1. C[1]

2. D[1]

3. C[1]

4. B[1]

5. D[1]

6. A[1]

7. D[1]

8. C[1]

9. A[1]

10. B[1]

11. (a) They act on the same body or do not act on different bodies (1)They are different types of, or they are not the same type of, force(1) 2

(b) As the passenger or capsule or wheel has constant speed (1)there is No resultant tangential force (acting on the passenger) (1) 2

(c) Friction between seat & person or push of capsule wall on person (1) 1[5]

12. (a) A baryon is a (sub-atomic) particle made up of 3 quarks(1) 1

(b) n (ddu) → (1)p (duu) (1) 2

[3]

13. (a) High frequency or high voltage(1)Alternating or square wave voltage(1) 2

(b) No electric field inside cylinders (due to shielding) (1)so no force (on electrons) (1) 2

(c) As speed increases (along the accelerator), (1)cylinders are made longer so that time in each stays the same(1) 2

[6]

14. The answer must be clear, use an appropriate style and be organised in alogical sequence (QWC)-particles fired at (named) metal (film) (1)in a vacuum (1)Most went straight through or suffered small deflections. (1)A few were reflected through large angles or some were reflected alongtheir original path (1)suggesting the mass or charge of the atom was concentrated in a very smallvolume (1) 5

[5]

15. (a) (i) Use of E = ½ CV2 (1)Answer [0.158 J] (1)

E = ½ CV2 = 0.5 × 2200 × 10–6 F × (12 V)2

E = 0.158 J 2

(ii) Correct substitution into Ep = mgh (1)Answer 0.75 [75%] (1)

Ep = 0.05 kg × 9.8 N kg–1 × 0.24 m [= 0.12 J]Efficiency = 0.12 ÷ 0.16 J = 0.75 [75%] 2

(b) (i) (t = CR) = 2200 × 10–6 (F) × 16 () = 35.2 (ms) (1) 1

(ii) Curve starting on I axis but not reaching t axis (1)

I0 = 1.6 V / 16 = 100 mA shown on axis (1)

Curve passing through about 37 mA at t = 35 ms (1) 3

(c) (i) The vibrations of the air particles (1)are parallel to the direction of travel of the wave (energy) (1) 2

(ii) T = 1/f = 50 ms (1)

Sensible comment related to time constant of 35 ms (1) 2[12]

16. (a) 3Li7 + 1p1 = 2He4 + 2He4

completing LHS (1)completing RHS(1) 2

(b) (i) Charge (1)(mass/) energy (1) 2

(ii) Mass of Li + p = 7.0143 u + 1.0073 u = 8.0216 u (1)Mass of 2 -particles = 2 × 4.0015 u = 8.0030 u (1)m = 8.0216 u – 8.0030 u = 0.0186 u

= 0.0186 × 1.66 × 10–27 kg = 3.09 × 10–29 kg (1)

E= c2m = (3.00 × 108 m s–1)2 × 3.09 × 10–29 kg = 2.78 × 10–12 J (1)

[Allow ecf from equation] 4

(iii)119

12

eV J 1060.1

J 1078.2

= 1.74 × 107 eV = 17.4 MeV (1)

The incoming proton has an energy of 300 keV = 0.30 MeV (1)So total energy = 17.4 MeV + 0.3 MeV = 17.7 MeV (1)The calculated energy differs by

%3%100MeV)2.177.17(

MeV 17.2-Mev 7.17

21

(1)

The experiment therefore provides strong evidence for Einstein’sprediction (1) 5

[13]

17. (a) Total (linear) momentum of a system is constant, (1)

provided no (resultant) external force acts on the system (1) 2

(b) The answer must be clear, use an appropriate style and be organised in alogical sequence

Use of a light gate (1)Use of second light gate (1)

Connected to timer or interface + computer (accept ‘log-it’) (1)

Cards on gliders (1)

Measure length of cards (1)Velocity = length ÷ time (1) 6

(c) Multiplies mass × velocity to find at least one momentum (1)

1560 g cm s–1 (0.0156 kg m s–1) before and after (1) 2[10]

18. (a) (i) Use of A = r2 leading to 0.87 (m) (1) 1

(ii) Correct use of = 2/t leading to 62.8 (rad s–1) (1) 1

(iii) Correct use of v = r = 55 m s–1 [allow use of show that value] (1) 1

(b) (i) Substitution into p = ½ Av3 (1)

3047 (W) (1) 2

(ii) Air is hitting at an angle/all air not stopped by blades (1)Energy changes to heat and sound (1) 2

(c) (i) Attempts to find volume per second (A × v) (1)

44 kg s–1 (1) 2

(ii) Use of F = mv/t (1)

F = 610 N (1) 2

(d) Recognises that 100 W is produced over 24 hours (1)Estimates if this would fulfil lighting needs for a day(1)Estimates energy used by low energy bulbs in day(1)Conclusion (2)The answer must be clear and be organised in a logical sequence

Example:

The 100 W is an average over the whole day. Most households would uselight bulbs for 6 hours a day in no more than 4 rooms, so this would meanno other energy was needed for lighting.4 low energy bulbs would be 44 W for 6 each hours so would require energyfrom the National grid.

[Accept an argument based on more light bulbs/longer hours that leads tothe opposite conclusion] 5

[16]

19. (a) Paths of alpha particlesPath A drawn less deflected than B (1)

Path A drawn as a straight line (1) 2

(b) (i) Why alpha source inside containerAlpha would be absorbed by [accept would not get through]container (material) (1) 1

(ii) Why the same kinetic energy?EitherTo restrict observation to two variables / closeness of approachand deflectionor so that speed / velocity / (kinetic) energy does not havean effect (on the observation / deflection /results /contact time) 1

(iii) Why an evacuated container?Eitherso that alphas do not get absorbed by / collide with / getdeflected by / stopped by / scattered by / get in theway of / ionise / lose energy to atoms / molecules (ofair) [Do not accept ‘particles’ of the air]

or so that all alphas reach the foil with the same (kinetic) energy 1[5]

20. Particle classification

Neutron: baryon and hadron (1)

Neutrino: lepton (1)

Muon: lepton (1) 3[3]

21. (i) Conservation laws

First reaction, Q: 0 + 0 ≠ 1 + 1 (1)

Second reaction B: 1 = 1 + 0 AND Q: –1 = –1 + 0 (1)

Hence only Ω– decay possible [based on B and Q conservation for thisdecay, accept simple ticks and crosses] (1) 3

(ii) Quark charges

Use of sss = –1 to show s = –⅓ (1)

Hence correct working (from baryons) to show u = ⅔ and d = –⅓ (1) 2[5]

22. (a) (i) Why speed is unchanged

Force/Weight [not acceleration] is perpendicular tovelocity/motion/direction of travel/instantaneous displacement[not speed]OR no component of force/weight in direction of velocity etc (1)

No work is doneOR No acceleration in the direction of motion (1) 2

(ii) Why it accelerates

Direction (of motion) is changing (1)

Acceleration linked to a change in velocity (1) 2

(b) Speed of satellite

Use of a = v2/r (1)

Correct answer [3.8 to 4.0 × 103 m s–1] (1)

Example calculation:

v = √(2.7 × 107 m × 0.56 m s–2)

[Allow 1 mark for ω = 1.4 × 10–4 rad s–1] 2[6]

23. (a) (i) W = QV (1)

Correct answer 3.2 nJ [3.2 × 10–9 J, etc.] (1)

Example of answer:

W = QV = 0.8 × 10–9 C × 4.0 V = 3.2 × 10–9 J 2

(ii) +0.8 (nC) on top plate and –0.8 (nC) on bottom plate (1)(both needed) 1

(b) Statement (E =) ‘Area’ or (E =) ½ QV (1)

See calculation ½ × 4.0 × 0.8 or ½ × base × height (1)

ORC found from graph (1)

Use of W = ½ CV2 (1)

Example of answer:

F 100.2V 0.4

C 100.8 109

V

QC

W = ½CV2 = 2

V) (4.0 F 100.2 210

= 1.6 × 10–9 J 2

(c) (i) Correct answer 0.2 nC (1) 1

(ii) Graph is straight and through origin (1)

ends at 3.0V and their Q (1) 2

(iii) Attempt to use C = Q/V or C = ∆Q/∆V (1)

Correct answer 0.067 nF / 67 pF (1)

Example of answer:

V 0.3

C 102.0 9

V

QC = 6.7 × 10–11 F 2

[10]

24. (a) (i) 1.2 keV = 1.2 × 103 × 1.6 × 10–19 JOR

Use of e∆V with e as 1.6 × 10–19 C and V as 1200 V (1)

Use of ∆(½mev2) with me as 9.1(1) × 10–31 kg. (1)

Correct answer 2.0 – 2.1 × 107 m s–1 (1) 3

(ii) 1200 × 8/100 = 96 (eV delivered per electron) (1)96/2.4 = 40 (1)

Or2.4 × 100/8 = 30 (incident eV needed per photon) (1)1200/30 = 40 (1)

Or1200 / 2.4 = 500 (photons per electron, ideally) (1)500 × (8/100) = 40 (1) 2

(b) Electrons on screen repel electrons in beam / force opposeselectron motion/decelerating force (1)

Electrons (in beam) decelerated /slowed /velocity reduced/ work done by electrons (against force) (1)

Electron (kinetic) energy reduced (not –shared–) (1)

Fewer photons (per electron, stated or implied) (1)

Trace less bright (1)

QoWC (1) Max 4[9]

25. (a) Scale interval is 0.1 (V) (1) 1

(b) (i) Use of ε = (–)N∆φ/∆t (1)

Correct answer 9.6 × 10–7 (Wb) / 0.96 (µWb) [ignore +/–] (1)

Example of answer:

= × N

t = 0.12 V ×

5000

s 1040 3 = 9.6 × 10–7 Wb 2

(ii) Use of ‘φ’ or ‘∆φ’ or ‘flux’ = BA, or B = ε ∆t/NA (1)

Correct answer 0.012 T / 0.013 T (1)

Example of answer:

= BA

B = 22

7

2

m 100.1

Wb106.9

A = 0.012 T

[N.B. φ = 0.96 µWb → 0.012T, φ = 1µWb → 0.013T] 2[5]

26. (a) pair of values of k.e. and υ2 read from graph / gradient (1)

υ2 > 5 × 1016 m s–2 (1)

mp = 1.62 – 1.69 × 10–27 (kg) to 3 s.f. (1) 3

(b) (i) (values 1.3 – 1.7, 3.1 – 3.5, 6.0 – 6.5) any two correct (1)(1)

(ii) ∆E = c2∆m / E = mc2 (1)

one value for ∆m (× 10–28 kg) (1)

use of mp from (i) [no mark]

one value of ∆m/mp : about 10%, 20%, 40% (1) 5[8]

27. (a) Show sum of quark charges in proton = +1+2/3 +2/3 –1/3 = (+) 1 (1)

Show sum of quark charges in neutron = 0+2/3 –1/3 –1/3 = 0 (1)

[ignore references to e] 2

(b) (i) • baryon (1)• meson (1) 2

(ii) baryon: 3 quarks (1)

meson: quark/antiquark (1)[1 for answers reversed or baryon/meson not specified] 2

(c) any 4 marks from the following examples:high speed means high energy/momentum (1)may need to overcome (electrostatic) repulsion (1)more energy available for creating particles (1)higher energy/momentum/speed means shorter wavelength (1)reference to λ = h/mv or λ = h/p (1)for diffraction/scattering (1)need λ approx equal to particle spacing/internal structure (1) max 4

(d) Speeds near the speed of light (1)[11]

28. Recall speed = s/t (1)Use of s = π D (1)Answer for speed (1)Conclusion (1)

ORUse of v = rωUse of ω = 2π × 20 000Answer for speedConclusion

v = s/ts = π × 8000 (m)v = π × 8000 × 20 000 (m/s)

v = 5 × 108 m/s

inaccurate/not possible since speed > c[4]

29. (a) Lines (1)[not crossing; minimum 2 lines starting from S pole of magnet]Correct arrow(s) (1)[minimum 1 arrow pointing towards S pole, any incorrect arrow scores 0] 2

(b) (i) Use of F = BIl rearranged to B = F/Il OR with two correct subs (1)Leading to correct answer (1)

B = F/Il = 0.008/(5.8 × 0.012) (T)B = 0.11 (T) 2

(ii) Assumption:parallel field/uniform field/constant field for 12 mm then falls to zero /assume wire perpendicular to field (at all points)/θ = 90° (where F=Bilsinθ given earlier)/force same at all points on the wire (1) 1

(c) Experimental value less because field divergesOR field strength decreases with distanceOR field could be 0.3 T at magnet surface and only 0.1 T at wire (1) 1

(d) Wire would levitate (again) (1)Two reversals cancel/applying FLHR (1)[wire moves downwards due to current OR field reversed scores 1] 2

[8]

30. (a) capacitors need d.c. (1)OR Mains is a.c. / mains current changes direction constantly

Charge given in one half of cycle is removed the next half (1)OR C charged then discharges 2

(b) (i) Voltage value for initial voltage × 1/e [or use of 37%] (1)

OR use 2 values where 2nd is 1/eth of firstOR draw tangent at time = 0 s

OR V = Vo e–t/RC with correct substitution of (t,V) from graph

T = 0.07 s [allow 0.065 s – 0.075 s] (1) 2

(ii) Recall time constant = CR (1)Answer for R [allow ecf for T] (1)

R = T/C = 0.07 s/(100 × 10–9 F)

R = 7 × 105 Ω 2

(c) (i) Recall Q = CV [equation or substitution] (1)Answer for Q (1)

Q = CV = 100 × 10–6 × 300Q = 0.03 C 2

(ii) Recall W = 1/2 CV2 OR W = 1/2 Q2/C (1)OR 2 correct subs into W = 1/2 QV [allow ecf]Answer (1)

Eg: W = 1/2 QV = 0.5 × 0.03 × 300 (J) = 4.5 J

OR W = 1/2 CV2 = 0.5 × 100 × 10–9 × 300 × 300 (J) = 4.5 J

OR W = 1/2 Q2/C = 0.5 × 0.03 × 0.03/(100 × 10–9) (J) = 4.5 J 2[10]

31. Any 8 marks from:Recall of p = mv (1)Use of momentum before collision = momentum after collision (1)Correct value for speed (1)

Example:1250 × 28.0 + 3500 x25.5 = (1250 + 3500) v

v = 26.16 m s–1

Recall of ke = 1/2 mv2 or ke = p2/2m (1)Total ke before (1)Total ke after (1)Loss in ke (1)Recall of work = force × distance (1)Correct answer for force to 2 SF (1)

Example:Total ke before = 1 138 000 J + 490 000 J = 1 627 938 JTotal ke after = 1 625 059 JLoss of KE = 2879 JBraking force = 2879/5 = 576 N Max 8

[8]

32. (a) 18 1 18 1 (1) O + p/H equals F + n (1)8 1 9 0 (1)[omitting the n with everything else correct = 1] 3

(b) Accelerated through 19 × 106 V / MVUsing linear accelerator / cyclotron / particle accelerator / (1)recognisable description (1) 2

(c) Time taken for half the original quantity/ nuclei /activity to decay (1)

Long enough for (cancer/tumour/body to absorb) and still beactive/detected (1)

Will not be in body for too long (1) 3

(d) Use of E = mc2 (1)Use of E = hf (1)Use of v = f λ (1)

λ = 2.4 × 10–12 m (1)

eg 9.11 × 10–31 × 9 × 1016 (×2)

f = 8.2 × 10–14 / 6.6 × 10–34 ecf

λ = 3 × 108 / 1.2 × 1020 ecf 4

(e) Conservation of momentum (1)

Before momentum = 0 (1)

so + for one photon and – for other (1) 2 max[14]

33. (a) Identify particle

Alpha (particle) / Helium nucleus/ He42 / 4

2He / 42 /

42

42

42 aalpha/alph/ / 1

(b) Momentum of particleMomentum equation [In symbols or with numbers] (1)

Either

Correct substitution into 2

1mv2 = energy (1)

Use the relationship to determine the mass [6.6 × 10–27 kg] (1)

Answer [9.3 × 10–20 (kg m s–1) Must be given to 2 sig fig. Nounit error] (1)

Or

Rearrangement of Ek = 2

1mv2 to give momentum ie

v

E2 K (1)

Correct substitution (1)

Answer [9.3 × 10–20 kg m s–1. Must be given to 2 sig fig. Nounit error] (1)

Eg 2

1m(1.41 × 107 m s–1)2 = 6.58 × 10–13 J

m = 217

13

)s m 1041.1(

J 1058.62

= 6.6 × 10–27 kg

momentum = 6.6 × 10–27 kg × 1.41 × 107 m s–1

= 9.3 × 10–20 (kg m s–1)

Or

Momentum = 17

13

s m 1041.1

J 1058.62

= 9.3 × 10 (kg m s–1) 4

(c) Consistent with the principle of conservation of momentum(Since total) momentum before and after (decay) = 0 (1)

State or show momentum / velocity are in opposite directions (1)[Values of momentum or velocity shown with opposite signswould get this mark]

Calculation ie 3.89 × 10–25 kg × 2.4 × 105 m s–1 = 9(.3) × 10–20 (kg m s–1) (1)

Eg 3.89 × 10–25 kg × 2.4 × 105 m s–1 = 9(.3) × 10–20 kg m s–1 3[8]

34. (a) Calculate the ratio the densities of the atom and the nucleusDensity equation [In symbols or numbers] (1)Show the relationship between density and radius. (1)[Candidates who start by stating that density is inverselyproportional to the radius cubed would get both these marks.Candidates who show an expression where the mass is

divided by 3

3

4r would set both these marks. Candidates who

write Ratio = (1/105)3 would get both of these marks.]

