct response (A) was the most common from candidates, however asignificant number of candidates appeared unsure of the impact of thedistance between plates and (incorrectly) selected response B.

W2

[1 mark]

17. A resistor of resistance R is connected to a fully charged cell of negligibleinternal resistance. A constant power P is dissipated in the resistor andthe cell discharges in time t. An identical cell is connected in series with twoidentical resistors each of resistance R.What is the power dissipated in each resistor and the time taken to discharge thecell?

MarkschemeB

Examiners reportThis question was not well answered, with fewer than 25 % of candidatescorrectly selecting response B. Furthermore, the discrimination index for thisquestion was remarkably low, suggesting this question would provide richclassroom discussion.

[1 mark]

18. Two currents of 3 A and 1 A are established in the same direction throughtwo parallel straight wires R and S.

What is correct about the magnetic forces acting on each wire?A. Both wires exert equal magnitude attractive forces on each other.B. Both wires exert equal magnitude repulsive forces on each other.C. Wire R exerts a larger magnitude attractive force on wire S.D. Wire R exerts a larger magnitude repulsive force on wire S.

MarkschemeA

Examiners reportThis question had the lowest difficulty index on the HL paper, with roughly 10 % of candidates selecting response A. Responses C and D were roughlyequally common candidate answers, with students not recognizing theapplicability of Newton’s 3rd law.

[1 mark]

19. A horizontal electrical cable carries a steady current out of the page. TheEarth’s magnetic field exerts a force on the cable.Which arrow shows the direction of the force on the cable due to the Earth’smagnetic field?

MarkschemeB

Examiners reportThe correct answer was well answered by candidates, with a relatively highdiscrimination index.

[1 mark]

20. In an experiment to determine the resistivity of a material, a studentmeasures the resistance of several wires made from the pure material.The wires have the same length but different diameters.Which quantities should the student plot on the -axis and the -axis of a graph toobtain a straight line?

MarkschemeC


x y

[1 mark]

21. A proton of velocity v enters a region of electric and magnetic fields. Theproton is not deflected. An electron and an alpha particle enter the sameregion with velocity v. Which is correct about the paths of the electron and thealpha particle?

MarkschemeD


22. Two copper wires X and Y are connected in series. The diameter of Y isdouble that of X. The drift speed in X is v. What is the drift speed in Y? A.

B. C. 2vD. 4v

MarkschemeA

v4v2

[1 mark]

[1 mark]


23. A wire of length L is used in an electric heater. When the potentialdifference across the wire is 200 V, the power dissipated in the wire is1000 W. The same potential difference is applied across a second similar wire oflength 2L. What is the power dissipated in the second wire? A. 250 WB. 500 WC. 2000 WD. 4000 W

MarkschemeB


[1 mark]

24. A combination of four identical resistors each of resistance R areconnected to a source of emf ε of negligible internal resistance. What isthe current in the resistor X?

A.

B.

C.

D.

MarkschemeC


ε5R

3ε10R

2ε5R

3ε5R

25. Two parallel wires are perpendicular to the page. The wires carry equalcurrents in opposite directions. Point S is at the same distance from bothwires. What is the direction of the magnetic field at point S?

[1 mark]

[1 mark]

MarkschemeA


26. A particle of mass m and charge of magnitude q enters a region ofuniform magnetic field B that is directed into the page. The particlefollows a circular path of radius R. What are the sign of the charge of the particleand the speed of the particle?

MarkschemeB

[1 mark]


27. Two parallel wires P and Q are perpendicular to the page and carry equalcurrents. Point S is the same distance from both wires. The arrow showsthe magnetic field at S due to P and Q.

What are the correct directions for the current at P and the current at Q?

MarkschemeA


[1 mark]

28. Two point charges Q and Q are one metre apart. The graph shows thevariation of electric potential V with distance from Q .

What is ?

A.

