topic 10 revision [147 marks] - peda.net · topic 10 revision [147 marks] 1. four identical,...

Topic 10 Revision [147 marks]

1. Four identical, positive, point charges of magnitude Q are placed at the vertices of asquare of side 2d. What is the electric potential produced at the centre of the square bythe four charges?

A. 0

B.

C.

D.

MarkschemeD

Examiners report[N/A]

4kQ

d

√2kQ

d

2√2kQ

d

2. The diagram shows 5 gravitational equipotential lines. The gravitational potential oneach line is indicated. A point mass m is placed on the middle line and is then released.Values given in MJ kg .

Which is correct about the direction of motion and the acceleration of the point mass?

–1

[1 mark]

[1 mark]

MarkschemeD


3. An electron of mass m orbits an alpha particle of mass m in a circular orbit of radius r.Which expression gives the speed of the electron?

A.

B.

C.

D.

MarkschemeA


e α

√ 2ke2

mer

√ 2ke2

mar

√ 4ke2

mer

√ 4ke2

mar

4. A moon of mass M orbits a planet of mass 100M. The radius of the planet is R and thedistance between the centres of the planet and moon is 22R.

What is the distance from the centre of the planet at which the total gravitational potential hasa maximum value?

A. 2R

B. 11R

C. 20R

D. 2R and 20R

[1 mark]

[1 mark]

MarkschemeC


5. The diagram shows the electric field and the electric equipotential surfaces between twocharged parallel plates. The potential difference between the plates is 200 V.

What is the work done, in nJ, by the electric field in moving a negative charge of magnitude 1nC from the position shown to X and to Y?

MarkschemeA


6. A positive point charge is placed above a metal plate at zero electric potential. Whichdiagram shows the pattern of electric field lines between the charge and the plate?

[1 mark]

[1 mark]

MarkschemeC


7. A satellite orbiting a planet moves from orbit X to orbit Y.

What is the change in the kinetic energy and the change in the gravitational potential energy asa result?

MarkschemeC


The mass of the Earth is M and the mass of the Moon is M . Their respective radii are

[1 mark]

8. The mass of the Earth is M and the mass of the Moon is M . Their respective radii areR and R .

Which is the ratio ?

A.

B.

C.

D.

MarkschemeC


E M

E M

escape speed from the Earthescape speed from the Moon

√ MMRM

MERE

√ MERE

MMRM

√ MERM

MMRE

√ MMRE

MERM

9a.

Hydrogen atoms in an ultraviolet (UV) lamp make transitions from the first excited state to theground state. Photons are emitted and are incident on a photoelectric surface as shown.

Show that the energy of photons from the UV lamp is about 10 eV.

[1 mark]

[2 marks]

MarkschemeE = –13.6 «eV» E = – = –3.4 «eV»

energy of photon is difference E – E = 10.2 «≈ 10 eV»

Must see at least 10.2 eV.

[2 marks]


1 213.6

4

2 1

9b.

The photons cause the emission of electrons from the photoelectric surface. The work functionof the photoelectric surface is 5.1 eV.

Calculate, in J, the maximum kinetic energy of the emitted electrons.

Markscheme10 – 5.1 = 4.9 «eV»

4.9 × 1.6 × 10 = 7.8 × 10 «J»

Allow 5.1 if 10.2 is used to give 8.2×10 «J».


–19 –19

−19

Suggest, with reference to conservation of energy, how the variable voltage source

[2 marks]

9c. Suggest, with reference to conservation of energy, how the variable voltage sourcecan be used to stop all emitted electrons from reaching the collecting plate.

MarkschemeEPE produced by battery

exceeds maximum KE of electrons / electrons don’t have enough KE

For first mark, accept explanation in terms of electric potential energy difference ofelectrons between surface and plate.

[2 marks]


9d. The variable voltage can be adjusted so that no electrons reach the collecting plate.Write down the minimum value of the voltage for which no electrons reach the collectingplate.

Markscheme4.9 «V»

Allow 5.1 if 10.2 is used in (b)(i).

Ignore sign on answer.

[1 mark]

[2 marks]

[1 mark]


9e.

The electric potential of the photoelectric surface is 0 V. The variable voltage is adjusted so thatthe collecting plate is at –1.2 V.

