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Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014
37
Pa
per
1 M
ark
sche
me
Que
stio
n nu
mbe
r
A
ccep
tabl
e an
swer
s
Add
ition
al g
uida
nce
Mar
k
1
D
1
2 A
1 3
D
1
4 D
1 5
B
1
6 C
1 7
D
1
8 D
1 9
C
1
10
A
1
(Tot
al fo
r M
ultip
le C
hoic
e Q
uest
ions
= 1
0 m
arks
)
Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014
38
Que
stio
n nu
mbe
r
A
ccep
tabl
e an
swer
s
Add
ition
al g
uida
nce
Mar
k
11
An
expl
anat
ion
that
mak
es re
fere
nce
to:
• M
ost a
lpha
par
ticle
s pas
s thr
ough
und
efle
cted
(1) O
R so
me
defle
cted
thro
ugh
a sm
all a
ngle
(1)
•
A v
ery
smal
l num
ber a
re d
efle
cted
thro
ugh
an a
ngle
gre
ater
than
90
0 (1
)
• Th
is su
gges
ts th
at th
e al
pha
parti
cles
are
def
lect
ed b
y a
char
ged
nucl
eus t
hat h
as a
ver
y sm
all d
iam
eter
com
pare
d to
that
of t
he
atom
rath
er th
an th
e ch
arge
bei
ng d
istri
bute
d th
roug
hout
the
atom
(1)
•
and
that
mos
t of t
he m
ass o
f the
ato
m is
con
cent
rate
d in
the
nucl
eus r
athe
r tha
n di
strib
uted
thro
ugho
ut th
e at
om (1
)
4
(Tot
al fo
r Q
uest
ion
11 =
4 m
arks
)
Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014
39
Que
stio
n nu
mbe
r
A
ccep
tabl
e an
swer
s
Add
ition
al g
uida
nce
Mar
k
12 (a
) •
Rec
ogni
ses t
hat w
eigh
t act
s at m
idpo
int o
f div
ing
boar
d 1.
8 (m
) fro
m X
(1)
• U
se o
f mom
ent =
per
pend
icul
ar fo
rce
x di
stan
ce (1
) •
Tota
l clo
ckw
ise
mom
ent =
315
0 N
m (1
) •
Rec
ogni
ses t
hat c
lock
wis
e m
omen
t = a
ntic
lock
wis
e m
omen
t (1
) •
F=35
00 N
(1)
Exam
ple
of c
alcu
latio
n:
Tota
l clo
ckw
ise
mom
ent =
(680
x3.6
)+(3
90x1
.8)
= 31
50 N
m
F =
3150
/0.9
= 3
500
N
5 12
(b)
• Th
e fo
rces
are
diff
eren
t typ
es (1
) •
The
forc
es a
ct o
n th
e sa
me
obje
ct (1
)
2
(Tot
al fo
r Q
uest
ion
12 =
7 m
arks
)
Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014
40
Que
stio
n nu
mbe
r
A
ccep
tabl
e an
swer
s
Add
ition
al g
uida
nce
Mar
k
13 (a
) •
The
num
ber o
f cha
rge
carr
iers
incr
ease
s with
tem
pera
ture
(1)
• So
this
low
ers t
he re
sist
ance
(des
pite
the
incr
ease
in la
ttice
vi
brat
ions
) (1)
Acc
ept n
umbe
r of e
lect
rons
. 2
13 (b
) •
R T =
0.7
– 0
.8 kΩ
[rea
d fr
om g
raph
] (1)
•
Use
of V
=IR
(with
3.5
V an
d R T
) to
find
I and
V=I
R (w
ith V
=
5.5
V) to
find
R (1
) •
R=11
00 –
130
0 Ω
(1)
Acc
ept u
se o
f ⎟⎟ ⎠⎞⎜⎜ ⎝⎛
+=
21
1 RRR
VsVo
O
r V
out/(
Vs –
Vou
t) x
RT
= R
Ex
ampl
e of
cal
cula
tion:
𝐼𝐼=3.5
750=0.0047 A
𝑅𝑅=
!.!
