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22

1–4 Particles and antiparticles

Grade booster1 (a) (i) electron

(ii) neutron

(b) (i) Examiner’s comment: The candidate has correctly calculated both numbers. Alpha particle has nucleon number 4 and proton number 2, so 229 = A + 4 and 90 = B + 2.

Mark = 2 out of 2

(ii) Ratio = mass of radium nucleus ____________ mass of a particle = 225 × 1.67 × 10 −27 ________________ 4 × 1.67 × 10 −27

= 56.25

(c) 0.511 MeV × 2 = 1.022 × 106 × 1.6 × 10−19

= 1.6352 × 10−13 J

Each photon has energy = 1.6352 × 10−13 ÷ 2

= 8.2 × 10−14 J (to 2 s.f.)

Exam-style questions

1 a Specific charge = 6 × 1.6 × 10 −19 _______________ 14 × 1.67 × 10 −27 (1)

= 4.1 × 107 (1)

C kg−1 (1)

b 4.8 × 107 C kg−1 = 6 × 1.6 × 10 −19 ______________ A × 1.67 × 10 −27 (1)

A = 6 × 1.6 × 10 −19 ____________ 4.8 × 10 7 × 1.67 × 10 −27

= 12 (1)

No of neutrons = 12 − 6 = 6 (1)

2 a They are annihilated / converted to γ rays / converted to a pair of photons. (1)

b The energy converted from rest mass is added. (1)

c Rest mass of proton = 938.257 MeV

Rest energy for proton and antiproton = 2 × 938.257 × 106 × 1.6 × 10−19

= 3.0024 × 10−10 J (1)

Total energy for both photons = 3.2 × 10−10 + 3.0024 × 10−10

= 6.2024 × 10−10 J (1)

Energy of each photon = 3.1012 × 10−10 J (3.1 × 10−10 J to 2 s.f.) (1)

5–8 Particle classification

Grade booster1 (a) proton number 7, nucleon number 14

(b) antineutrino

(c) baryons: proton, neutron; leptons: electron, electron antineutrino

Examiner’s comment: The student has incorrectly included electrons in the group of baryons and so has lost both marks as both groups are incorrect.

Mark = 0 out of 2

2 (a) mesons

(b) 0

(c) K−

(d) negative (e) (i) Classification K+ vµ µ+

Lepton

Charged particle

Hadron

Meson

(ii) Neutral pion, as kaons decay into pions and charge must be the same on both sides of equation.

Exam-style questions1 a hadrons contain quarks / hadrons feel the strong force (1)

b neutron = udd; neutral pion = uu− / dd− (2)

c proton (1)

d Using the Data Booklet values for the quarks uus

i charge = 2 _ 3 e + 2 _ 3 e − 1 _ 3 e = +1e (1)

ii baryon number = 1 _ 3 + 1 _ 3 + 1 _ 3 = 1 (1)

iii strangeness = 0 + 0 −1 = −1 (1)

e 13 25 A l → 12

25 M g + 1 0 β + + v e (2)

9–12 Fundamental interactions

Grade booster1 (a) Neutron is udd. (1)

+ 2 _ 3 e − 1 _ 3 e − 1 _ 3 e = 0 (2)

(b) beta-plus decay (1)

(c) Examiner’s comment: The answer is incorrect; β+ decay occurs through the weak interaction.

Mark = 0 out of 1

(d) p → n + β+ + νe (1)

Baryon number: 1 → 1 + 0 + 0 (1)

Lepton number: 0 → 0 −1 +1 (1)

Charge: +1 → 0 +1 + 0 (1) (e)

pW+

β

υe

n

Exam-style questions1 a 0

1 n → 1 1 p + − 1

0 β− + v e (1)

b Q: 0 → 1 −1 + 0 (1)

B: 1 → 1 + 0 + 0 (1)

L: 0 → 0 +1 −1 (1)

All are conserved, therefore decay is possible.

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33

c

nW+

p

v

e–

(4)

2 Q: 0 + 1 → 1 + 0 (1)

L: 0 −1 → 0 −1 (1)

B: 1 + 0 → 1 + 0 (1)

All conservation laws are obeyed so the reaction is possible.

3 a i us− (1)

ii K+ → π+ + π° (1)

Q: 1 → 1 + 0 (1)

B: 0 → 0 + 0 (1)

S: 1 → 0 + 0 (1)

Yes, the decay is possible, as charge and baryon number are conserved, and strangeness is allowed to vary by +1 in the weak interaction. (1)

b The weak interaction − (1)

strangeness is not conserved / it is varying by +1 (1)

4 a 3 quarks and 1 strange quark (2)

b i B: 1 → 0 + 1 (1)

Q: 1 → 1 + 0 (1)

S: −1 → 0 + 0 (1)

Possible via weak interaction as strangeness can vary by −1 (1)

ii Proton as it is the only stable baryon. (1)

5 Interaction Group of particles affected

Exchange particle

Strong hadrons pions [and gluons]

(1)

Weak hadrons and leptons

W + and W− bosons

(1)

Electromagnetic charged particles photons (1)

13–16 Wave–particle duality

Grade booster1 (a) (i)

(ii) Light travels as photons with energy hf. Above the threshold frequency, a photon has enough energy to overcome the work function, that is, to liberate an electron.

(b) (i) E = hf = hc _ λ 

E = 6.63 × 10 −34 J s × 3 × 10 8 _____________________ 560 × 10 −9

= 3.55 × 10−19 = 3.6 × 10−19 J (to 2 s.f.)

(ii) hf = Ek + φ 3.55 × 10−19 = Ek + 1.40 × 10−19

Ek = (3.55 − 1.40) × 10−19 = 2.15 × 10−19

= 2.2 × 10−19 J (to 2 s.f.)

(iii) Ek = 1 _ 2 mv2

2.15 × 10−19 J = 1 _ 2 × 9.11 × 10−31 kg × v2

v2 = 2 × 2.15 × 10 −19 ______________ 9.11 × 10 −31 = 4.72 × 1011

v = 6.87 × 105 = 6.9 × 105 m s−1 (to 2 s.f.)

(iv) λ = h _ mv

= 6.63 × 10 −34 ____________ 9.11 × 10 −31 × 6.87 × 10 5

= 1.06 × 10−9 = 1.1 × 10−9 m (to 2 s.f.)

(c) New evidence must be provided that is repeatable and must have has been checked by other scientists in the same field.

Examiner’s comment: Good answer.

Mark = 2 out of 2

Exam-style questions1 a no. of photons in one second on 1 mm2

= total energy

________________ energy of 1 photon

= 3 × 10 −10 _ 1.13 × 10 −18 (2)

= 2.66 × 108 = 2.7 × 108 (to 2 s.f.) (2)

b i minimum energy = hfo = φ = 6.63 × 10−34 × 5.8 × 1014 = 3.85 × 10−19 J (1)

hf = φ + EK(max), so at f = 2 × f0 = 1.16 × 1015 Hz

(6.63 × 10−34 × 1.16 × 1015) = 3.85 × 10−19 + EK(max)

(1)

EK(max) = (7.69 − 3.85) × 10−19 = 3.84 × 10−19 = 3.8 × 10−19 J (to 2 s.f.) (1)

ii eVs = Ek(max)

Vs = 3.8 × 10−19 / 1.6 × 10−19 = 2.4 V (1)

iii stopping potential increases (1)

since photon energy increases (1)

and so does the maximum kinetic energy (1)

c current doubled = 3.0 µA as twice as many photons incident and releasing twice as many electrons (2)

2 a i λ = h _ mv = 6.63 × 10 −34 ____________ 9.11 × 10 −31 × 5.20 × 10 6 (1)

= 1.4 × 10−10 m (to 2 s.f.) (1)

ii For diffraction, λ needs to be approximately same as crystal spacing (1)

speed is in inverse proportion to λ in de Broglie equation,

v = h _ mλ ~ 7 × 10 −34 ______________ 9 × 10 −31 × 10 −10 ~ 8 × 106 m s−1 (1)

17–20 Electron transitions

Grade booster1 (a) ionisation energy = 13.6 eV

(b) (i) Examiner’s comment: The student has correctly realised that 12.10 eV would excite electrons from the ground state to n = 3. Photons are emitted as the electron returns to the ground state. There are

E

f

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two routes back, either n = 3 to n = 2 and then n = 2 to n = 1, or n = 3 straight to n = 1, and the energy differences have been calculated correctly, but the arrows must point downwards. Arrows downwards from n = 3 to n = 2 (1.90 eV), from n = 3 to n = 1 (12.10 eV) and from n = 2 to n = 1 (10.20 eV).

Mark = 3 out of 4

(ii) Least energy = E1 − E2 = −1.50 − (−3.40) = 1.90 eV

Convert to joules = 1.90 × 1.6 × 10−19 = 3.04 × 10−19 J

E = hf = hc _ λ  3.04 × 10−19 = 6.63 × 10 −34 × 3 × 10 8 ___________________ λ 

λ = 6.54 × 10−7 m = 654 nm (to 3 s.f.)

(c) (i) λ = 486 nm

E = hf = hc _ λ  = 6.63 × 10 −34 × 3 × 10 8 ___________ 486 × 10 −9 = 4.09 × 10−19 J

= 4.09 × 10 −19 _ 1.6 × 10 −19 = 2.56 eV

(ii) n = 4 to n = 2

Exam-style questions1 a i An electron (1)

is given energy (1)

to move to a higher energy level within the atom (1)

ii A photon is emitted when an electron falls (1)

from the higher fixed energy level to a fixed lower level (1)

giving the photon a discrete amount of excess energy, and thus a specific wavelength. (1)

b i E = hf

f = 9.92 × 10 −19 _ 6.63 × 10 −34 (1)

= 1.50 × 1015 Hz

λ = c __ f

= 3 × 10 8 __________ 1.50 × 10 15 (1)

= 2.0 × 10−7 m (to 2 s.f.) (1)

ii energy (in eV) = 9.92 × 10 −19 _ 1.6 × 10 −19

= 6.20 eV (1)

transition from n = 2 to n = 1 (1)

iii transition from n = 4 to n = 1 (1)

2 a i KE = 4.1 × 10 −18 _ 1.6 × 10 −19

= 25.6 eV (26 eV to 2 s.f.) (1)

ii λ = h _ mv

= 6.63 × 10 −34 ____________ 9.11 × 10 −31 × 3.0 × 10 6 (1)

= 2.43 × 10−10 m (2.4 × 10−10 m to 2 s.f.) (1)

b hf = E1 − E2

E1 − E2 = (0.87 − 0.18) × 10−18

f = (0.87 − 0.18 ) × 10 −18

_________________ 6.63 × 10 −34 (1)

= 1.04 × 1015 Hz

λ = c _ f

= 3 × 10 8 _ 1.04 × 10 15 (1)

= 2.88 × 10−7 m (2.9 × 10−7 m to 2 s.f.) (1)

21–24 Progressive and stationary waves

Grade booster1 (a) Speck will move vertically upwards to the maximum

positive (upward) displacement, then downwards through the equilibrium position to the maximum negative displacement and then upwards back to the equilibrium position.

(b) String is 2.5 wavelengths long, so λ = 0.2 m

f = v _ λ = 150 _ 0.2 = 750 Hz

(c) The resultant displacement when two waves are added together as vectors.

(d)

(e) Distance between nodes = λ _ 2 so λ = 1.28 m

Examiner’s comment: The student’s answer is correct for their (wrong) diagram so an ‘error carried forward’ mark is awarded.

Mark = 1 out of 1 with error carried forward

(f) f = 1 _ 2l √ __

T __ µ     

= 1 _ 2 × 0.64 √

_________

70 _ 0.9 × 10 −3

= 1 _ 1.28 √ _______

77777.7

= 218 Hz

Exam-style questions1 a i f ∝ √

_ T means f = k √

_ T

Using values from Table 1

For f = 100 Hz, k = 100 ____________ √

__________ (0.4 × 9.81) = 50.5

For f = 120 Hz, k = 120 ____________ √

__________ (0.6 × 9.81) = 49.5

For f = 140Hz, k = 140 ____________ √

__________ (0.8 × 9.81) = 49.97 (1)

k has the same value of 50 to 2 s.f. so f α   √ __

T (1)

ii if m = 1.1 kg f = 50 × √ __________

(1.1 × 9.81 (1) = 164 Hz (1)

b µ = ‘mass of a 1 metre length =

f = 1 _ ( 2l ) √

_

( T _ µ )

Using values from Table 1:

140 = 1 _ 4 × √

_____________

[ ( 0.8 × 9.81 _ µ ) ] (1)

Squaring both sides

1402 = 0.252 × ( 0.8 × 9.81 ) _ µ

µ = 0.25 2 × 0.8 × 9.81 _______________ 140 2 (1)

= 2.5 × 10−5 kg m−1 (1)


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c Conditions:

– Wave and reflected wave or waves travelling in opposite directions meet or interact or overlap or cross or pass through

– waves have same wavelength / frequency

Features:

– Node = point of minimum or no or zero amplitude / displacement / disturbance

– Antinode = point of maximum amplitude / displacement

How features formed:

– Node is where two waves cancel / destructively interfere or are 180° out of phase / in antiphase. Note: not peak meets trough

– Antinode is where two waves reinforce /constructively interfere or are in phase

Other relevant points:

– Two waves superpose. Note: not superimpose

– Energy is not transferred

Marking:

6 marks: Two conditions for formation of stationary waves + description of oscillation at antinode + three other points from the list

5 marks: Two conditions for formation of stationary waves + three other points from list

4 marks: One condition for formation of stationary waves + three other points from list

3 marks: Any three points from list

2 marks: Any two points from list

1 mark: Any one point or reference to nodes and antinodes

d Points between two adjacent (1) nodes are in phase. (1)

25–28 Interference and diffraction

Grade booster1 (a) Light from two slits superposes. Constructive

interference produces bright fringes and destructive interference produces dark fringes.

(b) They are illuminated by the same source so the light passing through them has the same frequency. They have the same path difference as the distance from the laser to S1 is the same as that from the laser to S2, so there is a constant phase relationship.

(c) Fringe spacing = 3.6 _ 4 = 0.90 mm

w = λD _ s

λ = ws _ D = 0.90 × 10 −3 × 0.56 × 10 −3 _____________ 0.80

= 6.30 × 10−7 m (6.3 × 10−7 to 2 s.f.)

(d) (i) Examiner’s comment: The graph correctly shows maxima of similar intensity to the central maximum and all fringes the same width as the central fringe.

Mark = 2 out of 2

(ii) Examiner’s comment: The student has not labelled the graph. A maximum should be labelled red and a minimum dark.

Mark = 0 out of 1

(e) The bright fringes will be closer together (1) since D and s do not change, only λ does. w = λD / s, so If λ decreases so will w. (1)

Exam-style questions1 a The path difference between the two waves (1) gives

rise to a phase difference between them (1) causing destructive interference. (1)

b Path difference = 0.710 − 0.656 = 0.054 m (1)

0.054 = nλ for maxima and this is the third maximum, so n = 2

0.054 = 2λ λ = 0.027 m (1)

f = c _ λ = 3 × 10 8 _ 0.027 = 1.11 × 10 10 Hz

(1.1 × 1010 Hz to 2 s.f.) (1)

c Intensity of the signal decreases with distance (1) and as one wave from slit S2 travels further than the other from S1 (1) the intensities of the two waves meeting at the minimum points between A and B are not equal so do not totally cancel out. (1)

d The signal decreases (1) and becomes zero at 90° (1) because the waves are polarised. (1)

2 a A diffraction grating is a piece of glass onto which many thousands of very thin parallel and equally spaced lines have been engraved. (1)

Each of these ‘slits’ causes diffraction so that what you observe on the screen is the combined effect of all of them. (1)

b i maxima more widely spaced (1)

ii maxima closer together (1)

c Central / zero-order bright white maximum. (1)

Other orders will be seen as spectra with blue on the inside and red on outside. (1)

Darkness in-between the orders. (1)

29–32 Refraction of light

Grade booster1 (a) nair sin θ1 = ncore sin θ2

(b) n = speed of light in air

________________ speed of light in core

1.55 = 3.00 × 108

_________ ccore

ccore = 3.00 × 108

_________ 1.55 = 1.94 × 108 m s−1

(c) Examiner’s comment: The student has just copied the answer and forgotten to use geometry.

Mark = 0 out of 1

Sum of angles in a triangle = 180° so 180° − 90° − 19.52° = 70.48°

(d) Total internal reflection occurs above critical angle θc

ncore sin θc = ncladding sin 90

1.55 sin θc = 1.45 sin 90 (1)

sin θc = 11.45__55_

= 0.9355

θc = 69.3° (1)


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Yes, the diagram is correct as the angle x is 70°, which is bigger than the critical angle for the core–cladding interface, and so total internal reflection occurs. (1)

Exam-style questions1 A (1)

Look at the marked angles: they are not angles between the ray and the normal.

n1 sin θe = n2 sin θ2

n1 is refractive index of air, 1.00

1.00 × sin (90° − 54°) = n2 sin (90° − 67°)

n2 = (sin 36°)

______ (sin 23°)

= 1.50

2 D (1)

n = c __ cs

cs = c _ n = 3.00 × 108 ________ 1.33 = 2.26 × 108 m s−1

convert units: 2.26 × 105 km s−1

3 air na = 1.00; ice ni = 1.31; glass ng = 1.50

a na sin θ1 = ni sin θ2

1.00 × sin 60° = 1.31 × sin θ2 (1)

sin θ2 = (sin 60°)

______ 1.31 = 0.6611

θ2 = 41.4° (to 3 s.f.) (1)

b θ3 same as θ2 = 41.4° (1)

c ni sin θ3 = ng sin θ4

1.31 × sin 41.4° = 1.50 × sin θ4 (1)

sin θ4 = 1.31 × 0.6611 _____ 1.50 = 0.5774

θ4 = 35.3° (to 3 s.f.) (1)

d ng sin θc = na sin 90°

1.50 × sin θc = 1.00 × sin 90° (1)

sin θc = 1.00 ____ 1.50 = 0.6667

θc = 41.8° (to 3 s.f.) (1)

e Ray continued into air from the point at which it meets the glass – air boundary and parallel with path of ray entering ice at the top (1)

f Total internal reflection occurs in an optically dense medium at the boundary with a less optically dense medium (1) when angle of incidence > critical angle. (1)

4 a nair sin θair = nA sin θA1

1.00 × sin 20° = 1.48 × sin θA1 (1)

sin θA1 = (sin 20°)

______ 1.48 = 0.2311

θA1 = 13.4° (to 3 s.f.) (1)

b Ray refracted towards normal in prism A (see Figure 5 in red) (1)

c nA sin θc = nB sin 90°

1.48 × sin θc = 1.38 sin 90° (1)

sin θc = 1.38 ____ 1.48 = 0.9324

θc = 68.8° (to 3 s.f.) (1)

d A

B

20°

45°

45°S A1=13.4°θ

as shown in green, angle between boundary and ray = 45° + 13.4° (or other geometric method) (1)

angle of incidence = 58.4° (1)

e nA sin θA2 = nB sin θB (use additional s.f. in value of θ1 to avoid rounding errors)

1.48 × sin 58.36° = 1.38 × sin θB (1)

sin θB = 1.48 × (sin 58.36°)

_________ 1.38 = 0.9131

θB = 65.9° (to 3 s.f.) (1)

f Ray marked away from normal into prism B (see Figure 5 in blue) (1)

33–36 Optical fibres

Grade booster1 (a) The difference in refractive index makes total internal

reflection possible at the core–cladding interface so that the core of the fibre can carry light rays.

(b) Examiner’s comment: The answer will not gain any marks. There is not enough detail and two reasons should be given.

Mark = 0 out of 2

Two from: the cladding prevents scratches to the core / provides strength / prevents crossover of light between fibres.

(c) Modal dispersion occurs when rays take different paths through the core. A ray passing straight down the axis of the core will arrive first at the other end; whereas other rays will arrive at different times depending on how many times they have been totally internally reflected, and thus how long their path was. Therefore, the final pulse is broader than the original and may merge with the next pulse, corrupting the data. A narrow core minimises the possible path differences.

