Mark Scheme (Results)

Summer 2014

Pearson Edexcel International A Level in Mechanics 1 (WME01/01)

99

Q

uest

ion

Num

ber

Sch

eme

M

arks

Not

es

1.

(a)

0.9

20.

60

0.6

2v

M1

Equ

atio

n w

ith

all t

he te

rms

– co

ndon

e “0

” m

issi

ng.

Ter

ms

mus

t be

of th

e fo

rm m

v, b

ut c

ondo

ne s

ign

erro

rs.

Con

done

g p

rese

nt a

s a

com

mon

fac

tor.

A

1 C

orre

ct u

nsim

plif

ied

equa

tion

1v

A

1

(3)

(b)

0.6

21.

8I

v

N s

or

0.9

21.

8I

N s

M

1 C

hang

e in

mom

entu

m o

f A

or o

f B.

Con

done

sig

n sl

ips

and

nega

tive

ans

wer

. N

o g.

A1

1.8

only

(or

exa

ct e

quiv

alen

t)

Fro

m c

orre

ct w

ork

only

.

(2)

[5]

W

atch

out

for

for

tuit

ous

answ

ers

in (

b);

5v

fro

m

(a)

used

in (

b) w

ill s

core

at m

ost M

1A0

in (

b)

100

Que

stio

n N

umbe

r S

che

me

M

arks

Not

es

2 (a

) 1

205

9.8

252

h

M1

Use

of

21 2

sut

at

to

fin

d h.

Mus

t quo

te th

e co

rrec

t

form

ula

and

be u

sing

20

& 5

, but

con

done

slip

s in

su

bstit

utio

n.

Acc

ept c

ompl

ete

alte

rnat

ive

solu

tion

s w

orki

ng v

ia

the

max

imum

hei

ght.

(max

ht 2

0.4.

.., ti

me

to to

p 2.

04...

) A

ccep

t com

plet

e al

tern

ativ

e m

etho

ds u

sing

oth

er

suva

t equ

atio

ns.

A

1 C

orre

ctly

sub

stit

uted

equ

atio

n(s)

Con

done

use

of

a pr

emat

ure

appr

oxim

atio

n.

22

.5h

A

1

(

3)F

inal

ans

wer

. A

ccep

t 22.

5 or

23.

Max

imum

3sf

. -2

2.5

is A

0.

N

B D

o no

t ign

ore

subs

eque

nt w

orki

ng if

they

rea

ch

22.5

and

then

mov

e on

to d

o fu

rthe

r w

ork.

(b)

22

202

9.8

22.5

V

O

R

20

(59.

8)V

M

1 F

irst

bal

l -

use

of su

vat t

o fi

nd V

or

V2

Fol

low

thei

r h.

(

284

1V

)

29

A

1 C

orre

ct o

nly

(con

done

-29

)

2

23

29.

822

.54

Vw

M1

Sec

ond

ball

-

suva

t equ

atio

n in

V (

or th

eir

V )

to f

ind

w. M

ust b

e us

ing

the

3 4.

2

984

12

9.8

22.5

16w

A1f

t C

orre

ctly

sub

stit

uted

equ

atio

n w

ith

thei

r V

and

thei

r h.

5.

66w

A

1 o

r 5.

7. A

nsw

er c

orre

ct to

2 s

.f. o

r to

3 s

.f.

(5)

[8

]

101

Que

stio

n N

umbe

r S

che

me

M

arks

Not

es

3

N

(a)

F

3

0o

1.5

g F

or e

quili

briu

m

M1

For

res

olut

ion

of f

orce

s pa

ralle

l or

perp

endi

cula

r to

th

e pl

ane.

Wei

ght m

ust b

e re

solv

ed.

Con

done

si

n/co

s co

nfus

ion.

R

pla

ne1.

5co

s30

Ng

A1

Cor

rect

equ

atio

n fo

r N

(12

.7)

R

pla

ne1.

5co

s60

Fg

A1

Cor

rect

equ

atio

n fo

r F

(7.

35).

Con

done

R

co

s60

0.57

7...

