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 l× l10–6 l ✓
0.22 lm l s–2 l ✓
3
(c) 0.10(3)m l(allow lecf lfrom l(i))✓
1
(d) Force l on la lparticle l= lmg land
lacceleration l= lF l/ lm lso lalways l=
lg✓
Time lto lfall l(given ldistance) ldepends l(only) lon lthe ldistance land lacceleration✓
OR:
g l= lGM l/ lr2 l ✓
Time lto lfall l= l√2s l/ lg
so lno lm lin lequations lto ldetermine ltime lto lfall✓
2
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, Cardinal Newman College
(e) Mass l is lnot lconstant lsince lparticle lmass lwill
lvary✓ lCharge lon la lparticle lis lnot lconstant✓
Acceleration l= lEq l/ lm lor l(V l/ ld) l(q l/ lm) lor lVq l/ ldm✓
E lor lV l/ ld lconstant lbut lcharge land lmass lare l‘random’ lvariables lso lq l/ lm
lwill lvary l(or lunlikely lto lbe lthe lsame)✓
4
[12]
M4.(a) (i) (Mass lchange lin lu=) 1.71× l10−3 l (u)
or l(mass lBe−7) l‒ l(mass lHe−3) l‒ l(mass lHe−4) lseen lwith lnumbers
C1
2.84 l× l10−30 l (kg)
or lConverts ltheir lmass lto lkg
Alternative l2nd lmark:
Allow lconversion lof l1.71 l× l10−3 l (u) lto lMeV lby
lmultiplying lby l931 l(=1.59 l(MeV)) lseen
C1
Substitution lin lE l= lmc2 condone ltheir lmass
ldifference lin lthis lsub lbut lmust lhave lcorrect lvalue lfor
lc2 l (3×108)2 l or l9×1016
Alternative l3rd lmark:
Allow ltheir lMeV lconverted lto ljoules l(× l1.6 l× l10−13) lseen
C1
2.55 l× l10−13 l (J) lto l2.6 l× l10−13 l (J)
Alternative l4th lmark:
Allow l2.5 l× l10−13 l (J) lfor lthis lmethod
A1
4
(ii) Use lof lE=hc l/ lλ ecf
C1
Correct lsubstitution lin lrearranged lequation lwith lλ
subject lecf
C1
7.65 l× l10−13 l (m) lto l7.8 l× l10−13 l (m)ecf
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, Cardinal Newman College
A1
3
(b) (i) Use lof lEp l formula:
C1
Correct lcharges lfor lthe lnuclei land lcorrect lpowers lof l10
C1
2.6(3) l× l10−13 l J
A1
3
(ii) Uses lKE l= l3 l/ l2 lkT: lor lhalves lKET, lKE= l1.3 l× l10−13 l (J)
seen lecf
C1
Correct lsubstitution lof ldata land lmakes lT lsubject ecf
Or luses lKET l value land ldivides lT lby l2
C1
6.35 l× l109 l (K) lor l6.4 l× l109 l (K) lor l6.28 l× l109(K) lor l6.3 l×
109 l (K) lecf
A1
3
(c) (i) Deuteron l/ ldeuterium l/ lhydrogen−2
B1
Triton l/ ltritium l/ lhydrogen−3
B1
2
(ii) Electrical lheating l/ lelectrical ldischarge l/ linducing la
lcurrent lin lplasma l/ luse lof le−m lradiation l/ lusing
lradio lwaves l(causing lcharged lparticles lto
lresonate)
B1
1
[16]
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, Cardinal Newman College
M5. l(a) force lbetween ltwo l(point) lcharges lis
proportional lto lproduct lof lcharges l✓
inversely lproportional lto lsquare lof ldistance lbetween lthe lcharges l✓
Mention lof lforce lis lessential, lotherwise lno lmarks.
lCondone l“proportional lto lcharges”.
Do lnot lallow l“square lof lradius” lwhen lradius lis lundefined.
Award lfull lcredit lfor lequation lwith lall lterms ldefined.
2
(b) V lis linversely lproportional lto lr l[or lV l∝ l (−)1 l/ lr l] l✓
(V lhas lnegative lvalues) lbecause lcharge lis lnegative
[or lbecause lforce lis lattractive lon l+ lcharge lplaced lnear lit
or lbecause lelectric lpotential lis l+ lfor l+ lcharge land l− lfor l− lcharge] l✓
potential lis ldefined lto lbe lzero lat linfinity l✓
Allow lV l× lr l= lconstant lfor l1st l mark.
max l2
(c) (i) Q(= l4πɛ0 l rV l) l= l4πɛ0 l × l0.125 l× l2000
OR lgradient l= lQ l/ l4πɛ0 l = l2000 l/ l8 l✓
(for lexample, lusing lany lpair lof lvalues lfrom lgraph) l✓
= l28 l(27.8) l(± l1) l(nC) l✓
(gives lQ l= l28 l(27.8) l±1 l(nC) l✓
2
(ii) at lr l= l0.20m lV l= l−1250V land lat lr l= l0.50m lV l=
l−500V lso lpd lΔV l= l−500 l− l(−1250) l= l750 l(V) l✓
work ldone lΔW l(= lQΔV) l= l60 l× l10−9 l × l750
= l4.5(0) l× l10−5 l (J) l(45 lμJ) l✓
(final lanswer lcould lbe lbetween l3.9 land l5.1 l× l10−5)
lAllow ltolerance lof l± l50V lon lgraph lreadings.