Exam-style questions1 a Modal dispersion: a ray travelling along the axis of the

fibre has the shortest path to the far end; rays that are totally internally reflected take varied and slightly longer paths (1) so arrive later and pulse is broadened. (1)

b Angle of incidence less likely to fall below critical angle so more light stays in core (1); prevents modal dispersion as smaller path difference between axial and non-axial rays. (1)

c Monochromatic light prevents material dispersion (1) as it has only one speed / as in contrast violet light is slower than red, causing emerging pulse to broaden. (1)

2 a Blue light travels slower than red due to the greater refractive index / red reaches end before blue, leading to material pulse broadening. (1)


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Blue: v = c __ n = (3.00 × 108 m s−1)

_____________ 1.47

Time = d __ v = 1500 m × 1.47 ____________ (3.00 × 108 m s−1) = 7.35 × 10−6 s

Red: v = c __ n = (3.00 × 108 m s−1)

____________ 1.46

Time = d __ v = 1500 m × 1.46 _____________ (3.00 × 108 m s−1) = 7.30 × 10−6 s (1)

Time difference = (7.35 − 7.30) × 10−6 = 5.0 × 10−8 s (1)

b Diagram to show broader (1) and lower pulse (1)

c Two from:

– use of monochromatic signal to reduce material dispersion / pulse broadening (1)

– use of cladding with a lower refractive index (alternative: core with a higher refractive index) to decrease the critical angle and thus increase the amount of total internal reflection – this reduces loss of intensity (1)

– use of cladding with a higher refractive index (alternative: core with a lower refractive index) to increase the critical angle so fewer paths with total internal reflection are possible reducing material dispersion (1)

– use of repeaters along fibre to reform signal before too much broadening has taken place. (1)

3 Application: medicine or a particular named application, for example endoscope or arthroscope (1). Evaluation: improved diagnosis leading to better recovery and better quality of life. (1)

Or

Application: communications. (1) Evaluation: faster and more secure transmission of data / faster internet. (1)

37–40 Scalars, vectors and moments


W

R1

R2

F

(ii) sin 30 = x _ 6.0

x = 6.0 sin 30 = 3.0 m

cos 30 = y _ 6.0

y = 6.0 cos 30 = 5.2 m

taking moments about base of ladder, TCM = TAM

W × 1.5 = R1 × 5.2

R1 = 310 × 1.5 _ 5.2 = 89 N (to 2 s.f.)

(iii) horizontal forces are equal

R1= F = 89.42 N

Examiner’s comment: the student has correctly equated the horizontal forces so an error carried forward mark is awarded.

Mark = 1 out of 1 with error carried forward

(b) Upward forces = downward forces

R2 = 700 + 310 = 1010 N

Exam-style questions1 a

820 N 820 N

W

horizontal component = 820 × cos 44 = 589.85 N = 590 N (to 2 s.f.) (1)

vertical component = 820 × sin 44 = 569.6 N = 570 N (to 2 s.f.) (1)

b weight of girder = 2 × 569.6 = 1139 N = 1100 N (to 2 s.f.) (1)

2 D (1)

3 A Moment of couple = F × d = 23 × 0.38 = 8.74 N m (1)

4 a Taking moments about elbow joint

TCM = TAM

(25 × 0.16) + (40 × 0.38) = FB × 0.04 (1)

19.2 = FB × 0.04

FB = 19.2 _ 0.04 = 480 N (1)

b upwards forces = downwards forces

480 = 25 + 40 + FE (1)

FE = 480 − 65 = 415 N (1)

41–44 Motion graphs and equations

Grade booster1 (a) v = u + at

0 = 5.2 −9.81 × t

t = 5.2 _ 9.81 = 0.53 s (to 2 s.f.)

(b) s = ut + 1 __ 2    at2 = 5.2 × 0.53 − 1 __ 2    × 9.81 × 0.532 = 1.38 m

= 1.4 m (to 2 s.f.)

Examiner’s comment: the student has not used negative g for upward motion. Another method would have been the use of v2 = u2 + 2as

Mark = 0 out of 2

(c) s = ut + 1 __ 2    at2

1.38 = 0 + 1 __ 2    × 9.81 × t2

t = √

___________

[ ( 1.38 × 2 ) _ 9.81 ] = 0.530 s

Total time = 1.1 s

Exam-style questions1 a Both gradients represent acceleration due to gravity ‘g’

so are equal. (1)

b height of the first bounce (1)

c changes direction now going upwards (1)

d dissipation of thermal energy due to friction at surface during bounce (1)


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2 a Displacement = 0 m (1)

b total distance = (30 s × 2.0 m s−1) × 2, total time = 50 s

average speed = 120 _ 50 = 2.4 m s−1 (1)

c v2 = u2 + 2as

9.42 = 0 + 2a × 38 (1)

a = 9.4 2 _ 76 = 1.16 m s−2 = 1.2 m s−2 (to 2 s.f.) (1)

3 A (1)

4 v2 = u2 + 2as

02 = 272 − 2 × 5 s (1)

s = 27 2 _ 10 = 72.9 m = 73 m (to 2 s.f.) (1)

45–48 Projectiles

Grade booster1 (a) Horizontal displacement is 2 m in each equal

time interval. This means there is no acceleration in the horizontal direction, the horizontal velocity must be constant, so no air resistance or another force is acting.

(b) The graph shows vertical displacement is increasing in each successive time interval. The ball is accelerating due to gravity speed increasing.

(c) (i) Examiner’s comment: Air resistance affects motion both horizontally and downwards, so the answer should be that horizontal displacement will be less.

Mark = 0 out of 1

(ii) Vertical displacement would be less and the effect would be more noticeable as time went on, as air resistance increases with speed.

Exam-style questions1 P and Q fall vertically at same rate

Horizontally, for P after 1 s, s = ut = 1 × 0.1 = 0.1 m (1)

distance apart = 0.3 − 0.1 = 0.2 m (1)

2 a i Horizontally, sh = 25 m, t = 1.4 s

vh = s h _ t

= 25 _ 1.4 (1)

= 17.86 = 18 m s−1 (1)

ii zero (1)

iii vertically, t = 1.4 s

v = u + at

0 = uv − 9.81 × 1.4 (1)

uv = 13.73 m s−1 (1)

b using Pythagoras’ theorem

v2 = vh2 + vv

2

= 17.862 + 13.732 (1)

= 507.5

v = 22.53 ms−1 = 23 ms−1 (to 2 s.f.) (1)

tan θ = v v _ v h

= 13.73 _ 17.86 (1)

= 37.6° (1)

Initial ball velocity = 22.5 m s−1 at an angle of 38° to the horizontal.

c Maximum KE occurs as ball is kicked as KE = 1 __ 2 mv2

ball has greatest speed as it is kicked and no GPE (2)

49–52 Forces and Newton’s laws

Grade booster1 (a)

Reactionforce

Driving force

Drag

Weight

(b) (i) a = ( v − u ) _ t

= ( 28.2 − 0 ) _ 4.5 = 6.267 = 6.3 m s−1 (to 2 s.f.)

Examiner’s comment: The student copied the time incorrectly as 4.0 s, so has no marks for the answer, but the number of significant figures is correct so one mark is awarded.

Mark = 1 out of 3

(ii) F = ma

= (360 + 75) × 6.267

= 2726 = 2700 N (to 2 s.f.)

(c) The forward force must be increased as the car accelerates, as air resistance increases with speed. The forward force must always be greater than the resistive force in order that the resultant force stays the same to give constant acceleration.

(d) Component along plane = mg sin θ = 435 × 9.81 × sin 30°

= 435 × 9.81 × 0.5 = 2134 N = 2130 N (to 3 s.f.)

Exam-style questions1 a Points to make:

– acceleration = gradient of velocity–time graph

– acceleration / gradient greatest initially then decreases

– terminal / constant / maximum velocity shown by gradient / graph flattening to become horizontal

– as speed increases air resistance increases

– so resistive force increases while driving force stays the same, causing decreasing acceleration as resultant force decreases (Newton’s second law)

– until resistive force = driving force, therefore no acceleration and lorry has reached its maximum speed. No resultant force, Newton’s first law.

Marking:

6 marks: gradient change + meaning in terms of acceleration at each section + forces acting in each section + appropriate reference to Newton’s laws

5 marks: gradient change + meaning in terms of acceleration at each section + forces acting in each section or appropriate reference to Newton’s laws

4 marks: gradient change + meaning in terms of acceleration at each section + forces acting in at least one section or appropriate reference to Newton’s laws

3 mark: gradient change + meaning in terms of acceleration at least one section + forces acting at least one section or appropriate reference to Newton’s laws


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2 marks: 2 relevant points

1 mark: 1 relevant point

b Distance travelled = area under graph (1)

as graph is curved have to count number of squares and × area of 1 square (1)

2

F

2100 kg800 kg

T T

a i a = ?, u = 0 m s−1, v = 8.0 m s−1, t = 60 s

v = u + at

8.0 = 0 + 60a (1)

a = 0.133 m s−2 (1)

considering caravan

F = ma

Tension T = 800 × 0.133

= 106.4 N = 110 N (to 2 s.f.) (1)

ii considering car + caravan

engine thrust F = ma

F = 2900 × 0.133 (1)

= 386.7 = 390 N (to 2 s.f.) (1)

b u = 25 m s−1, v = 0, s = 50 m, a = ?

v2 = u2 + 2as

02 = 252 − 2a × 50 (1)

a = 25 2 _ 100 = 6.25 m s−2 (1)

F = ma

= 2100 × 6.25 = 13 125 N = 13 100 N (to 3 s.f.) (1)

c u =10 m s−1, v = 0, s = 0.3 m, m = 21 00 kg, a = ? F = ?

v2 = u2 + 2as

02 = 102 − 2 × a × 0.3 (1)

a = 100/0.6 = 166.67 m s−2

F = ma

= 2100 × 166.67 (1)

= 35 007 = 3.5 × 105 N (to 2 s.f.) (1)

Alternative answer using equations from pages 57–60

KE lost = work done against friction

1 __ 2 mv2 = F × d

F = ½ mv 2 _ d = 0.5 × 2100 × 10 2 ______________ 0.3 = 3.5 × 105 N

3 D (1)

53–56 Momentum

Grade booster1 (a) Total momentum before impact = total momentum

after impact provided no external forces are acting

(b) m1u1 + m2u2 = (m1 + m2) v

(8000 × 2.5) + (12 000 × 0) = 20 000 v

v = 20 000 _ 20 000 = 1.0 m s−1

(c) Examiner’s comment: The numerical answer is correct, but the units are incorrect (should be kg m s−1)

Change in momentum = 12 000 kg m s−1

Mark = 1 out of 2

(d) Kinetic energy after the collision

= 1 __ 2 × (8000 + 12 000) × 1.02 = 10 000 J

Collision is inelastic as there is a loss in kinetic energy of 15 000 J.


1 B F = m(u − v)

________ t

= 0.16 ( 35 − ( − 30 ) ) ___________ ( 52 × 10 −3 )

= 0.16 × 65 _ 52 × 10 −3

= 200 N (1)

2 Converting eV to J:

E = 6.1 × 106 × 1.6 × 10−19 = 9.76 × 10−13 J (1)

mα = 4.0u = 4.0 × 1.66 × 10−27 = 6.64 × 10−27 kg

mTl = 208u (1)

E = 1 __ 2 mv2

9.76 × 10−13 = 1 __ 2 × 6.64 × 10−27 × v2

v2 = 9.76 × 10 −13 _ 3.32 × 10 −27

v = 1.7 × 107m s−1 (1)

By law of conservation of momentum:

mTl vrecoil = mαv (1)

208u × vrecoil = 4v × 1.7 × 107

vrecoil = 4u × 1.7 × 10 7 ____________ 208u

= 3.3 × 105 m s−1 (1)

3 a Momentum:

– momentum is conserved because there are no external forces acting on probe + capsule as it’s in free space

– in free space large masses are so remote that gravitational forces are negligible

– during the explosion, there are equal and opposite forces, acting within the system, between the probe and the capsule

– momentum is a vector and is conserved so the capsule must continue to move along the same line of movement

Energy:

– total energy is always conserved in any physical process because energy can be neither created nor destroyed

– energy may be converted from one form to another

– the probe is already moving and has kinetic energy

– during the explosion, some chemical energy is converted into kinetic energy and some thermal energy is dissipated in the surroundings

– after the explosion the probe and capsule have more kinetic energy than just the probe had originally, because some kinetic energy has been released by the explosion

6 marks momentum conserved + detailed reasons why conditions satisfied + energy conserved during explosion + transfers including initial KE as probe moving.


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5 marks momentum conserved + reasons why conditions satisfied + energy conserved during explosion + transfers but may not include consideration that probe already moving.

4 marks momentum conserved + limited reasons why conditions satisfied and energy conserved during explosion.

3 marks momentum conserved + limited reasons why conditions satisfied or energy conserved during explosion + transfers.

2 marks momentum and energy conserved with limited explanation.

1 mark momentum or energy conserved with little or no explanation.

b momentum before explosion = momentum after explosion

(600 × 150) = − (180 × v) + (420 × 220) (1)

90 000 = − 180v + 92 400

180v = 92 400 − 90 000

v = 2400 _ 180 = 13.3 m s−1 backwards (1)

KE before explosion = 1 __ 2 × 600 × 1502 = 6 750 000 J (1)

KE after explosion = 1 __ 2 × 420 × 2202 + 1 __ 2 × 180 × 13.32

= 10 179 920 J (1)

Energy released by explosion = 10 179 920 J − 6750 000 J

= 3 429 920 J = 3.4 × 106 J (1)

57–60 Work, power and energy

Grade booster1 (a) mass = ρV = 1.2 × 3.5 × 105

= 420 000 = 4 × 105 kg (to 1 s.f.)

(b) Ek = 1 __ 2 mv2 = 1 __ 2 × 420 000 × 112 = 2.54 × 107 J

Examiner’s comment: the student has been able to access the marks here by using the answer quoted in part (a). The mark scheme will give a range of answers (2.42 × 107 − 2.54 × 107) to accommodate this possibility.

Mark = 2 out of 2

(c) Efficiency = output power

____________ input power × 100 %

= 10 × 10 6 _ 2.42 × 10 7 × 100

= 41.3%

(d) (i) P = KE ___ t = 1 __ 2 mv2

________________ t = 1 __ 2 mv2 × ρ × A × L × v2

________________ t

= 1 _ 2 A ρ v3

so the hypothesis seems to be correct. (1) However, looking at the graph, it appears that if you double the velocity from 5 to 10 m s−1 you don’t get 23 = 8 times as much power. It is more like 5 × power. (1) Wind turbines are only about 40% efficient. The wind engineer’s equation is the input to the turbine. The graph is the output power. (1)

(ii) Density of air is less at higher levels. (1) Colder air is also denser. (1) A cooler low elevation with wind speeds of about 10 ms−1 would be ideal. (1)

Exam-style questions1 GPE at top = 1.5 × 9.81 × 3.0 = 44.145 J

(vertical height 3.0 m needed here) (1)

KE at base of slope = 1 __ 2 mv2 = 1 __ 2 × 1.5 × 42 = 12 J (1)

Work done against friction = 44.145 − 12 = 32.145 J (1)

Work done = Fs

32.145 = F × 5.0 (length of slope = 5.0 m needed here)

F = 32.145 _ 5 = 6.429 = 6.4 N (to 2 s.f.) (1)

2 C (1)

3 Output power = Fv = weight × v = 10 × 9.81 × 0.5 = 49.05 W (1)

Efficiency = output power

____________ input power × 100 %

= 49.05 _ 90 × 100 (1)

= 54.5% (1)

4 a i GPE = 65 × 9.81 × 4.5 (1)

= 2869.425 = 2900 J (to 2 s.f.) (1)

GPE = KE = 1 __ 2 mv2

2869.425 = 1 __ 2 × 65 × v2 (1)

v = √ ______________

( 2 × 2869.425 ____________ 65 ) = √ _____

88.29 = 9.396

= 9.4 m s−1 (to 2 s.f.) (1)

ii u = 0, v = 9.396, s= 4.5, a = g = 9.81, t =?

v = u + at

9.396 = 0 + 9.81 t (1)

t = 9.396 _____ 9.81 = 0.958 = 0.96 s (to 2 s.f). (1)

b GPE = KE so mgh = 1 __ 2 mv2 (1)

m cancels, so v depends only on height. (2)

61–64 Properties of materials

Grade booster1 (a) (i) Load is directly proportional to extension so the

spring is obeying Hooke’s law.

(ii) Load is no longer proportional to extension. The spring is now deforming plastically.

(b) k = gradient of graph = 2.5 _ 0.002 = 1250 N m−1

Examiner’s comment: The student should have used the straight-line section only and converted to m. The units are N m−1 (not nM, N m or nM−1). Capital and small letters must be correct in units.

Mark = 0 out of 3

(c) Work done = area under loading graph = area of triangle + area under curve

= ( 1 __ 2 × 0.002 × 2.5)

+ (number of squares × area of 1 square)

= ( 1 __ 2 × 0.002 × 2.5) + (11 × 0.5 × 0.0005)

= (2.5 + 2.75) × 10−3 = 5.25 × 10−3 J

(d) Point R shows the spring has a permanent extension as the unloading line does not return to 0.

Extension = OR = 0.8 mm


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Exam-style questions1 D (1)

2 a i 30 − 19 = 11 (m) (1)

ii F = kΔL − mg

= (380 × 11) − (75 × 9.81) (1)

= 3444 N (1)

b forces balanced

mg = kΔL

75 × 9.81 = 380 × ΔL

ΔL = 75 × 9.81 _ 380 (1)

= 1.936 = 1.9 m (to 2 s.f.) (1)

c loss in GPE = gain in elastic strain energy + gain in KE

mg Δh = 1 __ 2 FΔL + 1 __ 2 mv2 (1)

75 × 9.81 × 25 = 1 __ 2 × 380 × 52 + 1 __ 2 × 75 × v2 (1)

18 394 = 4750 + 37.5 v2

v2 = 18 394 − 4750 ____________ 37.5 (1)

= 363.84

v = 19.1 m s−1 (1)

d At the point when the rope begins to stretch, loss in GPE = gain in KE (1)

If k is greater, F _ ΔL is greater so stops in a shorter distance (1)

Increases force on jumper and increasing the risk of injury (1)

e Force Loading

Unloading

Extension (2)

The graph shows that the change in length for loading > change in length for unloading (1)

Plastic deformation, as the polymer does not return to its original length. (1)

3 A ΔL = F _ k = 2 _ 16 = 0.125 m

Energy = 1 __ 2 FΔL = 1 __ 2 × 2 × 0.125 = 0.125 J (1)

65–68 The Young modulus

Grade booster1 (a) Measurement: Diameter using micrometer, length using

metre rule, extension using Vernier callipers or pointer to show original length plus mm ruler

Calculations: Area = π ( d __ 2 ) 2, YM = FL _ AΔL , graph of F

vs ΔL

Accuracy: At least 8 sets of values of e and ΔL, repeats loading and unloading, micrometer × 5 and take mean, Vernier callipers or second parallel wire with Vernier scale

Safety: Care with toes (not drop masses) and eyes (goggles in case wire snaps)

Level 5–6: relevant information, using specialist vocabulary, written coherently and legibly, with

spelling, grammar and punctuation making meaning clear.

6 marks: 2 measurement + all calculation + 3 × accuracy + 1 safety

5 marks: 2 measurement + all calculation+ 2 × accuracy + 1 safety

Level 3–4: relevant information, less specialist vocabulary written legibly with spelling, punctuation, grammar and style allowing meaning to be understood.

4 marks: 2 measurement + 2 calculation (not graph) + 1 accuracy + 1 safety

3 marks: 1 measurement + Young modulus equation + 1 accuracy + 1 safety

Level 1–2: some relevant information but little specialist vocabulary, with legibility, spelling punctuation and grammar allowing some meaning to be conveyed.

2 marks: 3 relevant points

1 mark: 1 or 2 relevant points

Examiner’s comment: Although the student has mentioned measurements, calculations, accuracy and safety, too many important details have been missed out. To access 5–6 marks the student needs to mention plotting a graph, all instruments used and include all steps in calculations. The student must also write coherently and use specialist vocabulary wherever possible.

3 marks are awarded: 1 for measurement + Young modulus equation, 1 for accuracy and 1 for safety

(b) A smaller diameter wire produces larger extensions. (1) Reducing the percentage uncertainty in extension and in result for Young modulus. (1) A smaller diameter means cross sectional area is smaller. (1) Increases percentage uncertainty in cross sectional area and in result for Young Modulus. (1)

Exam-style questions1 a i For steel, ΔL = FL ________ YM × A

= 120 × 1.5 ___________ 2.0 × 10 11 × 2.4 × 10 −7 (1)

= 180 _ 4.8 × 10 4

= 37.5 × 10−4 m = 3.75 mm (1)

ii Extension for brass = 7.50 mm (2 × that of steel as

Young modulus of brass = 1 __ 2 Young modulus of steel) (1)

b No longer horizontal; the LH end will be lower than the RH end. (1)

2 B kg m−2 s−2 (1)

69–72 Electricity investigations

Grade booster1 (a) Resistivity ρ = RA _ L (where R = resistance, A =

cross-sectional area and L = length) Units are Ωm

(b) (i) Measure the diameter of the wire using a micrometer at five different places and find the mean value.


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(ii) R = ρL

___ A

y = mx + c (1)

gradient = ρ __ A

        ρ = gradient × A (1)

(c) R = V _ I = 0.3 _ 0.6 = 0.5 Ω

ρ = RA _ L

= 0.5 × 0.5 × 10 −7 _____________ 1.5

= 1.67 × 10−8 Ωm

= 1.7 × 10−8 Ωm (to 2 s.f.)

