Cardinal Newman College
Paper 2 – Booklet A - Mark Schemes
Electric Field 1
M1. A
[1]
M2.C
[1]
M3.(a) t= or 4.5 = × 9.81 × t 2 ✓
t = 0.96 s✓
2
(b) Field strength = 186000V m–1✓
Acceleration = Eq / m
or 186 000 × 1.2 × 10–6 ✓
0.22 m s–2 ✓
3
(c) 0.10(3)m (allow ec𝘧 𝘧rom (i))✓
1
(d) Force on a particle = mg and
acceleration = F / m so always = g✓
Time to 𝘧all (given distance) depends (only) on the distance and acceleration✓
OR:
g = GM / r2 ✓
Time to 𝘧all = √2s / g
so no m in equations to determine time to 𝘧all✓
2
Page 1
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(e) Mass is not constant since particle mass will vary✓
Charge on a particle is not constant✓
Acceleration = Eq / m or (V / d) (q / m) or Vq / dm✓
E or V / d constant but charge and mass are ‘random’ variables so q / m will
vary (or unlikely to be the same)✓
4
[12]
M4.(a) (i) (Mass change in u=) 1.71× 10−3 (u)
or (mass Be−7) ‒ (mass He−3) ‒ (mass He−4) seen with numbers
C1
2.84 × 10−30 (kg)
or Converts their mass to kg
Alternative 2nd mark:
Allow conversion o𝘧 1.71 × 10−3 (u) to MeV by
multiplying by 931 (=1.59 (MeV)) seen
C1
Substitution in E = mc2 condone their mass
di𝘧𝘧erence in this sub but must have correct value
𝘧or c2 (3×108)2 or 9×1016
Alternative 3rd mark:
Allow their MeV converted to joules (× 1.6 × 10−13) seen
C1
2.55 × 10−13 (J) to 2.6 × 10−13 (J)
Alternative 4th mark:
Allow 2.5 × 10−13 (J) 𝘧or this method
A1
4
(ii) Use o𝘧 E=hc / λ ec𝘧
C1
Correct substitution in rearranged equation with λ
subject ec𝘧
C1
7.65 × 10−13 (m) to 7.8 × 10−13 (m) ec𝘧
Page 2
, Cardinal Newman College
A1
3
(b) (i) Use o𝘧 Ep 𝘧ormula:
C1
Correct charges 𝘧or the nuclei and correct powers o𝘧 10
C1
2.6(3) × 10−13 J
A1
3
(ii) Uses KE = kT: or halves KET, KE= 1.3 × 10−13 (J)
seen ec𝘧
C1
Correct substitution o𝘧 data and makes T subject ec𝘧
Or uses KET value and divides T by 2
C1
6.35 × 109 (K) or 6.4 × 109 (K) or 6.28 × 109(K) or 6.3 ×
109 (K) ec𝘧
A1
3
(c) (i) Deuteron / deuterium / hydrogen−2
B1
Triton / tritium / hydrogen−3
B1
2
(ii) Electrical heating / electrical discharge / inducing a
current in plasma / use o𝘧 e−m radiation / using
radio waves (causing charged particles to resonate)
B1
1
[16]
Page 3
, Cardinal Newman College
M5. (a) 𝘧orce between two (point) charges is
proportional to product o𝘧 charges ✓
inversely proportional to square o𝘧 distance between the charges ✓
Mention o𝘧 𝘧orce is essential, otherwise no marks.
Condone “proportional to charges”.
Do not allow “square o𝘧 radius” when radius is unde𝘧ined.
Award 𝘧ull credit 𝘧or equation with all terms de𝘧ined.
2
(b) V is inversely proportional to r [or V ∝ (−)1 / r ] ✓
(V has negative values) because charge is negative
[or because 𝘧orce is attractive on + charge placed near it
or because electric potential is + 𝘧or + charge and − 𝘧or − charge] ✓
potential is de𝘧ined to be zero at in𝘧inity ✓
Allow V × r = constant 𝘧or 1st mark.
max 2
(c) (i) Q(= 4πɛ0 rV ) = 4πɛ0 × 0.125 × 2000
OR gradient = Q / 4πɛ0 = ✓
(𝘧or example, using any pair o𝘧 values 𝘧rom graph) ✓
= 28 (27.8) (± 1) (nC) ✓
(gives Q = 28 (27.8) ±1 (nC) ✓
2
(ii) at r = 0.20m V = −1250V and at r = 0.50m V = −500V
so pd ΔV = −500 − (−1250) = 750 (V) ✓
work done ΔW (= QΔV) = 60 × 10−9 × 750
= 4.5(0) × 10−5 (J) (45 μJ) ✓
(𝘧inal answer could be between 3.9 and 5.1 × 10−5)
Allow tolerance o𝘧 ± 50V on graph readings.