B. C. 4D. 16

MarkschemeD


1 2x 1

Q1

Q2

116

14

[1 mark]

29. Three resistors are connected as shown. What is the value of the totalresistance between X and Y?

A. 1.5 ΩB. 1.9 ΩC. 6.0 ΩD. 8.0 Ω

MarkschemeA


30. A liquid that contains negative charge carriers is flowing through a squarepipe with sides A, B, C and D. A magnetic field acts in the direction shownacross the pipe.On which side of the pipe does negative charge accumulate?

MarkschemeA

[1 mark]

[1 mark]


31a.

A proton is moving in a region of uniform magnetic field. The magnetic field isdirected into the plane of the paper. The arrow shows the velocity of the proton atone instant and the dotted circle gives the path followed by the proton.

Explain why the path of the proton is a circle. [2 marks]

Markschememagnetic force is to the left «at the instant shown»ORexplains a rule to determine the direction of the magnetic force ✔force is perpendicular to velocity/«direction of» motionORforce is constant in magnitude ✔force is centripetal/towards the centre ✔NOTE: Accept reference to acceleration instead of force


31b.

The speed of the proton is 2.0 × 10 m s and the magnetic field strength B is0.35 T.

Show that the radius of the path is about 6 cm.

Markscheme✔

OR 0.060 « m »

NOTE: Award MP2 for full replacement or correct answer to at least 2significant figures


6 –1

qvB = mv2

R

R = 1.67×10−27×2.0×106

1.6×10−19×0.35

[2 marks]

31c. Calculate the time for one complete revolution.

Markscheme✔

« s » ✔

NOTE: Award [2] for bald correct answer


T = 2πRv

T = « = 1. 9 × 10−72π×0.062.0×106

31d. Explain why the kinetic energy of the proton is constant.

MarkschemeALTERNATIVE 1work done by force is change in kinetic energy ✔work done is zero/force perpendicular to velocity ✔ NOTE: Award [2] for a reference to work done is zero hence E remainsconstant ALTERNATIVE 2proton moves at constant speed ✔kinetic energy depends on speed ✔NOTE: Accept mention of speed or velocity indistinctly in MP2

k

[2 marks]

[2 marks]


32a.

An electron is placed at a distance of 0.40 m from a fixed point charge of –6.0 mC.

Show that the electric field strength due to the point charge at theposition of the electron is 3.4 × 10 N C .

Markscheme ✔

OR ✔

NOTE: Ignore any negative sign.


8 –1

E = k×q

r2

E = 8.99×109×6.0×10−3

0.42 E = 3. 37 × 108 «N C−1»

[2 marks]

32b. Calculate the magnitude of the initial acceleration of the electron.

Markscheme OR ✔

✔

NOTE: Ignore any negative sign. Award [1] for a calculation leading to Award [2] for bald correct answer


F = q × E F = 1. 6 × 10−19 × 3. 4 × 108 = 5. 4 × 10−11 «N»a = « = » 5. 9 × 1019« m s−2»5.4×10−11

9.1×10−31

a = « m s−2»

32c. Describe the subsequent motion of the electron.

[2 marks]

[3 marks]

Markschemethe electron moves away from the point charge/to the right «along the linejoining them» ✔decreasing acceleration ✔increasing speed ✔NOTE: Allow ECF from MP1 if a candidate mistakenly evaluates the force asattractive so concludes that the acceleration will increase


33a.

X has a capacitance of 18 μF. X is charged so that the one plate has a charge of48 μC. X is then connected to an uncharged capacitor Y and a resistor via an openswitch S.

Calculate, in J, the energy stored in X with the switch S open.

Markscheme OR ✔

✔

E = 12

Q2

CV = Q

C

E = « = » 6. 4 × 10−5 «J»12

(48×10−6 )

18×10−6

[2 marks]


33b.

The capacitance of Y is 12 μF. S is now closed.

Calculate the final charge on X and the final charge on Y.