(d) Examiner’s comment: The student has not noticed that the 3.5 line is just visible on the barrel and has read the scales as 3.0 + 0.21 = 3.21 instead of 3.5 + 0.21 = 3.71. The units should be mm.

Mark = 0 out of 2.

(e) A superconductor has zero resistivity at or below the critical temperature.

Exam-style questions1 B ρ = RA _ L = VA _ IL (1)

2 a 0.01 mm (1)

b D (1)

3 a Current / A

Reverse bias

Forward bias

Voltage / V

Thresholdvoltage = 0.6V

(3)

b There is infinite resistance when reverse biased. (1) When forward biased, resistance begins to drop at approximately 0.6 V and decreases rapidly as shown by almost vertical line of the I–V characteristic. (1)

4 a circuit diagram to show: putty, ammeter, battery, variable resistor and switch in series (1)

voltmeter across the putty (1)

b ρ = RA ___ L

Vol. of cylinder V = AL

A = V __ L

Therefore

ρ = R __ L × V __ L

= RV ___ L2

c i Measurements: (2)

– length measured with a ruler

– thickness / diameter measured with Vernier callipers / micrometer

– voltage and current measured with voltmeter and ammeter.

Errors: (3)

– putty is not rigid so as it is measured may change shape / get thinner and longer.

– check still cylindrical when measuring length

– do not tighten the micrometer / Vernier calliper more than needed when measuring diameter

– check for zero errors on both meters and beware of fluctuations in readings.

ii Effect on ρ (2)

Using derived equation from b:

– volume remains same even when squashed as same amount of putty

– if length is larger ρ will be smaller

d ρ = RA ___ L

= (R × πr2)

________ L

= (20 × π × (10 ×

10−3)2

__________________ 60 × 10−3

= 0.1047 = 0.10 Ω m (3)

73–76 Electrical circuit analysis

Grade booster1 (a) lamp + resistor R = R1 + R2 = 8 + 4 = 12 Ω parallel

1 _ R T = 1 _ R 1 + 1 _ R 2

= 1 _ 12 + 1 _ 6 = 3 _ 12

RT = 4 Ω R for whole circuit = 6 + 4 = 10 Ω (b) The current supplied by the battery

I = V _ R = 20 _ 10 = 2 A (1)

pd across AB = 20 − (2 × 6) = 8 V (1)

I in 6 Ω resistor = V _ R = 8 __ 6 = 1.3 A (1)

I in lamp = 2 − 1.3 = 0.7 A (1)

(c) (i) Power = I 2R = 0.72 × 8 = 3.92 W

(ii) Mark = 2 out of 2 with error carried forward

Student has correctly calculated the power from the battery and been awarded error carried forward from (c) (i) for the power of the lamp in the % calculation. Correct answer is 10%.

Power from battery = IV = 2 × 20 = 40 W

% Power dissipated = 3.92 _ 40 × 100 % = 10 %

(d) The total resistance of the whole circuit will be less, so that the current in the whole circuit will be more. This means there will be less voltage across AB and as there are now 3 branches, the current through the lamp branch will be less, so the brightness of the lamp will be less.


1 a i I = V _ R

total R = 60 + 5 + 25 = 90 kΩ (1)

I = 12.0 _ 90 000 = 1.33 × 10−4 A (1)

ii V = IR = 1.33 × 10−4 × 5000 (1) = 0.67 V (1)


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b Resistance of LDR decreases. (1)

Voltmeter reading increases because greater

proportion

__________ share of the voltage across R. (1)

c I = 0.8 _ 5000 = 1.6 × 10−4 (A) (1)

Resistance LDR = R so pd across LDR = 0.75 V as well

pd across variable resistor = 12.0 − 0.75 − 0.75 = 10.5 (V) (1)

R = 10.5 ________ 1.6× 10−4 = 65 625 Ω = 66 000 Ω (1) (to 2 s.f.)

77–80 EMF and internal resistance

Grade booster1 (a) Across r, V = I × r

= 4.8 × 1 = 4.8 V (1)

(b) pd = emf − lost volts = 12 − 4.8 = 7.2 V (1)

I = V __ R = 7.2 _ 3.0 = 2.4 A (1)

(c) I = 4.8 − 2.4 = 2.4 A (1)

R = 7.2 _ 2.4 = 3.0 Ω (1)

(d) Resistors in parallel

1 _ R T = 1 _ R 1 + 1 _ s 2

= 1 _ 3 + 1 _ 3 = 2 _ 3 (1)

RT = 3 _ 2 = 1.5 Ω R of whole circuit = 1.0 + 1.5 = 2.5 Ω (1)

Examiner’s comment: Candidate did not use the value given for R so error carried forward cannot be awarded.

Mark = 0 out of 2

(e) (i) The battery converts chemical energy into electrical energy, which is then dissipated in the internal resistance (1) and through the two external resistors. (1)

(ii) In the internal resistance P = IV = 4.8 × 4.8 = 23.04 W (1)

In the 3.0 Ω resistor P = IV = 2.4 × 7.2 = 17.28 W (1)

(f) Electrical energy from cell / second = IV

= 12 × 4.8 = 57.6 W (1)

Energy dissipated in resistors / second

= 23.04 + 17.28 + 17.28 = 57.6 W (1)

Energy input per second = energy output per second

Therefore, energy has been conserved. (1)

Exam-style questions1 C

ε = I(R + r)

1.5 = (0.245 × 5.8 )+ (0.245 × r) (1)

1.5 = 1.42 + (0.245r)

r = 0.08 _ 0.245 = 0.327 Ω

2 a i ε = V + Ir

V = Ir + ε (1)

Comparing y = mx + c where m = gradient and c = intercept on y-axis (1)

Intercept = ε = 2.0 V (1)

Gradient = −r so r = 2 _ 3.33 = 0.6 Ω (1)

ii ε = V + Ir

2 = 1.6 + I × 1 (1)

I = 0.4 A

R = V _ I = 1.6 _ 0.4 = 4 Ω (1)

or from graph when V = 1.6 V, I = 0.4 A (1)

R = 1.6 _ 0.4 = 4 Ω (1)

b i E = IVt

= 0.85 × 4 × 60 × 60 (1)

= 12 240 J (1)

ii ε = I(R + r)

4 = I(20 + 1.2) (1)

I = 4 _ 21.2 = 0.189 A (1)

V = 0.189 × 20 = 3.77 V (1)

c Internal resistance limits current flowing (1) hence battery can provide higher current (1)

or energy is wasted in internal resistance of battery (1) so less energy wasted with lower internal resistance (1)

or charges quicker (1) as current higher or less energy wasted (1)

or lower internal resistance means higher terminal pd / voltage (1) as less pd / across internal resistance or mention of lost volts (1)

81–84 Circular motion


R

W

Examiner’s comment: The student has drawn arrow correctly but should have labelled upward arrow normal reaction R and downward arrow weight W.

Mark = 1 out of 2

(ii) mg – R = centripetal force F (1)

Centripetal force F = mv 2 _ r

mg − R = mv 2 _ r (1)

R = mg − mv 2 _ r

= m ( g − v 2 _ r ) (1)

(iii) R = 1500 × 9.81 − 1500 × 11 2 _ 23 (1)

= 6824 N = 6800 N (to 2 s.f.) (1)

(iv) R = 0 N when loses contact (1)

mg = mv 2 _ r m cancels

v = √ ___

rg (1)

= √ ________

23 × 9.81 = 15.02 m s−1 (1)


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(b) ω = 3 rad s−1 F = 200 N m = 20 kg r = ?

F = mω2r

200 = 20 × 32 × r (1)

r = 200 _ 20 × 9 = 1.1 m (1)

Exam-style questions1 B a = ω 2r (1)

ω = √ _

( a _ r )

T = 2π _ ω 

= 2π  √ _

( r _ a )

10T = 20π √ _

( r _ a )

2 a Vertically R cosθ = mg (1) (1)

Normal reaction R

R sin

R cosθ

Weight W

θ

θ

θ

Horizontally R sinθ = mv 2 _ r (2) (1)

Dividing equation (2) by equation (1)

R sinθ _ R cosθ = mv 2 _ mrg (1)

Cancelling:

tanθ = v 2 _ rg (1)

v2 = rg tan θ

b v = √ _________________

( 35 × 9.81 × tan 25 )

= 13 m s−1 (1)

3 a i T = 24 hrs = 24 × 60 × 60 = 86 400

ω = 2π _ T (1)

= 2π _ 86 400

= 7.3 × 10−5 (1) rad s−1 (1)

ii T = 365 days = 365 × 24 × 60 × 60 = 31 536 000 s

f = 1 _ 31 536 000 = 3.17 × 10−8 Hz (1)

ω = 2πf = 2 × π × 3.17 × 10−8

= 1.99 × 10−7 (1)

v = rω = (1.5 × 1011) × (1.99 × 10−7) = 29.9 × 103 m s−1 = 30 km s−1 (1)

b T = 30 mins = 30 × 60 =1800 s

f = 1 _ 1800 = 0.00055 (1)

ω = 2πf = 2 × π × 0.00055 = 3.49 × 10−3 (1)

a = rω2

= 65 × (3.49 × 10−3)2

= 7.9 × 10−4 ms−2 (1)

c i ω = 2π _ T = 2π _ 97 × 60

= 1.08 × 10−3 = 1.1 × 10−3 (to 2 s.f.) rad s−1 (3)

ii F = mω2r = (1.1 × 104) × (1.08 × 10−3)2 × (6.99 × 106) (1)

= 8.97 × 104 = 9.0 × 104 N (to 2 s.f.) (1)

d The centripetal force for both is the Earth’s gravity (1) but the building is on the surface where g = 9.81 m s−2 (1) whereas the Hubble telescope is above the surface / further out so the centripetal acceleration is less than g. (1)

85–88 SHM equations and graphs


Time

Displacement

Velocity

Acceleration

Time

Time

(2)

(2)

(ii) π _ 2 (1)

(iii) Mark = 0 out of 1

Examiner’s comment: The student has correctly identified the velocity as being π __ 2 out of phase in the previous question but not the acceleration. It should be π radians out of phase or in antiphase as a = − ω2x

(b) (i) a

x

(1)

(ii) The graph has a gradient = −ω2 (1) as a = − ω2x and it is a straight line. It is negative as the displacement x is measured outwards from the fixed point and the acceleration a is inwards towards the fixed point (i.e. in the opposite direction). (1)

(c) (i) 5 mm (1)

(ii) ω = 20π T = 1 _ f = 2π _ ω  = 2π _ 20π = 1 _ 10 = 0.1 s (1)

Time from max to centre = 0.1 _ 4 = 0.025 s (1)

Exam-style questions1 C (1)

2 a 0.11 m (1)

b T = 0.7 s ω = 2πf = 2π _ T = 2π _ 0.7 = 8.98 rad s−1 (1)

Max acceleration amax= ω2A = 8.982 × 0.11 (1) = 8.86 m s−2 (1) (to 3 s.f.)


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c restoring force F = ma = 0.05 × 8.86 (1) = 0.443 N (1)

d vmax = ωA = 8.98 × 0.11 (1) = 0.9878 = 0.99 (1) m s−1

e v = ± ω √ _

( A 2 − x 2 ) = ± 8.98 √ _____________

( 0.11 2 − 0.05 2 ) (1) = 0.88 m s−1 (1) (to 2 s.f.)

89–92 SHM in action

Grade booster1 (a) (i) In SHM the acceleration is always proportional

to the displacement (1) and is in the opposite direction to the displacement. (1)

(ii) Examiner's comment: The student has not made the link with air resistance so has not realised that at the equilibrium position, the bob is moving fastest, and air resistance increases with speed.

Marks: 0 out of 2

Answer: At the centre where the bob moves fastest since air resistance increases with speed.

Exam-style questions1 A

Period of a pendulum is inversely proportional to √ __

g so T decreases when taken to the Moon whereas period of a mass-spring system does not depend on g so does not change. (1)

2 C

Pendulums X and C are the same length so have the same natural frequency so resonance occurs (1)

3 T1 = 2π  √

_

( l _ g )

T2 = 2π  √

_

( l − 600 _ g )

T2 = 0.5T1 (1)

2π  √

_

( l − 600 _ g ) = 0.5 × 2π  √

_

( l _ g )

Cancel π and square both sides:

4 ( l − 600 ) _ g = l _ g

4l −2400 = l

3l = 2400

l = 800 mm (1)

93–96 Thermal energy transfer

Grade booster1 (a) Specific latent heat of vaporisation of water is the

energy required to convert 1 kg of water to steam.

(b) (i) Examiner’s comment: The student has correctly calculated the heat energy given out by the copper, but has forgotten to quote the answer to the correct number of significant figures.

Mark = 2 out of 3

Answer: 1.6 × 104 J

(ii) thermal energy gained by water and copper container = mw cw Δθ + mc cc Δθ

= 0.050 × 4200 × (100 − 84) + 0.020 × 390 × (100 − 84)

= 3485 J

energy for steam = 1.6314 × 104 − 3485 = 1.28 × 104 J

(iii) Q = mL

1.25 × 104 = m × 2.3 × 106

m = 1.25 × 104

_________ 2.3 × 106

= 0.0054 kg

Exam-style questions1 B

Heat taken in by ice cube in melting + heat taken in by melted ice = heat lost by water

(mI × L) + mI × cw (T − 0) = mw × cw × (15 − T)

(0.01 × 3.4 × 105) + 0.01 × 4200 × (T − 0) = 0.1 × 4200 × (15 − T)

3400 + 42T = 6300 − 420T

462T = 6300 − 3400

T = 2900 _ 462

= 6.3 oC (1)

2 a Energy supplied = heat taken in by lead as temp rises + heat taken in by lead in melting

Q = (mL × cL ×   Δθ) + (mL × L)

= 0.75 × 130 × (327.5 ‒ 21) + 0.75 × 23 000 (1)

= 29 884 + 17 250

= 47 134

= 4.7 × 104 J (to 2 s.f.) (1)

b Electrical energy supplied = P × t

t = 47 134 _ 200

= 236 s (1)

97–100 Ideal gases and kinetic theory

Grade booster1 (a) (i) pV

T / ºC0–273

(1)

(1)

(ii) pV = nRT, comparing to y = mx +c, gradient m = nR

(b) (i) Examiner’s comment: The student converted oC to kelvin by adding 273 and has chosen to use pV = nRT as this is the most relevant equation to the values given, along with the value of R from the Data Booklet.

Mark = 2 out of 2

(ii) mass = 130.5 × 0.043 = 5.61 kg

density = mass ____ volume

= 5.61 _ 0.200 = 28.1 kg m−3


1616

Exam Skills Builder

Physics

(iii) V2 = p1V1 ____ T1

= p2 V2 _____ T2

V2 = 1.6 × 10 6 × 0.200 × ( 273 − 50 ) _______________ 3.6 × 10 4 × ( 273 + 22 )

= 6.72 m3

Mass remaining = 5.61 × 0.20 _ 6.72

= 0.167 kg = 0.17 kg (to 2 s.f.)

Or

n = PV _ RT

= 3.6 × 10 4 × 0.20 ________________ ( 8.31 × ( 273 − 50 ) ) = 3.885 mol

Mass remaining = 3.885 × 4.3 × 10−2 = 0.17 kg

(to 2 s.f.)

Exam-style questions1 a p1 = 1.1 × 105 Pa p2 = 1.9 × 105 Pa

V1 = 8.5 × 10−5 m3 V2 = 5.8 × 10−5 m3

T1 = 20 = 293 K T2 = ?

p 1 V 1 _ T 1

= p 2 V 2 _ T 2

T2 = 1.9 × 10 5 × 5.8 × 10 −5 × 293 _____________ 1.1 × 10 5 × 8.5 × 10 −5 (1)

= 345 K (1)

= 345− 273 = 72 °C (1)

b pV = nRT

1.1 × 105 × 8.5 × 10−5 = n × 8.3 × 293 (1)

n = 1.1 × 10 5 × 8.5 × 10 −5 ___________ 8.3 × 293 (1)

= 3.84 × 10−3 (1)

c After compression, particles are closer together (1) and moving faster as the temperature is higher. (1) More collisions between particles and walls of gas syringe (1) larger change of momentum, increasing the force on the walls (1) and hence the pressure in the syringe. (1) (Max 4)

2 a No net flow of (thermal) energy (between two or more bodies) (1)

bodies at same temperature (1)

(kinetic) energy is exchanged in molecular collisions (1)

until average kinetic energy of all molecules is the same (1) (Max 3)

b i 1 _ 2 mcms2 = 3RT _ 2 N A

crms = √ 

_

( 3RT _ mN A ) and mNA = M = molar mass

= √ ___________

( 3 × 8.31 × 300 ) _____________ 4 × 10 −3 (1)

= 1367 m s−1 (1)

Or 1 _ 2 mcms2 = 3kT _ 2

m = M _ N A = 4 × 10 −3 _ 6.02 × 10 23 = 6.645 × 10−27 (1)

crms = √ _

( 3kT _ m )

= √ ____________________

( 3 × 1.38 × 10 −23 × 300 ) ___________________ 6.645 × 10 −27

= 1367 m s−1 (1)

ii Average KE of nitrogen molecules = average KE of helium molecules as temperature equal (1)

Average KE = 3 _ 2 kT

= 3 _ 2 × 1.38 × 10−23 × 300

= 6.21 × 10−21 J (1)

iii pV = NkT so p ∝ no. of molecules (1)

He:N = 2:1

PHe = 2 _ 3 × 125 = 83.33 kPa = 83 kPa (1)

101–104 Gravitational fields

Grade booster1 (a) (i) Gravitational field strength = force per unit mass.

It is a vector quantity.

(ii) Gravitational force on mass m, F = mg

Gravitational force on mass m, F = GMm _ ( R + h ) 2

mg = GMm _ ( R + h ) 2 (1)

g = GM _ ( R + h ) 2 (1)

(b) F = GMm _ ( R + h ) 2

= ( 6.67 × 10 −11 ) × ( 5.97 × 10 24 ) × 2550 _________________ ( 6.37 × 10 6 ) + ( 1.38 × 10 7 ) 2

= ( 6.67 × 10 −11 ) × ( 5.97 × 10 24 ) × 2550 _________________ ( 2.017 × 10 7 ) 2

= 2496 N = 2500 N (to 3 s.f.) (3)

(c) Examiner’s comment: The student has not shown that g is proportional to 1 _ r 2 . As the student has plotted g against r2 the graph should show an inverse proportionality not a straight line. A graph of g against 1 _ r 2 would be a straight-line graph.

Mark = 0 out of 2

Exam-style questions1 A (1)

Equal masses so g is equal for both stars and so g at point X is zero as resultant force is zero.

2 C (1)

F = GMm _ r 2

= ( 6.67 × 10 −11 ) × ( 9.11 × 10 −31 ) × ( 1.67 × 10 −27 ) ______________________ ( 5 × 10 −11 ) 2

(Values from the Data Booklet)

= 4.06 × 10−47 N

3 C (1)

Density = mass _ volume

Mass A = ρ × 4 π r A 3

_ 3

Mass B = ρ × 4 π ( 2 r A )

3 ___________ 2 × 3 =

ρ4 ( 4π r A 3 ) _ 3 = 4 × mass A

gA = GM A

_ r A 2

gB = G 4 M A

_ ( 2 r A )

2 =

G M A _ r A 2 = gA = 13.4 N kg−1


1717

Exam Skills Builder

Physics

4 a i Mass of larger sphere ML = 4 _ 3 πr3ρ

= 4 _ 3 π × (0.100)3 × 11.3 × 103 (1)

= 47.3 kg (1)

ii Gravitational force F = G M L M s _ x 2

= 6.67 × 10 −11 × 47.3 × 0.74 ____________ 0.125 2 (1)

= 1.5 × 10−7 N (1)

iii For the spheres, mass M = 4 _ 3 πr3ρ (1)

Mass of each sphere would be 23 or 8 × greater (378 kg, 5.91 kg) (1)

This would make the force 8 × 8 = 64 × greater (1)

but separation would be doubled causing force to be 22 = 4 × smaller (1)

Net effect would be to make the force ( 64 _ 4 ) = 16 × greater (1)

(that is 2.38 × 10−6 N)

b Me = 81Mm ge = 9.8 = GM e _ r e

2 gm = 1.7 = GM m

_ r m 2

g m

_ g e =

GM m r e 2 _ GM e r m 2 cancel G

1.7 _ 9.8 = M m r e

2 _ 81 M m r m 2 cancel Mm (1)

r e

2 _ r m 2 = 1.7 × 81 _ 9.8

re : rm = 3.7 (1)

105–108 Gravitational potential

Grade booster1 (a) (i) VS = − GM _ R

= − 6.67 × 10 −11 × 5.97 × 10 24 ___________ 6.37 × 10 6 (1)

= − 6.25 × 107 (1) J kg−1 (1)

(ii) 0

–6.25 × 107 Jkg–1 (1)

6.37 × 106 m (1)0

V

r

(1)

(iii) Gravitational potential at infinity is defined as zero. (1)

Work has been done against the field to reach infinity so potential at all points closer than infinity are negative. (1)

Gravitational potential is the work done per unit mass in bringing an object from infinity to the point. (1)

(iv) Examiner’s comment: The answer is incomplete as

g = − ΔV _ Δr . The student should have said ‘the magnitude of the gravitational field strength at any point’.