0.6

cos3

0

F N

e

quil

ibri

um

M1

Use

of

max

FN

and

com

pare

wit

h F,

or f

ind

the

valu

e of

thei

r F N

and

com

pare

with

A

1

(5)

Rea

ch g

iven

con

clu

sion

cor

rect

ly. T

hey

mus

t mak

e so

me

com

men

t, ho

wev

er b

rief

.

AL

T f

or f

irst

3 m

ark

s:

R

esol

ve v

erti

cally

co

s30

cos6

01.

5N

Fg

M

1A1

Res

olve

hor

izon

tally

co

s60

cos3

0N

F

A

1

A

LT

for

last

2 m

ark

s:

m

ax0.

612

.73

7.63

7.35

F

P

is a

t res

t

M1

A1

C

andi

date

s w

ho th

ink

that

the

diag

ram

app

lies

to (

a) w

ill s

core

no

thin

g in

(a)

but

if th

ey c

arry

thei

r re

sult

s fo

rwar

d in

to (

b) th

en

thei

r w

ork

can

scor

e th

e m

arks

ava

ilab

le in

(b)

.

If th

e ca

ndid

ate

has

give

n th

e eq

uati

on o

f m

otio

n fo

r th

e pa

rtic

le m

ovin

g do

wn

the

plan

e th

en

A1

for

1.5

sin

301.

5g

Ra

T

o sc

ore

mor

e th

ey n

eed

to c

omm

ent c

orre

ctly

on

thei

r an

swer

: a

= -

0.19

impo

ssib

le

M

1 C

oncl

ude

that

the

part

icle

can

not b

e m

ovin

g.

A1

102

Que

stio

n N

umbe

r S

che

me

M

arks

Not

es

N

F

(b)

X

30o

1

.5g

R

pla

ne1.

5co

s30

cos6

0N

gX

M1

Req

uire

s al

l 3 te

rms.

C

ondo

ne s

in/c

os c

onfu

sion

and

sig

n er

rors

.

R

pla

neco

s30

1.5

cos6

0X

gF

M

1 R

equi

res

all 3

term

s.

Con

done

sin

/cos

con

fusi

on a

nd s

ign

erro

rs.

A1

Bot

h eq

uati

ons

corr

ect u

nsim

plif

ied.

co

s60

1.5

cos3

01.

5co

s60

0.6

cos3

0N

gg

N

D

M1

Use

0.

6F

N

to f

orm

an

equa

tion

in N

or

in X

. D

epen

dent

on

the

two

prev

ious

M m

arks

co

s60

cos6

01

0.6

1.5

cos3

01.

5co

s60

cos3

0co

s30

Ng

g

OR

: 0.

6(co

s60

1.5

cos3

0)1.

5si

n30

cos3

0X

gg

X

(i

)

26 o

r 2

6.0

N

(N

) A

1 F

irst

val

ue f

ound

cor

rect

ly. (

N o

r X)

(i

i)

1.5

cos3

0co

s60

XN

g

D

M1

S

ubst

itut

e th

eir

N (

or X

) to

fin

d X

(or

N)

Dep

ende

nt o

n th

e pr

evio

us M

mar

k.

26or

26

.5X

A

1 (

7)

Sec

ond

valu

e fo

und

corr

ectly

.

[1

2]

A

lt:

co

s30

cos6

01.

5,

co

s30

0.6

cos6

01.

5N

Fg

NN

g

M1,

R

esol

ve v

erti

call

y. C

ondo

ne s

in/c

os c

onfu

sion

. M

ust h

ave

all t

erm

s.

DM

1 U

se

0.6

FN

A

1 C

orre

ct u

nsim

plif

ied

equa

tion

1.5

26 o

r 26

.0co

s30

0.6

cos6

0

gN

A

1

cos3

0co

s60,

0.6

cos3

0co

s60

XF

NN

M

1,

Res

olve

hor

izon

tally

. F

ollo

w th

eir

N.

Mus

t hav

e al

l ter

ms.

Con

done

sin

/cos

con

fusi

on.

DM

1 S

ubst

itut

e fo

r F

and

N

26or

26

.5X

A

1

103

Que

stio

n N

umbe

r S

che

me

M

arks

Not

es

4 (a

) (i

)

M3

13

24

52

CD

Rg

gg

M

1

e.g.

