Example - Amazon S3 · Acceptable answers Additional guidance Mark 1(a) An explanation that makes reference to the following points: • Comment on the data (1) • Correct consequent

Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014

133

Pa

per

3 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(a)

A

n ex

plan

atio

n th

at m

akes

refe

renc

e to

the

follo

win

g po

ints

: •

Com

men

t on

the

data

(1)

• C

orre

ct c

onse

quen

t con

clus

ion

(1)

4.43

is a

n an

omal

ous v

alue

So

the

mea

n va

lue

is to

o lo

w

Acc

ept

data

is c

onco

rdan

t so

mea

n va

lue

is c

orre

ct

2

1(b)

A

n ex

plan

atio

n th

at m

akes

refe

renc

e to

the

follo

win

g po

ints

: •

Ligh

t gat

es c

an re

cord

shor

t tim

es a

ccur

atel

y (1

) O

R w

ith sm

alle

r unc

erta

inty

(1)

• B

ecau

se h

uman

reac

tion

time

is n

ot in

volv

ed (1

)

2 (T

otal

for

Que

stio

n 1

= 4

mar

ks)


134

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

2 •

Sens

ible

est

imat

e of

unc

erta

intie

s fro

m re

adin

gs g

iven

(1)

• A

dds p

erce

ntag

e un

certa

intie

s (1)

•

Hen

ce c

alcu

late

s unc

erta

inty

in sp

eed

(1)

• C

andi

date

’s c

oncl

usio

n m

ust b

e su

ppor

ted

by th

eir e

stim

ate

of th

e un

certa

intie

s (1)

Exam

ple

of c

alcu

latio

n:

%U

in L

is (0

.1/2

5.6)

x 1

00 %

= 0

.4 %

%U

in F

is (1

/320

) x 1

00 %

= 0

.3 %

%

U in

spee

d is

0.7

%

328

x 0.

007

= 2

Spee

d =

328

± 2

All

thre

e re

sults

are

with

in th

e ca

lcul

ated

un

certa

inty

so c

oncl

udes

stud

ent B

is c

orre

ct

4

(Tot

al fo

r Q

uest

ion

2 =

4 m

arks

)


135

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

3 (a

)(i)

A d

escr

iptio

n th

at m

akes

refe

renc

e to

the

follo

win

g po

ints

: C

ircui

t dia

gram

show

ing:

• C

ell,

varia

ble

resi

stor

and

am

met

er in

serie

s and

vol

tmet

er

in p

aral

lel w

ith c

ell (

1)

• R

ecor

ding

pai

rs o

f rea

ding

s of t

erm

inal

p.d

. and

cur

rent

(1)

•

Use

the

varia

ble

resi

stor

to o

btai

n 5

othe

r pai

rs o

f rea

ding

s (1

)

Shou

ld b

e be

twee

n 5

and

10 o

ther

pai

rs

3 3

(a)(

ii)

A d

escr

iptio

n th

at m

akes

refe

renc

e to

the

follo

win

g po

ints

: •

Plot

a g

raph

of t

erm

inal

pot

entia

l diff

eren

ce o

n th

e y-

axis

an

d cu

rren

t on

the

x-ax

is (1

) •

Inte

rcep

t on

the

y-ax

is e

qual

s e.m

.f. (1

) •

And

gra

dien

t = -r

(1)

3

3(b)

•

From

gra

ph: a

fter 5

00 c

harg

ing

cycl

es in

tern

al re

sist

ance

of c

ell

is 3

27 mΩ

(1)

• U

se o

f V =

ε -

Ir (1

)

• U

se o

f %

100

0500×

VV

%99.6

%100

0500

=×

VV

(1)

• So

man

ufac

ture

r’s c

laim

is c

orre

ct (1

)

Exam

ple

of c

alcu

latio

n:

V3.352

V0.248

V3.6

Ω0.310

A0.800

V3.6

0

=−

=

×−

=V

V3.338

V0.262

V3.6

Ω0.327

A0.800

V3.6

500

=−

=

×−

=V

%99.6

%100

V3.352V

3.338

%100

0500

=×

=×

VV

Th

is la

st m

ark

is a

war

ded

only

if th

e co

nclu

sion

is

corr

ectly

supp

orte

d by

the

calc

ulat

ion.