Mark = 0 out of 1

(b) (i) V2 = − GM ____ R = − 6.67 × 10 −11 × 5.97 × 10 24 ___________ ( 6.37 × 10 6 + 850 × 10 3 )

= − 5.52 × 107 J kg−1 (1)

Work done = m ΔV

= m (V2 − V1)

= 2100 (−5.52 − (−6.25)) × 107 (1)

= 1.5 × 1010 (1) J (1)

(ii) If a satellite is orbiting:

Gravitational force = centripetal force

GMm _ r 2 = m v 2 _ r

mv2 = GMm _ r

KE =    1 __ 2  mv2 = GMm _ 2r

At 850 km KE = 6.67 × 10 −11 × 6 × 10 24 × 2100 _______________ 2 × ( 6.4 × 10 6 ) + ( 850 × 10 3 )

= 5.796 × 1010 J (1)

At 700 km KE = 6.67 × 10 −11 × 6 × 10 24 × 2100 _______________ 2 × ( 6.4 × 10 6 ) + ( 700 × 10 3 )

= 5.918 × 1010 J (1)

Change in KE = (5.918 − 5.796) × 1010 = 1.22 × 109 J (1)

(iii) Speed increases. (1)

(iv) GPE decreases as it is a lower height. (1) At lower altitude the GPE decreases as the KE increases. (1)

Exam-style questions1 D (1)

2 B

V ∝ 1 _ R (1)

3 a −63 = GM e _ R (1)

VM = G 0.1 M e _ 0.5 R =

GMe _ 5R = − 63 _ 5 = −13 MJ kg−1 (1)

b i

Mars

correct shape (1) arrows to show direction (1)

ii

Mars

(1)


1818

Exam Skills Builder

Physics

iii g = ΔV _ Δr = gradient of a V against r graph (1) (1)

rV

c ΔW = mΔV = 6 (−15 − (−30)) = +90 J kg−1 (1)

d PE = mΔ V Vs = − GM e _ R VH = −

GM e _ 2R

PE = m(VH − VS)

= m ( GM e _ 2R ) (equation 1) (1)

But g = GM e _ R 2

GMe = gR2 (equation 2) (1)

Sub from equation 2 into equation 1

PE = m ( g R 2

_ 2R )

= mgR

_ 2 (1)

109–112 Orbits of planets and satellites

Grade booster1 (a) It has the same orbital period as the earth i.e. 24 hours /

86 400 s. The satellite orbits in the plane of the equator or it is always above the same location on the Earth.

(b) Examiner’s comment: The student has remembered to add the radius of the Earth to the height of the satellite. Mark = 1 out of 1

(c) v = rω = 2πr _ T

= 2 × π × 4.237 × 10 7 ________________ 24 × 60 × 60

= 3081 m s−1

KE = 1 _ 2 mv2

= 1 _ 2 × 1200 × 30812

= 5.70 × 109 J

(d) Vs = − GM _ r

= − 6.67 × 10 −11 × 5.97 × 10 24 ____________ 6.37 × 10 6

= − 6.25 × 107 J kg−1

Vo = − GM _ r

= − 6.67 × 10 −11 × 5.97 × 10 24 ____________ 4.23 × 10 7

= − 9.41 × 106 J kg−1

Ep =   ΔV × m = (Vo − Vs) × m = (−9.41 × 106) − (−6.25 × 107) × 1200

= 6.37 × 1010 J

(e) Radius of orbit increases so the velocity gets smaller as

from circular motion v2 is proportional to 1 _ r . Kepler’s

3rd law (R3 ∝ T2)

(f) Communications. The satellite stays at the same position above the Earth so the satellite dish does not need to turn to track it.

Exam-style questions1 a i GMm ______ r 2 = mv 2 _ r

v2 = GM _ r = ( 6.67 × 10 −11 ) × ( 1.99 × 10 30 ) ______________ 1.5 × 10 11

= 8.8489 × 108 (1)

v = 2.975 × 104 m s−1 (3.0 × 104 to 2 s.f.) (1)

ii R M 3

_ T M 2 = R E 3

_ T E 2

RM3 =

( 1.5 × 10 11 ) 3 × 88 2 _______________ 365 2 (1)

RM = 5.81 × 1010 m (1)

b GMm _ r 2 = mv 2 _ r

v2 = GM _ r

v = √ _______

  ( GM _ r )

ω = v _ r

= √ _

( GM _ r 3 ) (1)

= √ ______________

( 6.67 × 10 −11 × 5.97 × 10 24 ____________ ( 4.24 × 10 7 ) 3 ) (1)

= 7.23 ×10−5 rad s−1 (1)

c v = √ ________

  ( 2GM _ r )

= √ ________________

2 × 6.67 × 10 −11 × 7.33 × 10 22 ______________ 1.74 × 10 6 (1)

= 2370 m s−1 (to 3 s.f.) (1)

2 GMm _ r 2 = mv 2 _ r

v2 = GM _ r

v = √ _

( GM _ r ) (1)

ω = v _ r

= √ _

( GM _ r 3 ) (1)

T = 2π ___ ω    = 2π √ _______

  ( r 3 _ GM ) (1)

Radial field: g decreases as r increases

V becomes less negative as r increases

Gradient decreases as r increases

Uniform field: g is constant

V becomes less negative as r increases

Gradient is constant

rV


1919

Exam Skills Builder

Physics

113–116 Electric fields and forces

Grade booster1 (a)

Neutralpoint

X+3 nC +6 nC

field lines outward from charges (1)

neutral point marked (1)

number of field lines double for 6nC (1)

neutral point closer to 3nC charge (1)

(b) Examiner’s comment: The student has missed out the 10−9 from the 6nC charge and so has been awarded the first mark with a power of ten error but not the second for the final answer.

(c) E = Q _ 4π ε o r 2

E3 = 3 × 10 −9 ________________ 4 × π × 8.85 × 10 −12 × ( 32 × 10 −3 ) 2

= 26 343 N C−1 to the right (1)

E6 = 6 × 10 −9 ________________ 4 × π × 8.85 × 10 −12 × ( 32 × 10 −3 ) 2

= 52 686 N C−1 to the left (1)

Resultant E = 52 686 − 26 343 = 26 343

= 2.6 × 104 (1) N C−1 (1) to the left (1)

Exam-style questions1 Units of E are normally quoted in Vm−1 or of NC−1 N =

Jm−1 so A is equivalent of NC−1 and D is the incorrect unit. (1)

2 a F = Q 1 Q 2 _ 4π ε o r 2

= 5 × 10 −9 × 10 × 10 −9 ________________ 4 × π × 8.85 × 10 −12 × ( 10 × 10 −2 ) 2 (1)

= 4.5 × 10−5 N (1)

b i correct directions for E5 (along direction line from +5.0 nC to P) and E10 (along line from P to −10.0 nC) (1)

E10 approx twice as long as E5 (1)

PR

Es

E10

ii correct direction of resultant R shown from completing parallelogram (1)

c E1 = Q _ 4π ε o x 2 E2 =

Q ____________ 4π ε o ( 10 − x ) 2

E1 = E2

5 × 10 −9 __________________________ 4 × π × 8.85 × 10 −12 × x2 =

10 × 10 −9 _______________________________ 4 × π × 8.85 × 10 −12 × ( 10 − x ) 2

5 _ x 2 = 10 _ ( 10 − x ) 2 (1)

Take square root of both sides:

( 10 − x ) √ _

5 = x √ _

10

10 √ _

5 = x ( √ _

10 + √ _

5 ) 22.36 = 5.389 x

x = 22.36 _ 5.398 = 4.14 cm (1)

117–120 Electric potential

Grade booster1 (a) (i) V is proportional to 1 _ r

(ii) Examiner’s comment: The student has correctly realised that electric potential is + for + charge and − for − charge. Alternatively ‘potential is defined to be zero at infinity’ would also be correct.

Mark = 1 out of 1

(b) (i) From graph: V = −1000 V 1 _ r = 4 m−1 gives r = 0.25 m

rearranging V = Q _ 4π ε o r

Q = 4πε0 rV = 4π × 8.85 × 10−12 × 0.25 × 1000

= 27.8 × 10−9 C

= 28 nC

(ii) E = Q _ 4π ε o r 2

= 27.8 × 10 −9 ____________ 4 π × 8.85 × 10 −12 × 0.3 2

= 2777.5 V m−1

(iii) At r = 0.33 m V = −750 V and at r = 1 m V = −250 V

So pd   ΔV = −250 − (−750) = 500 (V)

Work done   ΔW (= Q  ΔV) = 50 × 10−9 × 500

= 2.5 × 10−5 J

Exam-style questions1 D

Vp + VQ = 0

4 _ 4π ε o x + ( − 6 ) ____________ 4π ε o ( 100 − x ) = 0

4(100−x) − 6x = 0

400 − 4x −6x = 0

400 = 10x

x = 40 mm (1)

2 A

Work done = Q ΔV

= 3 × 10−9 (475−350)

= 3 × 10−9 × 125

= 375 × 10−9

= 3.75 × 10−7 J (1)


2020

Exam Skills Builder

Physics

3 V from 3 charges = 3Q _ 4π ε o x

3Q _ 2π ε o x

− 3Q _ 4π ε o x

= 3Q _ 4π ε o x

so the other charge = 3Q C (1)

4 Max 3 from:

– Forces are both proportional to product of the mases or charges and inversely proportional to square of distance apart / inverse-square law.

– A spherical object acts as a point mass or charge at centre of sphere.

– Field strengths are defined as a force per unit mass or charge.

– Field strengths are both proportional to 1 _ r 2 .

– Field strengths are proportional to the magnitude of mass or charge producing it.

– Both definitions of potential involve work done per unit mass or charge (work done in moving a mass or charge from infinity to a point).

– Both types of potential are proportional to 1 _ r .

– Potential is proportional to the mass or charge producing them.

– The work done in moving a mass or charge across a potential difference is calculated by multiplying the mass or charge by the potential difference.

Differences Max 1 from

– Gravitational forces always attractive. Electric forces attractive or repulsive.

– Gravitational field is always directed towards the mass producing it. Electric field is directed towards a negative charge but away from a positive charge.

– Gravitational potential is always a negative quantity but electric potential is negative for negative charges and positive for positive charges.

121–124 Capacitance

Grade booster1 (a) 1 C of charge is stored for every 1 V of pd between

the plates of the capacitor.

(b) (i) straight line graph through the origin; y-intercept = 3.5 V; x-intercept = 10.5 µC

(ii) For a small increase in charge ΔQ, V is almost constant

Work done ΔW = V ΔQ (1)

= area of narrow strip under graph

Energy stored = total work done (W )

= sum of areas of all strips

= total area under graph (triangle) (1)

= 1 __ 2 × base × height

= 1 __ 2 QV (1)

(iii) Examiner’s comment: The student has incorrectly substituted 3.5 V for the battery voltage and the maximum voltage it could reach instead of 1.5 V. Physics error so no marks awarded.

Mark = 0 out of 2

E = 1 _ 2 CV 2 = 1 _ 2 × 3.0 × 10−6 × 1.52

= 3.38 × 10−6 J

(c) C = A ε o ε r _ d

23 × 10−12 = π r 2 × 8.85 × 10 −12 × 1 ___________ 1.8 × 10 −3

r2 = 23 × 10 −12 × 1.8 × 10 −3 ___________ π × 8.85 × 10 −12

= 1.49 × 10−3 m2

r = 0.0386 m

Exam-style questions1 C

C = Q

_ V C = 400 × 10 −6 _ V C = 200 × 10 −6 _ V − 50

400 × 10 −6 _ V = 200 × 10 −6 _ V − 50

400V − 20000 = 200V

200V = 20000

V = 100 V

C = 400 × 10 −6 _ 100 = 4 × 10 −6 F (1)

2 A

Q = CV = (500 × 10−6) × 5 = 2500 × 10−6

I = Q

_ t = 2500 × 10 −6 _ 20 = 125 µA (1)

3 C

E = 1 _ 2 CV 2

1600 × 10−6 = 1 _ 2 × C × V2

400 × 10−6 = 1 _ 2 × C × (V−2)2

Combining equations and cancelling:

1600 × 10 6 _________ 400 × 10 6 = 1 _ 2 × C × V 2

______________ 1 _ 2 × C × ( V − 2 ) 2

Rearranging:

1600 _ V 2 = 400 _ ( V − 2 ) 2 (eqn 1)

Take square root of both sides of equation (1):

40 _ V = 20 _ V − 2

40V − 80 = 20V

20V = 80

V = 4

E = 1 _ 2 QV

1600 × 10−6 = 1 _ 2 × Q × 4

Q = 800 µC

C = Q

_ V = 800 × 10 −6 _ 4 = 200 µF (1)

4 B

C = ε 0 ε r x 2

_ d C = ε 0 4 x 2

_ 2d

ε r x 2

____ d = 4x 2 ___ 2d

εr = 2 (1)


2121

Exam Skills Builder

Physics

125–128 Capacitor charge and discharge

Grade booster1 (a) (i) 10 –15 ms

(ii) Examiner’s comment: The student has stated that data logging is ‘automatic’ for 1 mark but has not gone on to explain clearly the difficulties of manual timing in this particular experiment. ‘Random errors are introduced with manual timing as it is very difficult to read and record a voltmeter and a stopwatch at exactly the same time’ could have been added for the second mark. Alternatively, simply ‘120 ms is too fast for human measurement’ would be sufficient in this experiment for the second mark. Mark = 1 out of 2

(b) (i) From the graph, V0 = 6 V

Q0 = CV0 = 3.3 × 10−6 × 6 = 19.8 × 10−6 C = 20 µC (2)

(ii) From the graph, at 70 ms, V = 2.0 V

E = 1 _ 2 CV 2

= 1 _ 2 × 3.3 × 10−6 × 2.02

= 6.6 × 10−6 J

(iii) Time constant is time taken for V to fall to Vo / e (1)

       ∴V must fall to 6 __ e = 2.2 V (1)

From graph t = time constant = 64 ms (1)

Exam-style questions1 D

t _ RC = 4.8 ___________ ( 20 × 10 3 × 120 × 10 −6 ) = 2

Q = Q 0 e −t _ RC

Q

_ Q 0 = e−2 = 0.135 (1)

2 D

Q = CV = 500 × 10−6 × 10 = 5 × 10−3 C

Q = Q 0 e −t _ RC RC = 100 × 103 × 500 × 10−6 = 50

2.5 × 10−3 = 5 × 10−3 e −t _ 50

Taking ln on both sides of the equation gives:

ln 2.5 _ 5 = − t ____________ ( 500 × 10 −6 × 100 × 10 3 )

−ln2 = − t _ 50

t = 50 ln2 = 34.65 s = 35 s (1)

3 C

V = V o e −t _ RC

1.5 = 6 e −92 ____________ ( 220×10−6 R )

ln 0.25 = − 92 _____________ ( 220 × 10 −6 R )

R = − 92 ___________ ( − 1.386 × 220 × 10 −6 )

= 301 718 Ω = 300 kΩ (1)

4 a Time constant = RC = 500 × 10−6 × 2 × 106 = 1000 s (1)

b C

Q0 = CV0

= 24 × 500 × 10−6

= 0.012 C (1)

Q = Q0 (1 − e −t

__ RC )

= 0.012 (1 − e −600 _ 1000 ) (1)

= 5.4 × 10−3 C (1)

c V = V0 (1 − e −t _ RC )

0.95 V0 = V0(1 − e −t ____ 1000 ) (1)

0.95 = 1 − e −t _ 1000

0.05 = e −t _ 1000

ln 0.05 = − t _ 1000 (1)

−2.996 = − t _ 1000

t = 2996 s (1)

129–132 Magnetic forces

Grade booster1 (a) (i) It is not possible as the force due to the magnetic

field is perpendicular to the direction of travel of the protons. (1)

(ii) It accelerates the particles as they cross the gap. Electric field reverses direction when the particle reaches the other side so that it is accelerated when travelling in the opposite direction. (2)

(iii) Examiner’s comment: The student has drawn a ‘spiral’ shape, although not very well, so first mark ‘just’ awarded. There should be arrows on the path in an anticlockwise direction to show that student has used Fleming’s LH rule correctly to determine the direction, so second mark not awarded.

Mark = 1 out of 2

(b) B Q v = m v 2 _ r (1)

BQ

_ m = v _ r

v = BQr

_ m

Time for semicircle t = distance _ speed = π r _ v (1)

Substitute for v

t = π m _ BQ (1)

which does not contain r and so is independent of r

(c) v = B Q r

_ m

= 0.44 × 1.6 × 10 −19 × 0.55 ____________ 1.67 × 10 −27 (1)

= 2.3 × 107 m s−1 (1)


BIL = mg

0.04 × 3.5 × 0.25 = m × 9.81

m = 3.57 × 10−3 kg = 3.57 g (1)

2 B

Fβ = Bev Fα = 3B × 2e × v _ 20 = 0.3Bev = 0.3F (1)


2222

Exam Skills Builder

Physics

3 D

Bev = mv 2 _ r t = 2πr _ v

v _ r = Be _ m = 2π _ t

t = 2πm _ Be (1)

4 C

F = BIL F2 = 3B × 0.3 I × 2L = 1.8BIL = 1.8F (1)

133–136 Electromagnetic induction

Grade booster1 (a) (i) maximum: 5π _ 6 radians; minimum: π _ 3 radians

(ii) 60o or π _ 3 radians

(b) (i) Examiner’s comment: The student has not been precise enough. The answer should be induced voltage or emf or rate of change of flux linkage. Voltage is insufficient. Mark = 0 out of 1

(ii) T = 20 ms

Frequency = 1 _ T = 1 _ ( 20 × 10 −3 ) = 50 Hz

No of revs per min = 50 × 60 = 3000

(iii) BAN = maximum flux linkage from graph

= 0.44 (Wb turns)

ω = 2π _ T = 2π _ 20 × 10 −3 = 314 rad s−1

peak emf = BANω = 0.44 × 314 = 138 V


Ε = BLv = 3.5 × 10−5 × 1.6 × 25 = 1.4 × 10−3 V (1)

2 ε =   NΔΦ _____ Δt = 6 × (16 − 8) × 10−3

_______________ 0.4 = 0.12 V (1)

3 a Φ = BA = 45 × 10−3 × π × (70 × 10−3)2 (1)

= 6.93 × 10−4 Wb (1)

b i NΔΦ = NBA − 0 = 850 × 6.93 × 10−4 (1)

= 0.589 Wb turns (1)

ii induced emf = NΔΦ _  Δt = 0.589 _ 120 × 10 −3 (1) = 4.91 V (1)

137–140 Transformers and ac

Grade booster1 (a) (i) The primary coil has more turns than the

secondary coil. (1) The input must be ac. (1)

(ii) The core directs the magnetic field round to the secondary coil so that more of the flux passes through the secondary coil.

(b) Examiner’s comment: Excellent answer, the student has mentioned changing magnetic flux passing through the secondary coil and Faraday’s law.

Mark = 1 out of 1

(c) (i) NP = N S V P

_ V S = 300 × 230 _ 20

= 3450 turns (3500 to 2 s.f.)

(ii) Efficiency = power out

_ power in = 65 × 100 _ 230 × 0.30 = 94%

Exam-style questions1 a Removal of top of core means less direction of field

lines to secondary coil (1) so the secondary voltage will be less. (1)

b A dc voltage on primary means the flux only changes as primary voltage is switched on and off. (1) A voltage will be recorded momentarily on the secondary voltmeter as power is switched on and off but no voltage while it is just ‘ON’. (1)

2 a Step-up transformer raises the transmission voltage. High voltage means low current. (1) Power loss = I2R (1) so a lower current means less power loss (1) during transmission along the cables.

b Ns = V s × N p

_ V p =

( 275 × 10 3 ) × 800 _______________ 25 × 10 3 (1)

= 8800 (1)

c P = IV

Irms = 2 × 10 6 _ 25 × 10 3 (1)

= 80 A (1)

d power output = 0.95 × power input = 0.95 × 2.0 × 106 = 1.9 MW (1)

Irms output = 1.9 × 10 6 _ 275 × 10 3 (1)

= 6.9 A

Ipeak = 6.9 × √ _ 2 (1)

= 9.76 A = 9.8 A (to 2 s.f.) (1)

141–144 Alpha, beta and gamma radiation

Grade booster1 (a)

Type of radiation

Typical range of travel in air / m

α 0 − 0.05 (1)

β 0 − 0.70 (1)

γ 0 − 5 (1)

(b) Examiner’s comment: (i) β particles are negative so will be attracted to the positive plate. Alpha particles are positive so will be attracted to the negative plate of an electric field. (ii) β particles might pass through the pipe but they would not be able to reach the detector above the ground, so gamma should be used.