Factor 10–15 established. [Some working must be shown forthis mark] (1)

Eg (Density)atom = 3atom3

4r

m

or Density

3

1

r

(Density)nucleus = 3nucleus3

4r

m

3

atom

nucleus

nucleus

atom )(Density)(

(Density)

r

r

= (10–5)3

Assumption – (entire) mass of the atom is concentrated in thenucleus[there must be a reference to the nucleus] (1)[eg mass of the atom =/approx – mass of the nucleus; most /majority of the atom’s mass is in the nucleus. The followingwould not be awarded marks; The atom is mostly emptyspace; mass of the electrons is negligible; the nucleus is avery dense.] 4

(b) ObservationA very small percentage of particles [accept ‘very few’ notjust ‘a few’. Do not accept ‘some’] are deflected throughangles greater than 90° / are back-scattered / deflected back. (1)[Allow; nearly all / most (alpha) particles pass through(the atom) without being deflected (showing the atom isvirtually empty space).][Accept ‘nearly all’, not ‘many’ for the word ‘most’.] 1

[5]

35. (a) Calculation of angular speed

Use of = 2/T (1)

7.27 × 10–5 [2 sig fig minimum] (1) 2

2/(24h × 3600 s h–1) = 7.27 × 105 rad s–1

(b) (i) Calculation of acceleration

Use of a = r2 OR v = r and a = v2/r (1)

0.034/031 m s–2 (1) 2

(6400 × 10J m)(7.27 × 10–5 rad s–1)2

= 0.034 m s–2

(ii) Direction of accelerationArrow to the left (1)[No label needed on arrow. If more than one arrow shown, nomark unless correct arrow labelled acceleration] 1

(iii) Free-body diagramArrow to left labelled Weight/W/mg/pull of Earth/gravitationalforce (1)Arrow to right labelled Normal reaction/N/R/push of Earth (ORground)/(normal)contact force (1)

[Don’t accept “gravity” as label][More than two forces max 1][Diagram correct except rotated gets 1 out of 2] 2

(iv) How the acceleration is producedN is less than W (1)Resultant (OR net OR unbalanced) force towards centre (1)

[Accept downward / centripetal for towards the centre, butnot as an alternative to “resultant”] 2

[9]

36. (a) (i) P.d. across capacitor

Use of VR = I × R (1)

[allow one error of 103 in individual substitutions; disallow ifVR value is 6V]

VC = 6.0 V – 4.0 V (= 2.0V) (1)[No ecf] 2

Example of answer:

VR = 20 × 10–6 A × 2.0 × 105 = 4.0 V

Hence Vc = 6.0 V – 4.0 V = 2.0 V

(b) Calculation of charge

Use of Q = C × V with 560 F & 2.0 V (1)

[Check correct equation is being used; allow power of 10 error incapacitance value. If capacitance value mis-transcribed, allow thisfirst mark only]

Answer 1.1 (2)mC (1120C) [no ecf] (1) 2

(c) Calculation of energy stored

Use of W = ½CV2 with given values, or W = ½ V with their Q,to get 1.1(2) mJ (1120J) or their correct answer. (1)[same numerical value as in (b)] 1

(d) Calculation of energy transferred

Use of E = QV, with their Q and V = 6.0 V, to get 6.7(2) mJ (6720J)or their answer [6 × value at part c] correctly found. (1) 1

(e) Main reason for energy difference

Energy is transferred to thermal / heat energy in / work is doneagainst, the resistance of the resistor in the circuit [NOT just ‘the resistance of the wires’, nor the ‘components’] (1)

[Do not credit vague reference to energy dissipation, nor ‘energyis lost to the surroundings’] 1

[7]

37. (a) Calculation of potential difference

Use of ½mpv2 with v = 2.77 × 105 m s–1

and mp = 1.67 × 10–27 kg (1)

Use of eV with e = 1.60 × 10–19 C (1)[beware confusion of v and V]

Answer = 400(.4) / 401 V (1)[If data used to 2 sf, 380V, 384V or 364V, allow 2/3] 3

Example of answer:

eV = ½ mpv2

V = C 106.12

)s m 1077.2( kg 1067.1

2 19

215272

e

vm p

= 400V[beware unit error of eV here]

(b) Add second path to diagram

Path at B stays equidistant from that at A [gauge by eye] (1) 1

(c) (i) Add path to diagram

Added path at A [allow through letter A] also curves upwards (1)

But is less curved than the original, straight beyond plates andcontinues to diverge from it (1) 2

(ii) Explanation

Charge on a is double that on proton / has 2 protons /force on a is double force on proton. (1)

Mass of a particle is (approx) 4 times / more than double that of theproton. (1)

[hence acceleration is approximately halved].

[Ignore reference to F = Bqv; do not credit reference

to He42 unless implication of numbers 4 and 2 is made clear]. 2

[8]

38. (a) (i) Direction of current

Position 1 = Q to P / anticlockwise / to the left (1)Position 3 = P to Q / clockwise / to the right [both needed; arrows added to diagram may givecurrent directions at 1 & 3]

Position 2 = no current (1) 2

(ii) Current calculation

Use of t

BA

t

)(

, or = Blv, = 2 × 10–2 T × 0.12 m × 0.05 m s–1 (1)

[ignore power of 10 errors in dimension and velocity values]

(Emf =) 1.2 × 10–4 V (1)

I = R

V or I =

R

E seen or used (1)

Answer = 6.0 × 10–5 A or 60A [ecf their emf] (1) 4

(b) Uniform acceleration?

QoWC (1)

Magnitude of current would be increasing as frame movesthrough position 1 (or position 3) (1)

Magnitude of current would be greater for position 3 than 1[Beware comparison of position 3 with position 2 here] (1)

Reference to increased rate of flux cutting / increased rate of flux change / increased area swept out per second (1)(Beware suggestion that B or flux density is changing)

So induced emf is greater (1)

Current for position 2 is zero [Do not credit equal and opposite currents cancelling] Both needed (1)

Since flux linkage is constant / (net) rate of flux cutting is zero / Emfs in PS and QR are equal and opposite Max 4

[10]

39. (a) recall of p = mv [eqn or sub] (1)answer (1)

p = mv

= 2 × 0.024 × 0.88 (N s) = 0.042(24) N s OR kg m s–1 2

(b) recall of KE = 1/2mv2 OR p2/2m [eqn or sub] (1)answer (1)

KE = 1/2mv2

= 0.5 × 2 × 0.024 × 0.88 × 0.88 (J) = 0.0185(856) (J) 2

(c) (i) provided no external force acts (1)OR balls do not interact with/transfer momentum to anything else 1

(ii) v = momentum/mass = 0.042(24)/0.072 = 0.5833 (0.5867) (m s–1) (1) 1

(iii) 0.5 × 0.096 × 0.442 OR 0.5 * B9 * C9 * C9 (1) 1

(iv) 3 points from:can’t be one ball as too much KE (1)collision pretty elastic/not much loss of energy (1)so won’t be 3 or 4 or 5 balls (1)2 balls gives same energy (1) Max 3

(d) 2 points fromkinetic energy is lost (as sound/through deformation/to heat) (1)OR collisions not perfectly elasticMomentum still conserved (1)as the total ke decreases (column D) more balls are in motion (1) Max 2

[12]

40. (a) use of Q = CV OR statement or use of W = CV2/2 OR Q2/2C (1)answer (1)

W = CV2/2= 0.5 × 2500 × 2 × 2 (J) = 5000 J 2

(b) 1 correct value (1)All correct values; 1.62, 1.39, 1.16 (1)(1 mark for one correct or inappropriate sig figs) 2

(c) graph of ln (y) v. time (x) (1)appropriate scales and both axes labelled fully (1)points plotted properly (+/– 1 mm) (1)best fit line drawn (1) 4

(d) recognise that gradient = (–)1/RC (1)evaluate gradient (1)conversion days to seconds (1)obtain appropriate value for R (1)

gradient = (–) 0.92/(40 (days))

R = 40 × 24 × 3600 (s) / 0.92 × 2500 (F)= 1500 Ω

(OR method using graph of V v. t)recognise that time to Vo/e = RC (1)this time estimated (42–45 days) (1)conversion days to seconds (1)obtain appropriate value for R (1) Max 4

[12]

41. (a) energy (of proton) converts to mass (1)7 TeV > 251 GeV, (so enough energy present to create Higgs particle) (1) 2

(b) (i) calculate rest–mass energy of proton in J (1)comparison with 7 TeV (1)

rest mass energy of proton – E = mc2 = 1.67 × 10–27 × c × c

= 1.5 × 10–10 J

= 1.5 × 10–10 / 1.6 × 10–19 (eV) = 9.4 × 108 (eV)much less than 7 TeV.

OR 7 TeV = 7 × 1012 × 1.6 × 10–19 (J)

= 1.12 × 10–6 (J)

>> 1.5 × 10–10 J 2

(ii) Appropriate use of 1.6 × 10–19 OR energy from above in J (1)Answer (1)

momentum = energy/c = 7 × 1012 × 1.6 × 10–19 (J)/(3 × 108 (m/s)) =

3.73 × 10–15 (kg m s–1) 2

(iii) Attempt to use r = p/Bq (1)two correct subs into formula OR rearrangement (1)circumference => radius (1)answer (1)

r = p/BqB = p/rQ

= 3.73 × 10–15 / [(27000/2_) × 1.6 × 10–19] (T)= 5.4 T 4

(iv) Yes (stated or clearly implied) (1)because motion and force both horizontal OR motion/force/B must allbe perpendicular (1) 2

[12]

42. (a) 1 equating PE and KE (1)2 recall of mv2/r (1)3 find centripetal force = 2mg (1)4 force on rider = centripetal force + weight OR force = 3mg (1)5 hence “g-force” = 3 (1) 5

(b) height not a factor, so B is correct (1)(some will reach this conclusion via much longer routes) 1

[6]

43. (a) Table

[Ignore crosses. If more than one tick in a line, no mark.]Top line: To the left (1)Bottom line: Downwards (1) 2

(b) Calculation of rotation period

Use of T = 2πr/v or T = 2π/ω and ω = v/r (1)Correct answer [0.084 s] (1)

e.g.

2π (0.28 m)/(21 m s–1)= 0.084 s 2

(c) (i) How the angular speed is affected

ω is increased, plus correct supporting argument in formula or words (1)i.e. Since v = rω / T decreases / f increases / wheel must turn faster 1

(ii) Speedometer reading

Speedometer reading is too high because frequency (1)

(OR ω OR revs per second OR rate of rotation of wheel)is increased[Allow ecf from “ω decreased” in c (i)] 1

[6]

44. (a) Show that

See ‘v = T

r2’ OR ‘ω =

T

r2’ (1)

Substitution of (60 × 60 × 24)s or 86400s for T (giving 7.27 × 10–5, no u.e.) (1)

Unit of ω

s–1/rad s–1 (1) 3

(b) Height above Earth’s surface

Statement / use of 2

2

2 OR

r

mGM

r

mv

r

mGM EE = mrω2 (1)

[Equations may be given in terms of accelerations rather than forces][Third mark (from below) may also be awarded here if (rE + h) is used for r]

Correct value for r, i.e. 4.2(3) × 107 m (1)Use of h = their r – RE (1)

Correct answer = (3.58 – 3.60) ×107 m [no ecf] (1)

Example of answer:

r

mv

r

mGM E2

2

rr

r

r

v

r

GM E 222

2

)(

GME = 2r3

3215

242211

32

)s 1027.7(

kg 1098.5kg m N 1067.6

EGMr

= 4.23 × 107 m

h = 4.23 × 107 m – 6.38 × 106 m

= 3.59 × 107 m 4[7]

45. (i) Add to diagram.

Arrows at A and B, both pointing directly away from the nucleus. (1)[Arrow end (head or tail) need not touch A /B, but direction must be correct.Gauge by eye, accept dotted construction lines as indication of intent] 1

(ii) Calculation of force

Use of F = 20

21

4 r

QQ

or F = 221

r

QkQ (1)

[ignore error/omission of ‘2’ and/or ‘79’ or ‘e’ or ‘1.6 × 10–19’ for this first mark,providing numerator clearly has a product of charges and denominator a distancevalue squared. Ignore power of 10 errors in values of Q or r]

2 × 1.6 × 10–19 C and 79 × 1.6 × 10–19 C seen (consequential mark, dependentupon correct use of equation previously) (1)

Correct answer = 1.6 – 1.7 N (1)

Example of answer:

213112

1919

20

21

)m 105.1(m F 1085.84

C) 106.12(C) 101679(

4

r

QQF

= 1.62 N 3

(iii) Effect on motion of α

Slows down [decelerates] and then speeds up again [accelerates].(both needed)[accept ‘slows down at A and speeds up at B] (1) 1

[5]

46. (a) (i) Direction of e.m.f.?Hub ‘–’ and Rim ‘+’.[Allow mark for either on its own, but not if contradicted.] (1) 1

(ii) Why a constant e.m.f.?

Reference to flux cutting / rate of change of flux / change offlux linkage due to spoke motion / spokes moving at rightangles to field / Reference to Faraday’s Law (1)

Constant rate of spin implies constant rate of flux cutting.[Link made clear] (1)[continuous process does not mean constant rate] 2

(iii) The time for one revolution

Use of ε = t

BA with ‘A’ recognisable as area of a circle (1)

[ignore power of 10 errors for e.m.f. and radius values, andinclusion of N = 24]

Correct substitution of all values [ but only N = 1 acceptable here] (1)

Correct answer 0.31 – 0.32 s (1)[t = 7.6s scores 1/3; t = 1.12s scores 0/3, t = 0.64s scores 1/3 here]

Example of answer:

BA

tt

BA

t

s 317.0V 1025

m) 10(30 T 108.26

225

t 3

Alternative answer

Use of ε = Blv with v = (mean) velocity of spoke. ((1))

→ v = 2.98 m s–1 ((1))

Hence rim velocity = 2.98 × 2 = 5.96 m s–1.

1s m 96.5

m 3.022

RIMv

rt = 0.316s. ((1))

[t = 0.63s scores 2/3 here]

(b) What effect?

(i) Reduced [accept ‘halved’] ANDRate of flux cutting is reduced / Fewer field lines are being cut /Component of Earth’s field perpendicular to the wheel is less /Flux through wheel is less / Area of wheel perpendicular tofield is less / Wheel is no longer perpendicular to the field (1)[do not credit answers suggesting changes in the field strength itself] 1

(ii) Increased / increasing ANDRate of flux cutting [etc.] would be increasing (1) 1

(iii) (Reduced to) zero [but not ‘very small’ / ‘negligible’, etc.] AND

No flux cut by spoke(s) / No component of the Earth’s fieldperpendicular to the wheel / No flux through wheel / Wheelis spinning parallel to the field / in plane of field (1)[but not just ‘∆Ф = 0’, nor ‘motion is not perpendicular to field’]

[Allow 1/3 for three correct statements of ‘ε’ outcome without anyexplanation, but only if score would otherwise be zero.]

[Disallow ‘breaking’ for ‘cutting’ on first occasion in entirequestion, but allow, ecf, thereafter] 3

[9]

47. (a) (i) Not matter/antimatter pair [stated or implied] (1)particle/antiparticle have same mass OR electron/proton not samemass OR other correct reason (eg electron is fundamental, protonis quarks) (1)antiparticle to proton is antiproton OR antiparticle to electron ispositron/antielectron (1) 3

(ii) Not matter/antimatter pair [stated or implied] (1)anti to up is anti-up OR anti to down is anti-down (1)up and down have different charge (1) any 5

(b) particles/antiparticles carry opposite charge (1)(component of) field perpendicular to travel (1)(magnetic/LH rule) forces act in opposite directions (1)some pairs uncharged so no separation/deflection (1)[not annihilation] any 2

(c) number = 5000 × 10–12 kg / 9.11 × 10–31 kg = 5.5 (5.488) × 1021 (1) 1

(d) (i) correct use of E = mc2 [subs] (1)correct use of E = hf and c=fλ [rearranged or subbed] (1)correct answer [ue] (1)

E = mc2 = 9.11 × 10–31 × (3 × 108)2 J (= 8.199 × 10–14 J) (1)E = hf = hc/λ λ = hc/E (1)

= 6.63 × 10–34 × 3 × 108/8.199 × 10–14 m

= 2.4 (2.426 or 2.42 or 2.43) × 10–12 m [ignore omissionof both factors of 2] (1)[factor of 2 wrong is a.e. = –1][use of λ = h/p scores 0] 3

(ii) this wavelength is not visible lightOR this is x-ray or gamma or high energy photon so need shielding (1) 1

[12]

48. (a) (i) arrow towards centre of curvature (1) 1

(ii) Use of formula with correct q OR v subbed (1)correct answer (1)

F = Bqv

= 0.5 × 1.6 × 10–19 × 800 000 N (correct q or v) (1)

= 6.4 × 10–14 N (1) 2

(iii) Use of formula: EITHER correct m subbed OR d identified with r (1)correct answer

r = p/Bq = 1.67 × 10–27 × 800 000/0.5 × 1.6 × 10–19 (m) (1)= 0.017 m (1)[Penalise factor 1000 error once only in question] 2

(iv) derive formula for T (1)correct answer (1)

T = π r/v (OR T = 2πr/v for (1)x) (1)= π × 0.017/800 000 (s) (ecf)

= 6.6 (6.5 – 6.7) × 10–8 s (1) 2

(v) correct statement of force = change of momentum/time (1)correct use of factor 2 (1)correct answer (1)

F = change of momentum/time (1)

= 2 × 1.67 × 10–27 × 800 000/6.7 × 10–8 (N) (ecf) (1)

= 4.1 (4.0) × 10–14 N [errors in m are self-cancelling] (1) 3

(b) Recall of formula (1)correct answer (1)

F = k q1 q 2/r2 OR F = q1 q 2/4πε0r2 OR k = 1/4πε0 (1)

= 1.6 × 10–19 × 1.9 × 10–6/4 × π × 8.85 × 10–12 × 5 × 5 (N)

= 1.1 × 10–16 N (1) 2[12]

49. (i) magnetic field changing (1)field cuts across conductor/flux linkage changes (1)Faraday/V induced (1) (any 3)V causes I (1) 3

(ii) Direction of induced current has an effect tending to cancel itscause OR [reasonable attempt at putting Lenz into words –not just “Lenz”] (1) 1

[4]

50. (a) (i) recall of formula (1)correct answer (1)

C = Q/V (stated or implied) [this way round] (1)= (appropriate pair of values, eg 4 C/4.8 V)= 0.83 (0.82 – 0.84) F (1) 2

(ii) strip width ∆Q (1)1∆W = V. ∆ Q (1)2add strips => area under graph (1)3area = 1/2Q V (1)4energy stored = work done (1)5showing 1/2QV has unit J/joule (1)6 (any 3)[integration answer – max (1)(1)][answer in words – max (1)(1)] 3

(iii) derive or recall E = 1/2 C V2 OR use correct Q value from graph (1)OR line across graph at 4 Vcorrect answer (1)