On the diagram, draw and label the equipotential lines at –0.4 V and –0.8 V.

Markschemetwo equally spaced vertical lines (judge by eye) at approximately 1/3 and 2/3

labelled correctly

[2 marks]


An electron is emitted from the photoelectric surface with kinetic energy 2.1 eV.

[2 marks]

9f. An electron is emitted from the photoelectric surface with kinetic energy 2.1 eV.Calculate the speed of the electron at the collecting plate.

Markschemekinetic energy at collecting plate = 0.9 «eV»

speed = «

» = 5.6 × 10 «ms »

Allow ECF from MP1

[2 marks]


√ 2×0.9×1.6×10−19

9.11×10−315 –1

10a.

A planet has radius R. At a distance h above the surface of the planet the gravitational fieldstrength is g and the gravitational potential is V.

State what is meant by gravitational field strength.

Markschemethe «gravitational» force per unit mass exerted on a point/small/test mass

[1 mark]

[2 marks]

[1 mark]


10b. Show that V = –g(R + h).

Markschemeat height h potential is V = –

field is g =

«dividing gives answer»

Do not allow an answer that starts with g = – and then cancels the deltas andsubstitutes R + h

[2 marks]


GM

(R+h)

GM

(R+h)2

ΔV

Δr

Draw a graph, on the axes, to show the variation of the gravitational potential V of the

[2 marks]

10c. Draw a graph, on the axes, to show the variation of the gravitational potential V of theplanet with height h above the surface of the planet.

Markschemecorrect shape and sign

non-zero negative vertical intercept

[2 marks]


10d. A planet has a radius of 3.1 × 10 m. At a point P a distance 2.4 × 10 m abovethe surface of the planet the gravitational field strength is 2.2 N kg . Calculatethe gravitational potential at point P, include an appropriate unit for your answer.

6 7

–1

[2 marks]

[1 mark]

MarkschemeV = «–2.2 × (3.1 × 10 + 2.4 × 10 ) =» «–» 6.0 × 10 J kg

Unit is essential

Allow eg MJ kg if power of 10 is correct

Allow other correct SI units eg m s , N m kg

[1 mark]


6 7 7 –1

–1

2 –2 –1

10e. The diagram shows the path of an asteroid as it moves past the planet.

When the asteroid was far away from the planet it had negligible speed. Estimate the speed ofthe asteroid at point P as defined in (b).

[3 marks]

Markschemetotal energy at P = 0 / KE gained = GPE lost

«mv + mV = 0 ⇒» v =

v = « =» 1.1 × 10 «ms »

Award [3] for a bald correct answer

Ignore negative sign errors in the workings

Allow ECF from 6(b)

[3 marks]


12

2 √−2V

√2 × 6.0 × 107 4 –1

10f. The mass of the asteroid is 6.2 × 10 kg. Calculate the gravitational forceexperienced by the planet when the asteroid is at point P.

12 [2 marks]

MarkschemeALTERNATIVE 1

force on asteroid is «6.2 × 10 × 2.2 =» 1.4 × 10 «N»

«by Newton’s third law» this is also the force on the planet

ALTERNATIVE 2

mass of planet = 2.4 x 10 «kg» «from V = – »

force on planet «» = 1.4 × 10 «N»

MP2 must be explicit

[2 marks]


12 13

25 GM

(R+h)

GMm

(R+h)213

11. A charge of −3 C is moved from A to B and then back to A. The electric potential at A is+10 V and the electric potential at B is −20 V. What is the work done in moving thecharge from A to B and the total work done?

MarkschemeC


A spacecraft moves towards the Earth under the influence of the gravitational field of

[1 mark]

12. A spacecraft moves towards the Earth under the influence of the gravitational field ofthe Earth.

The three quantities that depend on the distance r of the spacecraft from the centre of theEarth are the

I. gravitational potential energy of the spacecraftII gravitational field strength acting on the spacecraftIII. gravitational force acting on the spacecraft.

Which of the quantities are proportional to ?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

MarkschemeC


1r2

13. An isolated hollow metal sphere of radius R carries a positive charge. Which graphshows the variation of potential V with distance x from the centre of the sphere?

MarkschemeB


[1 mark]

[1 mark]

14a.