l[Alternative lfor l1st l mark:
ΔV l= (or lsimilar lsubstitution lusing
l60 lnC
instead lof l27.8 lnC:
use lof l60 lnC lgives lΔV l= l1620V) l]
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 l× l10–6 l ✓
0.22 lm l s–2 l ✓
3
(c) 0.10(3)m l(allow lecf lfrom l(i))✓
1
(d) Force l on la lparticle l= lmg land
lacceleration l= lF l/ lm lso lalways l=
lg✓
Time lto lfall l(given ldistance) ldepends l(only) lon lthe ldistance land lacceleration✓
OR:
g l= lGM l/ lr2 l ✓
Time lto lfall l= l√2s l/ lg
so lno lm lin lequations lto ldetermine ltime lto lfall✓
2
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, Cardinal Newman College
(e) Mass l is lnot lconstant lsince lparticle lmass lwill
lvary✓ lCharge lon la lparticle lis lnot lconstant✓
Acceleration l= lEq l/ lm lor l(V l/ ld) l(q l/ lm) lor lVq l/ ldm✓
E lor lV l/ ld lconstant lbut lcharge land lmass lare l‘random’ lvariables lso lq l/ lm
lwill lvary l(or lunlikely lto lbe lthe lsame)✓
4
[12]
M4.(a) (i) (Mass lchange lin lu=) 1.71× l10−3 l (u)
or l(mass lBe−7) l‒ l(mass lHe−3) l‒ l(mass lHe−4) lseen lwith lnumbers
C1
2.84 l× l10−30 l (kg)
or lConverts ltheir lmass lto lkg
Alternative l2nd lmark:
Allow lconversion lof l1.71 l× l10−3 l (u) lto lMeV lby
lmultiplying lby l931 l(=1.59 l(MeV)) lseen
C1
Substitution lin lE l= lmc2 condone ltheir lmass
ldifference lin lthis lsub lbut lmust lhave lcorrect lvalue lfor
lc2 l (3×108)2 l or l9×1016
Alternative l3rd lmark:
Allow ltheir lMeV lconverted lto ljoules l(× l1.6 l× l10−13) lseen
C1
2.55 l× l10−13 l (J) lto l2.6 l× l10−13 l (J)
Alternative l4th lmark:
Allow l2.5 l× l10−13 l (J) lfor lthis lmethod
A1
4
(ii) Use lof lE=hc l/ lλ ecf
C1
Correct lsubstitution lin lrearranged lequation lwith lλ
subject lecf
C1
7.65 l× l10−13 l (m) lto l7.8 l× l10−13 l (m)ecf
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, Cardinal Newman College
A1
3
(b) (i) Use lof lEp l formula:
C1
Correct lcharges lfor lthe lnuclei land lcorrect lpowers lof l10
C1
2.6(3) l× l10−13 l J
A1
3
(ii) Uses lKE l= l3 l/ l2 lkT: lor lhalves lKET, lKE= l1.3 l× l10−13 l (J)
seen lecf
C1
Correct lsubstitution lof ldata land lmakes lT lsubject ecf
Or luses lKET l value land ldivides lT lby l2
C1
6.35 l× l109 l (K) lor l6.4 l× l109 l (K) lor l6.28 l× l109(K) lor l6.3 l×
109 l (K) lecf
A1
3
(c) (i) Deuteron l/ ldeuterium l/ lhydrogen−2
B1
Triton l/ ltritium l/ lhydrogen−3
B1
2
(ii) Electrical lheating l/ lelectrical ldischarge l/ linducing la
lcurrent lin lplasma l/ luse lof le−m lradiation l/ lusing
lradio lwaves l(causing lcharged lparticles lto
lresonate)
B1
1
[16]
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, Cardinal Newman College
M5. l(a) force lbetween ltwo l(point) lcharges lis
proportional lto lproduct lof lcharges l✓
inversely lproportional lto lsquare lof ldistance lbetween lthe lcharges l✓
Mention lof lforce lis lessential, lotherwise lno lmarks.
lCondone l“proportional lto lcharges”.
Do lnot lallow l“square lof lradius” lwhen lradius lis lundefined.
Award lfull lcredit lfor lequation lwith lall lterms ldefined.
2
(b) V lis linversely lproportional lto lr l[or lV l∝ l (−)1 l/ lr l] l✓
(V lhas lnegative lvalues) lbecause lcharge lis lnegative
[or lbecause lforce lis lattractive lon l+ lcharge lplaced lnear lit
or lbecause lelectric lpotential lis l+ lfor l+ lcharge land l− lfor l− lcharge] l✓
potential lis ldefined lto lbe lzero lat linfinity l✓
Allow lV l× lr l= lconstant lfor l1st l mark.
max l2
(c) (i) Q(= l4πɛ0 l rV l) l= l4πɛ0 l × l0.125 l× l2000
OR lgradient l= lQ l/ l4πɛ0 l = l2000 l/ l8 l✓
(for lexample, lusing lany lpair lof lvalues lfrom lgraph) l✓
= l28 l(27.8) l(± l1) l(nC) l✓
(gives lQ l= l28 l(27.8) l±1 l(nC) l✓
2
(ii) at lr l= l0.20m lV l= l−1250V land lat lr l= l0.50m lV l=
l−500V lso lpd lΔV l= l−500 l− l(−1250) l= l750 l(V) l✓
work ldone lΔW l(= lQΔV) l= l60 l× l10−9 l × l750
= l4.5(0) l× l10−5 l (J) l(45 lμJ) l✓
(final lanswer lcould lbe lbetween l3.9 land l5.1 l× l10−5)
lAllow ltolerance lof l± l50V lon lgraph lreadings.
l[Alternative lfor l1st l mark:
ΔV l= (or lsimilar lsubstitution lusing
l60 lnC
instead lof l27.8 lnC:
use lof l60 lnC lgives lΔV l= l1620V) l]
2
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