E = 1/2 Q V= 1/2 × 3.3 (3.3–3.35) C × 4 V (1)= 6.6 (6.6–6.7) J (1)

OR E = 1/2 C V2 (1)

= 1/2 × 0.83 F × (4 V)2 = 6.6 J (1) 2

(b) (i) Q decreases V decreases OR I decreases (1)mention of P = VI (1) 2

(ii) 125–145 s (1) 1[10]

51. B in accelerators:changes direction of motion of charged particles ORforce/B perpendicular to motion of charged particlesOR ref to LHR (1)1(moving) charged particles stored in circles/circularpath/spirals (1)2

Bqv = mv2/r (1)3cyclotron: T = 2 π m/Bq (1)4fixed frequency voltage for acceleration (1)5diag/construction detail [probably on diag] (1)6synchrotron: r fixed, B adjusted as needed (1)7 (up to 4)

B in detectors:charged particles (detectable) curved paths (1)8find sign of charge from sense of curvature (1)9find momentum/speed/energy/mass from r (= p/Bq) (1)10

[5]

52. (i) two correct arrows [ignore labelling] (1) 1

(ii) Some use of mv2/r with v correctly subbed OR

mrω2 with v correctly used (1)[subbing may happen later in answer]

T cos θ = mg mg

OR T sin θ = mv2/r [either gains (1)] (1)

tan θ = v2/rg (1)

r = v2/g tan θ

= 30 × 30/9.81 × tan 20 m

= 252 m (1) 4[5]

53. (a) Expression for gravitational force:

F = GMm/r2 (1) 1

(b) Expression for gravitational field strength:

g = force on 1 kg, so g = GM/r2, or g = F/m so g = GM/r2 1

(c) Radius of geostationary orbit:Idea that a = g, and suitable expression for a quoted [can bein terms of forces] (1)substitution for velocity in terms of T (1)algebra to obtain required result (1) 3

Example of derivation:

g = v2/r or g = ω2rand v = 2πr/T or ω = 2π/T

so (2πr/T)2/r = GM/r2 or (2π/T)2r = GM/r2, leading to expression given

(d) Calculation of radius:Substitution into expression given (1)

Correct answer [4.2 × 107 m] (1) 2

Example of calculation:

r3 = 6.67 × 10–11 N m2 kg–2 × 6.0 × 1024 kg × (24 × 60 × 60 s)2 /4π2

= 7.6 × 1022 m3

So r = 4.2 × 107 m

(e) (i) Satellite with greater mass:Yes – because, in geostationary orbit, r constant soacceleration remains the same, regardless of mass (1)

(ii) Satellite with greater speed:No + suitable argument (1) 2[e.g. for geostationary orbit, T and r are fixed, so v cannotincrease (v = 2πr/T)]

(f) Why satellite must be over equator:Idea that centre of satellite’s orbit must be the centre of theEarth (can be shown on diagram) (1)there must be a common axis of rotation for the satellite andthe Earth / the satellite’s orbit must be at right angles to thespin axis of the Earth (1) 2

[11]

54. (a) Momentum at impactp = mv seen or used (1)

Answer [11 kg m s–1] (1) 2

eg momentum = 0.42 kg × 27 m s–1

= 11.34 kg m s–1

(b) Momentum at releaseMinus (1)

8.4 kg m s–1 (1) 2

(c) (i) Average force(ecf momenta values)

Use of F = t

p

ie for using a momentum value divided by (1)

0.22Adding momentum values (1)

Answer [88.0 N – 89.8 N] (1)

F = s 22.0

s m kg 3.11s m kg 4.8 11

F = (–) 89.5 N

Or

Use of F = ma (1)Adding velocities to calculate acceleration (1)Answer [88.0 N – 89.8 N] (1)

Eg acceleration = s 22.0

s m 27s m 20 11 (= –213.6 m s–2)

Force = 0.42 kg × –213.6 m s–2 = (–)89.7(2) N 3

(ii) Direction of force on diagramRight to left (1)[Accept arrow drawn anywhere on the diagram. Label not required] 1

(d) Difference and similarityDifference: opposite direction / acts on different object (1)

Similarity: same type of force / same size / acts along same line / (1)act for same time / same size impulse[‘Magnitude’ and ‘size’ on their own is sufficient. ‘They are equal’is OK. Accept; they are both contact forces; they are bothelectrostatic forces] 2

[10]

55. (a) EK of helium nucleus

Use of EK = 2

1mv2 (1)

Answer [ 3.1 × 10–15 J. No ue. Min 2 sig fig required] (1) 2

eg EK = 2

1 × 6.65 × 10–27 kg × (9.65 × 105 m s–1)2

= 3.096 × 10–15 J

(b) (i) Loss of EK of proton [ecf their value for EK of helium nucleus]

3 × 10–15 J or 3.1 × 10–15 J (1) 1

(ii) Speed of proton after collision[ecf their value for loss of EK of proton, but not if they havegiven it as zero]Calculation of initial EK of proton (1)

Subtraction of 3.1 × 10–15 J [= 1.7 × 10–15 J] (1)

Answer [(1.40 – 1.50) × 106 m s–1] (1)

eg EK = 2

11.67 × 10–27 kg × (2.4 × 106 m s–1)2 (= 4.8 × 10–15 J)

EK after collision = 4.8 × 10–15 J – 3.1 × 10–15 J (= 1.7 × 10–15 J)

v =

kg 1067.15.0

J 107.127

15

= 1.43 × 106 m s–1

OrUse of the principle of conservation of momentum. (1)Correct expression for the total momentum after the collision (1)

Answer [ (1.40 – 1.50) × 106 m s–1] (1)

Eg 1.67 × 10–27 kg × 2.4 × 106 m s–1

= 6.65 × 10–27 kg × (9.65 × 105 m s–1) + 1.67 × 10–27 kg × V

V = –1.44 × 106 m s–1

[For both these solutions allow the second marking point to 3candidates who incorrectly write:

the mass of the proton as 1.6 × 10–27 kg or 1.7 × 10–27 kg,

or the mass of the helium as 6.6 × 10–27 kg or 6.7 × 10–27 kg

or the velocity as 9.6 × 105 m s–1 or 9.7 × 105 m s–1]

(c) Other factor conservedMomentum / mass / charge / total energy (1) 1

[7]

56. Deductions

(a) (i) The atom is mainly empty space (1)[The atom must be referred to. The words ‘empty’ and ‘space’ mustbe qualified eg ‘there is a large amount of space in the atom’ is notsufficient]

(ii) Within the atom there is an area / the nucleus which is positive/ charged or more massive than the alpha particle[If they choose to describe only the mass it must be a comparison ie‘the nucleus is (much) more massive than the alpha’. ‘The atom hasa dense centre,’ ‘the nucleus has a large mass’ are both insufficient.] (1) 2

(b) Explain(Deflection could have been) repulsion from positive nucleus (1)(Deflection could have been) attraction towards negative nucleus (1)

[The words repulsion and attraction can be described eg ‘α deflectedaway from positive nucleus’, ‘α is deflected towards a negativenucleus’] [Diagrams showing the path of an alpha deflected by botha negatively and a positively charged nucleus would get both marks] 2

(c) Value of n (4 – 6) (1)[Allow minus values] 1

[5]

57. (a) (i) Recall of Q = CV or W = 1/2 CV2 (1)Correct calculation of W or V or C (1) Conclusion [must be consistent] (1)

eg W = 1/2 CV2

C = 2 W/V2 = 2 × 0.045 / (30,000)2 (F)

= 1 × 10–10 (F) = 100 (pF) ( NOT COMPATIBLE)

or W = 1/2 × 10 × 10–12 × (30,000)2 (J) = 0.0045 (J) ( NOT COMPATIBLE) 3

[no mark for conclusion; but ue for saying 100pF ~ 10 pF]

(ii) Sub of one appropriate value into Q = CV or W = 1/2QV (1)Correct value (1)

eg Charge = 1 × 10–10 × 30,000 (C)

= 3 × 10–6 C 2

(b) (i) Use of E = V/d [Rearranged or subbed into] (1)Correct value (1)

eg d = V/E = 30,000/3 × 106 (m)= 0.01 m 2

(ii) Use of E = F/q [Rearranged or subbed into – any charge value] (1)Correct value (1)

eg F = Eq = 3 × 106 × 1.6 × 10–19 (N)

= 4.8 × 10–13 N 2

(e) Correct use of 1 mm in W = Fd or V = Ed [ecf from(b)(ii)] (1) 3000 (V or eV)) (1) correct value (1)

eg W = Fd = 4.8 × 10–13 N × 0.001 m (= 4.8 × 10–16 J) 3000 (eV)3000/35 = 85/86/85.7

or V = Ed = (3 × 106 V/m) × 0.001 (m) 3000 (V)3000/35 = 85/86/85.7 3

[12]

58. (a) Any 2 from:momentum conserved (1)initial momentum zero (1) (Any 2)

final momentum zero (1)[opposite charges repel xx] 2

(b) 0.140 GeV/c2 (1)

– 1.6 × 10–19 C (1)anti-u, d (1) 3

(c) Meson (1) 1

(d) [(1) for 0.14 (alone) or correct use of 109] (1)

Minimum energy = 1.4 × 108 (eV) or 0.14 × 109 (eV) (1) 2[0.14 G is (1)x]

(e) Particles have K.E. (as well as mass) (1) 1

(f) Use of ∆E = c2∆m [rearrangement OR one correct line subbed) (1)correct value (1)

eg ∆m = ∆E / c2 = 0.14 × 109 × 1.6 × 10–19 J / (3 × 108 m s–1)2

Mass loss = 2.5 × 10–28 kg 2[ecf from (d)]

[11]

59. (a) R drawn [10° to vertical] (1)D drawn [10° to horizontal] (1)drag force D = 140 – 155 N [147.6 N by calc is OK] (1) 3

(b) Resolve vertically (1)correct value (1)

eg P cos 40° = 850 N P = 1100 N 2

(c) (i) velocity not constant / direction changing (1)[NOT “if no force, goes straight”]acceleration (towards centre of circle) (1) (Any 2)F = ma (1) 2

(ii) P/push of ice (on sled) (1)horizontal component (1) 2[“additional centripetal force” = 0]

(d) Recall circular motion formula (1)resolve horizontally (1)correct value (1)[incorrect force is eop]

[also possible: W.tan40 = mv2/r (1)(1)]

eg F = P sin 40 = 713 (643) (N) [formula or value]

R = mv2/F

= 87 × 352 / 713 (643) (m)radius = 149 (166) m 3

[12]

60. (a) emf/voltage (1)induced / created / caused by flux change (1) 2

(b) Lenz (1)effect opposes change producing it (1) 2

(c) dynamo generates emf (1)lights off no current (1)lights on current flowing (1) Any 4If current, then force on dynamo rotor/F = BIl (1)[or field acting against field in dynamo]This force opposes rotation (1) 4

[8]

61. (a) Shape [lines not crossing] (1)arrow(s) (1) 2

(b) [reference to] changing B field/ flux cuts coil / changing flux

(linkage) (1)1induces emf or current [NOT “output”] /EM induction (1)2emf α rate of change / Faraday’s law stated (1)3output is gradient of flux graph (1)4signal +ve while increases / –ve while decreases (1)5max emf for max d / dt / steepest gradient (1)6 (Any 5)Emf 0 when gradient = 0 (1)7[(1)5 (1)6 (1)7 can be gained by annotations on graph] 5

(c) (Binary / 1 or 0 / 210

maximum number =) 1024 (1) 1

(d) 10010 (1) 1

(e) Attempt to calculate circumference (formula or numbers) (1)

dividing by 0.83 × 10–6 (m) (1)correct value (1)

eg C = πd = π × 0.089 m (= 0.2796 m)

No of bits along circumference = C ÷ (0.83 × 10–6 m) 3

(= 3.37 × 105)

Rate = 3.37 × 105 × 120 (7200 revs/min = 120 Hz)

= 4.04 × 107 s–1

[2.43 × 109 min–1 is OK][12]

62. (a) Newton’s Second Law of Motion

(The) force (acting on a body) is proportional/equal to the rate of (1)change of momentum (1)and acts in the direction of the momentum change[accept symbols if all correctly defined for the first of these marks][ignore any information that is given that is not contradictory] 2

(i) Calculate the mass

Correct calculation for volume of air reaching tree per second[Do not penalise unit error or omission of unit] (1)Correct value for mass of air to at least 3 sig fig [246 kg. No ue.] (1)[If 1.23 × 10 × 20 = 246 kg is seen give both marks.Any order for the numbers]

Example

20 ms–1 × 10m2 = 200 m3

1.23 kg m–3 × 200 m3 = 246 kg

(ii) Calculate the momentum

Answer: [ (246 kg × 20 m s–1 =) 4920 kg m s–1]

[Accept (250 kg × 20 m s–1 =) 5000 kg m s–1. Accept 4900 kg m s–1. (1) 2Ecf value for mass. Ignore signs in front of values.]

(iii) Magnitude of the forceAnswer: [F = 4920 N or 5000 N or 4900 N.][Ecf value from b(ii). Ignore signs in front of values] (1) 2

[6]

63. β – decay equations

(i) n = udd and p = uud (1)

– and v have no quarks / are leptons / are fundamental (1) 2

(ii) p → n (1)

β+ and ν [on RHS, allow e+] (1) 2[4]

64. Antihydrogen

(i) Antiproton [or anti–up quark, anti–down quark] and positron (1) 1

(ii) p = –1 and e+ = +1 [accept correct u , d charges for p ] (1)

u u d (e+ fundamental / no quarks) [ecf from (b), credit if in (i)] (1) 2

(iii) zero / neutral (1) 1

Antimatter storage

(iv) Annihilates (1)(On contact) with matter / container / protons / HOR Not charged: not affected by magnetic fields (1) 2

[6]

65. Explanation

There is a resultant (or net or unbalanced) force (1)

Plus any 3 of following:–

Direction of motion is changing (1)Velocity is changing (1)Velocity change implies acceleration (1)Force produces acceleration by F = ma (or N2) (1)Force (or acceleration) is towards centre / there is a centripetal (1)force (or acceleration) / no force (or acceleration) parallel to motionNo work done, so speed is constant (1) Max 3

[4]

66. (a) Formula in words

(The force between two charged particles is directly) proportional tothe product of their charges [plural] and (1)

inversely proportional to the square of their separation [not just ‘radius’]. (1)

OR Either equation for F*, with valid word replacements for Q1, Q2 (1)

and r or r2 symbols. One mark for numerator, one for denominator. (1)

4in or Fin wordsi.e.*

20

212

21

r

QQ

r

QkQ

[If equation given in symbol form, followed by a key to the symbol 2meanings, then 1/2.]

(b) Base units of constant

[Either k or (4)0, be sure which][ecf from part a if power of Q or r wrong]

F = 221

r

QkQ or F = 2

0

21

4 r

QQ

[OR using k units N m2 C–2]

Q1Q2 (or C2) → A2s2 (1)

F (or N) → kg m s–2 (1)

→ (units of) k = kg m3 A–2 s–4 OR (units of) ε0 = kg–1 m–3 A2 s4 (1)

OR using ε0 units F m–1:

C = As and either F = CV–1 or V = JC–1 (1)

J = kg m2 s–2 or N = kg m s–2 (1)

→ (units of) ε0 = kg–1 m–3 A2 s4 (1) 3[5]

67. (a) Electron speed

Substitution of electronic charge and 5000V in eV (1)

Substitution of electron mass in ½ mv2 (1)

Correct answer [4.2 (4.19) × 107 (m s–1), no ue] to at least 2 sf (1) 3[Bald answer scores zero, reverse working can score 2/3 only]

Example of answer:

v2 = (2 × 1.6 × 10–19C × 5000 V)/(9.11 × 10–31 kg) = 1.76 × 1015

v = 4.19 × 107 m s–1

(b) (i) Value of E

Correct answer [2.80 × 104 V m–1/N C–1 or 2.80 × 102 V cm–1] (1) 1

Example of answer:

E = V/d = 1400 V / 5.0 × 10–2

= 28 000 V m–1

(ii) Value of force F

Correct answer [4.5 × 10–15 N, ecf for their E] (1) 1

Example of answer:

F = Ee = 2.80 × 104 V m–1 × 1.6 × 10–19 C

= 4.48 × 10–15N

(c) Calculation of h 4

See a = their F / 9.11 × 10–31 kg (1)

[→ a = 4.9 × 1015 m s–2]

See t = 12 (× 10–2) m / 4 × 107 m s–1 (or use 4.2 × 107 m s–1) (1)[t = d/v, with d = plate length; 12 cm]

[→ t = 3.0 × 10–9 s, or 2.86 × 10–9 s]

See substitution of a and t values [arrived at by above

methods] into ½ at2 (1)

Correct answer [h = 0.020 m – 0.022 m] (1)

[Full ecf for their value of F if methods for a and t correct andtheir h 5.0cm]

Example of answer:

h = ½ a t2

= ½ × 4.9 × 1015 m s–2 × (2.86 × 10–9 s)2

= 2.0 × 10–2 m

(d) (i) Path A of electron beamLess curved than original (1) 1

(ii) Path B of electron beamMore curved than original, curve starting as beam enters field[started by H of the Horizontal plate label] (1) 1

[For both curves: ignore any curvature beyond plates after exit new path must be same as original up to plates]

[No marks if lines not identified, OK if either one is labelled][11]

68. (a) (i) Additional force

Correct answer [3.9 × 10–3 N ] (1) 1Example of answer:

0.4 × 10–3 kg × 9.81 N kg–1 = 4 × 10–3 N

(ii) ExplanationQuality of written communication (1)(Current produces) a magnetic field around the rod (1)[Do not accept in the rod]

There is an interaction between the two magnetic fields / fieldscombine to give catapult field (1)

Fleming’s Left Hand Rule/ Fleming’s Motor Rule (1)The rod experiences an upward force (1)

Using Newton 3 → downward force on magnet Max 4

(b) (i) DiagramLower pole labelled North/N and upper pole labelled South/S (1) 1

(ii) Calculation of current in rod

Use of F = BIl. (Ignore 10x. F is their force and l is 5cm) (1)See conversions; mT to T and cm to m (1)Correct answer [2.6/2.7 A ] (1) 3

Example of answer:

I = 3.9 × 10–3 N / (30 × 10–3 T × 5 × 10–2 m) = 2.6 A

(iii) New reading on the balanceValue < 85g [not a negative value] (1)84.6 g (1) 2