4

(Tot

al fo

r Q

uest

ion

3 =

10 m

arks

)


136

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

4 (a

)

Any

thre

e fr

om:

• In

cons

iste

nt p

reci

sion

for e

xten

sion

(1)

• La

ck o

f pre

cisi

on o

n m

ass,

shou

ld b

e sh

own

to 3

DP

(1)

• N

o ev

iden

ce o

f rep

eat r

eadi

ngs

(1) O

R th

ere

shou

ld b

e m

ore

read

ings

to c

ompe

nsat

e fo

r rep

eat r

eadi

ngs b

eing

inap

prop

riate

(1

) •

Inco

nsis

tent

inte

rval

s bet

wee

n re

adin

gs (1

)

Unc

erta

inty

sugg

este

d by

1 sf

is fa

r gre

ater

than

that

ex

pect

ed in

pra

ctic

e

3 4

(b)

A d

escr

iptio

n th

at m

akes

refe

renc

e to

two

of th

e fo

llow

ing

poin

ts:

• us

e of

fici

dual

mar

k (1

) •

eye

clos

e to

liqu

oric

e la

ce to

avo

id p

aral

lax

erro

rs (1

) •

Fixe

d m

etre

rule

clo

se to

lace

(1)

• U

se o

f set

squa

re to

ens

ure

rule

ver

tical

(1)

2

(Tot

al fo

r Q

uest

ion

4 =

5 m

arks

)


137

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

5 •

Dia

met

er is

onl

y ha

lf a

divi

sion

on

the

scal

e (1

) OR

dia

met

er is

m

easu

red

to o

nly

1 sf

(1)

• H

ence

ther

e is

a la

rge

perc

enta

ge u

ncer

tain

ty in

mea

sure

men

t of

diam

eter

of o

il dr

op (1

) •

Sinc

e th

e vo

lum

e of

the

drop

is c

alcu

late

d by

taki

ng (d

iam

eter

)3 , th

e pe

rcen

tage

unc

erta

inty

in v

olum

e be

com

es v

ery

larg

e (3

× %

un

certa

inty

in d

iam

eter

) (1)

•

Sugg

estio

n fo

r im

prov

emen

t: us

e a

larg

er o

il dr

op, u

se a

(v

erni

er) s

cale

cap

able

of r

eadi

ng to

nea

rest

0.1

mm

, pro

ject

im

age

of d

ropl

et to

larg

er si

ze (1

) •

Dro

p w

ill n

ot sp

read

out

as a

n ex

actly

circ

ular

are

a, so

dia

met

er

read

ing

may

be

inac

cura

te (1

) •

Sugg

estio

n fo

r im

prov

emen

t: th

e di

amet

er o

f spr

ead-

out o

il dr

op

shou

ld b

e ta

ken

a nu

mbe

r of t

imes

acr

oss a

num

ber o

f diff

eren

t di

rect

ions

and

a m

ean

calc

ulat

ed (1

)

Allo

w fo

r ide

ntifi

catio

n of

any

oth

er v

alid

pro

blem

s an

d im

prov

emen

ts b

ased

on

good

phy

sics

, for

ex

ampl

e pl

ace

met

re ru

le a

cros

s tra

y so

that

it is

cl

ose

to th

e su

rfac

e.

6 (T

otal

for

Que

stio

n 5

= 6

mar

ks)


138

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

6 (a

)

An

expl

anat

ion

that

mak

es re

fere

nce

to th

e fo

llow

ing

poin

ts:

• Th

e cu

rren

t pro

duce

s a m

agne

tic fi

eld

arou

nd th

e al

umin

ium

rin

g (1

) •

The

dire

ctio

n of

the

ring

field

opp

oses

the

chan

ge p

rodu

cing

it

(1)

• Th

e fie

lds r

epel

, pro

duci

ng a

forc

e (1

) •

The

elec

trom

agne

tic fo

rce

is e

qual

and

opp

osite

to th

e w

eigh

t of

the

ring

so it

rem

ains

in p

ositi

on sh

own

(1)

4

6 (b

)(i)

•

Use

of ½

mv2 =

mgh

(1)

• v

= 2.