MarkschemeALTERNATIVE 1

✔

✔

solving to get ✔ ALTERNATIVE 2

✔ ✔

✔ NOTE: Award [3] for bald correct answer


QX + QY = 48

=QX

18QY

12

QX = 29 «μC» QY = 19 «μC»

48 = 18 V + 12 V ⇒ V = 1. 6 «V»QX = «1. 6 × 18 = » 29 «QX = 1. 6 × 18 = 29 «μC»QY = «1. 6 × 12 = » 19 «μC»

[3 marks]

33c. Calculate the final total energy, in J, stored in X and Y.

MarkschemeALTERNATIVE 1

✔

✔

ALTERNATIVE 2

✔

✔

NOTE: Allow ECF from (b)(i)Award [2] for bald correct answerAward [1] max as ECF to a calculation using only one charge


ET = +12

(29×10−6 )2

18×10−612

(19×10−6 )2

12×10−6

= 3. 8 × 10−5«J»

ET = × 18 × 10−6 × 1. 62 + × 12 × 10−6 × 1. 6212

12

= 3. 8 × 10−5«J»

33d. Suggest why the answers to (a) and (b)(ii) are different.

[2 marks]

[2 marks]

Markschemecharge moves/current flows «in the circuit» ✔thermal losses «in the resistor and connecting wires» ✔NOTE: Accept heat losses for MP2


34a.

A girl rides a bicycle that is powered by an electric motor. A battery transfersenergy to the electric motor. The emf of the battery is 16 V and it can deliver acharge of 43 kC when discharging completely from a full charge.The maximum speed of the girl on a horizontal road is 7.0 m s with energy fromthe battery alone. The maximum distance that the girl can travel under theseconditions is 20 km.

Show that the time taken for the battery to discharge is about 3 × 10 s.

Markschemetime taken «= 2860 s» = 2900«s» ✔Must see at least two s.f.

Examiners reportThis question was generally well answered. Candidates should be reminded onquestions where a given value is being calculated that they should include anunrounded answer. This whole question set was a blend of electricity andmechanics concepts, and it was clear that some candidates struggled withapplying the correct concepts in the various sub-questions.

–1

3

2.0×104

7

[1 mark]

34b. Deduce that the average power output of the battery is about 240 W.

Markschemeuse of E = qV OR energy = 4.3 × 10 × 16 «= 6.88 × 10 J» ✔power = 241 «W» ✔Accept 229 W − 241 W depending on the exact value of t used from ai.Must see at least three s.f.


3 5

34c. Friction and air resistance act on the bicycle and the girl when theymove. Assume that all the energy is transferred from the battery to the

electric motor. Determine the total average resistive force that acts on the bicycleand the girl.

[2 marks]

[2 marks]

Markschemeuse of power = force × speed OR force × distance = power × time ✔«34N» ✔Award [2] for a bald correct answer.Accept 34 N – 36 N.


34d.

The bicycle and the girl have a total mass of 66 kg. The girl rides up a slope that isat an angle of 3.0° to the horizontal.

Calculate the component of weight for the bicycle and girl acting downthe slope.

Markscheme66 g sin(3°) = 34 «N» ✔

[1 mark]

Examiners reportMany candidates struggled with this question. They either simply calculatedthe weight, used the cosine rather than the sine function, or failed to multiplyby the acceleration due to gravity. Candidates need to be able to apply free-body diagram skills in a variety of “real world” situations.

34e. The battery continues to give an output power of 240 W. Assume thatthe resistive forces are the same as in (a)(iii).

Calculate the maximum speed of the bicycle and the girl up the slope.

Markschemetotal force 34 + 34 = 68 «N» ✔3.5 «ms »✔If you suspect that the incorrect reference in this question caused confusionfor a particular candidate, please refer the response to the PE.Look for ECF from aiii and bi.Accept 3.4 − 3.5 «ms ».Award [0] for solutions involving use of KE.Award [0] for v = 7 ms .Award [2] for a bald correct answer.