[11]

69. (a) Quality of written communication (1)Protons drift/move uniformly inside tubes (1)Accelerate between the tubes/in the gaps (1)Alternating p.d. reverses while p is in tube (1)The tubes must get longer as p speeds up (1)For time inside tube to be constant or to synchronisemovement with the pd (1) Max 5

(b) (i) Multiply by 419 or 420 (1)

Multiply by 1.6 × 10–19 (1)

Correct answer to at least 2 sf (1)

[5.36/5.38/5.4 × 10–11 (J)] [no ue]

m = energy (9.0 × 1016 m2 s–2) (1)

[ecf their energy or 5 × 10–11] (1)

m 1.01 × 1.66 × 10–27kg [ecf their m] (1)

Correct answer (1)

[0.36 or 36%] [Use of 5 × 10–11 gives 33%] (1) 6

[Accept routes via m in u and mp in J]

(ii) Use of 1/f (1)

time down linac = 420 3.9 × 108 s–1

or 210 ÷ 3.9 × 108 s–1 (1)

[t = 1.07/1.08/1.1 × 10–6(s) or 0.54 × 10–6 (s)] 2

(c) (i) Fixed target:Large(r) number of /more collisions or more likely to get collisions[not easier to get collisions] (1)

Other particle beams produced (1)

(ii) Colliding beams:More energy available for new particles (1)p = 0 so all energy available (1) Max 2

[15]

70. Calculation of voltage

Use of E=V/d [could be rearrangement] (1)

Correct answer [1.5 × 109 V] (1) 2

Example of calculation:

V = Ed = 3 × 105 × 5000 V

V = 1.5 × 109 V

Calculation of capacitance

Recall Q = CV(1)

Correct answer [2.7 × 10–8 F](1) 2

Example of calculation:

C = Q / V

= 40 C / 1.5 × 109 V

= 2.7 × 10–8 F

Resistance

Use of RC = 20 ms, or an appropriate time (eg, 20 ms ÷ 5 = 4 ms)(1)OR attempt to find current from I = Q/t

Correct answer [7.5 × 105 Ω (1.5 × 105 Ω)](1)

Example 1:

⇒ R = 20 ms ÷ 2.7 × 10–8 F (4ms ÷ 2.7 × 10–8 F)

= 7.5 × 105 Ω (1.5 × 105 Ω)

Example 2:

I = 40 C / 20 ms = 2000 A(1)

⇒ R = A2000

V105.1 9

I

V

⇒ 7.5 × 10 5 Ω (1) 2

[Also give credit for using “average” pd which is less than 1.5 GV

say V/2 3.75 × 105 Ω]

Drawing of electric field region

–

G ro u n d

5 k m

≥ 2 radial lines (1)

Arrow(s) (1) 2

Electric field

Recall E = 204 r

q

OR k = 04

1

= 9 × 109 m F–1) (1)

Correct answer [1.44 × 104 V m–1] (1) 2

[Use of d

V scores 0]

Example of calculation:

E = 204 r

q

212– 50001085.84(

40

V m–1

= 1.44 × 104 V m–1

Lightening strike

Field stronger near cloud OR Greater force/acceleration on chargesOR Mention of force on charges OR Mention of ionising atoms by collision (1) 1

[11]

71. Speed of sphere

Momentum conserved [stated or implied] (1)

Correct subs L.H.S or R.H.S of conservation of momentum equation (1) 3

Correct answer [ν = 1.43(m s–1)] (1)

Example of calculation:

54 × 2.57 (+ 0) = 54 × ν + 29 × 2.12 (g m s–1)⇒ 138.78 = 54 × ν + 61.48

⇒ ν = 1.43(m s–1)

Elastic or inelastic collision

Recall K.E: 1/2mν2(1)

Correct values for both KEs [178(mJ), 120(mJ), no ue](1)

Conclusion consistent with their results for KE(1) 3

[max 1 if only words used and inelastic ≡ energy lost implied]

Example of calculation:

= ½ × 54 × 2.572 = 178 mJ

Final total K.E: ½ × 29 × 2.122 + ½ × 54 × 1.432 mJ= 65 mJ + 55 mJ= 120 mJ⇒Inelastic

Average speed of the spheres

Recall ν = 2πr / t (1)

Correct answer [2.9 m s–1] (1) 2

Example of calculation:

ν = 2πr / t = π × 0.17 × 2 m / 0.37 s

= 2.9 m s–1 or 290 cm/s

Calculation of centripetal force

Recall F = mν2 / r OR mrω2 OR m ν ω (1)Correct answer [1.43 N, ecf for their ν ] (1) 2

Example of calculation:

F = mν2 / r

= 0.029 × 2.92 / 0.17 N [watch out for 29 twice]= 1.43 N [ecf]

Tension

Weight of sphere (= mg = 0.029 × 9.81 N = )0.28 N (1)

T = F – W OR F = T + W [using their values for F and T] (1) 2

Example of calculation:

= 1.43 – 0.28 (N) T = 1.15 N

[12]

72. “The standard model”

Everyday matter/atoms: p,n, e [maybe in two places] (1)

Protons / neutrons are made from quarks (1)

p: uud and n:udd (1)

show charge of either [p: u(+2/3) u(+2/3) d(–1/3) ⇒ + 1 OR n: u(+2/3) d(–1/3) d(–1/3) ⇒ 0] (1)

All baryons have three quarks (1)

Hadrons contain quarks (1)

Electron is fundamental/leptons are fundamental (1)

Electron-neutrino created during β-decay (1) Max 6[6]

73. Calculation of voltage

Use of ∆E = c2∆m (1)

Use of eV (1)

Correct answer [4.1 × 109 (V)] [no ue] (1)

Example of calculation:

∆E = c2∆m = eV

⇒ V = c2∆m / e = 9 × 1016 × 8000 × 9.1 × 10–31 / 1.6 ×10–19 V

= 4.1 × 109 V 3

Role of magnets

Field deflects/bends/curves the path (1)

Field is at curved parts / field at AD and BC / no field on straight parts (1)

Field perpendicular to page / velocity (1)

Force perpendicular to velocity or field (1)

Force is centripetal / towards centre (1) Max 4

Calculation of field strength:

r = p / Bq rearranged to B = p/rq (1)

correct substitution of either p OR of r and q (1)

Correct answer [0.124(T), no ue] (1)

Example of calculation:

r = P / Bq ⇒ B = P / qr

= 8000 × 9.1 × 10–31 × 3.0 × 108 / 110 × 1.6 × 10–19 T= 0.124 T 7

[10]

74. Smoke detectorsRecognition that mass alpha = 4 (1)Idea of - 4 to find resulting nucleus mass [237] (1)

Mudaughter product = mvalpha or momentum equations in context (1)

valpha = 59(.25) udaughter product [allow ecf incorrect masses eg 4 and 241 : 60.25] (1)

Use of ½ mv2 To give ratio = 59(.25) [allow ecf as long as rounds (1)

to 60 ; must have speeds sub ; valid use of p2/2m]

Energy:

E = mc2 / energy must have come from mass (1)

Total mass after is a (little) less than before/mass loss/mass defect/binding energy (1) max 6

[6]

75. (a) From what height?

Use of mgh and ½ mv2 (1)[ignore power of 10 errors]

mg h = ½ mv2 (1)

[shown as formulae without substitution, or as numbers substitutedinto formulae]

Answer [0.8(2) m] (1)

[It is possible to get 0.8 m by a wrong method:

If v2 = u2 + 2as is used, award 0 marks

If you see v2/a then apply bod and up to 2/3 marks – the 2nd and 3rd 3

marks. Note that v2/g is correct and gains the first 2 marks, with the

3rd mark if 0.8 m is calculated]

80 × (10–3) kg × 9.81 N kg–1 × h = ½ × 80 (× 10–3) kg × (4 m s–1)2

h = 13

213

kg N 18.9kg1080

)ms4(kg10805.0

= 0.8(2) m

(b) (i) Law of conservation of linear momentum

Provided no external [other/resultant/outside] force acts (1)The total momentum (of a system) does not change / total momentum (1)before(collision) = total momentum after (collision) 2[Total seen at least once] [Ignore all references to elastic and inelastic][Do not credit simple statement that momentum is conserved]

(ii) Speed of trucks after collision

Any correct calculation of momentum (1)

Use of conservation of momentum leading to the answer 1.3(3) m s–1 (1)

80 × (10–3) kg × 4 m s–1 = 240 × (10–3) kg × u, giving u = 1.3(3) m s–1 2

(c) Time for trucks to stop

[Do not penalise candidates for using a total frictional force of 0.36 N.3/3 possible]

Either

Correct use of power = f × v and ½ mv2 (1)[Do not penalise power of 10 errors or not dividing by 2 in f × V equation]

Use of energy divided by power (1)

Answer in range 2.6 s to 2.7 s (1)[ecf their value for u]

P = 0.12 N × 233.1 m s–1 = 0.08W

KE = ½ × (3) × 80 × (10–3) kg × (1.33 m s–1)2 = 0.21 J

W0.08J 21.0

powerEnergy

t = 2.6(5) s

[accept 2.6 or 2.7 as rounding] 3

OR

Use of F = ma (1)

Use of either v = u + at i.e or a = tv (1)

Answer in range 2.6 s to 2.7 s (1)

(–)0.12 N = (3) × 80 × (10–3) kg × a (a = (–)0.5 m s–2)

0 = 1.33 m s–1 – 0.5 m s–2 × t or (–)0.5 m s–2 = t

–1s m )1.33(

t = 2.6(6) s

OR

Select Ft = p (1)

Substitution (–)0.12t = (–3) × 80 × (10–3) kg × 1.33 m s–1 (1)Allow omission of any bracketed value]Answer in range 2.6 s to 2.7 s (1)

[10]

76. Conservation laws

Baryon (1)

– 1 (1)

Q: (–1) + (+1) = (0) + (+1) + (X) (1)

B: (0) + (+1) = (0) + (0) + (X) (1) 4

Quark content

uud (1)

su (1) 2[6]

77. (a) Radius of circular path

Correct use of v = T

r2 (allow substitution of their T) (1)

Radius = 70 – 80 m (74.48 m) (1) 2

(b) Resultant force

rmvF

2

[seen or used] (1)

Force = 0.08 N (0.077 N) [Allow ecf of their radius.] (1)

Towards the centre of the circular path / towards hub. (1) 3

(c) Forces on the man

(i) Force P : Normal contact/reaction force / EM force / push of (1)capsule or floor on man

Force Q : Pull of Earth on man / weight / gravitational pull (1) 2

(ii) Resultant force (to centre) (1)(at A provided by) friction (1) 2

(iii) at B resultant provided (by force Q being greater than P) (1) 1[10]

78. Conserved quantities

Momentum, charge, (mass-)energy, lepton number (1) (1) 2

[2 right gets 1 mark; all 3 right get 2 marks][Do not credit kinetic energy]

Charge of the pentaquark

31

312

322

= (+) 1(e) (1) 1

Charge on X

Positive since pentaquark was positive, neutron neutral [ecf] (1) 1

[Reasoning needed]

Possible quark composition for X with explanation

u s (1)

Left behind (after removing neutron/udd) (1) 2

Mass of pentaquark

Conversion from GeV to J or substitution of c2 (1)

answer [no ue] (1)

1.54 × 109 × 1.6 × 10–19/(3 × 108)2 2

= 2.7 × 10–27kg[8]

79. How electron gun creates beam of electrons

Any four from:

1. hot filament (1)

2. thermionic emission / electrons have enough energy to leave (1)

3. anode and cathode / ± electrodes [identified] (1)

4. E–field OR force direction OR cause of acceleration (1)

5. collimation [eg gap in anode identified as causing beam] (1)

6. need for vacuum (1) Max 4

Speed of electrons

(eV =) ½ mv2 (1)

Use of eV [ie substituted or rearranged] (1)

Answer [1.09 × 107 m s–1] (1)

1.6 × 10–19 × 340 (J) = ½ × 9.11 × 10–31 (kg) × v2

v =1.09 × 107 m s–1 3

Definition of term electric field

Region/area/space in which charge experiences force (1) 1ertical acceleration of electrons due to field[Bald answer =0]

Use of equation E = V/d (1)

E = V/d = 2500 V 0.09 m = 28 (kV m–1)

Rearranged equation E = F/q or substitution into it (1)

F = Eq = 28 000 × 1.6 × 10–19 (N) 4.4 × 10–15 (N)

Equation F = ma seen or substitution into it (1)

A = F/m = )(1011.9

)(104.431

15

hg

N

= 4.9 × 1015 (m s–2) (1) 4

[at least 2 sig fig needed] [No u.e.] [Reverse calculation max 3][12]

80. Direction

Force centre/perpendicular to velocity/motion (1) 1[Accept sideways/inwards]

Why force required

2 of:

Changing direction /charging velocity (1)

Acceleration (1)

Reference to or explanation in terms of NI or NII (1) Max 2

What provides force

Friction between tyres and road surface (1) 1

Maximum speed

F = mv2/r [Accept F = mrω2] (1)

v = mFr OR

16014470 (1)

= 6.4 m s–1 (1) 3

Why skid occurs

Smaller r (1)

F(mv2/r) (required) increases / use of F = mv2/r to deduce v decreases (1)

Only 470 N (available) / the force is the same (1) 3

Explanation

(m increased ) F (needed) increases (1)

EITHER only 470 N available or the force is the same NO OR friction increases as mass increases( YES) (1) 2

[12]

81. Approximate energy of alpha particle in MeV

1. r = 0.09 (m) [accept in range 0.07 – 0.12] (1)[must have unit if given in cm]

2. q = 2 × 1.6 × 10–19 (C) (1)

3. m = 4 × 1.7 × 10–27(kg) (1)

4. r = p/Bq p = rBq or v = rBq/m (1) [see equation or substitution]

[p = 0.09 × 3.7 × 3.2 × 10–19 N s]

5. =1.07 × 10–19 (Ns) OR v = 1.6 × 107 (m s–1) (1)

6. E = p2/2m or use of ½ mv2 (1)

[E = (1.07 × 10–19)2 / (2 × 4 × 1.67 × 10–27)J]

7. 8.6 × 10”13J (1)

(5.4 MeV/5.4 × 106 eV) 7[7]

82. (a) Advantage of avoiding metal contacts

Any one from:

makes possible a sealed unit

avoids electrocution

stops corrosion (by water)

water cannot enter/short contacts (1) 1

(b) How arrangement is able to charge the battery

Any six from:

1. current (in X) produces magnetic field

2. field links second coil

3. metal = iron

4. metal core increases field

5. field changes/alternates

6. changing /B or d/dt or Faraday induces/causes V

7. V causes I

8. diode needed (or a.c. so won’t charge)

9. field penetrates plastic

10. like a transformer / X is a primary and Y is a secondary

11. electromagnetic induction Max 6[7]

83. Base units of eV

(i) Reference to joule (1)

Useful energy equation / units shown [e.g. ½mv2, mgh, mc2, Fd, not (1)QV or Pt]

Algebra to J = kg m2 s–2 shown (e.g. kg (m s–1)2 or kg m s–2 m) (1) 3

(ii) Energy released

146 shown or used (1)∆m calculation [1.9415, ecf] (1)

Multiply by 930 [allow E = mc2 with mass in kg] (1)1800 MeV [no ue] (1) 4

[7]

84. (i) Decay numbers

11 p and 1

0 n (1)

01 + and 0

0 (1) 2

(ii) Tick the boxes

Proton: baryon and hadron only (1)

neutron: baryon and hadron only (1)

β+: lepton and antimatter only (1)

ν: lepton only (1) 4

[only penalise once for including meson] [if both baryon correct but no hadrons 1 mark out of 2 and vice versa]

[6]

85. (a) Resultant force required

The direction of speed OR velocity is changing (1)There is an acceleration/rate of change in momentum (1) 2

(b) (i) Angular speed

Use of an angle divided by a time (1)

7.3 × 10–5 rad s–1 OR 0.26 rad h–1 OR 4.2 × 10–3 o s–1 OR 15o h–1 (1) 2

(ii) Resultant force on student

Use of F = mrω2 OR v = rω with F = r

mv 2

(1)

2.0 N (1) 2

(iii) Scale reading

Evidence of contact force = mg − resultant force (1)Weight of girl = 588 (N) OR 589 (N) OR 60 × 9.81 (N) (1)Scale reading = 586 N OR 587 N [ecf their mg − their F] (1) 3

[9]

86. (a) Direction of field lines

Downwards (1) 1

(b) (i) Calculation of force

Use of V/d i.e. 250 V/0.05 m [if 5 used mark still awarded] (1)

Use of d

V e [Mark is for correct use of 1.6 × 10–19 C] (1)

= 8.0 × 10–16 N (1) 3

(ii) Direction and explanation

(Vertically) upwards / towards AB (1)

No (component of ) force in the horizontal direction OR because (1) 2(the force) does no work in the horizontal direction

(c) Calculation of p.d.