43 m

s-1 (1

) Ex

ampl

e of

cal

cula

tion:

v

= 2𝑔𝑔ℎ

= 2×9.81×0.30

= 2

.43

m s-1

2

6 (b

)(ii)

•

Use

of i

mpu

lse

= ch

ange

in m

omen

tum

(1)

• R

ecog

nise

s ini

tial v

eloc

ity is

zer

o (1

) •

Hen

ce F

= 0

.923

N (1

) •

Use

of l

= π

d (1

) •

Equa

tes c

alcu

late

d va

lue

of F

with

BIl

(1)

• H

ence

I =

191

A (1

)

Exam

ple

of c

alcu

latio

n:

Ft =

mv

– m

u w

here

u =

0

So F

= (0

.019

kg ×

2.43

m s-1

)/0.0

5 s =

0.9

23 N

l =

π ×

0.0

48 m

= 0

.151

m

I = 0

.923

N/(0

.032

T ×

0.1

51 m

) = 1

91 A

6 (T

otal

for

Que

stio

n 6

= 12

mar

ks)


139

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

7 (a

)(i)

A d

escr

iptio

n th

at m

akes

refe

renc

e to

the

follo

win

g po

ints

: •

Rec

ord

nT (w

here

n is

at l

east

5) (

1)

• D

ivid

e m

easu

rem

ent b

y n

(1)

2

7 (a

)(ii)

•

Ano

mal

ies c

an b

e sp

otte

d (1

) •

Red

uce

the

effe

ct o

f ran

dom

err

or (1

)

2

7 (b

)(i)

• B

FL is

smoo

th a

nd th

in w

ith a

def

inite

min

imum

an

d m

inim

um is

in ra

nge

0.26

m –

0.2

8 m

(1)

1

7 (b

)(ii)

•

Val

ues r

ead

corr

ectly

from

can

dida

te’s

line

(1)

• h

to 3

sig

fig a

nd T

to 4

sf (1

) V

alue

s fro

m th

eir c

urve

to w

ithin

1 sm

all s

quar

e w

ith

no u

nit p

enal

ty.

2

7 (c

) A

des

crip

tion

that

mak

es re

fere

nce

to th

e fo

llow

ing

poin

ts:

• Pl

ot T

2 h

agai

nst h

2 (1)

• C

is in

terc

ept o

n T2

h ax

is (

1)

OR

C is

the

valu

e of

T2 h

whe

n h2 is

zer

o (1

) •

Uni

t is m

s2 (1)

3

(Tot

al fo

r Q

uest

ion

7 =

10 m

arks

)


140

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

8 (a

) A

n ex

plan

atio

n th

at th

e st

uden

t’s c

oncl

usio

n is

inco

rrec

t bec

ause

: •

The

popp

er re

turn

s to

its o

rigin

al sh

ape,

eve

n th

ough

ther

e is

a

time

dela

y (1

) •

Elas

tic m

ater

ial r

etur

ns to

its o

rigin

al sh

ape

whe

n th

e de

form

ing

forc

e is

rem

oved

(1)

• B

ut a

pla

stic

mat

eria

l wou

ld su

ffer

a p

erm

anen

t def

orm

atio

n (1

)

3

8 (b

) A

des

crip

tion

that

mak

es re

fere

nce

to th

e fo

llow

ing

poin

ts:

• R

efer

to v

2 = u

2 + 2

as (1

) •

Whe

re s

is h

eigh

t rea

ched

, v is

zer

o, a

= -g

(1)

• So

u =

√2g

s (1)

Allo

w a

rgum

ent ½

mv2 =

mgh

to g

et th

e sa

me

resu

lts.

3 8

(b)(

ii)

• A

ir re

sist

ance

will

act

on

the

popp

er…

(1)

• …

As a

dec

eler

atin

g fo

rce

(1) O

R…

dis

sipa

ting

ener

gy (1

) •

So th

e in

itial

spee

d w

ill b

e lo

wer

than

in th

e ab

senc

e of

air

resi

stan

ce, s

o th

e su

gges

tion

is n

ot c

orre

ct (1

)

3 (T

otal

for

Que

stio

n 8

= 9

mar

ks)


141

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

9 (a

) •

Use

s wei

ght o

f dis

plac

ed a

ir = ρV

g (1

) •

Find

s res

ulta

nt fo

rce

= up

thru

st –

wei

ght (

1)

• U

ses F

= m

a to

find

acc

eler

atio

n (1

) •

Acc

eler

atio

n =

0.16

1 m

s−2 (1

)

Exam

ple

of c

alcu

latio

n:

Wei

ght o

f air

disp

lace

d = ρV

g =

1.20

kg

m−3×2

880

m3 ×

9.8

1m s−

2 = 3

3 90

3N

Res

ulta

nt u

pwar

d fo

rce

= 33

903

N –

(340

0 kg

×

9.81

m s−

2 ) = 5

49 N

A

ccel

erat

ion

= 54

9 N

/340

0 kg

= 0

.161

m s−

2

4 9

(b)*

Th

is q

uest

ion

asse

sses

a st

uden

t’s a

bilit

y to

show

a c

oher

ent

and

logi

cally

stru

ctur

ed a

nsw

er w

ith li

nkag

es a

nd fu

lly-

sust

aine

d re

ason

ing.