-1

-1

-1

[2 marks]

34f. On another journey up the slope, the girl carries an additional mass.Explain whether carrying this mass will change the maximum distancethat the bicycle can travel along the slope.

Markscheme«maximum» distance will decrease OWTTE ✔because opposing/resistive force has increasedORbecause more energy is transferred to GPEORbecause velocity has decreasedORincreased mass means more work required «to move up the hill» ✔

Examiners reportThis question was well answered in general, with the vast majority ofcandidates specifying that the maximum distance would decrease. This is an“explain” command term, so the examiners were looking for a detailed reasonwhy the distance would decrease for the second marking point. Unfortunately,some candidates simply wrote that because the mass increased so did theweight without making it clear why this would change the maximum distance.

[2 marks]

34g.

The bicycle has a meter that displays the current and the terminal potentialdifference (pd) for the battery when the motor is running. The diagram shows themeter readings at one instant. The emf of the cell is 16 V.

Determine the internal resistance of the battery.

MarkschemeV dropped across battery OR R = 1.85 Ω ✔so internal resistance = = 0.62«Ω» ✔For MP1 allow use of internal resistance equations that leads to 16V − 12V(=4V).Award [2] for a bald correct answer.


circuit4.06.5

[2 marks]

34h.

The battery is made from an arrangement of 10 identical cells as shown.

Calculate the emf of one cell.

Markscheme = 3.2 «V» ✔


165

34i. Calculate the internal resistance of one cell.

[1 mark]

[2 marks]

MarkschemeALTERNATIVE 1:2.5r = 0.62 ✔r = 0.25 «Ω» ✔ALTERNATIVE 2:

= 0.124 «Ω» ✔r = 2(0.124)= 0.248 «Ω» ✔Allow ECF from (d) and/or e(i).


0.625

35a.

Three identical light bulbs, X, Y and Z, each of resistance 4.0 Ω are connected to acell of emf 12 V. The cell has negligible internal resistance.

The switch S is initially open. Calculate the total power dissipated in thecircuit.

[2 marks]

Markschemetotal resistance of circuit is 8.0 «Ω» ✔

P = =18 «W» ✔

Examiners reportMost candidates scored both marks. ECF was awarded for those who didn’tcalculate the new resistance correctly. Candidates showing clearly that theywere attempting to calculate the new total resistance helped examiners toaward ECF marks.

122

8.0

35b. The switch is now closed. State, without calculation, why the current inthe cell will increase.

Markscheme«a resistor is now connected in parallel» reducing the total resistanceORcurrent through YZ unchanged and additional current flows through X ✔

Examiners reportMost recognised that this decreased the total resistance of the circuit. Answersscoring via the second alternative were rare as the statements were often fartoo vague.

power dissipated in Y with S open

[1 mark]

35c. The switch is now closed. Deduce the ratio .

Markschemeevidence in calculation or statement that pd across Y/current in Y is the sameas before ✔so ratio is 1 ✔

Examiners reportVery few gained any credit for this at both levels. Most performed complicatedcalculations involving the total circuit and using 12V – they had not realisedthat the question refers to Y only.

power dissipated in Y with S openpower dissipated in Y with S closed

36a.

Three identical light bulbs, X, Y and Z, each of resistance 4.0 Ω are connected to acell of emf 12 V. The cell has negligible internal resistance.

The switch S is initially open. Calculate the total power dissipated in thecircuit.

[2 marks]

[2 marks]

Markschemetotal resistance of circuit is 8.0 «Ω» ✔

«W» ✔

Examiners reportMost candidates scored both marks. ECF was awarded for those who didn’tcalculate the new resistance correctly. Candidates showing clearly that theywere attempting to calculate the new total resistance helped examiners toaward ECF marks.

P = = 18122

8.0

36b. The switch is now closed. State, without calculation, why the current inthe cell will increase.