Use of ∆EK = ½ mυ2 / ½ 9.11 × 10–31 (kg) × (1.3 × 107)2 (1)

Use of Ve / V × 1.6 × 10–19 (C) (1)

= 480 V (1) 3

(d) Beam of electrons

Diagram showing:

Spreading out from one point (1)fastest electrons labelled (1)

2[11]

87. (a) (i) QOWC (1)Link track to bubbles (1)Which reflects light / are illuminated (1)(produced as) the electron / it ionises liquid / particles / (1)H2 / air 4

(ii) Mention of B–field/F = Bqυ/ F = Beυ/ FLHR (1)B is perpendicular to υ/ direction of motion / in or out (1)of pageElectron loses energy/slows down (1)Colliding with / interacting with / ionising liquid particles / H2 (1) 4

(b) (i) & (ii)

r/m r/mm p/kg m s–1 m/kg

P 62 – 67 × 10–3 62 – 67 1.2 – 1.3 × 10–20 4.0 – 4.3 × 10–29

Q 43 – 48 × 10–3 43 – 48 0.83 – 0.92 × 10–20 2.8 – 3.1 × 10–29

R 28 – 33 ×10–3 28 – 33 0.54 – 0.63 × 10–20 1.8 – 2.1 × 10–29

Values for r in range above [ignore 10n and units] (1)

p = Ber any one correct p [ignore 10n but must have (1)unit] [ecf] (1) 3

All ps correct numerically [no ue] (1)p = mυ m = p/υ (1)

Any one correct m [ignore 10n but must have unit]

EITHER

Comment [e.c.f.]: any reference to 9 × 10–31 kg/rest (1)mass (of electron) / electron massBecause electron is moving close to / at the speed oflightOR(effective) mass (of electrons) is decreasing (1)

reference to E = mc2 / ∆E = c2∆m / mass–energy (1)conservation 4

[15]

88. Speed of car 2

2 − u2 = 2as

(−)u2 = (−) 2 × 3.43 × 23.9 [substitution or rearrange] (m s–1) (1)

u = 12.8 (m s–1) [value] (1) 2

Magnitude of the momentum of car 2

p = m (1)

= 1430 × 12.8 (13) (kg m s–1)

= 18 300 (18 600) N s or kg m s–1 (1) 2

Calculation of easterly component of momentum

Component = momentum × cos θ (1)

Car 1: 23 800 Ns × cos 45

= 16 800 N s (1)

Car 2: 18 300 (18 600) N s × cos 30 = 15 800 (16 100) N s (1) 3

Whether car 1 was speeding before accident

(Sum of two easterly components) ~ 33 000 N s [ecf] (1)

(÷ mass of car 1) ~ 16.8 m s–1 [ecf] (1)

Conclusion related to speed limit (17.8 m s–1) (1) 3

Explanation of how investigator could use conservation law

Any two from:

Momentum conservation After collision there is significant northerly momentum Before collision car 1 had no northerly momentum/only car 2

had northerly momentum (1) (1) 2[12]

89. Explanation

energy gained by electron accelerated through 1 V/W = QV (1)

W = 1.6 × 10−19 C × 1 V = 1.6 × 10−19 J (1) 2

Unit of mass

∆E = c2∆m so ∆m = ∆E/c2 (1)

GeV is energy GeV/c2 is mass (1) 2

Mass of Higgs boson

m = 115 × [109] × 1.6 × 10–19/(3 × 108)2 (1)

= 2.04 × 10–25 kg (1) 2

Antiparticle

Same mass and opposite charge (1)[Accept Particle and its antiparticle annihilate (→ photons)] 1

Explanation of need for a magnetic field and why it can be small

Force deflects particles/force produces circular motion (1)

Force is perpendicular to motion/force provides centripetal force (1)

r is large or curvature is small/gentle (1)

reference to B = p/rQ to show why small B is needed (1) 4[11]

90. Minimum charge on balloon

0 .5 m

1 .8 m

M g

F

T

any 2 forces correct (1)

T 3rd force correct (1) 2

F = kq1q2/r2/F = kq2/r2 (1)

mg = T cos θ/T = 1.8 × 10−2 N (1)

F = T sin θ/F = mg tan θ/F = 4.6 × 10−4 N (1)

r = 0.5 + 2 × 1.8 sin 1.5° (= 0.594 m) (1)

q2 = Fr2/k = Fr2 × 4πε0 (1) Max 3

= 0.0018 × 9.81 × tan 1.5° × 0.5942 × 4π ε0

q = 1.36 × 10–7 C (1)

[NB 2 for diagram, maximum 3 for intermediate parts, and final 1for the answer] 1

[6]

91. Momentum of neutron

Use of p = m (1)

p = 5.03 10–20 N s/kg m s–1 (1) 2

Speed of nucleus

Total mass attempted to be found (1)Conservation of momentum used (1)

= 2.01 106 m s–1 [ecf from p above only] (1) 3

Whether collision was elastic

Use of k.e. = ½ m2 (1)

ke = 7.45 10–13 (J) / 5.06 10–14 (J) (ecf) (1)

A correct comment based on their two values of ke. (1) 3[8]

92. Particle X

Positive (1) 1

Is a baryon (1) 1

Quark compositions

Proton uud; neutron udd BOTH (1) 1

Explanation and deduction of identity of X

Quality of written communication (1)

Strong / not weak interaction (1)

One strange quark on each side / no flavour change (1)

X is a proton (1) 4[7]

93. Resultant force

Direction of travel changing (1)

Velocity changing/accelerating (1)

Force is towards centre of circle (1) 3

Why no sharp bends

Relate sharpness of bend to r (1)

Relate values of , r and F (1) 2

[e.g. if r large, can be large without force being too large/if r small, must be small to prevent force being too large]

Bobsleigh

Ncos = mg (1)

Nsin (1)

= m2/r or ma(1)

Proof successfully completed [consequent on using correct formula] (1) 4

Calculation of angle

77 78° (1) 1[10]

94. Results of experiments and conclusions

Most pass straight through/undeflected (1)

A few deflect/reflect (at large angles) (1)

Small nucleus/mostly empty space (1)

Concentrated mass and/or positive charge (1) 4

How to determine x graphically

Plot log N v. log (sin / 2) [OR ln on both sides] [Any base] (1)

Gradient = x (1) 2

Meaning of numbers in the symbol for the gold nucleus

Bottom number: 79 protons (1)

Top number: 197 ns + ps )

OR )

197 nucleons ) (1)

OR )

197 79 = 118 ns ) 2

Mass of alpha particle

Mass of alpha particle 4 mp

= 4 1.67 10–27 = 6.7 [or 6.68] 10–27 kg (1) 1

Calculation of electric force

F = kq1q2/r2 OR q1q2/4o r2 (1)

q1 = 79 1.6 10–19 C and q2 = 2 1.6 10–19 C (1)

[stated or subbed]

F = 14.56 N (1) 3[12]

95. Situation to which equation refers

F = force on particle (1)

B = (m ag n e tic ) flu x d en sity /fie ld s tren g th= v e lo c ity /sp eed o f p a rtic leq = ch a rg e o f p a rtic le

= an g le b e tw een B an d /m o tio n/cu rren t

(1 )

(1 )

(1 )

F is perpendicular to B and (1)

[Some of these may be shown by diagram] Max 4

Description of situation modelled by equation

Curved/circular motion of particle (1)

p = momentum (1) 2

Why path of a particle is curved

Charged particles (1)

with (component of) motion perpendicular to field (1)

Force perpendicular to motion/ Fleming’s L.H. rule (1) Max 2

Why spiralling path decreases as it nears North Pole

Nearer pole field stronger (1)

Reference to r = p(m)/Bq OR r ∝ 1/B

OR B increasing centripetal/inward F increases

Alternative: ↓ due to resistive force (1)

Reference to r = p(m)/Bq OR r ∝ p/ 2[10]

96. Discussion of type of collision

Inelastic (1)

Energy heat at impact/plastic deformation (1) 2

Momentum vector diagram

Diagram [right–angle triangle with arrows on two perpendicular sides] (1)

Labelling (1) 2

M ass: 5 .8 × 1 0 k g × 3 5 0 0 0 m s (= 2 .0 × 1 0 N s)

C asta lia1 .2 × 1 0 k g × 2 5 0 0 0 m s(= 3 × 1 0 N s)

– 1 2

1 6

– 1

6 – 1

11

Calculation of change in direction

/sin /tan = 00025102.1

00035108.512

6

= 6.77 10–6 (1)

= 6.8 10–6 rad OR 3.9 10–4 degrees OR 1.4 seconds (1) 2

Formula for net force in terms of momentum

F = d / d t (m) OR words (1) 1

Calculation of number of rockets required

N F t = 2 1011 OR N = 2 1011 / (7 106 130) (1)

N = 220 [must be a whole number] (1) 2[9]

97. How properties of particles and antiparticles compare

Same mass/properties, opposite charge (1) 1

Energy

E = mc2 = 1.67 10–27 (3 108)2 J [m or c subbed correctly] (1)

= 1.503 10–10 J [u.e. if comparison made here]

= 1.503 10–10/109 1.6 10–19 GeV (1)

= 0.94 GeV (1) 3

[jump to “≈ 1 GeV” omitting last line scores (1)(1)]

Survival of anti-atom

Anti–proton meets proton OR positron meets electron OR (anti-atom)meets atom (1)

(leads to) annihilation (1) 2

Table 2

Meson Baryon Lepton

proton (1)

antiproton

electron (1)

positron

Quark structure

Antiproton: 2 2/3 (anti u) + 1 + 1/3 (anti d) (1)

= 1 (e not needed) (1)

[3 d –1 scores ] 2[10]

98. Explanation of why resultant flux in iron core is zero

Same current (in both coils) OR same turns (1)

Wound opposite ways (1) 2

OR leading to cancelling of magnetic effects

Explanation of how RCCB breaks circuit

Any five from:

Different currents give different (noncancelling) effects (1)

net B OR /B ≠ 0 (1)

Faraday/changing /B (1)

V induced in third coil [“I induced” is 4th (1) only] (1)

I in third coil/relay coil (1)

relay coil magnetized (1)

relay contact opens (1) Max 5[7]

99. What happens in circuit after switch closed then opened again

Any seven from:

S closed C charges (1)

up to VS (1)

Instantly/very quickly (1)

S open: discharge starts (1)

Exponential discharge (1)

(Vc = Vs et/RC )

¾ Vs = Vs et/RC (1)

ln ¾ = t/RC (1)

t = 29.7 s OR RC = 103 s [if no other calculation] (1)

Buzzer sounds for 29.7 s [ecf] (1) Max 7

[Marks 1-5 and mark 9 are available via appropriate graph. For mark 5graph must have axes labelled with a V/Q/I and same t, and a recognisableexponential curve.]

100. Rutherford scattering experiment

Most went (nearly) straight through (1)

A small proportion deflected through large angles (1) 2

Arrows to diagram

Two arrows directed away from N (1) 1

Sketch graph

Speeds equal at A and B (1)

A non-zero minimum at P (1) 2

Shape of graph

A to P: Force (component) against velocity so decelerates (1)

P to B: Force (component) in direction of velocity so accelerates (1) 2

Add to diagram

Same initial path but deflected through larger angle (1) 1

Observations

More alpha particles deflected/ alphas deflected through largerangles/fewer pass straight through (1) 1

[9]

101. Comparison between antiparticle and its particle pair

Similarity: same mass as its particle pair (magnitude of charge) (1)

Difference: opposite charge/baryon number/(Iepton number / spin) (1) 2

Quark composition

ū ū –

d [OR anti-down etc] (1) 1

Baryon number

–1 (1) 1

Why difficult to store antiprotons

As soon as they contact protons/matter (1)

they annihilate (1) 2

Maximum possible mass

×2 (1)

÷ 0.93 or equivalent [OR by using E = mc2 to 1.6 × 10–25 kg] (1)96 (u) OR 97 (u) (1) 3[48u x (1) (1)]

Two reasons why interaction cannot take place

Q/charge not conserved (1)B/baryon number not conserved (1) 2

[11]

102. Explanation

Diffraction (1)Molecular/atomic separation 1nm/de Broglie wavelength (1) 2

Kinetic energy

Use of = h/mv (1)

Use of k.e. = 1/2mv2 OR p2/2m (1)

k.e. = 9.1-9.2 × 10–23 J [no ecf] (1) 3[5]

103. Speed of rim of drum

v = r or v = 2r/T [either used] (1)

= s60

min rev 8002 –1 OR T = 1–min rev 800

s60 (1)

= 18.4 m s–1 [3 sf min.] (no ue) (1) 3

Acceleration

Use of a = r 2 OR a = v2 /r (1)

1.5 × 103 ms–2 (1) 2

Addition of arrow and explanation

Arrow labelled A towards centre of drum (1)Push of drum on clothing/normal contact exerted by drum on clothing (1) 2[Normal reaction accepted]

Arrow of path

Arrow labelled B tangential to drum, from P, in anticlockwise direction (1) 1[8]

104. Meaning of uniform magnetic field

Magnetic flux density constant / magnetic field lines parallel / (1)magnetic field strength is constant/ does not vary 1

Sizes and directions of forces on LM and NO

Force on LM: 2.4 × 10–4 N/ 0.24 mN (1)Direction:Downwards/into (paper) (1)

Force on NO: )

2.4 × 10–4 N / 0.24 mN [No unit penalty] ) Must have both (1) 3

Direction: )Upwards / out of (paper) )

Why no forces on MN and OL

Wires/current and B field directions are parallel [allow ‘samedirection’] / field due to current and B field of magnet areperpendicular to each other (1) 1

The effect on the square

A (turning) moment will be applied / it will (begin) to turn / spin / rotate (1) 1

Moving the pole pieces further apart

Reduces the size of the forces, (1)Because the flux density is reduced/ magnetic field (strength)reduced / B (field) reduced (1) 2

[8]

105. Charge on capacitor

220 F × 5 V [use of CV ignore powers of 10] (1)= 1100 C (1) 2

Energy on capacitor

2

220F × (5 V)2 /

2

1100C ×5 V / F

2202

C 1100 22

[ignore powers of 10] (1)

= 2750 J (2.8 × 10–3 J) (1) 2

Experiment

Method 1 (constant current method):

Circuit (1)

For a given V record time to charge capacitor at a constant rate (1)

for a range of values of V (1)

Use Q = It to calculate Q (1)

Plot Q V– straight line graph through origin / sketch graph /dive Q/V and obtain constant value (1)

Method 2:

Circuit (1)

For a given value V measure I and t (1)

Plot I t find area under graph Q (1)

Repeat for a range of values of V (1)

Plot Q V for straight line graph through origin/ sketch graph /dive Q/V and obtain constant value (1)

Method 3 (joulemeter method):

Circuit (1)

Record V and energy stored (1)

For range of V (1)

Determine Q from ½ QV or C

Q

2

2

(1)

Plot Q V– straight line graph through origin / sketch graph /divide Q/V and obtain constant value (1) 5

[Coulombmeter (will not work with this value of capacitor)circuit (1) ; record charge Q on colombmeter (1); for a range of values of V (1); PlotQ V for straight line through origin (1) – Max 3]

[9]

106. Lenz’s law

The direction of an induced current/emf/voltage is such as (1)to oppose the change (in flux) that produces it (1) 2

Polarity at top of coil

North (1)Direction of current

Only ONE arrow required (1) 2

Graph

Magnet is moving faster / accelerating (under gravity) (1)

(Rate of) change/ cutting of flux is greater (1)

Induced emf is greater (1) Max 2[6]

107. Explanation of what has happened in circuit

Charging process (1)

Plates oppositely charged OR charge moves from one plate to another (1)

Charge flows anticlockwise OR electrons flow clockwise OR leftplate becomes positive OR right plate becomes negative (1)

Build up of Q/V reduces flow rate (1) Max 3

Explanation of what would have been seen

Same as ammeter 1 (1)

Reason: Same I everywhere OR series circuit OR same I/Q in eachcomponent (1) 2

Estimate of charge

Attempt to find area under correct region of graph (1)

= 52 C (1) 2

[Allow 45 – 65 C]

Estimate of capacitance

p.d. across resistor at t = 10 s = 100 × 103 × 3 × 10–6 A = 0.3 V (1)

(hence p.d. across capacitor = 1.5 V – 0.3 V = 1.2 V)

C = V

Q =

V2.1

C105 5– (equation or sub) [ecf] (1)

C = 42 F [If 1.5 V is used to obtain C = 33 F, then 2/3] (1) 3

Alternative method using e –t/RC

Correct answer appropriate to set of values (1)

Correct ln line (1)

Correct answer (40–44uF) (1)

Alternative method using T = RC

Using T = RC (1)

Appropriate T value (1)

correct answer (1)

Observations

Same picture as before (1)

since same V (1) 2

[OR C now carries twice the previous charge][12]

108. Meaning of E

Voltage/e.m.f. (1)

Induced/caused/created (when magnetic field/flux changes) (1) 2

Definition of

= BA (1)

(Magnetic) flux OR magnetic field lines (1) 2

Additions to diagram of paths of currents

Joined up, and one each side and wholly on disk (1) 1

Explanations

(i) Disc cuts B/ [relative motion implied] ( V/I induced) (1)

(ii) B + I [or two interacting magnetic fields] force (1)

Lenz or LH rule ( opposing force) OR energy argument (1) 3

Explanation of reasoning

Mention of F = BIl (1)

B Ib (1)

I (in disc) B/ (1) 3[11]

109. Addition to diagram

2 lines or more, vertical (1)

Arrow downwards (1) 2

N io b iu m sp h e re

To p p la te

B o tto m p la te

Electric field strength

E = V/d OR 2000 ÷ (0.8 × 10–2) (1)

= 250 000 V m–1 OR N C–1 [OR 2500 V cm–1] (1) 2

(2.5 × 105)

Magnitude of charge

F = ma OR 1.8 × 10–7 × 3.0 × 10–7 (N) OR 5.4 × 10–14 (N) (1)

(F = Eq ) q = F/E OR 5.4 × 10–14/2.5 × 105 (C) (1)

= – 2.16 (2, 2.2) × 10–19 C (1) (1) 4

[NB 1 mark for –, 1 mark for rest of answer]

Why a vacuum

Air/gas/molecules would alter acceleration OR provide anotherforce OR collide with niobium sphere (1) 1

[9]

110. Comparison of positron with electron

Same mass (1)

Opposite charge (1) 2

Minimum energy

Use of E = mc2 (1)

= 2 × 9.11 × 10–31 × (3 × 108)2 J

= 1.6398 × 10–13 J (1)

[Factor 2 omitted: lose second tick]

= 1.6398 × 10–13/1.6 × 10–19 (× 106) MeV= 1.02 MeV (1) 3

How process releases energy

Annihilation (1) 1

Any two from:

em radiation/photon(s)

2 photons

0.51 MeV each (1) (1) Max 2[8]

111. Alpha particle scattering experiment

Quality of written communication (1)

Most alpha went straight through/deflected very little (1)

A tiny minority were deflected through large angles / > 90° (1)

Atom had a dense/massive nucleus (1)

Most of the atom was empty space/small nucleus (1)[5]

112. Mass of head of malletSelecting density x volume (1)Correct substitutions (1)Mass = 1.15 (kg) [3 significant figures, minimum] (1) 3

Momentum change

p = m used (1)

p = 1.15 or 1.2 kg (4.20 + 0.58) m s–1 (1)

= 5.50 / 5.74 kg m s–1/N s (1) 3

Average force

Their above / 0.012 s (1)

F = 458/478 N [e.c.f. p above] (1) 2

Value for force

Handle mass/weight/ head weight/force exerted by user (handle)neglected (1) 1

Effectiveness of mallet with rubber head

t goes up/p goes up (1)

less force, less effective/more force, more effective [consequent] (1) 2[11]