There is a proposal to power a space satellite X as it orbits the Earth. In this model, X isconnected by an electronically-conducting cable to another smaller satellite Y.

Satellite X orbits 6600 km from the centre of the Earth.

Mass of the Earth = 6.0 x 10 kg

Show that the orbital speed of satellite X is about 8 km s .

Markscheme« » =

7800 «m s »

Full substitution required

Must see 2+ significant figures.


24

–1

v = √ GME

r√ 6.67×10−11×6.0×1024

6600×103

–1

14b.

Satellite Y orbits closer to the centre of Earth than satellite X. Outline why

the orbital times for X and Y are different.

[2 marks]

[1 mark]

MarkschemeY has smaller orbit/orbital speed is greater so time period is less

Allow answer from appropriate equation

Allow converse argument for X


14c. satellite Y requires a propulsion system.

Markschemeto stop Y from getting ahead

to remain stationary with respect to X

otherwise will add tension to cable/damage satellite/pull X out of its orbit


The cable between the satellites cuts the magnetic field lines of the Earth at right

[2 marks]

14d. The cable between the satellites cuts the magnetic field lines of the Earth at rightangles.

Explain why satellite X becomes positively charged.

Markschemecable is a conductor and contains electrons

electrons/charges experience a force when moving in a magnetic field

use of a suitable hand rule to show that satellite Y becomes negative «so X becomespositive»

Alternative 2

cable is a conductor

so current will flow by induction flow when it moves through a B field

use of a suitable hand rule to show current to right so «X becomes positive»

Marks should be awarded from either one alternative or the other.

Do not allow discussion of positive charges moving towards X


Satellite X must release ions into the space between the satellites. Explain why

[3 marks]

14e. Satellite X must release ions into the space between the satellites. Explain whythe current in the cable will become zero unless there is a method for transferringcharge from X to Y.

Markschemeelectrons would build up at satellite Y/positive charge at X

preventing further charge flow

by electrostatic repulsion

unless a complete circuit exists


14f. The magnetic field strength of the Earth is 31 µT at the orbital radius of thesatellites. The cable is 15 km in length. Calculate the emf induced in the cable.

Markscheme«ε = Blv =» 31 x 10 x 7990 x 15000

3600 «V»

Allow 3700 «V» from v = 8000 m s .

–6

–1

[3 marks]

[2 marks]


14g.

The cable acts as a spring. Satellite Y has a mass m of 3.5 x 10 kg. Under certaincircumstances, satellite Y will perform simple harmonic motion (SHM) with a period T of 5.2 s.

Estimate the value of k in the following expression.

T =

Give an appropriate unit for your answer. Ignore the mass of the cable and any oscillation ofsatellite X.

Markschemeuse of k = « »

510

N m or kg s

Allow MP1 and MP2 for a bald correct answer

Allow 500

Allow N/m etc.


2

2π√ mk

=4π2m

T 24×π2×350

5.22

–1 –2

Describe the energy changes in the satellite Y-cable system during one cycle of the

[3 marks]

14h. Describe the energy changes in the satellite Y-cable system during one cycle of theoscillation.

MarkschemeE in the cable/system transfers to E of Y

and back again twice in each cycle

Exclusive use of gravitational potential energy negates MP1


p k

15. An electric field acts in the space between two charged parallel plates. One plate is atzero potential and the other is at potential +V.

The distance x is measured from point P in the direction perpendicular to the plate.

What is the dependence of the electric field strength E on x and what is the dependence of theelectric potential V on x?

[2 marks]

[1 mark]

MarkschemeB


16. A satellite at the surface of the Earth has a weight W and gravitational potential energyEp. The satellite is then placed in a circular orbit with a radius twice that of the Earth.

What is the weight of the satellite and the gravitational potential energy of the satellite whenplaced in orbit?

MarkschemeC


17. Two point charges are at rest as shown.

At which position is the electric field strength greatest?

MarkschemeB

[1 mark]

[1 mark]


18. A positive charge Q is deposited on the surface of a small sphere. The dotted linesrepresent equipotentials.

A small positive point charge is moved from point P closer to the sphere along three differentpaths X, Y and Z. The work done along each path is W , W and W . What is a correctcomparison of W , W and W ?