Markscheme«a resistor is now connected in parallel» reducing the total resistanceORcurrent through YZ unchanged and additional current flows through X ✔

Examiners reportMost recognised that this decreased the total resistance of the circuit. Answersscoring via the second alternative were rare as the statements were often fartoo vague.

power dissipated in Y with S open

[1 mark]

36c. The switch is now closed. .

Markschemeevidence in calculation or statement that pd across Y/current in Y is the sameas before ✔so ratio is 1 ✔

Examiners reportVery few gained any credit for this at both levels. Most performed complicatedcalculations involving the total circuit and using 12V – they had not realisedthat the question refers to Y only.

Deduce the ratio power dissipated in Y with S openpower dissipated in Y with S closed

[2 marks]

36d. The cell is used to charge a parallel-plate capacitor in a vacuum. Thefully charged capacitor is then connected to an ideal voltmeter.

The capacitance of the capacitor is 6.0 μF and the reading of the voltmeter is 12 V.Calculate the energy stored in the capacitor.

Markscheme« » ✔

Examiners reportMost answered this correctly.

E = « CV 2 = × 6 × 10−6 × 122 = »4.3 × 10−412

12 J

36e.

When fully charged the space between the plates of the capacitor is filled with adielectric with double the permittivity of a vacuum.

Calculate the change in the energy stored in the capacitor.

[1 mark]

[3 marks]

MarkschemeALTERNATIVE 1capacitance doubles and voltage halves ✔since energy halves ✔so change is «–»2.2×10 «J» ✔ ALTERNATIVE 2

✔ capacitance doubles and charge unchanged so energy halves ✔so change is «−»2.2 × 10 «J» ✔

Examiners reportBy far the most common answer involved doubling the capacitance withoutconsidering the change in p.d. Almost all candidates who did this calculated achange in energy that scored 1 mark.

E = CV 212

–4

E = CV 2 and Q = CV so E =12

Q2

2C

−4

36f. Suggest, in terms of conservation of energy, the cause for the abovechange.

Markschemeit is the work done when inserting the dielectric into the capacitor ✔

Examiners reportVery few scored on this question.

[1 mark]

37a.

A proton moves along a circular path in a region of a uniform magnetic field. Themagnetic field is directed into the plane of the page.

Label with arrows on the diagram the magnetic force F on the proton.

MarkschemeF towards centre ✔

Examiners reportExaminers were requested to be lenient here and as a result most candidatesscored both marks. Had we insisted on e.g. straight lines drawn with a ruler ora force arrow passing exactly through the centre of the circle very few markswould have been scored. For those who didn’t know which way the arrowswere supposed to be the common guesses were to the left and up the page.Some candidates neglected to label the arrows.

[1 mark]

37b. Label with arrows on the diagram the velocity vector v of the proton.

Markschemev tangent to circle and in the direction shown in the diagram ✔

Examiners reportExaminers were requested to be lenient here and as a result most candidatesscored both marks. Had we insisted on e.g. straight lines drawn with a ruler ora force arrow passing exactly through the centre of the circle very few markswould have been scored. For those who didn’t know which way the arrowswere supposed to be the common guesses were to the left and up the page.Some candidates neglected to label the arrows.

6 -1

[1 mark]

37c.

The speed of the proton is 2.16 × 10 m s and the magnetic field strength is0.042 T.

For this proton, determine, in m, the radius of the circular path. Giveyour answer to an appropriate number of significant figures.

Markscheme« ✔

R = 0.538«m» ✔R = 0.54«m» ✔

Examiners reportThis was generally well answered although usually to 3 sf. Common mistakeswere to substitute 0.042 for F and 1 for q. Also some candidates tried toanswer in terms of electric fields.

6 -1

qvB = ⇒ »R = /mv2

RmvqB

1.673×10−27×2.16×106

1.60×10−19×0.042

[3 marks]

37d. For this proton, calculate, in s, the time for one full revolution.

Markscheme ✔

«s» ✔

Examiners reportThis was well answered with many candidates scoring ECF from the previouspart.

T = /2πRv

2π×0.542.16×106

T = 1.6 × 10−6

[2 marks]

38a.