113. Homogeneity

p = mass × velocity (1)

p units N s or kg m s–1 [This alone implies above mark] (1)

E unit (J) N m or kg m 2 s–2 (1)

c unit m s–1 (1) 4[4]

114. Classification of particles

– is a baryon (1) is a baryon (1)

– is a meson (1) 3[Allow bbm]

Charge of strange quark

Show that –1 = –1/3(d) + –1/3(s) + –1/3 (s) (1) 1 particle

is neutral (1)+2/3 + –1/3 + –1/3 = 0 and udsOR charge conservation (–1) = 0 + (–1) (1) 2

[6]

115. Magnitude of F

F = m2 /r (1)Towards the centre (1) 2

Calculation

(i) 9.07 × 10 3 N (1)

(ii) R = mg – m 2 / r (1)Substitutions (1)

5.37 × 103 N

[Calculation of m 2 / r max 1] (1) 4

Explanation

Required centripetal force > mg (so cannot be provided) (1) 1

Critical speed

Use of (m)g = (m) 2 / r (1)

15.7 m s–1 (1) 2

Apparently weightlessThis means no force exerted on/by surroundings OR R = 0 OR onlyforce acting is weight (1)When car takes off it is in free fall [consequent] (1) 2

[11]

116. F proportional to I Quality of written communication (1)Any two from:[In words or on diagram]

Method of producing and measuring a varying direct current

Wire perpendicular to B field

Method of measuring/detennining forces, e.g. moments / acceleration (1) (1)

Graph of F - I straight line through origin for F = added weight (1) 4

[OR correct straight line if F is total weight OR I

F constant]

Calculation of Initial acceleration

F = BIl

= 0.20 T × 4.5 A × 5.0 × 10–2 m

= 4.5 ×10–2 Na = F/m

= kg1050

N104.53–

–2

= 0.90 m s–2

Recall/state/use F – ma and F = BIl (1)

Use 5.0 × (10–2 m) for length (1)

Conversion of g to kg 50 × 10–3 (1)

a = 0.90 m s–2 (1) 4[8]

117. E.m.f.Motion of magnet (1)produces changing magnetic field over the coil (1) 2ORField lines (of magnet) cut across coilORProduces changes in flux linkage between coil and magnetsDiagramX at both ends of path (1)X in middle of path (1) 2

Rate of change of flux

500

))(10(3 3– V

t

(1)

= 6.0 ×10–6 wb s–1/V/T m2 s –1 (1) 2

Changes to apparatusAny three from:

more coils stronger magnet [Accept ‘more powerful’] decrease length of suspension ) [Not just ‘increase’ larger amplitude ) speed of magnet] larger cross sectional area of coil iron core within coil (1) (1) (1) 3

[9]

118. Diagram

Electric pattern:

Straight, parallel, reasonably perpendicular to plates and equispaced[Minimum 3 lines] (1)Correct direction labelled on one line [Downwards arrow] (1) 2Equipotential lines:Any two correct equipotentials with any labelling to identifypotentials (rather than field lines) (1) 1[Arrows on electric field lines – none on equipotential beingsufficient labelling]

Force

E = m1025

30003–

V [Correct substitution] (1)

Use of F = Ee even if value of “e” is incorrect (1)

F = 120 ×(103) V m–1 × 1.6 × 10–19 C

= 1.9 (2) × 10–14 (N) (1) 3

Graph

Straight horizontal line [Even if extending beyond 25 mm] (1)Value of F marked [e.c.f. their value] provided graph begins onforce axis and is marked at this point (1) 2

SpeedUse (1)

eV = ½ mv2

v2 = 2 eV/mv2 = 2

m

Fs

Fd = ½ mv2

v2 = 2Fd/m

Substitution (1)

V2 =

kg109.11

V3000C101.6231–

–19

= 2 kg109.11

N)10(1.9231–

–14

× 25 10–3 m

= kg109.11

m1025N101.92231–

–3–14

Answer: V = 3.2 × 107 ms–1 (1) 3

[If F= 2 × 10–14 N, then V= 3.3 × 107 ms–1][11]

119. Formula for magnitude of force

F = Eq (1) 1

Direction

Down page (1) 1

Calculation

Eq = Bq (1) = E/B (*)

= 1.2 × 104/0.4 ms –1 (*)(*) [Equation or substitution] (1)

= 3 × 104 ms–1 (1) 3

Explanation

mg << eE and/or Bq (1) (1)

[OR gravity force << E and/or B force 2 marks

OR m very small 1 mark only

OR gravity is a weak force 1 mark only 2

OR ion moving fast 1 mark only][7]

120. Corrected errors

line 3 Mesons are made from q and antiq (1) (1)OR leptons are fundamental/not made from smaller etc.

line 4 as line 3 [only one (1) for same correction made twice]OR quarks, leptons, neutrinos, and others (1)

line 6 Neutron is made from 3 q s (1) (1)OR meson is made from q and antiq [with restriction as in line 4]

line 10........... energy .... [instead of momentum] (1) (1) Max 6[6]

121. Direction of centripetal acceleration

Towards centre/downwards/inwards (1) 1

Explanation

F=maa and F in same direction (1) 1

Resultant force

= T

r2 =

5.4

82 [Equation OR substitution] (1)

= 11.2 m s –1 (1)

F =r

mv 2

= 8

2.1160 2 [Equation OR substitution] (1)

= 936 N (1)

OR w = T

2 =

5.4

2 [Equation OR substitution] (1)

= 1.396 rad s–1 (1)

F = mrw2 (*)

= 60 × 8 × 1.42 (*) [(*) Equation OR substitution] (1)= 936 N (1) 4

Calculation of weight

W = mg= 60 × 9.81= 589 N (1) 1

Calculation of magnitude of push

Fnet = W + PP = F – W= 936 – 589= 347 N (1) 1

Diagram

W (1)

P (1) 2[10]

122. Deceleration of Earth

F = m / t = 7 × 3 × 104 N [Equation or substitution] (1)

=2.1 × 105 N (1)

a = F/m = 2.1 × 105/6.0 × 1024 [Equation or substitution] (1)

= 3.5 × 10–20 m s–2 (1) 4

OR

in 1 second: for debris = 3 × 104 m s–1

a = 3 × 104 m s–2

F = ma = 7 kg × 3 × 104 ms–2

= 2.1 × 105 N (1) (1)

F on Earth = F on debris

aearth = F/m = kg106

N101.224

5

= 3.5 × 10–20 m s–2 (1) (1)OR Conservation of momentum (1)

or m1u1 + m22 = m33 etc

6 × 1024 × 3 × 104 (+ 7 × 0) = (6 × 1024 + 7) × 3 (1)

OR An answer (3.5 × 10–20 m s–2) calculated from any specific (1) (1) (1)arbitrary time] (1)Comment: negligible (1) 1[OR energy loss negligible]

[5]

123. Momentum and its unit

Momentum = mass × velocity 1

kg m s–1 or N s 1

Momentum of thorium nucleus before the decay

Zero 1

Speed of alpha particle/radium nucleus and directions of travel

Alpha particle because its mass is smaller/lighter 1

So higher speed for the same (magnitude of) momentum OR N3 argument 1

Opposite directions/along a line 1[6]

124. Table

(i)pa rtic le

(ii)q u a rk c o n te n t

(iii)a n tip ar tic le

(iv )q u a rk c o n te n t

p ro to n

K

u u d

d u

d s

p

K

u u d (2 )

u d (1 )

sd (2 )

–

– – – –

–

–

– +

0 0

Shaded boxes show answers: circled terms count as one.

Proton is uud 1

antiproton or p is uud [allow

p or p-bar ] 1

+ 1

Anti K0 is 0

K [ allow K0-bar] 1

Quark composition is dsanddu 1[5]

125. Angular speed

Use of = 2/T 1

= 1.2 × 10–3 [min 2 significant figures) [No ue as units given] 1

Free-body force diagram

Pull of Earth/Weight/mg/Gravitational Pull 1

Why satellite is accelerating

Resultant/Net/Unbalanced force on satellite must have an acceleration OR F = ma. 1

Magnitude of acceleration

Use of a = 2 r OR 2 ÷ r 1

a = 9.36-9.42 OR 6.5 m s–2 1

[Depends on which value used][6]

126. Alpha particle: diagram

Curving path between plates 1

Towards 0 V plate 1

Emerging from plates and carrying on straight 1

Calculation

Electric field = m)10(10

V20003–

Substitution 1

Force = EQ

3–1010

2000Vm–1 × (2) × 1.6 × 10–19 C

Substitution [ecf their E] 1

= 6.4 × 10–14 N

Correct answer 1[6]

127. Horizontal component

4.8 × 10–5 T × cos 66°

= 1.95 [2.0] × 10–5 T

Use cos 66°/sin 24° 1

Answer 1

Calculation of induced voltage

Speed after 2 seconds = 9.81 m s–2 × 2 s 1

[ecf their B]

Induced e.m.f. = 1.95 × 10–5 T × 2.5 m × [9.81 m s–2 × 2 s] 1

= 9.6 × 10–4V 1

North-south rod

Induced emf = 0 (V) 1

Rod does not cut magnetic field lines/no flux cutting/no change in flux 1[7]

128. (i) Protons are positively charged / like current 1

refer to Fleming or motor rule / Rev / Bqv / perpendicular F and 1

[not right hand rule ]

(ii) υυ2

Ber

m mr2 = Be 2

[accept q for e ]

t

r

T

r π/

π2υ

tT

π/

π2ω 1

(iii) Quality of written communication 1

Each time it crosses gap/between dees it accelerated / is attracted / is given E 1

Idea that p.d. between the dees reverses while the proton completes half a revolution / c.e.p. 1

As energy becomes large the mass/inertia of the proton increases 1[not protons hit edge ]

so it cannot exceed the speed of light [i.e. ref to c ]/synchronousproperty breaks down/formula no longer gives constant f 1

(iv) E = (1.6 × 10–19 C) (12000 V) [allow × 12] 1

= 1.9/1.92 x 10–15 (J) [no e.c.f.] 1

(v) r2 = 2m × k.e. ÷ B2e2r = same

Substitute 1.66 / 1.7 × 10–27 kg /1860 me /2000 me and

1.6 × 10–19 C 1

Use of k.e. = (1.9 × 10–15J) × 850 1

[e.c.f. for 1.9 × 10–15 J e.g. 2 × 10–15J 1.7 × 10–12J]

r = 0.575 m /57.5 cm 1

[2 × 10–15J 0.59 m]

[9.1 × 10–31 kg 0.0137 m e.o.p. max 1/3][15]

129. Direction of travel of electron

Anticlockwise (1) 1

Why track is curved and spiral

Force perpendicular v (1)

Force perpendicular B (1)

electron loses energy or v (1) 3

Measurements from photograph

rmax = 108 (100 - 120) mm (1)

onsubstitutiORequation)108(108.0106.14 19

BQrp (1)

= 6.9 (6.4 – 7.7) × 10-20 (kg ms-1) [no u.e.] (1) 3

Consistency of tracks

Any two from:

photon leaves no track

opposite/different directions of curvatures/spiral [NOT oppositedirections without reference to curve/spiral]

similar/same curvatures/radii/shape

no evidence of any other particles/two tracks only (2) 2

[Symmetrical scores once under bullet point 2 or 3]

How event obeys two conservation laws

Naming two laws e.g. charge and momentum (1)

Any one from:

Charge O +1 + – 1 Momentum initial electron + positron Momentum electron = positron Energy of photon = mass + energy of electron and positron (1) Max 2

[11]

130. Why person moving in a circle must have an acceleration

Acceleration due to changing directionORIf not it would continue in straight line (1) 1

Centripetal acceleration

a= 22

OR rwr

v (1)

)srad(103.7

OR

)ms(465606024

104.622

15

16

w

T

πrv

(1)

a = 0.034 (m s-2) [ no u.e.] (1) 3

Which force is the larger

mg is larger than R/R is smaller than mg (1)

mg – R / centripetal/accelerating/resultant force acts towards centre (1) 2

Differing apparent field strength

(0.034 ÷ 9.81) × 100%

= 0.35%

OR (0.03 ÷ 9.81) × 100% = 0.31 % [NOT 0.3%] (1) 1[6]

131. Energy stored in a capacitor

Justify area: W = QVORwork/area of thin strip = V × Q (1)

Area under graph (1) 2

Energy stored when capacitor charged to 5000 V

W= 21 QV= 2

1 × 0.35 × 5000 J

= 875 J (1) 1

Time constant for circuit

5000/e or 3 = 1840/1667 V (1)

T.C = 3.3 m s [3.1 – 3.6 m s] (1)

OR

Initial tangent t-axis (1)

Accept between 3.5 and 4.0 m s (1) 2

[Also allow use of exponential formula with appropriate substitutionof correct V and t, e.g. 2000 and 3 ms]

Capacitance

C =R

T or as numbers (1)

3.3 m s 7.0 × 10-5 F [Allow e.c.fs.]

4.0 m s 8.5 × 10-5 F (1) 2

[OR using graph: C = Q/V (1)= 0.35/5000 = 7.0 × 10-5 F (1)]

Energy left in capacitor

At 2 ms, V = 2700 V [2600 – 2800] (1)

E = 21 CV2 OR 2

1 QV

= 255 J [e.c.f, depends on method] (1) 2

Energy setting

Energy leaving capacitor = (875 – 255) J

= 620 J [e.c.f ] (1)

Energy delivered = 620 × 60/100 J

= 372 J

380 J setting [Allow e.c.f] (1) 2[11]

132. Electric field

)m(10300

)V(1006

(1)

= 3.3 × 105 V m-1 (1) 2

Force

F = Eq = 3.3 × 1.6 × 10-19 (N)

= 5.3 × 10-14 N [Allow e.c.f] (1) 2

Why force has this direction

Vertical line (1)

directionfieldoftermsinOR

platepositivetoAttracted (1) 2

How much energy hole gains

W = F × d = 5.3 × 10-14 × 2.8 × 10-10 (J) (1)

= 1.5 × 10-23 J [Allow e.c.f] (1) 2[8]

133. How torch works and factors which affect brightness

At least two lines leaving N and S pole, diverging and crossingwires (1)

Arrow leaving N pole/towards S pole (1)

Field/flux lines cut wires (1)

changing B/ OR td

d OR Faraday's law causes V or I (1)

If causes V followed by causes I (1)

Any two of rotation, field strength, number of coils (1)

Appropriate direction e.g. faster rotation brighter/more V/I (1)

R brighter (1)[Max 6]

134. Explanation of Pelton wheel

Quote of F = m/t [or in words] (1)

Negative moment/velocity after (1)

Increased (twice) momentum / velocity change/m - mu comparedwith falling off plane paddle (1)

Idea of doubled (1) Max 3

Percentage efficiency of station

Energy available = mgh (1)

Power input = 270 × 9.8 × 250 (1)

Efficiency = 500 000 × 100 / 661 500 = 76 (%) (1) 3

Other desirable properties

For example:

Hard – does not wear/scratch/dent (2)

Tough – can withstand dynamic loads/plastic deformation (2)

Strong – high breaking stress/force (2)

Smooth – low friction surface (2)

Durable – properties do not worsen with time (2) Max 4

Time to repay initial investment

Each hour worth 500 × 0.0474 = £23.70 OR total no of kW hrequired 1.000000/0.0474 = 2.1 × 107

No. of hours to repay = 1 000 000 /23.70 OR 2.1 × 107 /500

24 × 365 = 4.8 years to repay debt (1)

Assumption

Constant power production / no interest charges / no repair costsno wages for employees (1) 4

[14]

135. Outline of evidence from Geiger’s and Marsden’s scattering experiment

Most alpha particles went (almost) straight through (1)

Some or a few deflected at larger angles/>90°/rebounded (1)

A tiny minority [e.g. 1 in 8000] were deflected at angles > 90° OR rebounded (1) 3

Suggestion

No large deflections/all go (almost) straight through (1)

Explanation

No concentrated charge/mass OR no massive object (to hit) no denseobject to hit [consequent] (1) 2

[5]

136. Fundamental particle

A particle which cannot be further divided/which has no “parts”inside it/one of the 12 particles of which all matter is made (1)

[Not “one which cannot decay to another particle”]

Circled fundamental particles in list (2)

Positron and muon

[If more than two circled, –1 for each extra one] 3

Explanation

Any three from:

Quality of written communication (1)

Mesons are composed of a q and an q (1)

These have charges ± 2/3 and ± 1/3 (1)

Shows all possibilities (+1, 0, – 1) OR other convincingarithmetic to show max +1 (1) Max 3

[6]

137. Angular speed

Conversion of 91 into seconds – here or in a calculation (1)

Use of T = 2/ allow T= 360/

= 1.15 – 1.20 × 10–3 rad s–1 /6.9 × 10–2 rad min–1 /0.066 deg s–1 (1) 3

Acceleration

Use of a = r2 / 2 / r (1)

Adding 6370 (km) to 210 (km)/ 6580 (km) (1)

a = 8.5 to 9.5 m s–2 [No e.c.f. for 210 missed but allow for in rad s–1] (1) 3

Resultant force

Recall/Use of F = ma (1)

F = 35 –39 N [Allow e.c.f their a above only] (1)

Towards (centre of the) Earth (1) 3[9]

138. Cathode Ray Tube

Electron emission

Heating effect (due to current) (1) (Surface) electrons (break free) because of energy gain (1) 2

[Thermionic emission scores both marks]

Electron motion towards anode

The electrons are attracted to/accelerated by the positive anode (1) 1

Energy

Electron energy = (10 × 103 V) (1.6 × 10–19 C)

=1.6 × 10–15 J

Correct use of 1.6 × 10–19 OR use of 10 × 103 (1)Answer (1) 2

Number of electrons per second

Number each second = J106.1

A105.119

3

9.4×1015s–1

Correct conversion mA AAnswer (1) 2

Rate

Energy each second = (9.4 × 1015 s–1) (1.6×10–15 J) (1)

= 15 Js–1 (W) / 14.4 Js–1 (1) 2

[ecf their energy][9]

139. (i) Tracks (of alphas) are the same length/alphas travel same orequal distance (1)

(ii) H/p + Li 2/2He (1)

Heandp 42

11 correctly labelled (1)

Li73 (1) 4

(iii) Mass defect = 0.01865u (1)

Either Or

Use of × 1.66 × 10–27 Use of × 930 (1)

Use of × 9.0 × 1016 Use of × 1.6 ×10–13 (1)