A. W > W > W

B. W > W = W

C. W = W = W

D. W = W > W

MarkschemeB


X Y Z

X Y Z

Z Y X

X Y Z

X Y Z

Z Y X

The graph shows the variation of the gravitational potential V with distance r from the

[1 mark]

19. The graph shows the variation of the gravitational potential V with distance r from thecentre of a uniform spherical planet. The radius of the planet is R. The shaded area isS.

What is the work done by the gravitational force as a point mass m is moved from the surfaceof the planet to a distance 6R from the centre?

A. m (V2 – V1 )

B. m (V1 – V2 )

C. mS

D. S

MarkschemeB


20. Four uniform planets have masses and radii as shown. Which planet has thesmallest escape speed?

[1 mark]

[1 mark]

MarkschemeC


21a.

The diagram shows the gravitational field lines of planet X.

Outline how this diagram shows that the gravitational field strength of planet Xdecreases with distance from the surface.

Markschemethe field lines/arrows are further apart at greater distances from the surface


The diagram shows part of the surface of planet X. The gravitational potential at the

[1 mark]

21b. The diagram shows part of the surface of planet X. The gravitational potential at thesurface of planet X is –3V and the gravitational potential at point Y is – V.

Sketch on the grid the equipotential surface corresponding to a gravitational potential of –2 V.

Markschemecircle centred on Planet Xthree units from Planet X centre


A meteorite, very far from planet X begins to fall to the surface with a negligibly small

[2 marks]

21c. A meteorite, very far from planet X begins to fall to the surface with a negligibly smallinitial speed. The mass of planet X is 3.1 × 10 kg and its radius is 1.2 × 10 m. Theplanet has no atmosphere. Calculate the speed at which the meteorite will hit the surface.

Markschemeloss in gravitational potential =

«= 1.72 × 10 JKg »

equate to v

v = 590 «m s »

Allow ECF from MP1.


21 6

6.67×10−11×3.1×1021

1.2×106

5 −1

12

2

−1

21d. At the instant of impact the meteorite which is made of ice has a temperature of 0 °C.Assume that all the kinetic energy at impact gets transferred into internal energy in themeteorite. Calculate the percentage of the meteorite’s mass that melts. The specific latent heatof fusion of ice is 3.3 × 10 J kg .5 –1

[3 marks]

[2 marks]

Markschemeavailable energy to melt one kg 1.72 × 10 «J»

fraction that melts is = 0.52 OR 52%

Allow ECF from MP1.

Allow 53% from use of 590 ms .


5

1.72×105

3.3×105

-1

22a.

The gravitational potential due to the Sun at its surface is –1.9 x 10 J kg . The following dataare available.

Mass of Earth = 6.0 x 10 kgDistance from Earth toSun = 1.5 x 10 m

Radius of Sun = 7.0 x 10 m

Outline why the gravitational potential is negative.

Markschemepotential is defined to be zero at infinity

so a positive amount of work needs to be supplied for a mass to reach infinity


11 –1

24

11

8

The gravitational potential due to the Sun at a distance r from its centre is V . Show

[2 marks]

22b. The gravitational potential due to the Sun at a distance r from its centre is V . Showthat

rV = constant.

MarkschemeV = so r x V «= –GM» = constant because G and M are constants


S

S

S − GMr S

22c. Calculate the gravitational potential energy of the Earth in its orbit around theSun. Give your answer to an appropriate number of significant figures.

MarkschemeGM = 1.33 x 10 «J m kg »

GPE at Earth orbit «= – » = «–» 5.3 x 10 «J»

Award [1 max] unless answer is to 2 sf.

Ignore addition of Sun radius to radius of Earth orbit.


20 –1

1.33×1020×6.0×1024

1.5×101133

[1 mark]

[2 marks]

22d. Calculate the total energy of the Earth in its orbit.

MarkschemeALTERNATIVE 1work leading to statement that kinetic energy , AND kinetic energy evaluated to be«+» 2.7 x 10 «J»

energy «= PE + KE = answer to (b)(ii) + 2.7 x 10 » = «–» 2.7 x 10 «J»

ALTERNATIVE 2statement that kinetic energy is gravitational potential energy in orbit

so energy « » = «–» 2.7 x 10 «J»

Various approaches possible.