!.!!"#=1170 Ω
3 (T
otal
for
Que
stio
n 13
= 5
mar
ks)
Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014
41
Que
stio
n nu
mbe
r
A
ccep
tabl
e an
swer
s
Add
ition
al g
uida
nce
Mar
k
14 (a
) E
ither
•
Cal
cula
te a
ccel
erat
ion
(1)
• U
se o
f 𝐹𝐹=𝑚𝑚𝑚𝑚
(1)
• F
= 38
N (1
) O
R
• C
alcu
late
cha
nge
in m
omen
tum
(1)
• U
se o
f 𝐹𝐹= ∆!"
∆! (1
) •
F =
38 N
(1)
Exam
ple
of c
alcu
latio
n:
𝐹𝐹= !.!" ! !"
!.!"
=37.5 N
3 14
(b)
• U
se o
f 𝑠𝑠=𝑢𝑢𝑢𝑢+ ! !𝑎𝑎𝑡𝑡
! (1
)
• U
se o
f 𝑠𝑠= ! !𝑎𝑎𝑡𝑡
! w
ith v
ertic
al c
ompo
nent
s to
find
t (1)
•
Use
of 𝑠𝑠
=𝑢𝑢𝑢𝑢
with
hor
izon
tal c
ompo
nent
s to
find
s (1)
•
Subt
ract
12
from
thei
r ans
wer
for h
oriz
onta
l dis
tanc
e (1
) •
Dis
tanc
e fr
om n
et =
6 m
(1)
• M
akes
con
clus
ion
whe
ther
the
ball
is w
ithin
the
requ
ired
rang
e of
the
net
(1)
Ans
wer
con
sist
ent w
ith c
alcu
late
d va
lue.
Ex
ampl
e of
cal
cula
tion:
𝑡𝑡 =√ ! ×!.!
!.!"
= 0
.714
s 𝑠𝑠=25 ×0.714=17.85
m
Dis
tanc
e fr
om n
et =
17.
85 –
12
= 5.
9 m
6 (T
otal
for
Que
stio
n 14
= 9
mar
ks)
Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014
42
Que
stio
n nu
mbe
r
A
ccep
tabl
e an
swer
s
Add
ition
al g
uida
nce
Mar
k
15 (a
)*
This
que
stio
n as
sess
es a
stud
ent’s
abi
lity
to sh
ow a
coh
eren
t an
d lo
gica
lly st
ruct
ured
ans
wer
with
link
ages
and
fully
-su
stai
ned
reas
onin
g.
Mar
ks a
re a
war
ded
for i
ndic
ativ
e co
nten
t and
for h
ow th
e an
swer
is st
ruct
ured
and
show
s lin
es o
f rea
soni
ng.
The
follo
win
g ta
ble
show
s how
the
mar
ks sh
ould
be
awar
ded
for i
ndic
ativ
e co
nten
t. N
umbe
r of
indi
cativ
e m
arki
ng
poin
ts se
en in
an
swer
Num
ber o
f m
arks
aw
arde
d fo
r ind
icat
ive
mar
king
poi
nts
6 4
5–4
3 3–
2 2
1 1
0 0
Gui
danc
e on
how
the
mar
k sc
hem
e sh
ould
be
appl
ied:
Th
e m
ark
for i
ndic
ativ
e co
nten
t sho
uld
be a
dded
to
the
mar
k fo
r lin
es o
f rea
soni
ng. F
or e
xam
ple,
an
answ
er w
ith fi
ve in
dica
tive
mar
king
poi
nts w
hich
is
parti
ally
stru
ctur
ed w
ith so
me
linka
ges a
nd li
nes o
f re
ason
ing
scor
es 4
mar
ks (3
mar
ks fo
r ind
icat
ive
cont
ent a
nd 1
mar
k fo
r par
tial s
truct
ure
and
som
e lin
kage
s and
line
s of r
easo
ning
).
If th
ere
are
no li
nkag
es b
etw
een
poin
ts, t
he sa
me
five
indi
cativ
e m
arki
ng p
oint
s wou
ld y
ield
an
over
all
scor
e of
3 m
arks
(3 m
arks
for i
ndic
ativ
e co
nten
t and
no
mar
ks fo
r lin
kage
s).
Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014
43
Que
stio
n nu
mbe
r
A
ccep
tabl
e an
swer
s
Add
ition
al g
uida
nce
Mar
k
15 (a
)*
(con
tinue
d)
The
follo
win
g ta
ble
show
s how
the
mar
ks sh
ould
be
aw
arde
d fo
r stru
ctur
e an
d lin
es o
f rea
soni
ng.
N
umbe
r of m
arks
aw
arde
d fo
r stru
ctur
e of
ans
wer
and
su
stai
ned
line
of
reas
onin
g
Ans
wer
show
s a
cohe
rent
and
logi
cal
stru
ctur
e w
ith li
nkag
es
and
fully
sust
aine
d lin
es o
f rea
soni
ng
dem
onst
rate
d th
roug
hout
2
Ans
wer
is p
artia
lly
stru
ctur
ed w
ith so
me
linka
ges a
nd li
nes o
f re
ason
ing
1
Ans
wer
has
no
linka
ges b
etw
een
poin
ts a
nd is
un
stru
ctur
ed
0
Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014
44
Que
stio
n nu
mbe
r
A
ccep
tabl
e an
swer
s
Add
ition
al g
uida
nce
Mar
k
15 (a
)*
(con
tinue
d)
Indi
cativ
e co
nten
t •
At t
erm
inal
vel
ocity
the
forc
es o
n th
e dr
op a
re b
alan
ced
OR
w
eigh
t = d
rag
• Th
e p.
d. c
reat
es a
n el
ectro
stat
ic fo
rce
actin
g up
war
ds o
n th
e dr
op
• Th
e el
ectro
stat
ic fo
rce
incr
ease
s as p
.d. i
ncre
ases
•
The
net u
pwar
d fo
rce
caus
es th
e dr
op to
hav
e a
nega
tive
acce
lera
tion
• A
s spe
ed d
ecre
ases
the
drag
dec
reas
es
• Th
e dr
op re
mai
ns st
atio
nary
whe
n th
e fo
rces
are
bal
ance
d O
R u
ntil
the
drop
rem
ains
stat
iona
ry w
hen
wei
ght =
el
ectro
stat
ic fo
rce
6
15 (b
) •
Equa
te th
e el
ectri
c fo
rce
and
the
grav
itatio
nal f
orce
(1)
• U
se o
f E=V
/d to
obt
ain
q =
mgd
/V (1
)
qE =
mg
q(V/
d) =
mg
q =
mgd
/V
2
15 (c
) A
n ex
plan
atio
n th
at m
akes
refe
renc
e to
: •
Elec
trost
atic
/upw
ard
forc
e (o
n dr
op) w
ould
be
grea
ter t
han
the
wei
ght/d
ownw
ard
forc
e (1
) •
So d
rop
wou
ld a
ccel
erat
e up
war
ds (1
)
Indi
catio
n of
whi
ch fo
rce
is g
reat
er, u
nbal
ance
d is
in
suff
icie
nt.
2
(Tot
al fo
r Q
uest
ion
15 =
10
mar
ks)
Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014
45
Que
stio
n nu
mbe
r
A
ccep
tabl
e an
swer
s
Add
ition
al g
uida
nce
Mar
k
16 (a
)*
This
que
stio
n as
sess
es a
stud
ent’s
abi
lity
to sh
ow a
coh
eren
t an
d lo
gica
lly st
ruct
ured
ans
wer
with
link
ages
and
fully
-su
stai
ned
reas
onin
g.
Mar
ks a
re a
war
ded
for i
ndic
ativ
e co
nten
t and
for h
ow th
e an
swer
is st
ruct
ured
and
show
s lin
es o
f rea
soni
ng.
The
follo
win
g ta
ble
show
s how
the
mar
ks sh
ould
be
awar
ded
for i
ndic
ativ
e co
nten
t. N
umbe
r of
indi
cativ
e m
arki
ng
poin
ts se
en in
an
swer
Num
ber o
f m
arks
aw
arde
d fo
r ind
icat
ive
mar
king
poi
nts
6 4
5–4
3 3–
2 2
1 1
0 0
Gui
danc
e on
how
the
mar
k sc
hem
e sh
ould
be
appl
ied:
Th
e m
ark
for i
ndic
ativ
e co
nten
t sho
uld
be a
dded
to
the
mar
k fo
r lin
es o
f rea
soni
ng. F
or e
xam
ple,
an
answ
er w
ith fi
ve in
dica
tive
mar
king
poi
nts w
hich
is
parti
ally
stru
ctur
ed w
ith so
me
linka
ges a
nd li
nes o
f re
ason
ing
scor
es 4
mar
ks (3
mar
ks fo
r ind
icat
ive
cont
ent a
nd 1
mar
k fo
r par
tial s
truct
ure
and
som
e lin
kage
s and
line
s of r
easo
ning
).