2.79 ×10–12 J 2.78 × 10–12 J (1)

Assume: proton has zero/very little k.e. (1) Max 4[8]

140. How diagram confirms pion is negatively charged

Any two from:

bends opposite way to proton reference to magnetic interaction/Fleming’s left-hand rule proton + pion – (1) (1) 2

Charge carried by lambda particle

Neutral (1)

because charge conserved OR +1 – 1 0 OR not ionising/no track (1) 2

Deduction

r pion < r proton / straighter / less curved (1)

since r = p / BQ (P pion < P proton) (1) 2

Scale drawing

2 straight lines lpr > lpi (1)Orientation of lines (49°) joined correct way (1)Answer 10 ± 1 kg m s–1 (1) 3

Classification of particles

baryon meson

pion (1)

lambda (1) 2

Charge of a down quark

–1 / 3 (e) (1) 1[12]

141. Reason

Even a very small resistance (in series) with 2000 A through it would generatemuch heat

[Answer must refer to where heating effect occurs]

OR

Such a big current would need thick wires in the meter design

OR

Have to break circuit to insert meter (1) 1

Explanation

Any four from:

I1causes flux (in iron ring and coil) (flux) in iron ring/coil changing I1 changing flux

V in coil induced / V = N dt

d /cutting flux /Faraday’s law in context/

I induced (1) (1) V in coil I I in coil (1) (1) 4

[5]

142. How ions are accelerated

Electric field exists between +, – electrodes (1)

force on ions / force acceleration (1) 2

Speed of xenon atom

eV = ½ m2/eV = E k (1)

= m/eV2 (1)

= 125

19

ms102.2

)2251060(106.12

(1)

= 4.3 ×104 m s–1 [No u.e.] (1) 4

Thrust on space probe

Force = rate of change of momentum (1)

= 2.1 × 10–6 × 43 000 N (1)

= 0.090 N (1) 3

[Using 4 × 104 m s–1 gives F = 0.084 N]

Reason for reduced thrust

Xenon ions attracted back OR similar (1) 1

Why ion drives maybe preferable

Any two from:

less fuel required in total for example, 66 kg for a year thrust provided for longer/fuel lasts longer/accelerates for longer lower payload for initial launch/ion drive lighter (1) (1) 2

[12]

143. Diagrams showing forces:

D iag ra m 1 D ia g ram 2

(1) (1) 2

[Each diagram, one arrow only]

Discussion re artificial gravity

Any four from:

centripetal force, in context forces felt by astronaut both “upward” so feels like weight/gravity a 2/r / F = m2 / r / a = 2r / F = m2r / F = m

= 2

2

2

22

10

210442 )/(

t

rr/

t

r

[Any one] or calculate or

= 2.0 m s–2 [No u.e.] artificial field varies with / radius (1) (1) (1) (1) 4

1.6 N kg–1 [to one significant figure] OR other justification [e.c.f] (1) 1[7]

144. Observations of circuits

Any six from:

capacitor is charged energy stored in C/goes to lamps... heat and light in lamp as I/Q passes through lamp/discharges through lamp E = ½ CV2

V × 2 E × 4 (hence) 4 lots of energy/4 lamps lit similarly 5 V across 1 lamp same Q through each lamp as before discussion of T = RC R same for both circuits flash is bright and dies exponentially (1) (1) (1) V × 2 Q × 2 (1) (1) (1) same Q or I as before alone each parallel branch

[6]

145. Experiment2 light gates (1)Gate gives time trolley takes to pass [ not just ‘the time’] (1)Speed = length of ‘interrupter’/time taken (1)

OR

2 ticker timers (1)Dots at known time intervals (1)Speed = length of tape section/time taken (1) 3[ruler + clock could obtain third mark only, specifying a length/time]

Total momentum of trolleysZero (1)It was zero initially or momentum is conserved [consequent] (1) 2

Speed v of A Use of momentum = mass × velocity (1)Use of mass × speed (A) = mass × speed (B) (1)

1.8 m s1 [ignore -ve signs] (1) 3[8]

146. Topic C – Nuclear and Particle Physics

SimilarlySame mass

DifferenceCharge OR baryon number OR uud quarks duu (1) 2

Any two lepton pairs from the following:

e– e+ (

– + ( NOT e.g. muon and antimuon/ μ

- +

ve v e v v 0R just v v (2) 2v v

CollisionParticle and antiparticle annihilate/produce a burst of energy/of photons/of gamma rays (1) 1

[5]

147. Speed of electronSelection of = h/p and p = m (1)

m = 9.11 10–31 (1)

7.2 – 7.3 106 m s–1 (1)

Kinetic energy

Use of Ek = 1/2 m2 (1)147 – 152 [ecf] (1) 5

High energy electronNucleus tiny/a lot smaller so very small (1) or p very large [consequent] (1) 2

[7]

148. Faraday’s law of electromagnetic inductionThe induced e.m.f. (1)in a conductor is equal to/proportional tothe rate of change of magnetic flux linkage (1)

OR

E = t

φ

d

d or E

t

N

Δ

Δ [Accept or d]

E – induced voltage

d – change of magnetic flux (1)dt – time

[All symbols defined] 2

Conversion of sound waves into electrical signals

Any four from:

quality of language (1)

sound waves make the diaphragm/coil vibrate/oscillate (1)

coil: change in flux linkage/coil cuts field lines (1)

induced voltage across coil (1)

frequency of sound wave is frequency of induced voltage/current/electrical wave (1) Max 4

[6]

149. Calculation of charge

6000 V 20 10–6 F (1)= 0.12 C (1) 2

Energy stored in capacitor

2

2CV

2

V)6000(C1002 26

(1)

= 360 J (1) 2

Resistance

A40

V6000= 150 (1) 1

Time to discharge capacitor

Time = A40

C12.0 /their Q (1)

= 0.0030 s / 3.0 10–3 s [e.c.f.] (1) 2

ReasonTime is longer because the rate of discharge decreases/ current decreases

with time (1) 1[8]

150. Free-body force diagram

Tension/pull of thread (1)F/push of charged sphere/electric force/electrostatic force (1)Weight/W/pull of Earth [Not mg, unless W = mg stated] (1) 3

Te n s io n /p u ll o f th re ad

F /p u sh o f ch arg ed sp h ere /e lec tric fo rc e / e le c tro s ta tic fo rce

W eig h t/ /p u ll o f E arth [N o t g , u n le ss = g s ta ted ]W m W m

Force equation

W = T cos ( ( (1)F = T sin (

Processing mark, e.g. F = cos

W sin OR tan =

cos

sin (1)

OR

F, T, W labelled (1)both angles labelled (1) 2

T

TW

Table

Distance r = 36 10–3 mF = 35.5/36 [No u.e.] (1)

Distance r = 27 10–3 m

Using any pair of values (1)Seeing correct constant for their pair of values (1) F = 63.1 [n.o u.e.] (1)

OR

Valid simple ratio calculation using a pair of values (1)stating produce Q1Q or kQQ2 constant (1) F = 63.1 [no u.e.] (1) 4

Measurements taken quickly because

Leakage/discharge of charge [Allow dissipation or description of process] (1) 1[10]

151. (i) Wc = ½ CV2 = ½ (0.0047 F) (25 V)2 [Ignore 10n] (1) = 1.5 J / 1.47 J [no e.c.f.] (1) 2

Quality of written communicationWc is (very) small (1)Even at 50 V it is only 6 J (1)Any T is difficult to measure/wire spread out/ (1)something like a thermocouple is needed (1)Wire (might) melt/fuse (1)Heat/energy loss to air/surroundings [not to connecting wires] (1) Max 4

(ii) Exponential (decay) (1)Radioactive decay/radioactivity [independent] (1)Use of one of five approved methods [Name it] (1)

Data off graph appropriate to method [ignore 10n] (1)Use of RC/use of R = V/I (1)R = 7.2 8.5 [no e.c.f.] (1)[7200 – 8500 gets 3/4] 6

[12]

Methods:M1 RC = time to Q0 e [35 – 39 ms]M2 RC ln2 = t1/2 [24 – 28 ms]M3 RC = where initial tangent hits t axis [32 – 40 ms]

M4 Use of RC in Q = Q0e–t/RC with numbers [ correct]M5 Calculation of T0 initial current from gradient [2.7 – 3.0 A]

152. Quarks: What is meant by “charge = + 2/3”

sign: +/positive/sign same as proton/sign opposite to electron (1)

size: 2/3 charge on a proton / electron (1) 2

Mass of strange quark in kilograms

m = 0.2 GeV/c2

= 0.2 × 109 × 1.6 × 10–19 (1)

/ 9 × 1016

= 3.6 × 10–28 kg (1) 2

Charge and mass of anti – particle to the charmed quark

Charge: –3

2 (1)

Mass: 1.3 GeV/c2 [No unit penalty for omitting GeV/c2] (1) 2

Prediction of top quark

Symmetry of the model / 3rd generation partner / other valid statement 1

Reason for length of time to find experimental evidence for top quark

High energy needed (to create it) / needs a big accelerator/other valid reason 1

Use of conservation law to explain prediction

Momentum (in context)

Total momentum = 0 OR mtt = mbb OR in words

mt >> mb t « b / greater mass ( lower velocity) 3[11]

153. How movement of magnet produces voltage shown on c.r.o screen

Any 4 from:

Boxes correct

Mention of Faraday’s law/equation/word description

Flux max when magnet vertical / box 1 / box 3

Flux zero when magnet horizontal / box 2 / box 4

When flux max, not changing, V = 0

When flux changing fastest, V max

Appropriate comment about sense of voltage, e.g.,when poles reversed, V reversed 4

Differences between figures (i) and (ii)

Qualitative points: (max 2)

(Faster turning, giving) td

d (1)

= V and f OR T (1)

OR

Quantitative points: (max 3)

(f × 2 =) td

d (max) × 2 (1)

= V × 2 (1)

f × 2 (OR T ÷ 2) (1) 3

Flux at each end of magnet

Area 1 big square = 100 (Vs) or 100 × 10–6 (Vs)

OR area of 1 little square = 4 (Vs) or 4 × 10–6 (Vs)

OR area = 32 little squares (29 – 35)

OR area = 4/3 big squares (1.2 – 1.4) (1)

Area = 130 × 10–6 (Vs) (120 – 140) (1)

Area / 2 × 240

= 2.7 × 10–7 Wb (2.5 – 2.9) (1) 3

Magnetic flux density at end of bar magnet

B = /A OR = BA OR A = 0.01 × 0.005 OR A = 5 × 10–5 m2 (1)

= 3.0 × 10–7 / 5.0 × 10–5

= 6.0 × 10–3 T (accept Wb m–2) (1) 2[12]

154. Magnitude of charges

Value of v or (v = 1023 ms–1, = 2.7 × 10 –6 s–1) (1)

Value of a (a = 2.7 × 10–3 ms–2) (1)

F = r

m 2 (1)

Value of F (F = 2 × 1020 N) (1)

F = 2

1 2

r

QkQ (1)

Charge = 5.7 × 1013 C (1) 6

[Use of 2r

GMm to calculate F:

Allow Me in range 1024 – 1025 kg without penalty, otherwise max 4 for question.]

155. Atom is neutral (1)Quark composition is duu (1)Antiproton is (2/3) + (2/3) + (+1/3) (= 1) (1) 3

Explanation:As soon as it touches the container/matter (1)(Matter and antimatter) annihilate (1)[Not “cancel”; not “react”] 2

Completion of table:

Quarks Charge

up charm TOP +2/3

down strange BOTTOM 1/3

1

[OR TRUTH & BEAUTY][Both needed for 1 mark]

(i) Neutral strange meson: ds OR d s (1)

(ii) Positive charmed meson: c d OR c s (1)

(iii) Neutral strange baryon: uss/css/uds/cds OR any of their antiparticles,e.g. u s s (1) 3

[9]

156. Conservation laws:

(i) Charge: (1) + (+1) = (0) + (1) + (+1) + (0) (1)Baryon number: (0) + (+1) = (+1) + (0) + (0) + (0) (1)[So possible, no mark]

(ii) Charge: (+1) + (+1) = (+1) + (+1) + (+1) + (1) (1)Baryon number: (+1) + (+1) = (+1) + (+1) + (+1) + (1) (1) 4[So possible, no mark]

[4]

157. Discussion:No equilibrium or there is a resultant force (1)Direction changing or otherwise would move in a straight line (or off at an tangent) (1)acceleration or velocity changing (1)Force towards centre or centripetal (1)The tension provides this force [consequent] (1)[OR for last 2 marks: weight of ball acts downwards (1)vertical component of tension balances it (1)] 5

Free-body diagram:

W/weight/mg/gravitational ‘attraction’ [not ‘gravity’] (1) 1[6]

158. Isotope of lead:

Pb20682 1

Other particles:

(82) electrons 1

How appropriate number of quarks can combine:

3 quarks involved (1)

2 × + 2/3 + 1 × –1/3 = + 1 (1) 2

Explanation:

High energy is needed/high temperature/high speed (1)

Mention of E m OR E = mc2 (1) 2

Description:

Relates to electron (1)

e.g. charge +1/antiparticle/annihilates with (1) 2[8]

159. Calculation of total momentum:

In 1 s, p = 1400 J /3 × 108 m s–1

= 4.7 × 10–6 N s (kg m s–1)

[No u.e.] 2

Force exerted on whole sail:

Momentum change/time [OR symbols] (1)

= 4.7 × 10–6 × 1.5 × 106

= 7.0 (7.1) N (1) 2

Explanation of why force is doubled:

Photons bounce back (1)

So their change of momentum is doubled (1) 2

Calculation of maximum increase in speed:

a = F/m = 7/1200 m s–2 [allow e.c.f.] (1)

(= 0.006 m s–2 )

= a t = 0.006 m s–2 × 604 800 s (1)

= 3500 m s–1 (1) 3[9]

160. Calculation of energy:

E = ½ CV2 (1)

= ½ × 100 × 10–6 F× (4 V)2

= 8 × 10-4 J (1) 2

Parts of circuit which will transfer energy to surroundings:

Ss / the wires between it and C 1

Discussion:

No – negligible energy 1

Completion of graph:

Convex curve up from 0 (1)

through (200, 4) (1)

Then drops to zero (1) 3

6

4

2

0 1 0 0 2 0 0 3 0 0 t /s

p .d ./V

Approximate value for R :

Time constant = RC (1)

R = 200 s/100F

= 2× 106 [Allow 1.6 – 2.2 M] (1) 2

Discussion of effect of increasing resistance, R :

R goes up T goes up (1)

T goes up means longer toasting (1) 2[11]

161. (a) The origin of the induced e.m.f:

Faraday’s law (1)

As conductor cuts field lines (1)

Electrons experience force along wire (1)

move to one end e.m.f. (1) 3

(b) Reduction in orbit height due to flow of current:

Current + field force OR Fleming L H rule (1)

Lenz’s law: (1)

Force opposes motion (1)

Orbiting craft lose energy/speed (1) 3[Max 5]

162. Calculation of minimum height x :

a = r

2 (1)

r

2 g at top (1)

At top: ½ m2 = mgx [OR mgh 1 mark] (2)

2 = 2gx (Note: derived from 2 = u2 + 2as = 0)

[For 2 = 2g(x + 1) (1)(speed at bottom)]

2gx/r g at top

x r/2

x 0.25 m

[ or =] (1)[5]

163. Principle of conservation of linear momentum:

[Not just equation – symbols must be defined]

Sum of momenta/total momentum remains constant (1)

[Equation can indicate]

[Not “conserved”]

If no (resultant) external force acts (1)

[Not “closed/isolated system”]

2

Laws of Motion:

2nd and 3rd laws (1)

1

Description:

Measure velocities/speeds before and after collision (1)

Suitable technique for measuring velocity

e.g. ticker tape/ticker timer

light gate(s)

motion sensor (1)

[Not stop clock or just datalogger]

How velocity is found from their technique

[Need distance ÷ time + identify distance and time. Could get with (1)stopclock method.] (1)

after = ½ before/ calculate m before and after/check e.g. m11 = (m1 + m2) 2

4

Reason for discrepancy:

Friction/air resistance (1)

[Ignore any reference to energy] (1)

Explanation:

The Earth (plus car) recoils (1)

With same momentum as the car had (1)

2[10]

164. Relationship between current and charge:

Current is the rate of flow of charge/rate of change of charge OR current is charge per second

OR I = Q/t (with or without d or ) but with symbols defined (1)

1

Explanation:

Since I is constant, Q o on capacitor (= It) increases at a steady rate OR charge flows at a constant rate (1)

Since V Q, V also increases at a steady rate (1)

OR

V = Q/C = It/C (1)

and V = (I/C) × t compared with y = (m) × x

2

Determination of current, using graph:

Use of Q = CV Attempt to get grad (1)

Use of I = Q/t Use of I = C × grad (1)

= 1.1 mA 1.1 mA (1)

3

Explanation:

Decrease [If increase, 0/3] (1)

As capacitor charges, VR decreases (1)

R must decrease because I = VR/R OR R must decrease to prevent I

falling (1)

3

Second graph:

Line added to graph showing:

Any curve getting less steep with time [from origin; no maximum] (1)

And with same initial gradient as original straight line (1)

2[11]

165. Calculation of potential difference:

Use of E = V/d [d in m or cm] (1)

V = 90 kV (1)

Calculation of maximum kinetic energy:

Use of × 1.6 × 10–19 [in E = qV e.c.f. value of V] 1.4 × 10–14 (J) (1)

[e.c.f. their V × 1.6 × 10–19] (1)

4

Maximum speed of one of these electrons:

Use of k.e. = ½ m2 with m = 9.1 × 10 –31 kg (1)

[Full e.c.f. their k.e. possible; make sure v is speed term]

= 1.8 × 108 m s–1 [u.e. but only once] (1)

Diagram:

2

At least 3 radial lines touching object (1)

Direction towards electron (1)

2

Expression for electric potential V:

V = r

19106.1

επ

OR re

04 OR r

91044.1

[not k unless defined]

definedunless

4 0

Qr

QNot

[With or without “–” sign] (1)

1[9]

166. Flux through closed window:

Flux = 20 × 10–6 T × (1.3 × 0.7) m2

[if two equations, must use (1.3 × 0.7) each time] (1)

BH chosen OR area correct (1)

= 1.8 × 10–5 Wb/T m2

2

Average e.m.f. induced:

E = s8.0

Wb108.1 5 [e.c.f.] (1)

= 2.3 × 10–5 V [5.6 × 10–5 V if Bv used] (1)

2

Effect on induced e.m.f. of converting window: (1)

Zero induced e.m.f. [Not “very small”]

No change in flux linkage OR no flux cut OR e.m.fs. in opposite sides cancel out 2/2 [Consequential] (1)

2[6]

167. Momentum of driver:

Correct use of p = m [OR with numbers] (1)

= 1500 N s OR 1500 kg m s–1 (1) 2

Average resultant force:

Correct choice of F × t = p OR F = ma (1)

F × 0.07 (s) = 1500 (N s) F = 50 × 429/50 × 30/0.07 (1)

= 21 kN = 21 kN (1) 3

[Ignore sign of answer]

Why resultant force is not the same as force exerted on driver by seatbelt:

Air bags /floor/friction/seat/steering wheel (1)[Named force other than weight/reaction] 1

[6]

168. Charge on strange quark = – 1/3 (1) 1

Conservation law:

Charge – (–1) + (+1) (0) + X/by charge conservation (1)

X is neutral (1) 2

Particle X is a meson (1)

Baryon number conservation (0) + (+1) (+1) + (0) (1) 2

OR discussion in terms of total number of q + q = 5 OR q – q = 3

Composition of X is s d [0/3 if not q q ](1)

Justify S quark:

This is not a weak interaction/only a weak interaction can change quark type/this isa strong interaction/strangeness is conserved/ quark flavour cannot change (1)

Justify d quark:

X neutral; s – 1/3; d + 1/3. [e.c.f. if s = – 1/3 in first line.]