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:

The

mar

k fo

r ind

icat

ive

cont

ent s

houl

d be

add

ed to

th

e m

ark

for l

ines

of r

easo

ning

. Fo

r exa

mpl

e, a

n an

swer

with

five

indi

cativ

e m

arki

ng p

oint

s whi

ch is

pa

rtial

ly st

ruct

ured

with

som

e lin

kage

s and

line

s of

reas

onin

g sc

ores

4 m

arks

(3 m

arks

for i

ndic

ativ

e co

nten

t and

1 m

ark

for p

artia

l stru

ctur

e an

d so

me

linka

ges a

nd li

nes o

f rea

soni

ng).

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).

6


142

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

9 (b

)*

(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

mar

ks

awar

ded

for

stru

ctur

e of

an

swer

and

su

stai

ned

line

of

reas

onin

g

Ans

wer

show

s a c

oher

ent a

nd

logi

cal s

truct

ure

with

link

ages

an

d fu

lly su

stai

ned

lines

of

reas

onin

g de

mon

stra

ted

thro

ugho

ut

2

Ans

wer

is p

artia

lly st

ruct

ured

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


143

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

9 (b

)*

(con

tinue

d)

Indi

cativ

e co

nten

t •

As p

arac

hute

ope

ns (a

t B) t

he u

pwar

ds fo

rce

incr

ease

s •

Alo

ng B

C th

e ve

loci

ty is

dec

reas

ing

at a

no

n-co

nsta

nt ra

te

• Th

e dr

ag is

gre

ater

than

wei

ght (

nega

tive

grad

ient

) •

The

drag

is d

ecre

asin

g (c

urve

d lin

e)

• Ev

entu

ally

the

drag

forc

e ba

lanc

es th

e w

eigh

t •

No

acce

lera

tion

so li

ne is

hor

izon

tal

(Tot

al fo

r Q

uest

ion

9 =

10 m

arks

)


144

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

10 (a

) A

des

crip

tion

that

mak

es re

fere

nce

to th

e fo

llow

ing

poin

ts:

• g

is d

irect

ly p

ropo

rtion

al to

r up

to R

0 (1)

•

and

then

g d

ecre

ases

with

incr

easi

ng r

(1)

• w

here

g is

pro

porti

onal

to th

e in

vers

e of

the

squa

re o

f r (1

)

3

10 (b

) •

Forc

e on

obj

ect =

mg

(loca

l g) (

1)

• Fo

rce

is p

ropo

rtion

al to

dis

plac

emen

t (1)

•

Forc

e ac

ts in

the

oppo

site

dire

ctio

n to

the

disp

lace

men

t (1)

•

Ther

efor

e w

e ca

n sa

y F

= -k

x, so

the

cond

ition

for S

HM

is m

et

and

the

pred

ictio

n is

cor

rect

(1)

4

10 (c

)(i)

Eith

er

• W

hen

x =

R0,

F =

GM

m/R

02 (1)

• F

= G

Mm

R0/R

03 so k

= mω

2 = G

Mm

/R03 (1

) •

Use

of T

=2π/ω

(1)

• T2 =

4π2 /ω

2 = 4π2 R

03 /GM

So

𝑇𝑇=2𝜋𝜋√(𝑅𝑅 !! /𝐺𝐺𝐺𝐺

) (1)

OR

• Fr

om g

raph

F=-

(g/R

o)r (

1)

• Fr

om w

hich

ω =√(𝑔𝑔/𝑅𝑅 !

) (1

) •

Use

of T

=2π/ω

(1)

• So

𝑇𝑇=2𝜋𝜋√(𝑅𝑅 !