Mark = 0 out of 2

(c) The radiation needs to pass through the body to the gamma camera outside the body to be detected. (1) If the half-life of the source is shorter, it will become too weak before the tracer reaches the kidneys. (1) If the half-life is longer it may remain in the body for a long time and could cause damage. (1)

(d) (i) Nuclear radiation which is ever present without a radioactive source nearby.

(ii) corrected count rate at 0.2 m = 2750 − 50

= 2700 (c min−1)

corrected count rate at x = 5500 − 50

= 5450 (c min−1)


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I 1 _ I 2

= ( x 2 )

2 _

( x 1 ) 2

x22 =

I 1 x 1 2 _ I 2

= 2700 × ( 0.2 ) 2 ___________ 5450

x2 = √ _

0.0198

= 0.141 m

Exam-style questions1 A = 215, Z = 84 (2)

2 a Z down 2, A down 4 (2)

b Gamma is more penetrating. (1)

c I 1 _ I 2

= ( x 2 )

2 _

( x 1 ) 2 = 1 2 _ 4 2 = 1 _ 16 (2)

3 a Thicker material absorbs more particles so count rate falls if material is thicker (1) and rises if it is thinner. (1) There is a feedback system which adjusts the rollers to restore to original thickness. (1)

b paper = strontium-90 (1) steel = Cobalt-60 (1)

c Strontium-90 chosen for paper as it emits beta particles and has a long half-life (1) so source does not need replacing too often. Americium-241 not chosen as alpha particles would be absorbed by paper (1) and Cobalt and technetium emit gamma radiation which would not be affected by the paper, it would pass straight through. (1)

145–148 Radioactive decay

Grade booster

1 (a)

No. of atoms × 1022

00

6

3

LeadUranium

2 4 6 8 10Time × 109 years

(b) Examiner’s comment: More working needed. In this question it is not possible to show more significant figures so all working must be shown.

Mark = 0 out of 1

If N = no of U atoms

N = 2 × (6 × 1022 − N)

= 12 × 1022 − 2N

3N = 12 × 1022

N = 4 × 1022 atoms

(c) λ = ln 2 _ 4.5 × 10 9 = 1.54 × 10−10

N = No e−λt

4 × 1022 = 6 × 1022 × e−λt

ln 4 × 10 22 _ 6 × 10 22 = (−1.54 × 10−10) × t

ln 0.67 ____________ − 1.54 × 10 −10 = t

t = 2.6 × 109 years

Exam-style questions1 1 _ 16 = ( 1 _ 2 )

4

Four half-lives have passed

64 _ 4 = 16 days C (1)

2 a λ = ln 2 ______________ 8.04 × 24 × 3600 (1) = 1.0 × 10−6 s−1 (1)

b A = λN

N = A _ λ = 6.0 × 10 4 _ 1.0 × 10 −6 (1) = 6.0 × 1010 (1)

c N = Noe−λt

6.0 × 104 = 6.6 × 104 × e−1 × 10−6t

dividing and taking logs of both sides

ln 6.0 _ 6.6 = − 1.0 × 10−6 t

−0.095 = − 1.0 × 10−6 t

t = − 0.095 _ − 1.0 × 10 −6

= 95 310 s (1)

= 95 310 _ 3600

= 26.5 h (1)

d Half-life is 8.04 days (1) so the patient is still going to be radioactive for many days (1) and will be advised to stay away from young children and pregnant women for a couple of weeks (1). The benefits are that the patient’s condition and life should improve (1). It is a beta emitter so high doses will kill the thyroid cells but lower ones could increase the risk of cancer (1). [Max 4]

3 a in 1 s source emits 1.2 × 1014 particles (1)

energy emitted in 1 s = 1.2 × 1014 × 5.2 × 1.6 × 10−13 (J) (1)

= 99.8 J

b T 1 _ 2 = ln2 _ λ

λ = 0.693 ____________ 90 × 365 × 24 × 60 × 60 (1)

= 2.44 × 10−10 s−1 (1)

c N = A _ λ = 1.2 × 10 14 _ 2.44 × 10 −10 (1)

= 4.91 × 1023

Molar mass of uranium contains Avogadro’s number of atoms

Mass of isotope = 4.91 × 10 23 × 0.238 ________________ 6.02 × 10 23 (1)

= 0.19 kg (1)

149–152 Nuclear radius and instability

Grade booster1 (a) (i) Straight through α particles show that most of

the atom is empty space. (1) Most of the mass and all of the positive charge are concentrated in the central nucleus (1) resulting in enough force to repel some particles back on themselves.

(ii) Examiner’s comment: The student is incorrect. It is the electrostatic / electromagnetic force that causes repulsion not the strong nuclear force. Mark = 0 out of 1


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(b) KE of α = 1.6 × 10−19 × 5.5 × 106 J

= 8.8 × 10−13 J

KE lost = PE gained = 8.8 × 10−13 J

Using V = Q _ 4π ε o r

and Ep = qV:

Ep = Qq

_ 4π ε o r

r = qQ _ 4π ε o E p

= 2 × 1.6 × 10 −19 × 79 × 1.6 × 10 −19 ________________ 4π × 8.85 × 10 −12 × 8.8 × 10 −13

r = 4.13 × 10−14 m

Exam-style questions1 a Straight line graph (1) through origin showing

proportionality. (1)

b gradient = Ro

c R F

_ R U = R o 56 ⅓

_ R o 235 ⅓ = 0.6199 (1)

RU = 4.35 × 10 −15 _ 0.6199 = 7.017 × 10−15 m (1)

2 Max 4 from:

– it only emits γ rays

– may be detected outside the body / weak ioniser and causes little damage

– it has a half-life of 6.01 h and will not remain active in the body after use

– it has a long enough half-life to remain active during diagnosis

– it may be prepared on site by β− emission from molybdenum-99 as 88% of the molybdenum nuclei will decay in this way.

153–156 Mass and energy

Grade booster1 (a) (i) Fission occurs at A values above the peak and

fusion occurs at A values below the peak (1) Fission is the splitting of a large nucleus (high A number) into two smaller nuclei and fusion is the joining of two smaller nuclei (low A number) to form a larger one. (1)

(ii) Fusion energy is greater as the increase in BE / A for fusion > increase in BE / A for fission as the gradient is steeper. Energy is released when the binding energy (per nucleon) is increased.

(b) (i) Δm = (26mp + 30mn) − Miron

= (30 × 1.00867) + (26 × 1.00728) − 55.92067

= 30.2601 + 26.18928 − 55.92067 = 0.52871u

= (0.52871 × 1.661 × 10−27) = 8.78 × 10−28 kg

E = m × c2 = 8.78 × 10−28 × (3.00 × 108)2 = 7.902 × 10−11 J

BE = 7.902 × 10 −11 ___________ 1.6 × 10 −13 = 494 MeV

(ii) Examiner’s comment: The student read binding energy per nucleon correctly as 8.8 MeV from the graph but forgot to multiply by 56 (the number of nucleons) to get 493 MeV. Mark = 1 out of 2

Exam-style questions1   Δm = Zmp + Nmn − MB

= (4 × 1.00728u) + (3 × 1.00867u) − 7.01473u

= 0.0404u

Binding energy = 0.0404 × 931.5

= 37.63 MeV

BE _ nucleon = 37.63 _ 7 = 5.4 MeV C (1)

2   Δm = Zmp + Nmn − ML

= 3 × 1.00728u + 4 × 1.00867u − 7.01436u

= 3.02184u + 4.03468u − 7.01436u

= 0.04216u D (1)

3 mass of nucleus = 63.92915 − (30 × 0.00055) = 63.91265 u

  Δm = (30 × 1.00728) + (34 × 1.00867) − 63.91265

= 0.60053u

BE = 0.60053 × 931.5 = 559.4 MeV

BE _ nucleon = 559.4 _ 64 = 8.74 MeV / nucleon A (1)

4 (a) (i) (144 × 8.3 + 8.6 × 89) − (7.5 × 236) (1) = 1960 − 1770 = 191 MeV (1)

(ii) Power = E __ t = 1.3 × 10–5 × (3 × 108)2

___________________ 3600 (1)

= 325 MW (1)

(b) 3.27 MeV = (3.27 × 106) × (1.6 × 10−19) = 5.232 × 10−13 J

m = E __ c2 = 5.232 ×10−13

___________ (3 × 108)2 (1) = 5.81 × 10−30 kg greater

before fusion. (1)

157–160 Nuclear reactors and safety

Grade booster1 (a) Examiner’s comment: A good answer in many respects

as the student was able to identify the main source of high-level waste and describe the various stages of treatment in the correct order. However, the student has only been awarded 4 marks because of the mention of the solution to only one problem. For 5–6 marks, at least two more problems and solutions should have been discussed.

• Left in cooling ponds as very hot for several years

• Screening of ponds to absorb radiation

• Vitrification prevents leakage of high-level liquid waste

Mark = 4 out of 6

Exam-style questions1 a i reduced (1)

ii unchanged (1)

b Each fission produces two or three neutrons one neutron per fission must go on to produce further fission. (1)

Some neutrons escape or are absorbed by U-238 without fission (1) control rods absorb sufficient neutrons to maintain steady rate of fission. (1)

c Neutrons need to pass through a moderator (1) to reduce speed to cause further fission.

Neutrons that leave one fuel rod and pass through the moderator are unlikely to re-enter the same fuel rod (1) so many are needed and makes it easier to replace the fuel in stages. (1)


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2 Max 5 from:

– Cumbria has been the dumping ground for the world’s radioactive waste with its very long half-lives.

– However, it has generated an income and provided employment.

– Transport of radioactive waste across the sea and land creates risk of accident.

– Recycling unused uranium is good as supplies of uranium are finite.

– Impact resistant flasks are used for transport to reduce risks.

– In the future Cumbria will not have as much employment, but will be less at risk from radiation.

– Other countries will be more at risk of radiation as they can’t send the waste elsewhere.

161–164 Practical skills

Grade booster1 (a) Examiner’s comment: The student has explained the

procedure and correctly taken logs to base e or ln.

Mark = 2 out of 2

(b) Comparing with y = mx + c:

gradient = − k

= − ( 3.40 − 2.975 ) ____________ 56 − 7 = − 0.0087 s−1

k = 0.0087 s−1

(c) As lnφ60 is below the x-axis of the graph you need to substitute in the equation

lnφ0 = 3.46 and k = 0.0087 s−1

lnφ60 = lnφ0 − kt

= 3.46 − (0. 0087 × 60)

= 3.46 − 0.522

= 2.94

R60 = k φ60

= 0.0087 × 18.88

= 0.164 oC s−1

(d) The results seem to verify Newton’s law but if the liquid were allowed to cool for a much longer time (1) and the experiment repeated with a much higher initial temperature θ (1) giving a larger initial temperature difference φ then the results would be more conclusive over a larger range of values. (1)

Exam-style questions1 a (1)

Mass m / kg Force F / N

Extension   ΔL / mm

2.0 19.62 0.923.0 29.43 1.384.0 39.24 1.855.0 49.05 2.326.0 58.86 2.787.0 68.67 3.268.0 78.48 3.72

b Graph (4)

0 13.00

3.05

3.10

F / N

∆L / mm

3.15

3.20

3.25

3.30

2 3 4

c Gradient = 74 − 10 _________________ ( 3.5 − 0.475 ) × 10 −3 (1)

= 21.16 × 103 N m−1 (1)

d YM = F × L _ A × ΔL = gradient × L

___________ A

= 21.16 × 10 3 × 1.85 ________________ π × ( 0.255 × 10 −3 ) 2 (1)

= 1.92 × 1011 Pa (1)

% uncertainty in YM = (2 × % uncertainty in r) + % uncertainty in L + % uncertainty in m + % uncertainty in ΔL

= 2 ( 0.001 × 100 ) ____________ 0.255 + 0.001 × 100 _____________ 1.850 + 1 + 0.01 × 100 _ 0.92 (1)

= 0.78 + 0.054 +1 + 1.086

= 2.92% (1)

Uncertainty in YM = 2.92 _ 100 × 1.92 × 1011 = ± 5.6 × 109

YM = (1.92 ± 0.06) × 1011 Pa (1)

e Diameter measurement should be taken at five different places along the wire and a mean diameter calculated. Extension measurement should be made with a Vernier scale. (2)

f Larger diameter wire will produce a smaller extension (1) for a given weight applied so will increase the % uncertainty in the extension and in the Young modulus. (1)

Larger diameter will reduce the % uncertainty in the diameter (1) and also in the cross-sectional area and in the Young modulus. (1)


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165–168 Telescopes

Grade booster1 (a) (i) Largest distance = 2.57 + 1 = 3.57 AU

3.57 AU = 3.57 × 1.5 × 1011 m

= 5.36 × 1011 m

(ii) Examiner’s comment: The student has correctly used the arc length equation from the mathematical formulae section of the Data Booklet to find the angle subtended by the asteroid in radians. The student has given the answer to more s.f. than quoted in the question to ‘show that’. Mark = 2 out of 2

Angle = s _ r

= 5.4 × 10 5 _ 1.73 × 10 11

= 3.12 × 10−6 rad

(b) (i) LH mirror should be labelled convex RH mirror concave/ parabolic

(ii) Min angle of resolution θ = λ _ D

D = 1 × 10 −6 _ 3.3 × 10 −7

= 3.0 m 2 s.f. needed

(c) Minimum angle of resolution is smaller than the size of the asteroid. 3m << 540000 m (1)

3.3 × 10−7 ≈ 1 _ 10 of 3.12 × 10−6

1 _ 10 of size of asteroid = 54 km so details of 50 km

can be seen (1)

Exam-style questions1 a Two sources will just be resolved if the central maximum

of the diffraction pattern of one coincides (1) with the first minimum of the diffraction pattern of the other. (1)

b θ  = λ _ D = 2.0 × 10 −6 _ 3.8 (1) = 5.26 × 10−7 rad (1)

c – Reflectors collect more light, as the mirrors can be larger than a lens.

– Larger diameters give better resolution.

– No chromatic aberration with mirrors as no refraction.

– No spherical aberration if mirror is parabolic. (4)

2 a D = λ _ θ  = 5.7 × 10 −7 _ 1.1 × 10 −5 (1) = 5.2 × 10−2 m (1)

b Max 5 from:

– Resolution of the eye varies across retina whereas in a CCD it is uniform.

– Overall eye resolution < CCD in centre.

– All information received by CCD is processed whereas optic nerve does not transmit all the resolved information.

– Quantum efficiency of eye = less than 10% but Quantum efficiency of CCD is greater than 70%.

– CCD can process more than just visible radiation.

– CCD can process faint parts of a bright image.

– CCD can be used for a long exposure time.

– CCD facilitates direct computer analysis.

– CCD can be used remotely.

169–172 Classification of stars (1)

Grade booster1 (a) PC = σ AcTc

4

= 5.67 × 10−8 × 4 × π × (2.5 × 6.96 × 108)2 × 96024

= 1.83 × 1028 W

(b) C has the smallest apparent magnitude of 0.03 and so appears the brightest on Earth.

(c) Closest of three stars is A.

m − M = 5 log ( d _ 10 )

0.76 − 2.22 = 5 log ( d _ 10 )

−1.46 = 5 log ( d _ 10 )

10 −1.46 _ 5 = d _ 10

0.51 = d _ 10

d = 5.1 pc

= 5.1 × 3.08 × 1016

= 1.57 × 1017 m

(d) A is the smallest. They all have approximately the same temperature although A has the lowest. A has a larger apparent magnitude and absolute magnitude than the others so appears less bright. It has a smaller power output and a smaller surface area.

(e) Examiner’s comment: The student has used Wien’s law correctly but has not given the answer in nanometres as requested. Answer should be 340 nm.

Mark = 1 out of 2 (f)

300 600 900 1200

Relativeintensity

Wavelength

n / m

(1) (1)

Exam-style questions1 a m − M = 5 log ( d _ 10 )

2.2 − (−4.6) = 5 log ( d _ 10 ) (1)

6.8 _ 5 = log ( d _ 10 )

101.36 = d _ 10

22.9 = d _ 10

d = 229 pc (1)

b λmaxT = 0.0029

= 0.0029 _ 12 000 = 2.42 × 10−7 m (1)

2 a distance = 250 × 9.46 × 1015 = 2.37 × 1018 m (1) and

= 250 _ 3.26 = 76.7 pc (1)

b m − M = 5 log ( d _ 10 ) (1)

m = −2.8 + 5 log ( 76.7 _ 10 ) = 1.6 (1)

c Elnath (1)


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Elnath is dimmer than Bellatrix absolute magnitude −2.8 is brighter than −1.4 (1)

but appears to have same brightness from Earth, so must be closer. (1)

173–176 Classification of stars (2)

Grade booster1 (a) (i) An object that produces a rapid increase in

brightness or a lowering in value of absolute magnitude.

(ii) Small and extremely dense and made of neutrons.

(iii) Rs = 2GM _ c 2

= 2 × 6.67 × 10 −11 × 2 × 1.99 × 10 30 ________________ ( 3 × 10 8 ) 2

= 5899 m (b) (i)

Time / days0

–11

–15Absolutemagnitude

–19

100 200 300

(ii) Examiner’s comment: Excellent definition.

Mark = 1 out of 1

(iii) All type 1a supernovae have same peak absolute magnitude (approximately −19). (1) Apparent magnitude can be measured at this peak. (1)

(iv) The measurements of supernovae do not agree with predictions from Hubble’s theory. They showed the galaxies were moving away even faster. Universe must be expanding at an increasing rate or accelerating. It was controversial as there is no known energy source for expansion so may be due to dark energy.

Exam-style questions1 Rs = 2GM _ c 2

= 2 × ( 6.67 × 10 −11 ) × ( 7 × 2 × 10 30 ) ________________ ( 3 × 10 8 ) 2 (1)

= 2.08 × 104 (m) (1)

2 a Star B is much brighter (1) as it has a lower value of apparent magnitude (1) compared to star A.

b Correct main sequence (1) correct Giants and White Dwarfs. (1)

c A: m − M = 5 log ( d _ 10 ) B: m − M = 5 log ( d _ 10 )

11 − M = 5 log ( 1.3 _ 10 ) 1 − M = 5 log ( 160 _ 10 )

M = 11 + (−4.43) = 15.43 (1) M = 1 − 6.02 = −5.02 (1)

d Correct position on diagram (spectral class M, abs magnitude −5). (1)

e Both temperature less than 3500 K as in spectral class M. (1)

f B is brightest (and at same temperature) (1) so has largest surface area and hence diameter. (1)

177–180 Cosmology

Grade booster1 (a) Big Bang theory states that the universe began as a

small hot dense singularity and has been expanding ever since. (1) Steady State theory says that the Universe is unchanging, the same now as it ever was. (1)

(b) Examiner’s comment: The student has remembered that the galaxies are ‘moving away’ but not said ‘where from’ so has not proved expansion. The added phrase ‘from each other’ or ‘from the Earth’ would have gained the other mark.

Mark = 1 out of 2

(c) Max 4 from:

– CMBR stands for Cosmological Microwave Background Radiation.

– It is the radiation coming from all parts of the Universe / the radiation is isotropic.

– It can be interpreted as the radiation left over from the Big Bang.

– The electromagnetic radiation has been red shifted into the microwave region as the Universe has expanded.

– The spectrum has a peak in the microwave region. It corresponds to a temperature of 2.7 K.

(d) Fusion of H to He when hot enough.

Universe expanded and cooled rapidly when the ratio of H:He is 3:1.

It was then not hot enough for any further fusion or creation of larger nuclei.

Exam-style questions1 a Δλ = λv _ c = 7.5 × 10 5 × 6.5647 × 10 −7 ____________ 3 × 10 8

= 1.64 × 10−9 m (1)

λobs = (6.5647 + 0.0164) × 10−7 = 6.58 × 10−7 m (1)

b d = 4.9 × 10 7 _ 3.26 × 10 6 (1)

= 15 Mpc

H = v _ d

= 750 _ 15 (1)

= 50 kms−1 Mpc−1 (1)

c – Hubble constant varies with time (1)

– Current accepted value is 65 kms−1 Mpc−1 (1)

– If galaxies are receding the distance between them is increasing (1)

– the further away galaxies are moving away faster so the distance between them may be changing at a different rate. (1)

2 a Δλ _ λ  = − v _ c (4)

660.86 − 656.28 ______________ 656.28 = − v _ 3 × 10 8

v = −2094 km s−1 (1) table completed (1)

b points plotted (1) straight-line graph through origin (1)

c gradient = H = v _ d = 70 km s−1 Mpc−1 (1)


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181–184 Discovery of the electron

Grade booster1 (a) Electrons collide with the helium atoms. Electron in

an atom is excited into a higher energy level and then emits a light photon when the electron de-excites and moves back to a lower energy level.