Ion-thrust engines can power spacecraft. In this type of engine, ions are created ina chamber and expelled from the spacecraft. The spacecraft is in outer spacewhen the propulsion system is turned on. The spacecraft starts from rest.

The mass of ions ejected each second is 6.6 × 10 kg and the speed of each ion is5.2 × 10 m s . The initial total mass of the spacecraft and its fuel is 740 kg.Assume that the ions travel away from the spacecraft parallel to its direction ofmotion.

Determine the initial acceleration of the spacecraft.

Markschemechange in momentum each second = 6.6 × 10 × 5.2 × 10 «= 3.4 ×10 kg m s » ✔

acceleration = « =» 4.6 × 10 «m s » ✔


–6 4 –1

−6 4−1 −1

3.4×10−1

740−4 −2

[2 marks]

38b.

An initial mass of 60 kg of fuel is in the spacecraft for a journey to a planet. Half ofthe fuel will be required to slow down the spacecraft before arrival at thedestination planet.

Estimate the maximum speed of the spacecraft.

MarkschemeALTERNATIVE 1:(considering the acceleration of the spacecraft)time for acceleration = = «4.6 × 10 » «s» ✔

max speed = «answer to (a) × 4.6 × 10 =» 2.1 × 10 «m s » ✔ ALTERNATIVE 2:(considering the conservation of momentum)(momentum of 30 kg of fuel ions = change of momentum of spacecraft)30 × 5.2 × 10 = 710 × max speed ✔max speed = 2.2 × 10 «m s » ✔


306.6×10−6

6

6 3 −1

4

3 −1

[2 marks]

38c. Outline why scientists sometimes use estimates in making calculations.

Markschemeproblem may be too complicated for exact treatment ✔to make equations/calculations simpler ✔when precision of the calculations is not important ✔some quantities in the problem may not be known exactly ✔


[1 mark]

38d.

In practice, the ions leave the spacecraft at a range of angles as shown.

Outline why the ions are likely to spread out.

Markschemeions have same (sign of) charge ✔ions repel each other ✔


[2 marks]

38e. Explain what effect, if any, this spreading of the ions has on theacceleration of the spacecraft.

Markschemethe forces between the ions do not affect the force on the spacecraft. ✔there is no effect on the acceleration of the spacecraft. ✔


38f.

On arrival at the planet, the spacecraft goes into orbit as it comes into thegravitational field of the planet.

Outline what is meant by the gravitational field strength at a point.

Markschemeforce per unit mass ✔acting on a small/test/point mass «placed at the point in the field» ✔

[2 marks]

[2 marks]


38g. Newton’s law of gravitation applies to point masses. Suggest why the lawcan be applied to a satellite orbiting a spherical planet of uniform

density.

Markschemesatellite has a much smaller mass/diameter/size than the planet «soapproximates to a point mass» ✔


[1 mark]

39a.

A lighting system consists of two long metal rods with a potential differencemaintained between them. Identical lamps can be connected between the rods asrequired.

The following data are available for the lamps when at their working temperature.Lamp specifications 24 V, 5.0 WPower supply emf 24 VPower supply maximum current 8.0 ALength of each rod 12.5 mResistivity of rod metal 7.2 × 10 Ω m

Each rod is to have a resistance no greater than 0.10 Ω. Calculate, in m,the minimum radius of each rod. Give your answer to an appropriate

number of significant figures.

–7

[3 marks]

MarkschemeALTERNATIVE 1:

OR ✔

r = 5.352 × 10 ✔5.4 × 10 «m» ✔ ALTERNATIVE 2:

✔r = 5.352 × 10 ✔5.4 × 10 «m» ✔


r = √ ρl

πR√ 7.2×10−7×12.5

π×0.1

−3

−3

A = 7.2×10−7×12.50.1

−3

−3

39b. Calculate the maximum number of lamps that can be connectedbetween the rods. Neglect the resistance of the rods.