If th
ere
are
no li
nkag
es b
etw
een
poin
ts, t
he sa
me
five
indi
cativ
e m
arki
ng p
oint
s wou
ld y
ield
an
over
all
scor
e of
3 m
arks
(3 m
arks
for i
ndic
ativ
e co
nten
t and
no
mar
ks fo
r lin
kage
s).
Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014
46
Que
stio
n nu
mbe
r
A
ccep
tabl
e an
swer
s
Add
ition
al g
uida
nce
Mar
k
16 (a
)*
(con
tinue
d)
The
follo
win
g ta
ble
show
s how
the
mar
ks sh
ould
be
awar
ded
for s
truct
ure
and
lines
of r
easo
ning
.
N
umbe
r of m
arks
aw
arde
d fo
r stru
ctur
e of
ans
wer
and
su
stai
ned
line
of
reas
onin
g
Ans
wer
show
s a c
oher
ent
and
logi
cal s
truct
ure
with
lin
kage
s and
fully
sust
aine
d lin
es o
f rea
soni
ng
dem
onst
rate
d th
roug
hout
2
Ans
wer
is p
artia
lly
stru
ctur
ed w
ith so
me
linka
ges a
nd li
nes o
f re
ason
ing
1
Ans
wer
has
no
linka
ges
betw
een
poin
ts a
nd is
un
stru
ctur
ed
0
Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014
47
Que
stio
n nu
mbe
r
A
ccep
tabl
e an
swer
s
Add
ition
al g
uida
nce
Mar
k
16 (a
)*
(con
tinue
d)
Indi
cativ
e co
nten
t •
The
supp
ly c
reat
es a
cha
ngin
g m
agne
tic fi
eld
in th
e iro
n co
re
• R
ate
of c
hang
e of
flux
in to
othb
rush
coi
l is e
qual
to ra
te o
f ch
ange
of f
lux
in c
harg
er c
oil (
for a
n id
eal t
rans
form
er)
• Th
e ch
angi
ng fl
ux li
nkag
e in
the
coil
of th
e to
othb
rush
in
duce
s an
e.m
.f. a
ccor
ding
to F
arad
ay’s
law
•
E =
– N
dφ/
dt so
to st
ep d
own
the
e.m
.f. th
ere
mus
t be
few
er
turn
s in
the
toot
hbru
sh c
oil
• Th
e e.
m.f.
in th
e to
othb
rush
coi
l mus
t be
larg
er th
an th
e to
othb
rush
bat
tery
•
Dio
de is
incl
uded
so b
atte
ry is
not
dis
char
ged
by th
e al
tern
atin
g e.
m.f.
Allo
w p
rovi
des d
c to
cha
rge
batte
ry o
r sim
ilar.
6 16
(b)(
i) R
= 47
.4 Ω
(1)
Exam
ple
of c
alcu
latio
n:
R =
2.7
V/0
.057
A =
47.
4 Ω
1
16 (b
)(ii)
•
Use
of Ɛ
=V+I
r or
corr
ect a
ttem
pt to
find
r (1
) •
r = 5
7.9 Ω
or f
ind
ratio
! ! (1)
•
Mak
es c
oncl
usio
n by
com
parin
g r a
nd R
, rec
ogni
sing
m
axim
um p
ower
supp
lied
whe
n ! !=1
(1)
Ans
wer
con
sist
ent w
ith c
alcu
late
d va
lue.