(b) eV = 1 _ 2 mv2

v2 = 2eV _ m

= 2 × 1.6 × 10 −19 × 2500 ___________ 9.1 × 10 −31

= 8.79 × 1014

v = 2.965 × 107 m s−1

(c) F = Bev = mv 2 _ R

e _ m = v _ Br

= 2.965 × 10 7 ___________ 3.1 × 10 −3 × 5.7 × 10 −2

= 1.68 × 1011 C kg−1

(d) Examiner’s comment: The student has realised that collisions will reduce the speed but not that the radius

of the circle gradually decreases as r = vm _ Be and so v is

proportional to r. The path will gradually

spiral inwards.

Mark = 1 out of 2

(e) Specific charge of electron > specific charge of hydrogen ion. (1) Thomson concluded electrons have a very small mass or could have a high negative charge. (1)

Exam-style questions1 a Electrons are negatively charged so beam is attracted

to positive plate. (1)

Beam does not spread out (1) so electrons are travelling at constant speed if speeds varied, faster electrons would be deflected less than slower electrons. (1)

b 1 _ 2 mv2 = eVA

v = √ 

_

( 2 eV A

_ m ) (1)

c i Magnetic force on electron = electric force on electron (1)

Bev = eV p

_ d so v = eVp

____ Bd (1)

ii Combining the two equations and cancelling e:

√ 

_

( 2 eV A

_ m ) = eV p

_ Bd (1)

Squaring both sides of the equation:

2 eV A

_ m = V p

2 _ B2d 2

e _ m = V p

2 _ 2 V A d 2 B 2 (1)

= 4600 2 __________________ 2 × 3900 × ( 5 × 10 −2 ) 2 × ( 2.4 × 10 −3 ) 2 (1)

= 1.88 × 1011 C kg−1 (1)

iii Value obtained is a little higher but of the same order of magnitude (1) as that obtained by Thomson and the accepted value of today.

Student could lower Vp (1) or increase VA or B very slightly assuming the plate separation is fixed within the tube.

2 a Magnetic force on each electron in the beam is perpendicular to velocity no work is done on each electron by (magnetic) force so ke / speed is constant. (1) Magnitude of magnetic force is constant because speed is constant. (1) Magnetic force is always perpendicular to velocity so is centripetal. (1) [Max 3 marks]

b rearranging r = mv ___ Be

e __ m = v ___ Br (1)

= 7.5 × 106

_________ 5.4 × 10−4 × 78 × 10−3 (1)

= 1.78 × 1011 C kg−1 (1) (to 2 s.f.) (1)

185–188 Wave–particle duality (1)

Grade booster1 (a) An electromagnetic wave is propagated through space

due to electric and magnetic fields at 90° to each other and perpendicular to the direction of travel of the wave. (3)

(b) An induced emf is detected by the meter in the loop and must be caused by the radio waves passing through the loop and causing a changing magnetic flux around the loop.

(c) (i) A stationary wave is set up as the transmitted and reflected waves superpose. The maxima are maximum amplitude and are at the antinodes where the two waves constructively interfere and are in phase. The minima are at nodes and are zero amplitude where the two waves destructively interfere and are in anti phase.

(ii) Examiner’s comment: The student has failed to recall that the distance between adjacent nodes in

a stationary wave = λ _ 2 but has been awarded the

middle mark with error carried forward. The final mark has not been awarded as it should be obvious the answer was incorrect as the speed of all em waves = 3 × 108 m s−1.

Mark = 1 out of 3

Distance between nodes = λ _ 2

λ = 0.66m

c = f × λ = 4.5 × 108 × 0.66

= 2.97 × 108 m s−1

Exam-style questions1 a waves are emitted by each slit (1) each slit diffracts

light (1) the two slits are coherent sources of light waves (1) bright fringes formed where light from one slit reinforces light from the other slit (or dark fringes formed where light from one slit cancels light from the other slit) (1) path difference to a bright fringe = whole number of wavelengths (or path difference to a dark fringe = (whole number + half) wavelengths)) (4)


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b Corpuscles pass through a slit in a straight line and form a bright fringe (1) two slits produce only two bright fringes according to corpuscular theory (1) many / more than two fringes are observed due to diffraction and interference (1) dark fringes (or cancellation) cannot happen with corpuscles (1) (Max 3)

c Newton’s scientific pre-eminence (or Huygens’ theory considered light waves as longitudinal and therefore could not explain polarisation). (1)

d – Laser light is coherent so single slit is not needed.

– Slit-screen distance as long as possible to reduce uncertainty. Must be greater than 1 m.

– Measure across at least 5 fringes and find the mean fringe spacing to reduce uncertainty.

– Use Vernier callipers to measure across the fringes and to measure the slit separation. (4)

189–192 Wave–particle duality (2)

Grade booster1 (a) Condenser lens makes the beam of electrons parallel

and uniform across the sample. Objective lens forms a magnified image of the sample. Projection lens forms a further magnified image on a screen

(b) Image would be brighter because a higher anode voltage will accelerate the electrons more, so more electrons reach the screen per sec. According to the

de Broglie equation λ = h _ mv , a higher speed means a

smaller de Broglie wavelength which means better resolution and more detail in the image.

(c) Lens aberrations: the blurring of the image, is caused by the electrons having a range of speeds and so they are focused to different points on the image by the lens.

or

A thicker sample can cause a loss of speed, causing the wavelength to change by different amounts and each wavelength will then diffract differently.

(d) (i) Ek = eV = 1.6 × 10−19 × 20 × 103 = 3.2 × 10−15 J (1)

1 _ 2 mv2 = Ek

v2 = 2 E k _ m = 2 × 3.2 × 10 −15 ______________ 9.11 × 10 −31 = 7.03 × 1015 (1)

v = 8.38 × 107 m s−1 (1)

(ii) Examiner’s comment: The student has correctly used the value from the Data Booklet for the speed of light but has used the value for G the gravitational constant (6.67 × 10−11) instead of Planck’s constant (6.63 × 10−34) so has not obtained the correct answer. Mark = 0 out of 2

λ = h ___ mv

= 6.63 × 10 −34 ____________ 9.11 × 10 −31 × 8.37 × 10 7 = 8.7 × 1012 m

Exam-style questions1 a i Graph; straight line with a positive gradient (1)

intercept on − y-axis. (1)

f0

Vs

f

A

ii Rearranging hf = φ + Ek(max) and eVs = Ek(max)

hf = φ + eVs (1)

VS = hf

_ E − φ

 _ e (1)

y = mx + c

Graph of VS against f is a straight line with gradient h _ e (1) and y-intercept =

− φ _ e (1)

b hf = hc _ λ

= 6.63 × 10 −34 × 3 × 10 8 ___________ 418 × 10 −9 = 4.76 × 10−19 J (1)

EKmax = eVS = 1.6 × 10−19 × 1.92 = 3.07 × 10−19 J (1)

φ = hf − EK(max)

= 4.76 × 10−19 − 3.07 × 10−19

= 1.69 × 10−19 (1) J (1) (or 1.06 eV)

c parallel line above the previous one (with a smaller but still negative intercept on the y-axis) (1)

d Max 3 from:

Wave model predicts an increase in the photocurrent (1)

Because energy is transferred into each electron increases over time / electrons can gain sufficient KE to escape / electrons can leave the surface with greater KE. (1)

Photon model predicts photocurrent remains at zero (1)

Because the energy of a photon depends on the frequency not the intensity / energy of each incident photon remains the same / KE of electrons leaving the surface does not change / electrons cannot escape. (1)

193–196 Special relativity

Grade booster1 (a) (i) t0 = 800 s

t = t0 ( 1 − v 2 _ c 2 ) − 1 _ 2

= 800 (1 − 0.9942) − 1 __ 2

= 7300 s

(ii) distance = 0.994 × c × t

= 0.994 × 3 × 108 × 7300

= 2.2 × 1012 m (2.18 × 1012 m)


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(b) Examiner’s comment: The student has remembered the ‘twins paradox’ and that the space twin is younger. However, the student has not used the time dilation equation to confirm that reasoning. To improve the answer the student needed to add ‘In space the twin’s travel time = proper time = t0 whereas the time on

Earth is given by t where = t 0 (1− v 2 _ c 2 ) − 1 _ 2

. (1− v 2 _ c 2 ) − 1 _ 2

is

fractional so t > t0.

Mark = 1 out of 3

Exam-style questions1 a v = distance _ time = 44 _ 153 × 10 −9

= 2.88 × 108 m s−1 (1)

t = 153 ns (1)

= t0 ( 1 − v 2 _ c 2 ) − 1 _ 2

to = t ( 1 − v 2 _ c 2 ) 1 _ 2

= 153 × 10−9 × √ ___________

( 1 − 2.88 2 _ 3 2 ) (1)

= 42.9 ns (1)

Difference in time = 153 − 42.9 = 110.1 ns (1)

Observed time >> proper time

b m = m0 ( 1 − v 2 _ c 2 ) − 1 _ 2

10 = 1 ( 1 − v 2 _ c 2 ) − 1 _ 2

(1)

( 1 − v 2 _ c 2 ) 1 _ 2

= 0.1

Squaring both sides of the equation:

1 − v 2 _ c 2 = 0.12

v2 = [1 − 0.12] c2

v = c √ _

[1 − 0.1 2 ] (1)

= 0.995 c

= 2.985 × 108 m s−1 (1)

2 Distance between detectors in rest frame of particles:

L = L0 ( 1 − v 2 _ c 2 ) 1 _ 2

= 25 × ( 1 − 0.98 2 ) 1 _ 2 = 5.0 m (1)

Time taken in rest-frame of particles = distance _ speed

= 5 _ 0.98c = 1.7 × 10−8 s (1)

Time taken to decrease by 1 _ 4 = 2 half-lives (1)

Half-life = 1.7 × 10 −8 _ 2 = 8.5 × 10−9 s (1)

3 a l = υt = 1.50 × 108 × 12 × 10−9 = 1.80 m (1)

= l0 [1 − ( v 2 _ c 2 ) ] 1 _ 2

1.8 = l0 [1 − ( 1.5 2 ____ 3 2 ) ] 1 __ 2

(1)

l0 = 1.8 _ 0.866

= 2.08 m (1)

Difference in length = 2.08 − 1.80 = 0.3 m (1)

b i m = m0 ( 1 − v 2 _ c 2 ) − 1 _ 2

= mo ( 1 − 1.5 2 _ 3 2 ) − 1 _ 2

= m o _ 0.75

= 1.3 mo (1)

KE = mc2 − moc2

= (1.3mo − mo)c2

= 0.3moc2 (1)

= 0.3 × 1.67 × 10−27 × (3 × 108)2

= 4.51 × 10−11 J (1)

ii Total KE = (107 × 4.51 × 10−11) = 4.51 × 10−4 J (1)

Power = energy

_ time = 4.51 × 10 −4 _ 12 × 10 −9

= 37 575 W (1)

4 Explanation of how shift expected

– Mirror 2 lies in the direction of the Earth's velocity.

– Speed of light different in the two directions.

– The time taken for light to travel from P to Mirror 2 and back to P would be greater than the time taken from P to Mirror 1 and back to P.

– If the speed of light depends on the Earth's velocity through the ether.

– Rotating the apparatus through 90° would cause the time difference to reverse / change.

– When rotated there would be a change in the phase difference between the waves (at each point in the fringe pattern).

Results compared with prediction

– The apparatus was capable of detecting shifts of 0.05 fringe.

– No shift was detected then or in later experiments when apparatus rotated.

Conclusions

– The experiment showed that there is no absolute motion.

– Ether did not exist so light travels without the need for a material medium.

– The Earth was dragging the ether with it.

– Speed of light does not vary in free space as Einstein’s theory states.

6 marks: discussion of most of the statements in points 1, 2, 3.

5 marks: discussion of most of the statements in points 1, 2, 3 with minor omissions or errors.

4 marks: discussion of two points from 1+3 and 1 point from 2.

3 marks: discussion of one point from 1+2+3.

2 marks: comments on two of bullet points 1, 2 or 3.

1 marks: relevant comment on at least one point.


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v

Answers

198–205 Paper 1

Guided questions1 a total momentum before collision = total momentum

after collision (1)

provided no external force acts (1)

b ρ = m _ V

m = ρ × V

= 7480 × 1.34 × 10−6 (1)

= 0.0100232 kg (1) must be quoted to more than 3 s.f.

c p = mu

= 10 × 10−3 × 150 (1)

= 1.50 kg m s−1 or N s (1)

Total mass after collision = mass of bullet + mass of block = 0.30 kg (1)

Total momentum before collision = Total momentum after collision

1.5 = 0.3 v

v = 5.0 m s−1 (1)

d Kinetic energy = 1 _ 2 × 0.3 × 5.02 (1)

= 3.75 J

KE lost = GPE gained = mgh (1)

3.75 = 0.3 × 9.81 × h

h = 3.75 _________ 0.3 × 9.81 (1)

= 1.27 m (1) (to 3 s.f.) (1)

e Assumption made is that all KE is converted to GPE (1), but in fact there will be friction as the bullet enters the block (1), so the block will get warmer. Some of the energy will be transferred from the kinetic energy store to the thermal energy store of the block (1) and not as much to the gravitational potential energy store, so the block will not rise as high. (1) There will also be some air resistance as the block rises reducing the height further. (1)

f In the second experiment the pellet rebounds, so the change in momentum of pellet greater and therefore the change in momentum of the block is greater. (1)

Initial speed of block is greater. (1)

Mass stays the same so the initial KE of block greater. (1)

Therefore, more GPE of block and height reached by steel block is greater than with the wooden block. (1)

2 a – Only hadrons (baryons and mesons) experience strong interaction. Exchange particle is the pion

– Hadrons (baryons and mesons) and leptons experience weak interaction. Exchange particle is W+, W− or Z boson

– All charged particles experience electromagnetic interaction. Exchange particle is the photon.

Examples: Weak β− decay Electrons repelling electromagnetic

1 mark 2 interactions or 1 interaction +1 property or an example.2 marks 2 interactions + 1 property for 1 interaction or an example.3 marks 2 interactions + 1 property for each + 1 example.4 marks 2 interactions + 1 property for each + 2 examples.5 marks 3 interactions +1 property for each + 2 examples.6 marks 3 interactions + 2 properties for each + 2 examples of interactions.

b (Energy of positron + rest energy of positron and electron) is transferred to energy of 2 photons

Energy of 2 photons = 2 + (2 × 511) keV (1)

= 1024 × 103 × 1.6 × 10−19 J

= 1.64 × 10−13 J (1)

Energy of 1 photon = 8.2 × 10−14 J (1)

c Kinetic energy = 1 _ 2 m v2 = 2000 × 1.6 × 10−19 (1)

v = √ ___________

2 ( 2000 × 1.6 × 10 −19 ) ___________ 9.11 × 10 −31 (1)

= 2.65 × 107 ms−1 (1)

3 a YM = FL _ AΔL

ΔL = 580 × 15 ___________ 1.7 × 10 −5 × 1.2 × 10 11 (1)

= 4.26 × 10−3 m (1)

= 4.3 × 10−3 m (to 2 s.f.) (1)

b 580 = Tcos 67

T = 580 _ cos 67

= 1484 N (1)

c (4)

– Wind produces a wave / disturbance that travels along the wire

– Wave is reflected at each end / waves travel in opposite directions

– Incident and reflected waves interfere / superpose

– Only certain frequencies as fixed ends have to be nodes.

d M = ρ × V

Mass per m of wire μ = ρ × A

= 8900 × 1.7 × 10−5

= 0.151 kg (1)

Using the equation for the first harmonic from the Data Booklet and doubling the frequency:

f = 1 _ L [ √ _____

  (  T __ μ ) ]

= 1 _ 15 × √ ________

  ( 580 _____ 0.151 ) (1)

= 4.13 (1) Hz (1)

e Diagram showing two approximately equally spaced loops (1)

f Copper may be stretched beyond elastic limit / may deform plastically (1)

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Paper_Worked solutions.indd 31 30-Oct-19 19:31:00

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Permanent deformation / Does not return to original length (1)

4 a curve of decreasing gradient (1) followed by horizontal straight line (1)

Time0

0

Velocity

b initial acceleration / increase of speed as gravity is acting weight > drag, (1) resistance forces increase as speed increases (1) so acceleration gradually decreases

until weight = drag, or resultant force = 0 (1) reaches terminal / constant speed / velocity acceleration = 0 Newton’s first law applies (1)

c

r / m × 10−3 t / s v / ms−1 log10 (r / m) log10 (v / m s −1 )

3.40 7.2 0.1180 −2.46 −0.928

4.12 4.9 0.1735 −2.38 −0.761

1 mark per correct row (2)

d axes labelled (1) suitable scales chosen (1) points plotted correctly (1) line of best fit (1)

–2.8

–2.7

–2.6

–2.5

–2.4

–2.3

e v = arb

log v = b log r + log a (1)

y = mx + c

b = gradient = − 1.5 − ( − 0.8 ) ______________ − 2.752 − ( − 2.4 ) = − 0.7 _ − 0.35 = 2 (1)

Using 2 points which are on the graph line:

−1.575 = 2 × (−2.79) + log a eqn (1)

−1.124 = 2 × (−2.56) + log a eqn (2)

Add eqns 1 and 2

−2.699 = −5.58 − 5.12 + 2 log a

10.70 − 2.699 = 2 log a

8.001 = 2 log a

log a = 4.0005

Intercept on y-axis = log a = 4.0005 (1)

a = 104.0005 = 10011 = 104 m−1 s−1 (1)

This means that the terminal velocity is proportional to (radius of the ball bearing)2. (1)

or v = 104r2

5 a P = V 2 _ R

R = V 2 _ P

= 144 _ 60 (1)

= 2.4 Ω (1)

ρ = RA _ L

= 2.4 × 1.8 × 10 −8 _____________ 5 × 10 −2 (1)

= 8.64 × 10−7 (1) Ω m (1)

b Work done per unit charge by the battery or electrical energy gained per coulomb of charge passing through the battery (2)

c ε = V + Ir

12 = 11.8 + 9.2r (1)

r = 0.2 _ 9.2

r = 0.022 Ω (1)

d Brightness decreases as the current in circuit and in the battery is increased (1) so the lost volts increases leading to a decreased pd across working bulb / decreased terminal pd as its resistance is constant (1)

6 a For each spring, change in force ΔF = kΔL

= 25 × 50 × 10−3 = 1.25 N (1)

resultant force = 2ΔF = 2.5 N (1)

b Acceleration a = F _ m = 2.5 _ 0.7 = 3.6 m s−2 (1) to

the right (1)

c Acceleration is proportional to displacement from equilibrium position. (1)

Acceleration is in opposite direction to displacement acceleration is towards a fixed point (or towards equilibrium position). (1)

d The period of a mass–spring system is T = 2π √ _

( m _ k )

For this system, T = 2π √ _

( m _ 2k ) (1)

e T = 2π √ _

( 1 × 10 −25 _ 2 × 200 )

= 2π √ ____________

   ( 2.5 × 10 −28 ) s (1)

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Frequency = 1 _ T = 1.0 × 1013 Hz (1)

vmax = 2π fA = 2π × 1013 × 10−11

= 628 m s−1 (1)

max EK = 1 _ 2 mvmax2

= 1 _ 2 × 1.0 × 10−25 × 6282

= 2.0 × 10−20 J (1)

(f) Max 2 from:

Interatomic spacing ≈ 10−11 m so the amplitude is the absolute maximum and KE is unlikely to be quite as big. (1)

Mass of ion ≈ 10−25 kg is a good estimate for an ion such as copper but for those with smaller atomic numbers the max KE will be less. (1)

At cooler temperatures the max KE will be less as the amplitude of vibration will be less. (1)

7 C Three protons creating the excess charge (1)

8 C Increasing intensity only increases the number of photons released (1)

9 C Speed less in glass (1)

10 B (1)

Exam-style questions1 a Set up circuit as shown in diagram, using a new cell (1)

as the emf reduces with use. Measure current with an ammeter in series and the terminal pd with voltmeter. (1) Use a variable resistor or a selection of resistors to vary the external resistance in the circuit and so vary the current. (1) Check for zero errors on the meters and correct if necessary (1) Repeat the experiment to average out any random fluctuations in the meters. (1)

AV

R

rε

b There is a pd across internal resistance or energy is lost due to the internal resistance (1) so that the pd across internal resistance (or lost volts) increases as current increases. (1)

c emf = y-intercept = 1.52 V ± 0.01 V (1)