[2 marks]

Markschemecurrent in lamp = «= 0.21» «A»ORn = 24 × ✔ so «38.4 and therefore» 38 lamps ✔


524

85

39c. One advantage of this system is that if one lamp fails then the otherlamps in the circuit remain lit. Outline one other electrical advantage of

this system compared to one in which the lamps are connected in series.

[1 mark]

Markschemewhen adding more lamps in parallel the brightness stays the same ✔when adding more lamps in parallel the pd across each remains the same/atthe operating value/24 V ✔when adding more lamps in parallel the current through each remains thesame ✔lamps can be controlled independently ✔the pd across each bulb is larger in parallel ✔the current in each bulb is greater in parallel ✔lamps will be brighter in parallel than in series ✔In parallel the pd across the lamps will be the operating value/24 V ✔ Accept converse arguments for adding lamps in series:when adding more lamps in series the brightness decreaseswhen adding more lamps in series the pd decreaseswhen adding more lamps in series the current decreaseslamps can’t be controlled independentlythe pd across each bulb is smaller in seriesthe current in each bulb is smaller in series in series the pd across the lamps will less than the operating value/24 VDo not accept statements that only compare the overall resistance of thecombination of bulbs.


40a.

Ion-thrust engines can power spacecraft. In this type of engine, ions are created ina chamber and expelled from the spacecraft. The spacecraft is in outer spacewhen the propulsion system is turned on. The spacecraft starts from rest.

The mass of ions ejected each second is 6.6 × 10 kg and the speed of each ion is5.2 × 10 m s . The initial total mass of the spacecraft and its fuel is 740 kg.Assume that the ions travel away from the spacecraft parallel to its direction ofmotion.

Determine the initial acceleration of the spacecraft.

Markschemechange in momentum each second = 6.6 × 10 × 5.2 × 10 «= 3.4 ×10 kg m s » ✔

acceleration = « =» 4.6 × 10 «m s » ✔


–6 4 –1

−6 4−1 −1

3.4×10−1

740−4 −2

[2 marks]

40b.

An initial mass of 60 kg of fuel is in the spacecraft for a journey to a planet. Half ofthe fuel will be required to slow down the spacecraft before arrival at thedestination planet.

(i) Estimate the maximum speed of the spacecraft.(ii) Outline why the answer to (i) is an estimate.

[3 marks]

Markscheme(i) ALTERNATIVE 1:(considering the acceleration of the spacecraft)time for acceleration = = «4.6 × 10 » «s» ✔

max speed = «answer to (a) × 4.6 × 10 =» 2.1 × 10 «m s » ✔ ALTERNATIVE 2:(considering the conservation of momentum)(momentum of 30 kg of fuel ions = change of momentum of spacecraft)30 × 5.2 × 10 = 710 × max speed ✔max speed = 2.2 × 10 «m s » ✔ (ii) as fuel is consumed total mass changes/decreases so accelerationchanges/increasesORexternal forces (such as gravitational) can act on the spacecraft soacceleration isn’t constant ✔


306.6×10−6

6

6 3 −1

4

3 −1

40c. Outline why scientists sometimes use estimates in making calculations. [1 mark]

Markschemeproblem may be too complicated for exact treatment ✔to make equations/calculations simpler ✔when precision of the calculations is not important ✔some quantities in the problem may not be known exactly ✔


40d.

In practice, the ions leave the spacecraft at a range of angles as shown.

Outline why the ions are likely to spread out.

Markschemeions have same (sign of) charge ✔ions repel each other ✔

[2 marks]

Printed for Jyvaskylan Lyseonlukio

© International Baccalaureate Organization 2020 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®


40e. Explain what effect, if any, this spreading of the ions has on theacceleration of the spacecraft.

Markschemethe forces between the ions do not affect the force on the spacecraft. ✔there is no effect on the acceleration of the spacecraft. ✔


[2 marks]

18ibtopic5 20200922 [119 marks]

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