GMm

2r33

33 33

= − 12

= answer to (b)(ii)2

33

22e. An asteroid strikes the Earth and causes the orbital speed of the Earth to suddenlydecrease. Suggest the ways in which the orbit of the Earth will change.

[2 marks]

[2 marks]

Markscheme«KE will initially decrease so» total energy decreasesOR«KE will initially decrease so» total energy becomes more negative

Earth moves closer to Sun

new orbit with greater speed «but lower total energy»

changes ellipticity of orbit


22f. Outline, in terms of the force acting on it, why the Earth remains in a circularorbit around the Sun.

Markschemecentripetal force is required

and is provided by gravitational force between Earth and Sun

Award [1 max] for statement that there is a “centripetal force of gravity” without furtherqualification.


What is the unit of Gε , where G is the gravitational constant and ε is the permittivity of

[2 marks]

23. What is the unit of Gε , where G is the gravitational constant and ε is the permittivity offree space?

A. C kg

B. C kg

C. C kg

D. C kg

MarkschemeB


0 0

–1

2 –2

2 2

24. Two parallel metal plates are connected to a dc power supply. An electric field forms inthe space between the plates as shown.

What is the shape of the equipotentials surfaces that result from this arrangement?

MarkschemeB


A satellite of mass 1500 kg is in the Earth’s gravitational field. It moves from a point

[1 mark]

[1 mark]

25. A satellite of mass 1500 kg is in the Earth’s gravitational field. It moves from a pointwhere the gravitational potential is –30 MJ kg to a point where the gravitationalpotential is –20 MJ kg . What is the direction of movement of the satellite and the change in itsgravitational potential energy?

MarkschemeA


–1

–1

26a. Explain what is meant by the gravitational potential at the surface of a planet.

Markschemethe «gravitational» work done «by an external agent» per/on unit mass/kg

Allow definition in terms of reverse process of moving mass to infinity eg “work done onexternal agent by…”.Allow “energy” as equivalent to “work done”

in moving a «small» mass from infinity to the «surface of» planet / to a point

N.B.: on SL paper Q5(a)(i) and (ii) is about “gravitational field”.

[1 mark]

[2 marks]


26b. An unpowered projectile is fired vertically upwards into deep space from the surface ofplanet Venus. Assume that the gravitational effects of the Sun and the other planetsare negligible.

The following data are available.

Mass of Venus = 4.87×10 kg Radius of Venus = 6.05×10 m Mass of projectile = 3.50×10 kgInitial speed of projectile = 1.10×escape speed

(i) Determine the initial kinetic energy of the projectile.

(ii) Describe the subsequent motion of the projectile until it is effectively beyond the gravitationalfield of Venus.

24 6 3

[5 marks]

Markschemeiescape speedCare with ECF from MP1.

v = « »

or 1.04×10 «m s »

or «1.1 × 1.04 × 10 m s »= 1.14 × 10 «m s »

KE = «0.5 × 3500 × (1.1 × 1.04 × 10 m s ) =» 2.27×10 «J»

Award [1 max] for omission of 1.1 – leads to 1.88×10 m s .Award [2] for a bald correct answer.

iiVelocity/speed decreases / projectile slows down «at decreasing rate»

«magnitude of» deceleration decreases «at decreasing rate»Mention of deceleration scores MP1 automatically.

velocity becomes constant/non-zero ORdeceleration tends to zero

Accept “negative acceleration” for “deceleration”.

Must see “velocity” not “speed” for MP3.


√( ) =2 GMR

√( )2×6.67×10−11×4.87×1024

6.05×1064 –1

4 -1 4 –1

4 –1 2 11

11 -1

27. A negative charge moves in an electric field. Equipotential lines for the field and fourpossible paths of the charge are shown. Which path corresponds to the largest workdone on the charge by the field?

[1 mark]

MarkschemeB


28. In an experiment, oil droplets of mass m and charge q are dropped into the regionbetween two horizontal parallel plates. The electric field E between the plates can beadjusted. Air resistance is negligible. Which is correct when the droplets fall vertically atconstant velocity?

A. E=0

B.

C.

D.

MarkschemeC


E < mg

q

E = mg

q

E > mg

q

A satellite orbits a planet. Which graph shows how the kinetic energy E , the potential

[1 mark]

29. A satellite orbits a planet. Which graph shows how the kinetic energy E , the potentialenergy E and the total energy E of the satellite vary with distance x from the centre ofthe planet?

