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
, 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
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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