[Alternative 𝘧or 1st mark:
ΔV = (or similar substitution using 60
nC
instead o𝘧 27.8 nC:
use o𝘧 60 nC gives ΔV = 1620V) ]
2
Page 4
Paper 2 – Booklet A - Mark Schemes
Electric Field 1
M1. A
[1]
M2.C
[1]
M3.(a) t= or 4.5 = × 9.81 × t 2 ✓
t = 0.96 s✓
2
(b) Field strength = 186000V m–1✓
Acceleration = Eq / m
or 186 000 × 1.2 × 10–6 ✓
0.22 m s–2 ✓
3
(c) 0.10(3)m (allow ec𝘧 𝘧rom (i))✓
1
(d) Force on a particle = mg and
acceleration = F / m so always = g✓
Time to 𝘧all (given distance) depends (only) on the distance and acceleration✓
OR:
g = GM / r2 ✓
Time to 𝘧all = √2s / g
so no m in equations to determine time to 𝘧all✓
2
Page 1
, Cardinal Newman College
(e) Mass is not constant since particle mass will vary✓
Charge on a particle is not constant✓
Acceleration = Eq / m or (V / d) (q / m) or Vq / dm✓
E or V / d constant but charge and mass are ‘random’ variables so q / m will
vary (or unlikely to be the same)✓
4
[12]
M4.(a) (i) (Mass change in u=) 1.71× 10−3 (u)
or (mass Be−7) ‒ (mass He−3) ‒ (mass He−4) seen with numbers
C1
2.84 × 10−30 (kg)
or Converts their mass to kg
Alternative 2nd mark:
Allow conversion o𝘧 1.71 × 10−3 (u) to MeV by
multiplying by 931 (=1.59 (MeV)) seen
C1
Substitution in E = mc2 condone their mass
di𝘧𝘧erence in this sub but must have correct value
𝘧or c2 (3×108)2 or 9×1016
Alternative 3rd mark:
Allow their MeV converted to joules (× 1.6 × 10−13) seen
C1
2.55 × 10−13 (J) to 2.6 × 10−13 (J)
Alternative 4th mark:
Allow 2.5 × 10−13 (J) 𝘧or this method
A1
4
(ii) Use o𝘧 E=hc / λ ec𝘧
C1
Correct substitution in rearranged equation with λ
subject ec𝘧
C1
7.65 × 10−13 (m) to 7.8 × 10−13 (m) ec𝘧
Page 2
, Cardinal Newman College
A1
3
(b) (i) Use o𝘧 Ep 𝘧ormula:
C1
Correct charges 𝘧or the nuclei and correct powers o𝘧 10
C1
2.6(3) × 10−13 J
A1
3
(ii) Uses KE = kT: or halves KET, KE= 1.3 × 10−13 (J)
seen ec𝘧
C1
Correct substitution o𝘧 data and makes T subject ec𝘧
Or uses KET value and divides T by 2
C1
6.35 × 109 (K) or 6.4 × 109 (K) or 6.28 × 109(K) or 6.3 ×
109 (K) ec𝘧
A1
3
(c) (i) Deuteron / deuterium / hydrogen−2
B1
Triton / tritium / hydrogen−3
B1
2
(ii) Electrical heating / electrical discharge / inducing a
current in plasma / use o𝘧 e−m radiation / using
radio waves (causing charged particles to resonate)
B1
1
[16]
Page 3
, Cardinal Newman College
M5. (a) 𝘧orce between two (point) charges is
proportional to product o𝘧 charges ✓
inversely proportional to square o𝘧 distance between the charges ✓
Mention o𝘧 𝘧orce is essential, otherwise no marks.
Condone “proportional to charges”.
Do not allow “square o𝘧 radius” when radius is unde𝘧ined.
Award 𝘧ull credit 𝘧or equation with all terms de𝘧ined.
2
(b) V is inversely proportional to r [or V ∝ (−)1 / r ] ✓
(V has negative values) because charge is negative
[or because 𝘧orce is attractive on + charge placed near it
or because electric potential is + 𝘧or + charge and − 𝘧or − charge] ✓
potential is de𝘧ined to be zero at in𝘧inity ✓
Allow V × r = constant 𝘧or 1st mark.
max 2
(c) (i) Q(= 4πɛ0 rV ) = 4πɛ0 × 0.125 × 2000
OR gradient = Q / 4πɛ0 = ✓
(𝘧or example, using any pair o𝘧 values 𝘧rom graph) ✓
= 28 (27.8) (± 1) (nC) ✓
(gives Q = 28 (27.8) ±1 (nC) ✓
2
(ii) at r = 0.20m V = −1250V and at r = 0.50m V = −500V
so pd ΔV = −500 − (−1250) = 750 (V) ✓
work done ΔW (= QΔV) = 60 × 10−9 × 750
= 4.5(0) × 10−5 (J) (45 μJ) ✓
(𝘧inal answer could be between 3.9 and 5.1 × 10−5)
Allow tolerance o𝘧 ± 50V on graph readings.
[Alternative 𝘧or 1st mark:
ΔV = (or similar substitution using 60
nC
instead o𝘧 27.8 nC:
use o𝘧 60 nC gives ΔV = 1620V) ]
2
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