/𝑔𝑔) (

1)

4


145

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

10 (c

)(ii)

Eith

er

Cen

tripe

tal f

orce

= m

v2 /R0 =

GM

m/R

02 (1)

• 4π

2 R02 / T

2 R0 =

GM

/R02 (1

) •

T2 = 4π2 /ω

2 = 4π2 R

03 /GM

So 𝑇𝑇=2𝜋𝜋√(𝑅𝑅 !! /𝐺𝐺𝐺𝐺

) (1)

O

R

• m

g =

mv2 /R

0 =

m ω

2 R0 (

1)

• So

ω =

√(𝑔𝑔/𝑅𝑅

!) (1

) •

T =

2π/ω

= 2𝜋𝜋√(𝑅𝑅

!/𝑔𝑔)

(1)

3

(Tot

al fo

r Q

uest

ion

10 =

14

mar

ks)


146

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

11 (a

) A

n ex

plan

atio

n th

at m

akes

refe

renc

e to

the

follo

win

g po

ints

: •

Res

ista

nce

incr

ease

s with

dec

reas

ing

inte

nsity

(1)

• A

s dis

tanc

e in

crea

ses l

ight

inte

nsity

dec

reas

es so

resi

stan

ce

incr

ease

s (1)

2

11 (b

) A

n ex

plan

atio

n th

at m

akes

refe

renc

e to

the

follo

win

g po

ints

: •

Show

s exp

ansi

on ln

R =

p ln

(d) +

ln(k

) (1

) •

Com

pare

s with

y =

mx+

c a

nd st

ates

that

the

grad

ient

is p

whi

ch

is c

onst

ant (

1)

2

11 (c

)(i)

•

Ln v

alue

s cor

rect

and

to 3

or 4

SF

(1)

d/m

R

/kΩ

ln

(d/m

) ln

(R/kΩ

)

1.00

1.

79

0.00

0 0.

582

1.20

2.

24

0.18

2 0.

806

1.60

3.

32

0.47

0 1.

200

2.00

4.

04

0.69

3 1.

396

2.20

4.

70

0.78

8 1.

548

2.60

5.

50

0.95

6 1.

705

•

Labe

ls a

nd u

nit (

1)

• Sc

ales

(1)

• Pl

ots (

1)

• Li

ne o

f bes

t fit

(1)

See

mar

king

gui

danc

e fo

r gra

ph p

lotti

ng

5


147

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

11 (c

)(i)

(con

tinue

d)

11 (c

) (ii)

•

Find

s gra

dien

t with

larg

e tri

angl

e –

at le

ast h

alf t

he p

lotte

d le

ngth

(1

) •

1.13

< p

< 1

.23

to 2

/3 S

F an

d no

uni

ts (1

) •

Obt

ains

k =

1.8

(1)

• St

ates

rela

tions

hip

betw

een

R an

d d

(1)

4

(Tot

al fo

r Q

uest

ion

11 =

13

mar

ks)


148

Que

stio

n nu

mbe

r

Acc

epta

ble

answ

ers

Add

ition

al g

uida

nce

Mar

k

12 (a

)(i)

•

use

of F

= Q

1Q2/4πε

0r2 (1

) •

use

of F

= G

m1m

2/r2 (1

) •

Expr

esse

s for

ces a

s a ra

tio (1

) OR

cal

cula

tes t

he in

divi

dual

fo

rces

Fe=

8.1

x 10

-8 N

Fg =

3.6

x 1

0-47 N

(1)

• R

atio

= 2

x 1

039 o

r 5 x

10-4

0 and

iden

tifie

s gra

vita

tiona

l for

ce a

s in

sign

ifica

nt (1

)

4

12 (a

)(ii)

•

use

of F

= m

v2 /r a

nd F

= Q

1Q2/4πε

0r2 (1

)

• to

der

ive 𝑣𝑣 =

! !! !

!!! !!"

(1)

• ve

loci

ty =

2.2×

106 m

s-1 (1

)

Exam

ple

of c

alcu

latio

n:

v =

! !! !

!!! !!"

v =

!.! ×!"

!!" C×!.! ×!"

!!" C

!!×!.!"

!!" F!!! ×!.! ×!"

!!!m×!.!×!"

!!" kg

v =

2.1

85 ×

106 m

s-1

3

12 (b

) •

Cal

cula

tes w

avel

engt

h λ

(circ

umfe

renc

e) (1

) •

Use

of p

= h

/λ (1

) •

v =

2.2 ×

106 m

s-1 (1

)

Exam

ple

of c

alcu

latio

n:

λ =

2πr =

2π ×

5.3 ×

10-1

1 m =

3.3

3 ×

10-1

0 m

λ =

h/m

v so

v =

h/m

. v

= 6.

