7/25/2019 Gravity M3 Answers

1/2

1. (a) F=2x

k

[kmay be seen as Gm1m2, for example] M1

Equating Fto mg at x= R, [mg =2

R

k

] M1

Conin!ing !ompletion [k= mgR2] to gie F=

2

2g

x

Rm

" #1 $

(b) Equation of motion% (m)a= (&) 2

2

2

2 )(

'

')(

)(

x

gRm

x

vvm

x

gRm=

M1;M1

Integrating: v2= x

gR2

( + c) or equivalent M1A1

Use of 2

$2 gRv =,x= Rto find c[ c=& *gR or use in def! int! M1

"u#stitutingx= $Rand finding V; V= 6gR

M1;A1 %[10]

Alternative in(b)

&or'energ (&)a

R x

mgR2

2

dx; = mv2& mu

2M1;M1

Integrating: [ R

mgR

x

mgR 22

= mv2

& m 2

g$ R

M1A1M1*inal 2 ar's as scee M1A1

2. (a) -2. ma=& 2x

cm

/1

221

'' vx =&

2x

c

21

v2 =A+ m

c

ignoreA M1 A1

v2= B+ mc2

x= R, v= U B= U2&Rc2

M1

.eading to v2= U

2+ 2c

Rx11

0 cso A1

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7/25/2019 Gravity M3 Answers

2/2

(#)

+=

RRcUmmU 1

212

21

21

21 22

M1 A1

.eading to c= 21

RU2

A1 $

[8]

3. (a) At surface

"22

mgRkmgR

k==

cso M1A1 2

(#) -2. 2

2

x

mgRxm =

$

22

2

2

2

1

'

'or

'

'

x

gRv

xx

gR

x

vv =

=

M1

== xx

gRvxx

gRvv '1

2

1or'

1'

$

22

2

2

M1

)(2

1 2

2 Cx

gRv +=

A1

x= 2R, v= C= 2gR

M1A1

v2=

gRxgR

2

2

Atx= R,gR

R

gRv =

22 2

M1

v= (gR) A1 %[9]

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