Ex
ampl
e of
cal
cula
tion:
𝑟𝑟=(6.0 𝑉𝑉−2.7 𝑉𝑉)
0.057 𝐴𝐴
−50
=57.9 Ω
3 (T
otal
for
Que
stio
n 16
= 1
0 m
arks
)
Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014
48
Que
stio
n nu
mbe
r
A
ccep
tabl
e an
swer
s
Add
ition
al g
uida
nce
Mar
k
17 (a
) •
Use
of:
AL
Rρ
=(1
)
• 5.0
=AB RR
[acc
ept 2
:4 o
r 1:
2 or
1/2
] (1)
Exam
ple
of c
alcu
latio
n:
2 21
2 12 dLd
LRR AB
=
()
5.042
2222
==
=d
LLd
RR AB
2
17 (b
) •
Cor
rect
tran
sfer
of d
ata
for g
radi
ent (
1)
• La
rge
trian
gle
used
(1)
• ρ
= 1.
1 ×
10-4
Ωm
(1)
Con
clus
ion:
•
1.1 ×
10-4
Ωm
is g
reat
er th
an 3
x 1
0-5 Ω
m, s
o re
sist
ivity
incr
ease
s w
hen
clay
is a
dded
(1)
Ans
wer
to b
e co
nsis
tent
with
cal
cula
ted
valu
e.
Exam
ple
of c
alcu
latio
n:
2-6Ωm
1016.5
Gradient
×=
=Lρ
ρ
= 16
.5 ×
10-6
Ωm
2 /0.1
5 m
= 1
.1 ×
10-4
Ωm
4
17 (c
) A
n ex
plan
atio
n th
at m
akes
refe
renc
e to
: •
A ri
se in
tem
pera
ture
cau
ses t
he a
mpl
itude
of t
he v
ibra
ting
ions
to
incr
ease
(1)
• Th
is c
ause
s the
num
ber o
f col
lisio
ns p
er se
cond
bet
wee
n th
e io
ns
and
the
mov
ing
elec
trons
to in
crea
se (1
) •
So th
e ra
te o
f flo
w o
f ele
ctro
ns d
ecre
ases
(cau
sing
the
resi
stan
ce
of th
e m
etal
to in
crea
se) (
1)
Allo
w re
duce
mea
n fr
ee p
ath
or d
rift v
eloc
ity o
f el
ectro
ns.
Acc
ept c
urre
nt d
ecre
ases
3
(Tot
al fo
r Q
uest
ion
17 =
9 m
arks
)
Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014
49
Que
stio
n nu
mbe
r
A
ccep
tabl
e an
swer
s
Add
ition
al g
uida
nce
Mar
k
18 (a
) •
Use
of F
= B
Il or
use
of F
=Bq
v (1
) •
Con
verts
N to
kg
m s-2
(1)
Exam
ple
𝐵𝐵=𝐹𝐹[kg m 𝑠𝑠!! ]
𝐼𝐼[A] 𝑙𝑙[m]
So u
nits
are
kg
A-1
s-2
2 18
(b)
An
expl
anat
ion
that
mak
es re
fere
nce
to:
• Th
e m
agne
tic fo
rce
on th
e el
ectro
ns a
cts a
t rig
ht a
ngle
s to
the
plan
e co
ntai
ning
B a
nd v
(1)
• H
ence
the
forc
e is
alw
ays t
owar
ds th
e ce
ntre
of t
he c
ircle
(1)
So p
rovi
ding
a c
entri
peta
l for
ce o
n th
e el
ectro
n or
a c
entri
peta
l ac
cele
ratio
n th
at m
aint
ains
circ
ular
mot
ion
(1)
3
18 (c
) •
Cal
cula
tes B
x r
(1)
• C
alcu
late
the
perc
enta
ge u
ncer
tain
ty (1
) •
Suita
ble
com
men
t on
diff
eren
ce fr
om e
xpec
tatio
n (1
) •
Wea
k co
nclu
sion
bec
ause
onl
y th
ree
read
ings
(1) O
R n
o re
peat
s (1)
OR
lim
ited
rang
e (1
)
Exam
ple
of c
alcu
latio
n:
%U
= (0
.06/
5.01
) x 1
00%
= 1
.2%
Rad
ius
/cm
M
agne
tic fl
ux
dens
ity/m
T
8.0
0.63
5.
04
9.5
0.52
4.
94
11.0
0.