63×1

0-34 J

s/(9

.1×1

0-31 k

g×3.

33×1

0-10

m)

v =

2.18

8 ×

106 m

s-1

3

(Tot

al fo

r Q

uest

ion

12 =

10

mar

ks)


149

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

13 (a

)

An

expl

anat

ion

that

mak

es re

fere

nce

to th

e fo

llow

ing

poin

ts:

• R

eson

ance

is o

ccur

ring…

(1)

• …

whe

n th

e dr

ivin

g fr

eque

ncy/

forc

ed v

ibra

tion

(at w

alki

ng

freq

uenc

y) m

atch

es th

e na

tura

l fre

quen

cy …

(1)

• …

ener

gy tr

ansf

er is

max

imum

(1)

• Su

ppor

ting

the

obse

rvat

ion

that

the

ampl

itude

rapi

dly

incr

ease

s (1

)

4

13 (b

)(i)*

Th

is q

uest

ion

asse

sses

a st

uden

t’s a

bilit

y to

show

a c

oher

ent

and

logi

cally

stru

ctur

ed a

nsw

er w

ith li

nkag

es a

nd fu

lly-

sust

aine

d re

ason

ing.

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:

The

mar

k fo

r ind

icat

ive

cont

ent s

houl

d be

add

ed to

th

e m

ark

for l

ines

of r

easo

ning

. Fo

r exa

mpl

e, a

n an

swer

with

five

indi

cativ

e m

arki

ng p

oint

s whi

ch is

pa

rtial

ly st

ruct

ured

with

som

e lin

kage

s and

line

s of

reas

onin

g sc

ores

4 m

arks

(3 m

arks

for i

ndic

ativ

e co

nten

t and

1 m

ark

for p

artia

l stru

ctur

e an

d so

me

linka

ges a

nd li

nes o

f rea

soni

ng).

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).

6


150

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

13 (b

)(i)*

(c

ontin

ued)

Th

e fo

llow

ing

tabl

e sh

ows h

ow th

e m

arks

shou

ld

be a

war

ded

for s

truct

ure

and

lines

of r

easo

ning

.

N

umbe

r of

mar

ks

awar

ded

for

stru

ctur

e of

an

swer

and

su

stai

ned

line

of

reas

onin

g

Ans

wer

show

s a c

oher

ent a

nd

logi

cal s

truct

ure

with

link

ages

an

d fu

lly su

stai

ned

lines

of

reas

onin

g de

mon

stra

ted

thro

ugho

ut

2

Ans

wer

is p

artia

lly st

ruct

ured

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


151

Que

stio

n nu

mbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

13 (b

)(i)*

(c

ontin

ued)

In

dica

tive

cont

ent

• D

eter

min

e th

e na

tura

l fre

quen

cy b

y di

spla

cing

the

tea

in th

e cu

p an

d m

easu

ring

the

time

for o

scill

atio

ns

• Ti

me

(5 to

10

or ‘s

uita

ble

num

ber’

if te

st ru

n m

entio

ned)

full

osci

llatio

ns a

nd d

ivid

e by

the

num

ber

• C

arry

the

tea

for a

kno

wn

volu

me

of te

a fo

r fix

ed n

umbe

r of

step

s at a

stea

dy p

ace

• D

eter

min

e th

e fr

eque

ncy

of th

e ga

it •

Mea

sure

the

quan

tity

of te

a re

mai

ning

•

Rep

eat f

or o

ther

wal

king

pac

es

13 (b

)(ii)

A d

escr

iptio

n th

at m

akes

refe

renc

e to

the

follo

win

g po

ints

: •

Plot

vol

ume

of re

mai

ning

tea

agai

nst w

alki

ng fr

eque

ncy

(1)

• D

eter

min

e w

heth

er th

ere

is a

rela

tions

hip

betw

een

step

fr

eque

ncy

and

spill

age

(1)

• If

ther

e is

, det

erm

ine

whe

ther

max

imum

spill

age

occu

rs a

t or

near

the

natu

ral f

requ

ency

(1)

3

(Tot

al fo

r Q

uest

ion

13 =

13

mar

ks)

Documents

Example - Amazon S3 · Acceptable answers Additional guidance Mark 1(a) An explanation that makes reference to the following points: • Comment on the data (1) • Correct consequent