MarkschemeB


K

P

30. Which of the following experiments provides evidence for the existence of matterwaves?

A. Scattering of alpha particles

B. Electron diffraction

C. Gamma decay

D. Photoelectric effect

MarkschemeB

[1 mark]

[1 mark]


31a. Outline what is meant by escape speed.

Markschemespeed to reach infinity/zero gravitational fieldORspeed to escape gravitational pull/effect of planet’s gravity

Do not allow reference to leaving/escaping an orbit.Do not allow “escaping the atmosphere”.


31b. A probe is launched vertically upwards from the surface of a planet with a speed

where v is the escape speed from the planet. The planet has no atmosphere.

Determine, in terms of the radius of the planet R, the maximum height from the surface of theplanet reached by the probe.

v = vesc34

esc

[1 mark]

[3 marks]

Markscheme«kinetic energy at take off» =

kinetic energy at take off + «gravitational» potential energy = «gravitational» potentialenergy at maximum height

OR

solves for r and subtracts R from answer =

Award [0] for work that assumes constant g.


×916

GMm

R

× − = −916

GMm

R

GMm

R

GMm

R

9R7

31c. The total energy of a probe in orbit around a planet of mass M is wherem is the mass of the probe and r is the orbit radius. A probe in low orbit experiences a smallfrictional force. Suggest the effect of this force on the speed of the probe.

Markschemeenergy reduces/lost

radius decreases

speed increases

Do not allow “kinetic energy reduces” for MP1


E = − GMm

2r

A negatively charged particle falls vertically into a region where there is an electric field.

[3 marks]

32. A negatively charged particle falls vertically into a region where there is an electric field.The equipotentials of this field are shown.

What is the path followed by the particle?

MarkschemeD


33a.

This question is in two parts. Part 1 is about electric fields and radioactive decay. Part 2 isabout waves.

Part 1 Electric fields and radioactive decay

An ionization chamber is a device which can be used to detect charged particles.

The charged particles enter the chamber through a thin window. They then ionize the airbetween the parallel metal plates. A high potential difference across the plates creates anelectric field that causes the ions to move towards the plates. Charge now flows around thecircuit and a current is detected by the sensitive ammeter.

On the diagram, draw the shape of the electric field between the plates.

[1 mark]

[2 marks]

Markschememinimum of three lines equally spaced and distributed, perpendicular to the plates anddownwards; edge effect shown; } (condone lines that do not touch plates)

Examiners reportField patterns were often negligently drawn. Lines did not meet both plates, edge effectswere ignored, and the (vital) equality of spacing between drawn lines was not considered.Candidates continue to show their inadequacy in responding to questions that demand acareful and accurate diagram.

33b.

The separation of the plates d is 12 mm and the potential difference between the plates is 5.2kV. An ionized air molecule M with charge is produced when a charged particle collideswith an air molecule.

Calculate the electric field strength between the plates.

Markscheme

Examiners reportThis sequence of calculations was often undertaken well with appropriate figures carriedthrough from part to part.

V+2e

4.3 × 105 (NC−1)

Determine the change in the electric potential energy of M as it moves from the

[1 mark]

33c. Determine the change in the electric potential energy of M as it moves from thepositive to the negative plate.

Markschemeor ;

;

negative/loss;

Examiners reportThis sequence of calculations was often undertaken well with appropriate figures carriedthrough from part to part. The only common error was the omission of a consideration of thegain or loss of the energy change in part (iii).

ΔEP = qΔV 3.2 × 10−19 × 5.2 × 103

1.7 × 10−15 (J)

33d.

Radium-226 decays into an isotope of radon (Rn) by the emission of an alpha particleand a gamma-ray photon. The alpha particle may be detected using the ionization chamber butthe gamma-ray photon is unlikely to be detected.

Construct the nuclear equation for the decay of radium-226.

Markscheme

or ;

numbers balance top and bottom on right-hand side;

Examiners reportAs one of the easiest questions on the paper this was predictably well done.