Tak

e m

omen

ts a

bout

D –

req

uire

s al

l 4 te

rms

of

the

corr

ect f

orm

, but

con

done

sig

n er

rors

. 1x

nee

d no

t be

seen

A1

Cor

rect

uns

impl

ifie

d eq

uatio

n

5 o

r 49

NCR

g

A

1

(i

i)

R4

23

CD

RR

gg

g

M

1 e.

g.R

esol

ve v

erti

cally

to f

orm

an

equa

tion

in R

C a

nd

R D, r

equi

res

all 5

term

s

A

1 C

orre

ct u

nsim

plif

ied

equa

tion

4or

39

or

39.

2NDR

g

A

1 (

6)

Alt

M

( A)

34

63

25

30C

Dg

gR

Rg

M

1A1

T

wo

equa

tion

s –

M1A

1 fo

r ea

ch

M(B

)

3

46

24

24D

Cg

gR

Rg

M

1A1

M3

22

14

43

DC

Rg

gg

M(c

entr

e) 3

32

23

CD

gR

Rg

5 o

r 49

NCR

g

,

4or

39

or

39.

2NDR

g

A

1,A

1 S

olve

sim

ulta

neou

sly

for

R C a

nd R

D

(b)

M3

810

CD

Rxg

gg

(

3

18CR

xg

) M

1 F

irst

equ

atio

n in

x a

nd R

(or

RC a

nd R

D)

– co

rrec

t te

rms

requ

ired

but

con

done

sig

n sl

ips.

R4

2C

DR

Rg

gxg

Alte

rnat

ives

:

M

412

12C

DB

RR

gg

M(

):2

56

34

CD

AR

Rxg

g

M:2

23

41

4D

Cg

Rxg

g

M1

A s

econ

d eq

uati

on, c

orre

ct te

rms

requ

ired

but

co

ndon

e si

gn s

lips.

2

183

6x

gx

g

D

M1

Use

C

DR

R

and

sol

ve f

or x

. (as

far

as

x =

…..)

Dep

ende

nt o

n th

e tw

o pr

evio

us M

mar

ks.

3.

6x

A

1 (

4)

[10

]

104

Que

stio

n N

umbe

r S

che

me

M

arks

Not

es

5 (a

) S

peed

=

22

22

13

(2)

or

32

13m

s

M

1

U

se P

ytha

gora

s

A1(

2)

Acc

ept 3

.6 o

r be

tter

(b)

Igno

re th

eir

diag

ram

if it

doe

s no

t sup

port

thei

r w

orki

ng

2

tan

3

,

33.7

O

R

3ta

n2

, 56

.3

OR

fin

d an

othe

r us

eful

ang

le

M1

Fin

d a

rele

vant

ang

le

A

1 T

heir

ang

le c

orre

ct (

seen

or

impl

ied)

B

eari

ng =

124

A1

(3)

C

orre

ct b

eari

ng. A

ccep

t 124

o or

awrt

124

/124

o A

ccep

t N 1

24 E

or

S 5

6 E

(c

)

10

32

Bt

r

ji

j

M1

Fin

d th

e po

siti

on v

ecto

r of

B o

r G

at t

ime

t

A1

Cor

rect

for

B

5

42

23

Gt

ri

ji+

j

A1

Cor

rect

for

G

5

34

3t

t

O

R

10

22

2t

t

DM

1 C

ompa

re c

oeff

icie

nts

of i

or o

f j

to f

orm

an

equa

tion

in t.

(i

)

3t

s

A1

Cor

rect

una

mbi

guou

s co

nclu

sion

.

(ii)

103

32

94

m

r

j+i

ji

j

OR

5

42

32

94

3

ri

ji

ji

jm

A

1 (

6)

Fin

al a

nsw

er. A

ccep

t wit

h no

uni

ts. D

o no

t ign

ore

subs

eque

nt w

orki

ng.

[1

1]

105

Que

stio

n N

umbe

r S

che

me

M

arks

Not

es

6(a)

1

81.