Gradient = r = 1.12 _ 2.5 (1)= 0.45 Ω (1) ± 0.02 Ω

d Same intercept (same emf) (1) and twice gradient (2r). Line must go through (1.25, 0.40) (1)

e Same intercept (same emf) horizontal line (gradient = 0) (1)

f Q = lt

= 0.89 × 20

= 17.8 (1) C (1)

g P = I2r

= 0.892 × 0.45 (1)

= 0.36 W (1)

2 a v = rω and ω = 2πf

= 2 π × 50 × 0.012 (1)

= 3.77 m s−1 (1)

a = v 2 _ r

= ( 3.77 ) 2 _ 0.012 (1)

= 1.2 × 103 m s−2 (1)

b Panel resonates because the motor frequency = natural frequency of panel (2)

c tighten it up if possible or add some insulation / padding (1) to damp the vibration (1)

3 a V across resistor = 3.0 − 2.2 = 0.8 V (1)

R = V _ I

= 0.8 _ 35 × 10 −3

= 22.9 Ω (1)

b Q in 1 s = 0.035 (C) (1)

No. of electrons (in 1 s) = 0.035 ÷ 1.6 × 10−19

= 2.19 × 1017 (1)

c E = hf = hc _ λ

= 6.63 × 10 −34 × 3 × 10 8 ___________ 635 × 10 −9 (1)

= 3.13 × 10−19 J (1)

d P = VI

= 2.2 × 0.035

= 0.077 W (1)

Maximum no. of photons emitted per sec. = 0.077 ÷ 3.13 × 10−19

= 2.46 × 1017 (1)

4 a v = h _ mλ

= 6.63 × 10 −34 ___________________ 9.1 × 10 −31 × 1.2 × 10 −9 (1)

= 6.1 × 105 m s−1 (1)

b Ek = 1 _ 2 mv2

= 1 _ 2 9.1 × 10−31 × (6.1 × 105)2 (1)

= 1.7 × 10−19 J (1)

Total E = Ek + eV = 1.7 × 10−19 + (− 1.6 × 10−19 × 2.8) (1)

= − 2.8 × 10−19 J (1)

c Energy of photon = hc _ λ

= 6.63 × 10 34 × 3 × 10 8 ___________ 650 × 10 −9 (1)

= 3.1 × 10−19 J (1)

d Electron can gain enough energy to escape (1) as energy of photon > total E. (1)

5 a Automatic timing and distance recording. Distance measurements made later from photograph. (1) Less uncertainty or reduces random errors (1).

b Average = 2.20 ± 0.05 m s−1 (1) larger values are more reliable / accurate (1)

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c t / s x / m y / m (

y _ t ) / ms−1

0.05 0.116 0.093 1.86

0.10 0.220 0.210 2.10

0.15 0.324 0.350 2.33

0.20 0.442 0.518 2.59

0.25 0.548 0.710 2.84

0.30 0.660 0.922 3.07

graph: straight line y _ t against t axes labelled correctly

(1) suitable scale chosen so that points fill more than half of both axes (non-zero origin needed) (1) points correctly plotted (1) best straight line (1)

1.8

1.6

2.0

2.2

2.4

2.6

2.8

3.0

3.2

0.10 t / s0.15 0.20 0.25 0.30

ms–1yt

0.05

d a = y-intercept = 1.62 ± 0.04 m s−1 read from graph (1)

b = large triangle (1)

gradient = 3.14 − 1.80 ______________ 0.3125 − 0.0375 (1)

= 4.87 ± 2.5 m s−2 (1)

e in vertical direction using SUVAT, y = ut + 1 _ 2 gt2

y _ t = u + 1 _ 2 gt

a: initial vertical component of velocity uv (1)

b = 1 _ 2 g (1)

f u2 = uh2 + uv

2 by Pythagoras theorem (1)

u = ( 1.62 2 + 2.2 2 ) 1 _ 2

= 2.7 ± 0.1 m s−1 (1)

g In horizontal direction, x = 0.77m, uh = 2.2 ms−1 t =?

Speed = distance _ time

Time to reach floor = distance _ speed

= 0.77 _ 2.2

= 0.35 s (1)

In vertical direction uv = 1.62 ms−1 t = 0.35 s

y = ut + 1 _ 2 gt2

= 1.62 × 0.35 + 1 _ 2 × 9.81 × 0.352 (1)

= 1.17 m (1)

6 D Leptons do not contain quarks, a muon is a lepton (1)

7 C Li-7 has 3 protons, 4 neutrons, 3 electrons so 7 hadrons. 7 baryons and 3 leptons (1)

8 C λ = v _ f = 30 _ 150 = 0.2 m = 20 cm 5 cm = 1 _ 4 of 20 cm

so 360 _ 4 = 90o (1)

9 A s = ut + 1 _ 2 at2 = 0 + 1 _ 2 × 10 × 9 = 45 m (1)

10 B Area under graph = F × t = 1 _ 2 × 10 × 40 = 200.

Ft = mv − mu. v = Ft _ m = 200 _ 0.5 = 400 m s−1 (1)

11 A v2 = u2 +2as = 100 + (2 × 1.6 × 120) = 484,

v = 22 ms−1 (1)

12 C Same material means YM is the same (1)

206–213 Paper 2

Guided questions1 a n = PV _ RT

= 3.1 × 10 5 × 2.1 × 10 −3 ___________ 8.31 × 286

= 0.274 mol (1)

b P2 = T 2 × P 1 _ T 1

= 295 × 3.1 × 10 5 _____________ 286 (1)

= 3.198 × 105 Pa (1) = 3.20 × 105 Pa (to 3 s.f.) (1)

c Both have random motion / range of speeds. (1)

At the higher temperature the mean kinetic energy / root mean square speed / frequency of collisions of gas molecules with the walls is greater. (1)

2 a E = Q _ 4 πε o r 2

= 29 × 1.6 × 10 −19 _________________ 4 × π × 8.85 × 10 −12 × ( 1.2 × 10 −10 ) 2 (1)

= 2.90 × 1012 (1) Vm−1 or NC−1 (1)

b V = − GM _ R

= − 6.67 × 10 −11 × 65 × 1.661 × 10 −27 _______________ 1.2 × 10 −10 (1)

= − 6.00 × 10−26 (1) J kg−1 (1)

c F = Q 1 Q 2 _ 4 πε o r 2

= ( 1.6 × 10 −19 ) 2 _________________ 4 × π × 8.85 × 10 −12 × ( 1.25 × 10 −15 ) 2 (1)

= 147 N (1)

F = GM 1 M 2 _ R 2

= 6.67 × 10 −11 × ( 1.67 × 10 −27 ) 2 ______________ ( 1.25 × 10 −15 ) 2 (1)

= 1.19 × 10−34 N (1)

Gravitational force is much weaker than the electrostatic force. (1)

d 120

Neutron no. N

Proton no. Z

80

N = Z line

40

00

20 40 60 80 100

correct N scale (1) correct Z scale (1) correct line (1)

e Max 5 from:

At low neutron number strong nuclear force acts on neutrons (1) strong and electrostatic forces act

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on protons (1) strong nuclear force > electrostatic repulsion (1) so nuclei are stable when N = Z (1)

At higher neutron number, N > Z, (1) extra neutrons separate the protons (1) and reduce the electrostatic repulsion (1)

3 a time for V to halve = 8 ms = 0.008 s (1)

T 1 _ 2 = 0.69 RC

C = 8 × 10 −3 _ 0.69 × 150 (1)

= 77 μF (1)

b (Max 3)

• As capacitor discharges, pd decreases

• Current through resistor decreases (as I α V )

• Rate at which charge leaves the capacitor decreases

(as I = ΔQ

_ Δt )

• Rate of change of charge is proportional to rate of change of pd (as V α Q)

c Energy stored = 1 _ 2 CV 2

Initial energy = 1 _ 2 × 77 × 10−6 × 1.42

= 75.5 μJ (1)

Final energy = 1 _ 2 × 77 × 10−6 × 0.382

= 5.6 μJ (1)

Energy loss = 75.5 −5.6 = 69.9 μJ (1)

d Charge released = Qo − Q0.015

= C (Vo − V0.015)

= 77 × 10−6 (1.4 − 0.38) (1)

= 79 μC (1)

e Charge required = 79 × 10−6 × 60 every minute (1)

= 79 × 10−6 × 60 × 60 × 24 × 365 every year

= 2491.34 C every year (1)

Battery charge available = 4 × 3600 C

= 14 400 C (1)

Battery life = 14400 _ 249 1.34

= 5.78 years (1)

Needs replacing every 5 years (1)

4 a u_

u_

d_ (1)

b ud_ (1)

c mv 2 _ r = Bev

r = mv _ Be

= 1.67 × 10 −27 × 1.5 × 10 7 ___________ 0.16 × 1.6 × 10 −19 (1)

= 0.98 m (1)

d v, Q and B are the same for pions and protons

Mass of proton > mass of pion

Using r = mv _ Be .

Radius of pion < radius of proton

Pion path more curved than proton path (1)

e Max 3 from:

v = BQr

_ m (1)

Magnetic field B must be increased to increase the centripetal force. (1)

B must increase as v increases to keep r constant. (1)

Otherwise protons would attempt to travel in a path of larger radius. (1)

5 a Energy to heat water to 100 oC, = 180 × 4200 × 81 = 61.2 MJ (1)

Energy to vaporise water = 180 × 2.3 ×106 = 414 MJ (1)

Energy transferred (per sec) = (414 + 61.2) MJ = 475 MJ (1)

b Mass of rocks = 5.0 × 106 × 3200 = 1.6 × 1010 kg (1)

Temperature fall of ∆T in 1 day,

Energy removed = 1.6 × 1010 × 850 × ∆T

= 1.36 × 1013 ∆T (1)

Energy transfer in one day = 475 × 106 × 3600 × 24 = 4.104 × 1013 J (1)

ΔT = 4.104 × 10 13 ___________ 1.36 × 10 13

= 3.02 K (1)

c Number of nuclei in 1 kg of U-238 = 6.02 × 10 23 _ 0.238

= 2.5(3) × 1024 (1)

Activity of 1 kg of U 238 = λN

= ln 2 _ T 1 _ 2 × 2.53 × 1024 (1)

= 0.693 × 2.53 × 10 24 _____________ 4.5 × 10 9 × 365 × 24 × 3600

= 1.24 × 107 s−1 (1)

Energy released per sec per kg of U 238

= 1.24 × 107 × 4.2 × 1.6 × 10−13

= 8.30 × 10−6 J (1)

Mass of U-238 needed = 480 × 10 6 _ 8.30 × 10 −6

= 5.78 × 1013 kg (1)

6 A

V = −2000 = − GM _ R ,

V2 = − GM _ 2R = − 2000 _ 2 = −1000 J kg−1

7 B

V = E × d = 2000 × 0.05 = 100,

W = QV = 1.6 × 10−19 × 100 = 1.6 × 10−17 (1)

8 B

F = Q 1 Q 2 _ 4 πε o r 2 = 2 × 1.6 × 10 −19 × 79 × 1.6 × 10 −19 ________________ 4 × π × 8.85 × 10 −12 × ( 4 × 10 −14 ) 2

= 22.73 N (1)

9 A

Q = Qoe

Q

_ Q 0 = e

= e−2

= 0.135 (1)

10 B

Z behaves the same as X as rubber will not create a back emf (1)

−t _ RC

−4.8 _________________ ( 120 × 10 −6 × 20 × 103 )

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11 C

V p

_ N p =

V s _ N s V s = 120 × 230 _ 2400 = 11.5 Vrms

Vpeak = 11.5 × √ _ 2

= 16.26 V

Peak power = V 2

___ R = (16.26)2

_______ 5 = 53 W (1)

12 A

alpha particles and the nucleus are both positive so must repel (1)

13 C (1)

14 B

4 × 1.00728u + 3 × 1.00867u = 7.05513u

Δm = 7.05513u − 7.01473u = 0.0404u

Binding energy = 0.0404 × 931.5 MeV

= 37.63 _ 7

= 5.37 MeV nucleon−1 (1)

15 D (1)

Exam-style questions1 a Energy produced per minute = 4.2 × 103 × 60

= 2.52 × 105 J (1)

Volume of gas = energy / min

___________ Energy / m 3 = 2.52 × 10 5 _ 35 × 10 6

= 7.2 × 10−3 m3 (1)

b ρ = m _ V

m = V × ρ = 7.2 × 10−3 × 0.72

= 5.18 × 10−3 kg (1)

Mass of one molecule = 1.6 × 10 −2 _ 6.02 × 10 23

= 2.66 × 10−26 kg (1)

Number of molecules = 5.18 × 10 −3 _____________ 2.66 × 10 −26 kg

= 1.95 × 1023 (1)

c Energy per minute = 2.52 × 105 J (1)

∆Q = mc∆θ

m = ΔQ

_ c Δθ

= 2.52 × 10 5 _____________ 990 × ( 35 − 15 ) (1)

= 12.7 kg (1)

d P = IV

I = P _ V

= 4200 _ 230

= 18.3 A (1)

Current required for 4.2 kW would exceed fuse rating (1)

2 a ρ = m _ V

m = ρ × V

= 1000 × 900 × 10−6

= 0.9 kg (1)

Energy required to reduce temperature of the water to 0° C = mc∆θ

= 0.9 × 4200 × (19 − 0)

= 71.8 × 103 J (1)

Energy required to freeze the water = ml

= 0.9 × 3.4 × 105

= 306 × 103 J (1)

Energy removed per sec = ( 71.8 + 306 ) × 10 3 J _________________ 1700 (1)

= 222 J s−1 (1)

b Energy transferred from mains = (25 × 1700) = 4.25 × 104 J (1)

Total energy transferred to surroundings

= 4.25 × 104 + 71.8 × 103 + 306 × 103 (1)

= 4.2 × 105 J (1)

3 Gravitational forces F:

1 The spacecraft experiences opposite gravitational attractions to both the Earth and the Moon during its journey.

2 Earth mass >> Moon mass, for most of the outward journey the FE > FM.

3 In the later stages of the outward journey FM > FE.

4 On the return journey FE > FM for most of the way

Gravitational field strength g:

1 During the outward journey gE becomes weaker and gM becomes stronger.

2 The resultant field strength is the vector sum of gE and gM separately.

3 A point X is reached at which gE and gM are equal and opposite, giving zero resultant. X is much closer to M than to E.

4 Once X has been passed, the spacecraft will be attracted to M by M’s gravitational field or on the return journey the spacecraft will ‘fall’ to E once it is beyond X.

Gravitational potential V:

1 VE increases (that is becomes less negative) as the spacecraft moves away from E.

2 The resultant gravitational potential is the (scalar) sum of VE and VM separately.

3 At X the gravitational potential reaches a maximum value before decreasing as M is approached.

4 In order to reach M on the outward journey, the spacecraft has to be given at least enough energy to reach X, and vice-versa for the return.

5 Much more work is needed to move the spacecraft from E to X than from M to X, as a larger force has to be overcome over a larger distance.

6 marks reference to all 4 force points + 4 field points + 5 potential points.

5 marks reference to all 4 force points + 4 field points + 4 potential points.

4 marks reference to 3 force points + 3 field points + 3 potential points.



1 mark reference to 2 force points + 1 field points + 0 potential points.

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4 a Max 2 from:

– use long-handled tongs

– use lead shielding

– keep source in a lead lined box when not in use

– warning signs in lab.

b Random error (1) resulting in a scatter of points around the line best fit. (1)

c Large triangle (1)

values read from graph:

5.21 − 4.19 _ 140 − 0 (1)

Gradient = − 0.00729 min−1 (1)

d Count rate Cc = activity of source (1)

Activity A = A0 e−λt

Taking logs of both sides:

ln A = −λt + ln A0 (1)

Comparing with y = mx + c:

Gradient = −λ (1) and intercept on ln A axis = ln A0

e Gradient = λ = 7.0 × 10−3 min−1

T 1 _ 2 = 0.693 _ λ 

= 0.693 _ 7 × 10 −3 (1)

= 99 mins (1)

f Uncertainty = √ ____

455 = 21.33 (1)

% uncertainty = uncertainty

_ value × 100 %

= 21.33 _ 455 × 100

= 4.7 % (1)

uncertainty = √ ___

80 = 8.94 (1)

% uncertainty = 8.94 _ 80 × 100

= 11.1 % (1)

Comparing % uncertainties the longer time gives a much lower % uncertainty / reduces the % uncertainty by half. (1)

g Absorption of radiation and spreading out due to the inverse square law (1) leads to less radiation reaching detector and a smaller count. (1) This will lead to a greater % uncertainty (1) and so less accurate results.(1) Background count has a greater effect on a smaller count. (1) (Max 3)

h Same gradient as same half-life. (1) Lower intercept as lower count rate due to smaller total count. (1) Points may not be as close to line of best fit. (1)

5 a 233 1 91 139

U + n → Kr + Ba + X n

92 0 36 56 (2)

b

142

141

140

13990 91 92 93 94

Proton number Z

143

233U

P

Q

92Neutron number N

(2)

i x = 4 (1)

ii LHS eqn = 232.98915u + 1.00867u = 233.99782 u (1)

RHS eqn = (90.90368 + 138.87810 + 4 × 1.00867)] u = 233.81646 u (1)

Δ m = LHS − RHS

= 0.18136 u (1)

Energy released = 0.18136 × 931.5 = 169 MeV (1)

6 A

1 _ 2 m crms2 = 3 _ 2 kT ; crms α √

__ T ) 17.11 _ 16.52 = 1.037 (1)

7 B

KE = PE = m ΔV = m × ( GM _ R ) (1)

8 D

FE = QE = 1.6 × 10−19 × 5000 = 8 × 10−16 (1)

FG = GMm _ r 2 = 6.67 × 10 −11 × 5.97 × 10 24 × 9.1 × 10 −31 __________________ ( 6.37 × 10 6 ) 2

= 8.93 × 10−30

FE : FG = 9.0 × 1013 (1)

9 D

Units can be NC−1 or Vm−1 the other three are equivalent to one of these two (1)

10 A

C = L 2 ε o _ D C 2 =

4 L 2 ε o × 2 _ D = 8C (1)

11 B

r = p _ Be r α =

2p _ B2e (1)

12 D

T = distance _ velocity = 2πr _ v v = rBe _ m T = 2πrm _ rBe (1)

13 B

P s _ P p

= 0.75 = 60 _ 230 × I p I p = 60 _ 230 × 0.75 = 0.35 A (1)

214–217 Paper 3A

Guided questions1 a Circuit diagram consisting of thermistor, power pack,

switch and ammeter in series with a voltmeter in parallel with the thermistor. Thermistor on long leads in beaker of water / ice. Thermometer in water and possibly a stirrer. (1)

Place thermistor in ice, allow time for temperature distribution, (1) record current on ammeter, voltage on voltmeter and temperature using thermometer. (1) Add some cold water to raise temp. to about 10 oC, stir to obtain even temp. distribution and wait before taking measurements. Continue adding cold water and / or warm water to obtain a range of measurements every 10 oC (1) up to about 70 oC (1).

b Max 4 from:

• Any zero errors (systematic) on meters or thermometer should be added or subtracted appropriately from the readings obtained.

• When water is added it takes time to achieve thermal equilibrium (an even temperature distribution), stirring the water will help so that recorded temperatures are as accurate as possible.

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• A water bath may be used to obtain an even temperature and reduce random error.

• Thermometers should be read at eye level to reduce parallax error.

• There may be fluctuations in the meter readings. (random errors) so readings should be repeated if possible.

c The student should have recorded the current and voltage data, not just the resistance values which is derived data not the original values (1) and has not given all the resistance values to the same number of decimal points. (1)

d R = ae b _ T

Taking logs of both sides:

lnR = b ( 1 _ T ) + ln a (1)


A graph of ln R against 1 _ T (1) will be a straight line of gradient b and intercept on the y-axis of ln a (1).

e 1 mark for each correct row (2)

θ / oC R / Ω T / K ( 1 _ T ) / 10−3-K−1 ln R / Ω

60 24.0 333 3.003 3.1870 20.1 343 2.915 3.00

f Graph scales chosen so that graph values occupy over half of each axis (1) axes labelled and with units (1) points plotted (1) line of best fit (1).

