Mark scheme 1 Paper - Amazon S3

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

)


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

)


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

)


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)


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)


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).


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


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)


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).


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


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

)


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

)


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

)


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)


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)

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