For the third mark accept any q q pair that creates a mesonof the charge deduced for X above. (1) 3

[The justification for both q and q can be done also by tracking individual quarks][8]

169. Momentum: mass × velocity [accept defined symbols] 1

Physical quantity:

(Net) force (1)

on lorry (1) 2

[“Rate of change of momentum” scores one only]

Magnitude:

s40

61Ns = 1.5 (3) × 104 N

Gradient measurements (1)

Correct calculation (1) 2

[Accept 1.4 – 1.7 to 2 s.f.]

Explanation of shape:

Force decreases as speed increases (1)

[Allow “rate of change of momentum”]

Any one of:

Air resistance increases

Transmission friction increases

Engine force reduces 2[7]

170. Velocity of protons:

p=Bqr => = m

Bqr(1)

= 27–

19–

1067.1

5.11060.12.0

= 2.9 × 107 ms–1 (1)

[must have 2.9]

10

103 8 (ms–1) (1) 3

Time for last semi-circle of orbit:

t = 71087.2

5.1

d

(1)

1.6(4) × 10–7 s (1) 2

Frequency of accelerating p.d.

f =t

1 = 3.0 MHz [allow ecf] (1) 1

[6]

171. Exponential shape (1)

Value at RC > 1.5 V [only if shape correct] (1)

Levels off at 3 V (1) 3

Why movement of diaphragm causes p.d:

No movement, no change in C, no signal (1)

OR moving diaphragm changes C

As C changes so V changes (1)

Vc + IR is constant (1)

Hence IR changes – signal (1) 4

OR for last 3 marks

As C changes Q changes

Q flows through R

hence V = IR for resistor as signal 7

172. Wavelength of photon:

2E (1)

= 135 × 106 × 1.6 × 10–19 (1)

E = 1.08 × 10–11 J (1)

E = hf =ch

(1)

= E

hc

= 1.84 × 10–14 m (1) 5[5]

173. Credit to be given for all good, relevant Physics

Examples of mark scoring points [each relevant formula is also worth 1 mark]:

Between plates field is uniform

Acceleration is constant

Energy gained = 2000e

All ions have same F or same energy

From hole to detector is zero field/force

Ion travels at constant speed

g negligible

time proportional to 1 /velocity

time proportional to 1 /mass

in a vacuum there are no collisions or friction forces[Max 7]

174. Demonstration of how statement leads to equation:

Momentum = mass × velocity (1)

Therefore force mass × rate of change of velocity (1)

Therefore force mass × acceleration (1)

Definition of newton or choice of units makes the proportionality constant equal 1 (1) 4

[Standard symbols, undefined, OK; “=“ throughout only loses mark 4. No marks for just manipulating units. If no (e.g. m/t),can only get marks 1 and 4]

Effect on time:

Time increases (1) 1

Explanation:

Acceleration smaller/momentum decreases more slowly/F = t

p

(1)

[Need not say p = constant]

So force is smaller (1) 2

[Independent mark, but must be consistent with previous argument][If no previous argument, this becomes fully independent mark] 7]

175. Explanation:

Changing direction/with no force goes straight on (along tangent) (1)

Acceleration/velocity change/momentum change (1) 2

Identification of bodies:

A: Earth [Not Earth’s gravitational field] (1)

B: scales [Not Earth/ground] (1) 2

Calculation of angular speed:

Angular speed = correct angle ÷ correct time [any correct units] (1)

= 4.4 × 10–3rad min–1 / 0.26 rad h–1/ 2rad day–1 etc (1) 2

Calculation of resultant force:

Force = mr2 (1)

= 55 kg × 6400 ×103 m × (7.3 × 10–5 rad s–1)2 (1)

= 1.9 N (1) 3

[No e.c.f here unless in rad s–1]

Calculation of value of force B:

Force B = 539N – 1.9N (1)

= 537 N (1) 2

[e.c.f. except where R.F = 0]

Force:

Scales read 537 N (same as B) [allow e.c.f.]

Newton’s 3rd law/force student exerts on scales (1) 1[12]

176. Estimation of charge delivered:

Charge = area under graph (1)

= a number of squares × correct calculation for charge of one square i.e. correct attempt at area e.g. single triangle (1)

= (3.5 to 4.8) × 10–3 C (A s, A s) (1)

[Limit = triangle from 41 A 300 s]

OR

Charge = average current × time (1)

= (something between 10 and 20 A) × 300 s (1)

= (3.5 to 4.8) × 10–3 C (1) 3

[But Q = It 0/3, e.g. 41 A × 300 s]

Estimation of capacitance

C = calculated charge/9.0V time constant 100 s (1)

= 390 to 533 F C = 100 s/220 k. = 450 F (1) 2[5]

177. Estimate of time constant, using graph:

V /3 O R V /e O R t

3 V 3 .3 V 1 6 s

2 6 s 2 3 .2 4 s 2 3 s

M e th o d

Valu e 2 3 2 6 s

1/2

(1 )

(1 ) 2

Calculation of resistance and hence capacitance:

R = i

V OR 3–1019.0

9

(1)

Resistance = 47 k [ue] (1)

Substitute in t = RC [e.c.f their t, their R] OR answer 300 F

Capacitance = 500 F (1) 3

Addition to graph of line showing how potential difference varies with time:

A curve of shape shown below, i.e. getting less steep (1)

Any convex curve ending at 7.5 V, crossing at 15 s (1) 2[7]

178. Calculation of e.m.f. induced across falling rod:

Correct use of E = Bl (1)

= 25 m s–1 (1)

e.m.f. = 7.3 – 7.4 × 10–4 V (1) 3

Explanation of why magnitude of vertical component is not required:

Earth’s field is parallel to direction of fall/body falls vertically (1)

Therefore no flux cut (1) 2[5]

179. Calculation of how long wheel takes to complete one revolution:

Time = 2 × 60 m/0.20 m s–1 (1)

= 1900 s/1884 s/31.4 min (1) 2

Change in passenger’s velocity:

Direction changes OR up (N) down (S) OR + – (1)

OR 180° (1) 2

0.40 m s–1

[0.40 m s–1 without direction = 2/2]

Calculation of mass:

(G)pe = mgh

m = 80 × 103 J/9.81 m s–2 × 120 m) (1)[This mark is for rearranging the formula; accept 10 instead (1)of 9.81 and 60 instead of 120 but do not e.c.f. to next mark]m = 68 kg (1) 3

Sketch graph:

G P E /k J G P E /k J

8 0 80

9 5 0 95 0t/s t/s

O R

Labelled axes and line showing PE increasing with timeSinusoidal shape (1)(950 s, 80 kJ) (1)[Accept half the time they calculated at start of question (1)instead of 950 s as e.c.f.][PE v h 0/3] 3

Whether it is necessary for motor to supply the gpe:

No, because passenger on other side is losing gpe (1)If wheel equally loaded OR balanced with people (1)

ORYes, because no other passengers (1)so unequally loaded (1) 2

[12]

180. Definition of linear momentum:

Mass × velocity [Words or defined symbols; NOT ft] (1) 1

Newton’s second law:

Line 3 only (1) 1

Newton’s third law:

Line 2 OR 1& 2 (1) 1

Assumption:

No (net) external forces/no friction/drag (1)

In line 3 (he assumes the force exerted by the other trolley is

the resultant force) [Only if 1st mark earned] (1) 2

Description of how it could be checked experimentally that momentum is conserved in a collision between two vehicles:

Suitable collision described and specific equipment tomeasure velocities [e.g. light gates] (1)

Measure velocities before and after collision (1)

How velocities calculated [e.g. how light gates used] (1)

Measure masses / use known masses/equal masses (1)

Calculate initial and final moment a and compare ORfor equal trolleys in inelastic collision, then 1 = ½ 2 (1) Max 4

[9]

181. Completion of nuclear equation:

One mark for top line all correct (1)

One mark for bottom line all correct

73 Li + 1

1 p 74 Be + 1

0 n 2

Calculation of energy transfer

P = V × I = 2.8 × 106 V × 2.0 × 10–3 A= 5.6 × 103 W OR 5.6 kW

One mark for value (1)

One mark for power of ten and unit (1) 2

Demonstration that energy is absorbed at rate of 17 GW per cubic metre:

23–6–

3

m101.2m10280

W105.6

Volume

Power

= 1.66 × 1010 W m–3

= 17 GW m–3

Substitution (1)Calculation (1) 2

Suggested problem:

Very hot/target overheats/vaporises/difficult to cool OR other good

relevant physics (1) 1[7]

182. Forces acting on molecule, shown on diagram A:

Forces not collinear and sense correct (1) 1

Explanation of why molecules align with field:

Forces not in same line (1)

Hence turning effect [OR torque] 2

Field lines shown on diagram B:

At least three lines drawn equidistant(1)

Direction correct (1) 2

Calculations of electric field strength:

m100.1

V5.15–

d

VE (1)

= 1.5 × 105 V m–1 (1) 2[7]

183. Demonstration that water must be thrown backwards at about 13 m s–1:

Force = t

waterofmomentum

)(

(1)

8 × 105 N = 6 × 104 × V (1)

V= 4

5

106

108

= 13 m s–1 (1) 3

Calculation of power expended:

P = F × = 8.0 × 105 N × 20 m s–1 (1)

1.6 × 107 W OR 15 MW (1) 2

Calculation of rate at which water gains kinetic energy:

½ × m/t × 2 = ½ 6 × 104 kg/s × (13 M s–1)2 (1)

= 5.07 × 106 W OR 5.1 MW (1)

[Allow 5.3 MW if 13.33 m s–l used] 2

Overall efficiency:

Power in = 1.6 s

l × 3.4 × 107

l

J = 5.44 × 107 W (1)

[Intermediate value not explicitly needed]

Power out = 16.0 × 106 + 5.4 × 106 = 21.4 × 106 (1)

Efficiency = 5.54

4.21 = 0.39 (39%) (1) 3

[10]

184. Capacitors and storage of energy:

E = ½ cV2 (1)

Ew = ½ × 68 × 10–3 F × (16 V)2 = 8.7 J (1)

E2 = ½ × 1 × 10–3 F × (400 V)2 = 80 J (1) 3

[Allow calculations in proportion using 2

V

V

F

C

Range of actual value:

900 F < C < 1500 F (1) (1) 2

(i) Calculation of charge:

Q=CV = 68 × 10–3 F × 16 V= 1.1 C (1) 1

(ii) Demonstration that maximum leakage is about 3000 A:

I = 0.003 × 10–6 A/F V × 68 000 F × 16 V (1)

= 3.26 × 10–3 A [3.3 mA] (1) 2

(iii) Estimate of time for capacitor to discharge with reasoning:

s334A1026.3

C09.13–

0

I

Q

[This is time constant] (1)

Numerical example such as: for less than 0.7% remaining t = 5 =

1670 s OR well-reasoned estimate showing t >> 300 s (1) 2[10]

185. Explanation of how it can be deduced that magnetic field acts out of the plane:

Current flow in opposite direction to e– movement/same as e+ movement (1)

(Force acts into spiral) hence Fleming’s left-hand rule (gives field out of

paper) (1) 2

Explanation of which e– moves faster:

(the “atomic” electron) since path is straighter so r larger and

BQ

mr

(1) 1

Calculation of momentum:

p = BQr = 5.4 × 10–3 T × 1.6 × 10–19 C × 0.048 m (1)

= 4.1 × 10–23 N s (1) 2

Explanation of why path of the positron is a spiral:

Positron continually losing speed/energy (by ionising) 1

Discussion of conservation of two properties:

Charge:

e+ and e– (1)

recoiling electron and stationary positive ion (1)

Energy:

e+ and e– creation (1)

since E = mc2 (1)

EK of recoiling electron (1)

EK of e+ and e– pair (1)

Ionisation energy (1)

Momentum:

Incoming photon momentum goes to recoil electron (mostly) (1)

After collision:

Momentum up = momentum down (1)

2 go up (one slightly) and only one goes down so down one is faster (1) Max 5[11]

186. Why large voltage is generated in secondary circuit:

Faraday’s Law in words including ‘flux linkage’

Current flow in primary (1)

causes magnetic flux in core (1)

Flux links secondary (1)

Opening switch S causes flux to reduce (1)

Changing flux in. secondary induces e.m.f (1)

Many turns on secondary means large flux linkage (1)

Hence rate of change of flux linkage is large

reduction time is short (1)

Hence induced e.m.f. is large (1) Max 6[6]

187. In this experiment alpha particles werescattered by thin films of metals such as gold.

The experiment led to the conclusion that the atom had a

positively charged nucleus of diameter approximately 10–15 m and containing

most of the mass of the atom[5]

188. Observations on voltmeter:

(a) Movement which implies brief or pulsed then V reads zero

(b) Negative reading with respect to direction above

(c) Alternating reading positive to negative and continuous[5]

189. Total magnetic flux through the loop when 30 mm from end of magnet:Flux = B × A= 1 × 10-3 T × 16 × 10-4 m2 (1)

[Substitution of 1, 16. Ignore × 10 here]= 1.6 × 10-6 Wb (1)

Total magnetic flux through the loop when 10 mm from end of magnet:Flux = 30 ×10–3 T × 16 × 10–4 m2

= 4.8 × 10–5 Wb [Unit penalty once only] (1) 3

Average speed of movement of the loop:E = /t

t = V1015

Wb104.466

6

(1)

= 3.1 s

Use of speed = distance time = 20 mm 3.1 s (1)= 6.5 mm s–1 (1) 3

Slow down nearer to the magnet (1) 1[Total 7 marks]

190. The following statements apply to a body orbiting a planet at constant speed and at constant height. Indicate whether each statement is true () or false (x).

Statement True/False

The body is travelling at constant velocity. x

The body is in equilibrium because the centripetal force is equal and opposite to the weight.

x

The only force acting on the body is its weight.

The body’s acceleration towards the planet equals the gravitational field strength at the position of the body.

(Total 4 marks)

191. State Lenz's law of electromagnetic inductionDirection of induced emf is such as to oppose the charge producing it (2)

(2 marks)

An exhibit at a science centre consists of three apparently identical vertical tubes, T1, T2 and T3, each about 2 m long. With the tubes are three apparently identical small cylinders, one to each tube.

When the cylinders are dropped down the tubes those in ~T, and ~T2 reach the bottom in less than I second, while that in ~T3 takes a few seconds.

Explain why the cylinder in T3 takes longer to reach the bottom of the tube than the cylinder in T1

In T3 magnetic flux cuts copper tube (1)induction occurs (1)current in copper tube (1)creates magnetic field (1)opposite to magnet’s which repels slows magnet

T1 is plastic so no induction/no current forms (1)

(5 marks)

Explain why the cylinder in T2 takes the same time to reach the bottom as the cylinder in T1

In T2 falling cylinder unmagnetised so no flux cut or no induction (1)

Both T1 and T2 have only force of gravity acting on them (1)

(2 marks)[Total 9 marks]

192. A light aluminium washer rests on the end of a solenoid as shown in the diagram.

I

I

A lu m in iu mw ash e r

S o len o id

A large direct current is switched on in the solenoid. Explain why the washer jumps and immediately falls back.

B field produced by solenoid (1)

Flux lines CUT washer (1)

Induced current/e.m.f. in washer (1)

B field of solenoid opposite to B field washer (1)

Repulsive force lifts washer (1)

Steady current so no changing of flux/no induction (1)

OR explain by force on current carrying conductor in B field (LH rule)

[Total 5 marks]

193. A satellite orbits the Earth once every 120 minutes. Calculate the satellite’s angular speed.Correct substitution into angle/time (1)

Answer with correct unit (1)

r.p.m. etc. not allowed

Angular speed = e.g. 0.052 rad min-1 180°h-1

(2 marks)

Draw a free-body force diagram for the satellite.

(1 mark)(If the Earth is shown, then the direction must be correct)

The satellite is in a state of free fall. What is meant by the term free fall? How can the height of the satellite stay constant if the satellite is in free fall?

Free fall – when gravitational force is the only force acting on an object (1)

Height – (1) for each clear and relevant physics statement (1) + (1)

(3 marks)[Total 6 marks]

194. Define capacitance

Capacitance = Charge / Potential difference.

(2 marks)

An uncharged capacitor of 200 F is connected in series with a 470 k resistor, a 1.50 V cell and a switch. Draw a circuit diagram of this arrangement.

470 k

200 F 1.50 V

(1 mark)

Calculate the maximum current that flows.Current = 1.5 V/470 k

Current = 3.2 A

(2 marks)

Sketch a graph of voltage against charge for your capacitor as it charges. Indicate on the graph the energy stored when the capacitor is fully charged.

Shaded areaequalsenergy stored

V

Q(4 marks)

Calculate the energy stored in the fully-charged capacitor.½CV2 = ½ (200 F) (1.5 V)2

Energy = 2.25 J

(2 marks)[Total 11 marks]