2.8 2.9 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.73.0

3.2

3.4

3.6

3.8

4.0

4.2

4.4

4.6

ln (R / Ω)

1 / T / 10–3 K–1

g Large triangle (1)

Gradient = 4.54 − 3.2 ___________ ( 3.65 − 3.005 ) × 10 −3 (1)

= 2.078 × 103 K (1)

h b = 2.078 × 103 K (1)

Using (ln R = 3.92, 1 __ T = 3.356 × 10−3) using values from the graph (1)

3.92 = ln a + (2.078 × 103 × 3.356 × 10−3)

Ln a = 3.92 – 6.972

= –3.052

a = 0.047 Ω (1)

i Yes it would be suitable for lower temperatures (1) as the change in resistance per K is greater between 0 oC and 10 oC than it is between 30 oC and 40 oC. (1)

2 a Measure from bench to ruler at both ends / in several positions. (1) Use set-square to adjust spring perpendicular to ruler (1) or Use a spirit level for ruler (1) and set-square for spring (1) or spirit level for both ruler and spring. (2)

b Fiducial marker / pin horizontal from clamp stand at same height as ruler’s rest position. (1)

c Begin counting as ruler passes horizontal fiducial marker, not at extremities of oscillation. (1) The fiducial marker precisely identifies the beginning and ending of an oscillation. (1)

d 20 oscillations to reduce % uncertainty in timing (1) due to reaction time and random error. (1)

e As shown for T (1) with heading and units for last column (1) and decimal places correct. (1)

Distance from hinge

Time for 20 oscillations Av. time for 20

Time period

a / m t1 / s t2 / s t3 / s t / s T / s T2 / s2

0.300 9.79 9.92 9.99 9.90 0.495 0.2450.400 11.20 11.07 11.21 11.16 0.558 0.3110.500 12.83 12.70 12.51 12.68 0.634 0.4020.600 13.54 13.67 13.71 13.64 0.682 0.4650.700 14.88 14.73 14.55 14.72 0.736 0.5420.800 15.60 15.71 15.79 15.70 0.785 0.6160.900 16.51 16.62 16.67 16.60 0.830 0.689

f Suitable graph scales so that graph covers more than half of the grid, (1) axes labelled with units and T 2 on y-axis (1) points plotted. (1)

Line of best fit drawn with equal numbers of points on either side of line. (1)

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

0.1

0.2

0.3

0.4

T2 / s2

0.5

0.6

0.7

a / m

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g Predicted graph is T 2 = 4 π 2 ma _ kX

It is a straight line through the origin so T 2 is proportional to a (1)

h Large triangle (1)

Gradient = 0.695 − 0.1 _ 0.9 − 0.13 (1) = 0.595 _ 0.77

= 0.77 (1) s2 m−1 (1)

i Comparing with y = mx + c the gradient

will be 4 π 2 m _ kX

k = 4 π 2 m ___________ gradient × X

= 4 × π 2 × 0.8 _ 0.77 × 0.9 (1)

= 45.6 kg s−2 (1)

j Uncertainty in t = 0.5 × range = 0.5 × 0.2 = 0.1 (1)

% uncertainty in t = 0.1 _ 9.9 × 100 = 1.01 % (1)

k % uncertainty in T is the same = 1.01 % (1)

l % uncertainty in a = 0.002 _ 0.300 × 100 = 0.6 % (1)

% uncertainty in X = 0.002 _ 0.900 × 100 = 0.2 % (1)

% uncertainty in T 2 = 2 (% uncertainty in T )

% uncertainty in k = % uncertainty in m + % uncertainty in a + % uncertainty in T 2 +

% uncertainty in X

= 2 + 0.6 + 2 (1.01) + 0.2 (1)

= 4.82 % (4.8 % to 2 s.f.) (1)

Exam-style questions1 a Laser light is coherent, so coherent light is needed to

illuminate the double slit. (1) Light from a single slit is coherent, that is same frequency and same phase relationship as the two light rays from the single slit travel the same distance to the double slit. (1)

b 3.65 cm across 20 fringe widths (1) mean fringe width 36.5 _ 20 = 1.825 mm or 1.825 × 10−3 m (1)

c w = λD _ s


The graph of w against D _ s will be a straight line (1) of gradient λ. (1)

s / 10−3 m D / m ( D __ s ) / 103 w / 10−3m

0.70 1.100 1.57 1.120.70 1.000 1.43 1.030.70 0.900 1.29 0.930.70 0.800 1.14 0.840.70 0.600 0.86 0.661.0 1.100 1.10 0.821.0 1.000 1.00 0.761.0 0.900 0.90 0.681.0 0.800 0.80 0.621.0 0.600 0.60 0.50

d Scales chosen so that values cover more than half of both axes (1) axes labelled and units (1) points plotted

to within 1 _ 2 mm square (1) line of best fit. (1)

0.20 0.4 0.6 0.8 1.0 1.2 1.4 1.6

0.25

0.5

0.75

1.0

1.25

w / 10–3 m

Ds

e large triangle (1) λ = gradient = ( 1.13 − 0.25 ) × 10 −3 _________________ ( 1.6 − 0.2 ) × 10 3 (1)

= 6.29 × 10−7 m (1)

f Fringe width = 0.125 mm from the graph

when D _ s = 0 = intercept on w-axis

or can be found by substituting known values from the

graph into w = mD _ s + c

w = 6.29 × 10−7 D _ s + (0.125 × 10−3) (1)

g A systematic error in the reading of w (1) so the Vernier calliper probably had a zero error of 0.125 mm (1) which the student didn’t notice.

h Uncertainty of Vernier calliper = ± 0.01 mm (1)

i Largest uncertainty in λ is for the smallest value of w, D and s

% uncertainty in λ = % uncertainty in w + % uncertainty in D + % uncertainty in s

= 0.01 _ 0.5 × 100 + 0.002 _ 0.6 × 100 + 0.02 _ 0.7 × 100 (1)

= 2 + 0.33 + 2.86

= 5.19 % (1)

Uncertainty in λ = ± 3.26 × 10−8 m (1)

j Error bars (1) line of worst fit (1) as shown in d.

k Gradient of line of worst fit = ( 1.16 − 0.25 ) × 10 −3 _________________ ( 1.6 − 0.29 ) × 10 3

= 6.95 × 10−7 m

% uncertainty in gradient = ( 6.95 − 6.29 ) × 10 −7 _________________ 6.29 × 10 −7 × 100

= 10.5% (1)

Uncertainty in λ = ± 6.60 × 10−8 m (1)

l The uncertainty in the wavelength from the gradient in k is larger than the uncertainty using w, D and s in i. (1) This is because the error bars only take into account w not D and s, (1) the worst gradient encompasses all the error bars from the top of the error bar of the point most above the line of best fit, to the bottom of the error bar of the point most below the line of best fit. (1)

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218–221 Paper 3BA

Guided questions1 a Diagram to show:

• both focal points coinciding and labelled, with fo > fe centre ray straight through objective (1)

• rays crossing at focal plane and proceeding to eyepiece (1)

• rays refracted at eyepiece and emerge parallel to construction line. (1)

b fo + fe = 3.5

f o _ f e

= 100

fo = 100fe

100fe + fe = 3.5

fe = 3.5 / 101 = 0.03465 m

fo = 3.5 − 0.03465 = 3.465 m

estimate fo ≈3.5 m (1) and fe ≈ 0.035 m (1)

c M = β

_α

α = 4 × 10 −3 _ 100

= 4 × 10−5 rad (1)

α = D _ r

D = 4 × 10−5 × 1.3 × 109 (1)

= 5.2 × 104 km (1)

d Max 4 from:

• no chromatic aberration – mirrors do not refract light

• no spherical aberration – use of parabolic mirror

• no distortion – mirror can be supported more strongly

• better resolving power or greater brightness – mirrors can be larger

• more light gets through (image brighter) – lens absorbs more light.

2 Apparent magnitude:

• means how bright a star appears on the Earth.

• brightness on Earth depends on how bright a star actually is and how far away it is. Brighter, far away stars can appear the same brightness as closer, dimmer stars.

• backward scale: he gave larger values of apparent magnitude to the dimmer stars.

• a difference of 1 on the apparent magnitude scale is associated with a difference in brightness of a factor of 2.51.

The Hipparcos scale:

• assumed the brightest stars have a magnitude of 1 and the dimmest a magnitude of 6.

• magnitude 1 stars are 100 times brighter than magnitude 6 stars.

• applying apparent magnitude to the Hipparcos scale means that some stars (the brightest) have a negative apparent magnitude.

5–6 marks: Clear description of what is meant by apparent magnitude, with discussion of logarithmic scale;

explanation of Hipparcos scale in terms of visibility of stars.

3–4 marks: Either apparent magnitude or Hipparcos scale discussed in detail, with other not discussed fully or containing errors.

1–2 marks: Either one discussed fully and other not at all, or both discussed with some errors.

3 a λmaxT = 0.0029 gives λ = 0.0029 _ 5800 (1)

= 5.0 × 10−7 m (1)

b values on axis : 0.5 1.0 1.5 2.0 (1) with 0.5 at the peak (1)

c for Arcturus, λ = 0.0029 _ 4300 = 6.74 × 10−7 m

giving a similar shaped curve with a lower peak (1) shifted to right at approx. 6.67 on λ-axis (1)

d Arcturus has the lower and more negative absolute magnitude, therefore is the brighter star (1)

difference in absolute magnitude ≈ 5 (1)

difference in actual brightness = 2.515 = 100 (1)

e Stefan’s law P = σAT 4

P A

_ P s = 100 =

σ A A T A 4 _ σ A s T s

4 (1)

A A

_ A s = 100 × 5800 4 _ 4300 4

AA = 331 As (1)

4 × π × rA2 = 331 × 4 × π rs

2

rA = 1.27 × 1010 m (1)

f Sun = G class (1) Arcturus = K class (1) g orange (1)4 a Apparent magnitude is the brightness of a star as seen

from Earth (1) Absolute magnitude is the brightness at a distance of 10 pc from the Earth. (1)

b Antares is much brighter (1) as it has a lower value of apparent magnitude. (1)

c Proxima Centauri:

m − M = 5 log ( d _ 10 )

11 − M = 5 log ( 1.3 _ 10 ) (1)

M = 11 + 4.43 = 15.4 (1)

Antares

m − M = 5 log ( d _ 10 )

1.0 − M = 5 log ( 160 _ 10 )

M = 1.0 − 6.02 = −5.02 (1)

d correct main sequence (1) correct Giants (1) and White Dwarfs (1)

e Sun: spectral class G abs mag = 5 (1) Proxima Centauri: spectral class M abs mag = 15 (1)

Antares: spectral class M, abs magnitude = –5 (1)

f Both in spectral class M (1) so have temperature less than 3500 K. (1)

g Antares is brightest at same temperature. (1) so has largest surface area and so diameter. (1)

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5 a c = fλ

Δf = 3 × 10 8 _ 540 × 10 −9 − 3 × 10 8 _ 561 × 10 −9 (1)

= (5.556 − 5.348) × 1014 (1)

= 2.08 × 1013 Hz (1)

b Δf

_ f = v _ c

v = c × Δf

_ f (1)

= 3 × 10 8 × 2.08 × 10 13 ___________ 5.556 × 10 14 (1)

= 1.12 × 107 ms−1 (1)

c d = v _ H

= 11.2 × 10 3 _ 65 (1)

= 172 Mpc

= 172 × 106 × 3.08 × 1016 (1)

= 5.30 × 1024 m (1)

Exam-style questions1 a Hole in mirror

Convexsecondary mirror

Concave / parabolicprimary mirror

(3)

b θ = λ _ D

= 2 × 10 −6 _ 3.8 (1)

= 5.26 × 10−7 rad (1)

Visible wavelengths are shorter than infrared (1) making resolving angle θ smaller and so giving a better resolving power. (1)

c rays

Rays close to axis and far from axis to different focal points (2)

d Use of parabolic mirror prevents spherical aberration (1)

e D = λ _ θ 

= 5.7 × 10 −7 _ 1.1 × 10 −5 (1)

= 5.2 × 10−2 m (1)

2 a Black hole: large gravitational field and is very dense (1) nothing can escape as the escape velocity v >> c. (1)

Quasar: large red shift and is very far away (1) bright, very powerful radio sources. (1)

Quasars are produced by supermassive black holes situated at the centre of galaxies. (1)

b Rs = 2GM _ c 2 = 2 × 6.67 × 10 −11 × 3 × 10 9 × 1.99 × 10 30 ___________________ ( 3 × 10 8 ) 2 (1)

= 8.8 × 1012 m (1)

3 a v = Hd so gradient = H (1)

1 ly = 9.46 × 1015 m

gradient = 68 × 10 6 ___________ 4 × 10 9 × 9.46 × 10 15 (1)

= 1.8 × 10−18 s−1 (1)

b Uncertainty is in the measurement of distance (1)

c v = d _ t (1) and v = Hd

d _ t = Hd

H = 1 _ t (1)

d Red shift = 469 − 393 nm

= 76 nm (1)

Δ λ _ λ = v _ c

v = 76 × 10 −9 × 3 × 10 8 ________________ 393 × 10 −9 (1)

= 5.8 × 107 m s−1 (1)

4 Big Bang:

• The Universe has expanded from a single hot dense point

• Approximately 13 billion years ago

• Replaced steady state theory where everything continues as it was.

Evidence comes from:

Hubble relationship

• observations of the red shift of distant galaxies according to Hubble’s law

• showing galaxies are moving outwards from a single common point.

Cosmological microwave background radiation:

• disproved the steady state theory

• in all directions / isotropic

• follows a black body radiation curve corresponding to a temperature of 2.7 K

• interpreted as the left-over radiation from the Big Bang,

% of Hydrogen and helium in the Universe:

• present in the ratio 3:1

• supports the idea that a very brief period of fusion occurred when the Universe was very young,

• then universe cooled and is consistent with the Big Bang theory.

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1–2 marks: one piece of evidence or meaning of Big Bang

3 – 4 marks: explains meaning of Big Bang reasonably + 2 pieces of evidence

5–6 marks: explains meaning of Big Bang + 3 pieces of evidence

222–224 Paper 3BD

Guided questions1 a Cathode rays (1)

b Electrons excite the gas atoms by collision (1) and photons of light are emitted as de-excitation occurs. (1)

c Electrons would collide with the gas atoms and lose energy or be scattered (1) and not have sufficient energy to cause excitation. (1)

d Cathode rays move from cathode toward anode. (1) The paddle wheel moves in that direction so it must have gained kinetic energy or momentum from cathode rays, so ‘particles’ could be hitting it. (1)

2 a At terminal velocity, viscous force = weight

6πηrv = mg = 4πr3 ρg

_ 3 (1)

r2 = 9ηv

_ 2ρg

r = ( 9 ηv

_ 2ρg ) 1 _ 2

(1)

b v = 2.40 × 10 −3 _ 27.6 = 8.7 × 10−5 m s−1 (1)

r = [ ( 9 × 1.8 × 10 −5 × 8.7 × 10 −5 ) _____________ ( 2 × 960 × 9.8 ) ]

1 _ 2

(1)

= 8.65 × 10−7 m

m = 4 _ 3 πr3 × ρ

= 4 _ 3 π (8.65 × 10−7)3 × 960 (1)

= 2.6 × 10−15 kg (1)

c Electric force = the droplet weight

QV

_ d = mg

Q = mgd

_ V

= 2.6 × 10 −15 × 9.81 × 7 × 10 −3 ______________ 370 (1)

= 4.8 × 10−19 C (1) to 2 s.f. (1)

d The charge on each droplet is a whole number × 1.6 × 10−19 C. (1)

A quantum of charge is 1.6 × 10−19 C / charge is quantised. (1)

e Experimental result matches Millikan’s conclusion (1)

4.8 × 10−19 = 3 × 1.6 × 10−19 = 3e (1)

3 a Stationary wave pattern is formed by incident and reflected waves (1) detector reading is zero at nodes due to destructive interference (1) nodes are at spacing equal to one-half wavelength. (1)

b λ = 0.36 × 2 = 0.72m

c = fλ = 4.2 × 108 × 0.72 (1)

= 3.024 × 108 m s−1 (1)

c Hertz used Maxwell’s equation c = 1 ________ √ _

( μ o ε o ) (1) that

predicted a value for em waves of 3 × 108 m s−1 to confirm that radio waves were electromagnetic in nature. (1)

4 a Matter particles have wave-like properties (1) and an

associated wavelength = h _ p where p is the momentum

of the particles. (1)

b EK = 0.023 eV = 0.023 × 1.60 × 10−19 or 3.68 × 10−21 J (1)

EK = 1 _ 2 mv2

v = ( 2 E K

_ m ) 1 _ 2

= ( 2 × 3.68 × 10 −21 ______________ 1.67 × 10 −27 ) 1 _ 2

= 2.099 × 103 m s−1

mv = 1.67 × 10−27 × 2.099 × 103 = 3.506 × 10−24 kg m s−1 (1)

λ = h _ mv = 6.63 × 10 −34 ___________ 3.506 × 10 −24

= 1.89 × 10−10 m (1) = 1.9 × 10−10 m (to 2 s.f.) (1)

c λ = h _ mv

Electron momentum (p) = neutron momentum (1)

But electron mass << neutron mass (1)

Velocity electron << neutron speed (1)

KE = 1 _ 2 mv2 = p 2

_ 2m

p is same for neutron and electron and electron has smaller mass.

KE electron << KE neutron

5 a The laws of physics are the same in all inertial frames of reference. (1)

The speed of light in free space is invariant. (1)

b 22 GeV = 22 × 109 × 1.6 × 10−19 or 3.52 × 10−9 J (1)

Energy = m 0 ( 1 − v 2 _ c 2 ) −1 _ 2

c 2

3.52 × 10−9 = 9.11 × 10−31 ( 1 − v 2 _ c 2 ) −1 _ 2

× (3 × 108)2

( 1 − v 2 _ c 2 ) 1 _ 2

= 9.11 × 10 −31 × ( 3 × 10 8 ) 2 ____________ 3.52 × 10 −9 (1)

= 2.33 × 10−5 (1)

c l = l0 ( 1 − v 2 _ c 2 ) 1 _ 2

= 3000 × 2.33 × 10−5 (1)

= 0.0699 = 0.070 m (to 2 s.f.) (1)

d • Starts at mo

• Shallow increase to no more than 2mo at 0.7c

• Then curves sharply upwards becoming greater than 0.9c at 6mo

• Never greater than 1.0 c / asymptote shown (2)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

5mo

0

10mo

0.8 0.9 1.0 1.1

Mass m

Speed c

Exam-style questions1 a Photon theory:

• light consists of photons (1)

• an electron in the metal absorbs a photon (1)

• an electron needs a minimum amount of energy to escape (1)

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• a blue photon has more energy than a red photon (1)

or hf > φ for blue photon, hf < φ for red photon (1)

Wave theory

• every electron would gain sufficient energy from the waves to escape in time (1)

• no matter what the frequency / colour / wavelength of the light (1)

Max 5 marks with at least 1 mark from wave theory for explanation + 1 mark for evaluation: wave theory does not explain the observation whereas the photon theory does.

b hc _ λ = φ + Ek max

6.63 × 10 −34 × 3 × 10 8 ___________ 4.7 × 10 −7 = φ + 2.3 × 10−19 (1)

φ = 4.23 × 10−19 − 2.3 × 10−19 = 1.93 × 10−19 J (1)

= 1.93 × 10 −19 _ 1.6 × 10 −19 = 1.2 eV (1)

c hf0 = φ

f0 = φ

 _ h

= 1.93 × 10 −19 _ 6.63 × 10 −34 (1)

= 2.9 × 1014 Hz (1)

d Threshold frequency is the minimum frequency of light (1) at which electrons are emitted from the metal surface. (1) Below this frequency, the incident photons do not have enough energy (E = hf) to release the electrons from the metal surface. (1) Different metals have different work functions and therefore different threshold frequencies. (1)

2 a Speed = distance _ time

Time = distance _ speed

= 10.8 × 10 3 _ 2.86 × 10 8

= 3.78 × 10−5 (s) (1)

No. of half-lives = 3.78 × 10 −5 _ 2.2 × 10 −6

= 17.165 or 17.17 (1)

b 2.6 × 108 × ( 1 _ 2 ) 17

(1) = 1983 = 1.98 × 103 (to 3 s.f.) (1)

c Any one from:

• when v ≈ c, less time passes in muon’s reference frame for the journey (so fewer decay)

• when v ≈ c, journey is shorter in length for the muon’s frame of reference (so fewer decay)

• when v ≈ c, muons are observed to travel further in a half-life (on Earth) than expected (so fewer decay during journey)

• when v ≈ c, muon’s half−life is observed to be longer (on Earth) (so fewer decay) (1)

d l = l 0 ( 1 − v 2 _ c 2 ) 1 _ 2

= 10.8 × 103 ( 1 − 2.86 2 _ 3 2 ) 1 _ 2

(1)

= 3260

to= l _ v

= 3260 _ 2.86 × 10 8 (1)

= 1.14 × 10−5 s (1)

e T 1 _ 2 = ln 2 _ λ

λ = 0.693 _ 2.2 × 10 −6

= 3.15 × 105 s−1 (1)

N = N0 e− λt

= 2.6 × 108 e− 315000 × 1.14 × 10−5 (1)

= 7.17 × 106 (1)

3 a Electrons are negatively charged so the beam is attracted to positive plate. (1)

b Beam would be spread out (1) with faster electrons deflected less than slower electrons. (1)

c The current raises the temperature of the filament (1) electrons gain sufficient KE to leave the metal / to overcome the work function. (1)

d Perpendicularly (1) into the plane of the diagram. (1)

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