(22688 Ra)

(22688 Ra →222

86 Rn +42 He +0

0 γ)22286 Rn 4

2He

[3 marks]

[2 marks]

33e. Radium-226 has a half-life of 1600 years. Determine the time, in years, it takes for theactivity of radium-226 to fall to 5% of its original activity.

Markscheme;

;

6900 (years);

Award [3] for a bald correct answer.

Award [2 max] for 2.18 10 (s).

Award [1 max] to a candidate who identifies time as about 4.3 half-lives but cannot getfurther or gives an approximate reasoned answer.

However award [3] if number n of half-lives is calculated from 0.05 = 2 (= 4.32 usuallyfrom use of log working) and time shown.

Examiners reportIn the past candidates have found calculation involving exponential change difficult. On thisoccasion, however examiners saw a large number of correct and well explained solutionsfrom candidates.

λ = = 4.33 × 10−4 (yr−1)ln21600

0.05 = e−λt

× 11

–n

2

An electric dipole consists of a positive and a negative charge separated by a fixed

[3 marks]

34. An electric dipole consists of a positive and a negative charge separated by a fixeddistance. The electric field due to the dipole is shown in the diagram below.

An electric force acts on an electron at point P. In which direction does this force act?

MarkschemeA


35. A particle has charge and mass. Which types of field cause a force to be exerted on theparticle when it is moving in the direction of the field?

A. Electric, gravitational and magnetic fields

B. Electric and magnetic fields only

C. Gravitational and magnetic fields only

D. Electric and gravitational fields only

[1 mark]

[1 mark]

MarkschemeD


36. An electron is held close to the surface of a negatively charged sphere and thenreleased. Which describes the velocity and the acceleration of the electron after it isreleased?

MarkschemeD


A particle of charge q is at point S in a uniform electric field of strength E. The particle

[1 mark]

37. A particle of charge q is at point S in a uniform electric field of strength E. The particlemoves a distance w parallel to the field lines and then a distance y perpendicular to thefield lines to reach point T.

What is the change in electric potential energy of the charge between S and T?

A. Eqw

B. Eqy

C. Eq (y + w)

D. Eq

MarkschemeA


√y2 + w2

38. The diagram shows two point charges P and Q. At which position is the electric fieldstrength equal to zero?

[1 mark]

[1 mark]

MarkschemeA


39. An electron is held close to the surface of a negatively charged sphere and thenreleased. Which describes the velocity and the acceleration of the electron after it isreleased?

MarkschemeD


Two spherical objects of mass M are held a small distance apart. The radius of each

[1 mark]

40. Two spherical objects of mass M are held a small distance apart. The radius of eachobject is r.

Point P is the midpoint between the objects and is a distance R from the surface of each object.What is the gravitational potential at point P?

A.

B.

C.

D. 0

MarkschemeB


− GM

(r+R)2

−2 GMr+R

− GMr+R

The diagram shows equipotential lines around two sources.

[1 mark]

41. The diagram shows equipotential lines around two sources.

Possible sources are

I. two equal masses

II. two equal charges of same sign

III. two equal charges of opposite sign.

What is/are the possible source(s) for the equipotential lines?

A. I and II only

B. I and III only

C. II only

D. III only

MarkschemeA


42a.

This question is in two parts. Part 1 is about momentum. Part 2 is about electric point charges.

Part 1 Momentum

State the law of conservation of linear momentum.

[1 mark]

[2 marks]

Markschemetotal momentum does not change/is constant; } (do not allow “momentum is conserved”)provided external force is zero / no external forces / isolated system;


42b. A toy car crashes into a wall and rebounds at right angles to the wall, as shown in theplan view.

The graph shows the variation with time of the force acting on the car due to the wall during thecollision.

The kinetic energy of the car is unchanged after the collision. The mass of the car is 0.80 kg.

[9 marks]

(i) Determine the initial momentum of the car.

(ii) Estimate the average acceleration of the car before it rebounds.

(iii) On the axes, draw a graph to show how the momentum of the car varies during the impact.You are not required to give values on the y-axis.

Markscheme(i) clear attempt to calculate area under graph;initial momentum is half change in momentum;

Award [2 max] for calculation of total change (1.92kg ms )

(ii) initial speed ;

or ;–15(ms ); (must see negative sign or a comment that this is a deceleration)

oraverage force =12 N;uses F=0.8×a ;–15(ms ); (must see negative sign or a comment that this is a deceleration)Award [3] for a bald correct answer.Other solution methods involving different kinematic equations are possible.