512

v

M

1 U

se o

f v

uat

or e

quiv

alen

t for

8

t

2

120.

820

v

M

1 F

ollo

w th

eir

12

2

28v

m s

-1

A1

(3)

(b)

B1

shap

e B

1ft

(2

)

nos:

8,2

8; 1

2,28

ind

icat

ed.

Fol

low

thei

r 12

, 28

(c)

firs

t 8 s

:

di

st =

1

812

482

M

1

Cor

rect

met

hod

for

dist

ance

for

the

tria

ngle

(0-

8) o

r th

e tr

apez

ium

(8-

28)

A

1ft

Fol

low

thei

r 12

ne

xt 2

0 s:

d

ist =

1

1228

2040

02

A1f

t F

ollo

w th

eir

12, 2

8

T

otal

dis

t = 4

48 m

A

1 (

4)

Cor

rect

ans

wer

onl

y (c

ao)

(d)

20

282

2.8s

M1

Fin

d ar

ea o

f ri

ght h

and

tria

ngle

or

an e

xpre

ssio

n in

T

for

the

trap

eziu

m (

rect

angl

e +

tria

ngle

).

2

2814

02

2.8

s

A1f

t F

ollo

w th

eir

28

44

814

028

2000

T

D

M1

For

m a

n eq

uati

on in

T f

or th

eir

16, 4

48 a

nd 1

40

20

0044

814

050

.428

T

A1

(4)

O

r be

tter

(50.

4285

7...)

Acc

ept 5

0.

[

13]

t8

28

v 28 12

106

Que

stio

n N

umbe

r S

che

me

M

arks

Not

es

7

(a)

33

gT

a

M1

Eqn

of

mot

ion

for

Q: m

ust h

ave

the

corr

ect t

erm

s bu

t con

done

sig

n er

rors

A1

Cor

rect

equ

atio

n

2

cos6

02

Tg

a

(

2

Tg

a

)

M1

Eqn

of

mot

ion

for

P: m

ust h

ave

the

corr

ect t

erm

s bu

t con

done

sig

n er

rors

. Wei

ght m

ust b

e re

solv

ed.

A1

Cor

rect

equ

atio

n

A

llow

M1A

1 fo

r 3

2co

s60

5g

ga

in p

lace

of

eith

er o

f th

ese

two

equa

tion

s

2

5g

a

2 5ga

*

D

M1

Use

an

exac

t met

hod

to s

olve

for

a (

i.e. n

ot th

e eq

uatio

n so

lver

on

thei

r ca

lcul

ator

). D

epen

dent

on

the

firs

t 2 M

mar

ks o

r th

e M

for

the

com

bine

d eq

uati

on.

A1

Giv

en a

nsw

er d

eriv

ed c

orre

ctly

fro

m e

xact

w

orki

ng.

2

92

55

gg

Tg

M

1

Use

giv

en a

ccel

erat

ion

to s

olve

for

T.

A

1 (

8)

acce

pt 1

8 or

17.

6

(b)

22

2.4

20.

65

5

gg

v

M1

Use

the

give

n ac

cele

rati

on to

fin

d th

e sp

eed

2

35

vg

o

e i

nvol

ving

g

A1

(2)

A

ccep

t 2.2

or

2.17

107

Que

stio

n N

umbe

r S

che

me

M

arks

Not

es

(c)

Str

ing

slac

k: a

ccel

of

P (u

p pl

ane)

=

1co

s60

2g

g

B

1

2.

40

5

ggs

M

1 U

se o

f 2

22

vu

as

or

equ

ival

ent f

or th

eir

acce

lera

tion

2 5g

2.

41

2.4

0.48

55

gs

g

A1

T

otal

dis

t = 1

.08

m

A1f

t

(4

) 0.

6 +

thei

r 0.

48

(d)

20

35

2gg

t

0

2.17

4.9t

M

1 U

se o

f vu

at

or

equ

ival

ent w

ith

thei

r

acce

lera

tion

2 5g

to

fin

d t.

4

30.

4426

5

gt

g

0.44

o

r 0

.443

A1

(2)

on

ly

[1

6]

108