46
5.06
4
(Tot
al fo
r Q
uest
ion
18 =
9 m
arks
)
Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014
50
Que
stio
n nu
mbe
r
A
ccep
tabl
e an
swer
s
Add
ition
al g
uida
nce
Mar
k
19 (a
) A
n ex
plan
atio
n th
at m
akes
refe
renc
e to
: •
Elec
trons
/cha
rge
trans
ferr
ed fr
om n
egat
ivel
y ch
arge
d pl
ate
to
posi
tivel
y ch
arge
pla
te th
roug
h th
e re
sist
or (1
) •
Hen
ce th
e ch
arge
on
capa
cito
r dec
reas
es (e
xpon
entia
lly) (
1)
• U
ntil
the
char
ge o
n th
e ca
paci
tor e
qual
s 0/n
eglig
ible
(1)
3
19 (b
) E
ither
•
Use
Q =
2.6
to re
ad ti
me
cons
tant
from
gra
ph (
1) O
R d
raw
ta
ngen
t to
curv
e at
t =
0 an
d ob
tain
tim
e co
nsta
nt fr
om
inte
rcep
t on
x ax
is (1
) •
t = 1
7 –
18 (m
s) (1
) •
Use
of 𝑇𝑇
=𝑅𝑅𝑅𝑅
with
thei
r T (1
) •
C =
0.0
19 -
0.02
1 m
F (1
) O
R
• Q
0 =
7 (m
C) r
ead
from
gra
ph (1
) •
Any
cor
resp
ondi
ng v
alue
s of Q
and
t re
ad fr
om g
raph
(1)
• U
se o
f Q=Q
0 e-t/
RC w
ith th
eir v
alue
s for
Q0,
Q a
nd t
(1)
• C
= 0
.019
5 –
0.01
96 m
F (1
) O
R
• Q
0 =
7 (m
C) r
ead
from
gra
ph (1
) •
Q=3
.5 (m
C) w
hen
T 1/2
= 1
2.3
(ms)
(1)
• U
se o
f T1/
2 =
RC
ln2
(1)
• C
= 0
.019
5 –
0.01
96 m
F (1
)
Exam
ple
of c
alcu
latio
n:
T =
19 (m
s)
C =
19
x 10
-3/9
00 =
0.0
21 m
F
4 (T
otal
for
Que
stio
n 19
= 7
mar
ks)
Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014
51
Que
stio
n nu
mbe
r
A
ccep
tabl
e an
swer
s
Add
ition
al g
uida
nce
Mar
k
20 (a
)(i)
• R
ecog
nise
that
for p
asse
nger
to re
mai
n in
thei
r sea
t nor
mal
re
actio
n R ≥
0 (1
) or
cen
tripe
tal f
orce
>=
wei
ght (
1)
• Eq
uate
cen
tripe
tal f
orce
and
wei
ght (
for R
=0) (
1)
• v
= 9.
1 m
s-1 (
1)
Exam
ple
of c
alcu
latio
n:
mg
rmv
=2
12
sm
9.13
sm
9.81
m8.5
−−=
×=
=rg
v
3 20
(a)(
ii)
• Eq
uate
dec
reas
e in
gra
vita
tiona
l pot
entia
l ene
rgy
to in
crea
se in
ki
netic
ene
rgy
at to
p of
loop
(1)
• A
dds t
his t
o 17
.0 (1
) • Δ
h =
21.3
m (1
)
Exam
ple
of c
alcu
latio
n:
2
21mv
mgh
=
()
m4.25
sm
9.81
2s
m9.13
222
12
=×
==
−
−
gvh
∆h =
17
+ 4.
3 =
21.3
m
3
20 (b
)(i)
• U
se o
f rva
2
= (1
) •
a =
6.1g
(1)
Exam
ple
of c
alcu
latio
n:
()
22
12
sm
59.6
m8.5s
m22.5
−−
==
=rv
a
a =
59.6
/9.8
= 6
.1 g
2
20 (b
)(ii)
A
n ex
plan
atio
n th
at m
akes
refe
renc
e to
: •
Rad
ius o
f cur
vatu
re sm
alle
st a
t the
top
of th
e lo
op (1
) OR
radi
us
larg
er a
t the
bot
tom
of t
he lo
op (1
) •
So a
ccel
erat
ion
at b
otto
m is
less
for t
he sa
me
spee
d (1
)
2
(Tot
al fo
r Q
uest
ion
20 =
10
mar
ks)