(iii) goes through t=0.08s and from negative momentum to positive / positive momentum tonegative;constant sign of gradient throughout;curve as shown;Award marks for diagram as shown.


( × × 24 × 0.16) = 0.96 (kgms−1)12

12

–1

= ( =) 1.2ms−10.960.8

a = 1.2−(−1.2)0.16 a = −1.2−1.2

0.16–2

–2

Two identical toy cars, A and B are dropped from the same height onto a solid floor

42c. Two identical toy cars, A and B are dropped from the same height onto a solid floorwithout rebounding. Car A is unprotected whilst car B is in a box with protectivepackaging around the toy. Explain why car B is less likely to be damaged when dropped.

Markschemeimpulse is the same/similar in both cases / momentum change is same;impulse is force × time / force is rate of change of momentum;time to come to rest is longer for car B;force experienced by car B is less (so less likely to be damaged);


42d.

Part 2 Electric point charges

Define electric field strength at a point in an electric field.

Markschemeelectric force per unit charge;acting on a small/point positive (test) charge;


Six point charges of equal magnitude Q are held at the corners of a hexagon with the

[4 marks]

[2 marks]

42e. Six point charges of equal magnitude Q are held at the corners of a hexagon with thesigns of the charges as shown. Each side of the hexagon has a length a.

P is at the centre of the hexagon.

(i) Show, using Coulomb’s law, that the magnitude of the electric field strength at point P due toone of the point charges is

(ii) On the diagram, draw arrows to represent the direction of the field at P due to point charge A(label this direction A) and point charge B (label this direction B).

(iii) The magnitude of Q is 3.2 µC and length a is 0.15 m. Determine the magnitude and thedirection of the electric field strength at point P due to all six charges.

kQ

a2

[8 marks]

Markscheme(i) states Coulomb’s law as or states explicitly q=1;states r=a;

(ii)

arrow labelled A pointing to lower right charge;arrow labelled B point to lower left charge;Arrows can be anywhere on diagram.

(iii) overall force is due to +Q top left and -Q bottom right / top right and bottom left andcentre charges all cancel; } (can be seen on diagram)force is therefore ;

2.6×106 (N C ) ;towards bottom right charge; (allow clear arrow on diagram showing direction)


kQq

r2 =Fq

kQ

r2

2kQ

a2

-1

Part 2 Motion of a rocket

43a.

Part 2 Motion of a rocket

A rocket is moving away from a planet within the gravitational field of the planet. When therocket is at position P a distance of 1.30×10 m from the centre of the planet, the engine isswitched off. At P, the speed of the rocket is 4.38×10 ms .

At a time of 60.0 s later, the rocket has reached position Q. The speed of the rocket at Q is4.25×10 ms . Air resistance is negligible.

Outline, with reference to the energy of the rocket, why the speed of the rocket ischanging between P and Q.

Markschemegravitational potential energy is being gained;this is at the expense of kinetic energy (and speed falls);


7

3 –1

3 –1

43b. Estimate the average gravitational field strength of the planet between P and Q.

[2 marks]

[2 marks]

Printed for Jyvaskylan Lyseon lukio

© International Baccalaureate Organization 2019 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markscheme;

gravitational field strength = acceleration of rocket (=2.17 N kg–1); } (allow g = a insymbols)

or

computes potential difference from KE per unit mass change (5.61 105),

computes distance travelled (0.259 Mm), uses ;g=( )2.17(ms );


(acceleration = = =) (−) 2.17 (ms−2)(v−u)t

4.25×103−4.38×103

60

×g = (−)ΔV

Δr

− -2

43c. A space station is in orbit at a distance r from the centre of the planet in (e)(i). A satelliteis launched from the space station so as just to escape from the gravitational field ofthe planet. The launch takes place in the same direction as the velocity of the space station.Outline why the launch velocity relative to the space station can be less than your answer to (e)(i).

Markschemethe satellite has velocity/kinetic energy as it is orbiting with the space station;


[1 mark]

topic 10 revision [147 marks] - peda.net · topic 10 revision [147 marks] 1. four identical,...

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