Chapter 3 Gravity and Satellites

8/12/2019 Chapter 3 Gravity and Satellites

http://slidepdf.com/reader/full/chapter-3-gravity-and-satellites 1/16

Heinemann Physics 12(3e)Copyright © Pearson Education Australia 2008 Teacher’s Resource and Assessment Disk(a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263

1

Worked solutionsChapter 3 Gravity and satellites

3.1 Newton’s law of universal gravitation

1 a Gravitational force F = GMm/ R2

= (6.67 ! 10 –11

! 0.100 ! 0.200)/(0.050)2

= 5.3 ! 10 –12

N

b Gravitational force F = (6.67 ! 10 –11

! 1000 ! 1000)/(10.0)2

= 6.7 ! 10 –7

N

c Gravitational force F

= (6.67 ! 10 –11

! 2.0 ! 104 ! 6.0 ! 10

24)/(7.0 ! 10

6)

2

= 1.6 ! 105 N

d Gravitational force F

= (6.67 ! 10 –11

! 7.3 ! 1022

! 6.0 ! 1024

)/(3.8 ! 108)

2

= 2.0 ! 1020

N

e Gravitational force F

= (6.67 ! 10 –11

! 6.0 ! 1024

! 60)/(6.4 ! 106)

2

= 5.9 ! 102 N

f Gravitational force F

= (6.67 ! 10 –11

! 1.67 ! 10 –27

! 9.11 ! 10 –31

)/(5.33 ! 10 –11

)2

= 3.61 ! 10 –47

N

2 a The term universal constant refers to the fact that the value of G (i.e. 6.67 ! 10 –11

N m2

kg –2

) has exactly the same value everywhere in the Universe. This means that Newton’s

law of gravitation is valid throughout the Universe.

b Another very famous universal constant is the speed of light in a vacuum.

3 a The gravitational force at any point is inversely proportional to R2. Since the new force is

one-quarter of the previous force, the distance from the centre of the Moon has doubled

i.e. two Moon radii.b F = 160/4

2 = 160/16 = 12.5 N

4 F Phobos/ F Deimos =2

D

D

2

P

P

R

M

R

M ÷

=2

PD

2

DP

R M

R M

=2615

2716

)104.9(101.08

)1035.2(101.08

!!!

!!!

= 38



Worked solutions Chapter 3 Gravity and satellites


2

5 a F = GMm/ R2

= (6.67 ! 10 –11

! 1.9 ! 1027

! 1000)/(7.15 ! 107)

2 = 2.48 ! 10

4 N

b The magnitude of the gravitational force that the probe exerts on Jupiter is equal to the

magnitude of the gravitational force that Jupiter exerts on the probe = 2.48 ! 104 N.

c Acceleration = " F /m = 2.48 ! 104/1000 = 24.8 m s

–2

d Acceleration = " F /m

= 2.48 ! 104/1.9 ! 10

27 = 1.3 ! 10

–23 m s

–2

6 F = GMm/R2

Using the inverse square relationship between F and R, for the force to be 1% (100

1) of its

strength at the surface, the distance must be 10 times greater; i.e. at 10 Earth radii (10 RE).

7 a F #/ F $ = M /0.01 M = 100

b Let x = the required distance

Then let 100( R – x)2/ x

2 = 8100

and x = 0.1 R

8 Let d = required distance

Then 1/d 2 = 0.01/( R – d )

2

and d = 0.91 R





3

3.2 Gravitational fields

1 Mercury: g = GM / R2

= (6.67 ! 10 –11

! 3.30 ! 1023

)/(2.44 ! 106)

2

= 3.70 N kg –1

Saturn: g = GM / R

2

= (6.67 ! 10 –11

! 5.69 ! 1026

)/(6.03 ! 107)

2

= 10.4 N kg –1

Jupiter: g = GM / R2

= (6.67 ! 10 –11

! 1.90 ! 1027

)/(7.15 ! 107)

2

= 24.8 N kg –1

2 a F g = mg = 80 ! 3.70 = 300 N

b F g = mg = 80 ! 10.4 = 830 N

c F g = mg = 80 ! 24.8 = 2.0 ! 103 N

3 While Saturn is about 100 times more massive than Earth, this is offset by Saturn having about

10 times the radius of Earth. Gravitational field strength increases with mass, but decreases with

distance from the centre of the planet.

4 g = 9.8 /4.02

= 0.61 N kg –1

5 Acceleration = 0.61 m s –2

6 Acceleration at 4.0 RE/acceleration at 2.0 RE = g(4R E)/g(2R E)

= (2.0/4.0)2 = 1:4 = 0.25

7 g = GM / R2

= (6.67 ! 10 –11! 3.0 ! 10

30)/(10 ! 10

3)

2

= 2.0 ! 1012

N kg –1

8 g = GM / R2

= (6.67 ! 10 –11! 3.0 ! 10

30)/(5.0 ! 10

6)

2

= 8.0 ! 106 N kg

–1

9 Acceleration = g = 8.0 ! 106 m s

–2

10 The gravimeter would respond to differences in the gravitational field strength in a particular

region due to either large deposits of low-density material (e.g. oil) or large deposits of high-

density material (e.g. iron ore).

11 a F = GMm/ R2

= (6.67 ! 10 –11

! 7.3 ! 1022

! 500)/(3.0 ! 106)

2 = 270 N

b " F = mg

a = g = " F /m = 270/500 = 0.54 m s –2

c F = GMm/ R2

= (6.67 ! 10 –11

! 7.3 ! 1022

! 50)/(3.0 ! 106)

2 = 27 N

d " F = mg

a = g = " F /m = 27/50 = 0.54 m s –2

e g = 0.54 N kg –1





4

12 Let x = required distance

Then 6.0 ! 1024

/ x2

= 7.3 ! 1022

/(3.8 ! 108 – x)

2

and x = 3.4 ! 108 m





5

3.3 Satellites in orbit

1 D is correct. The gravitational field strength at this altitude is less than 9.8 N kg –1

, so a satellite

will orbit with an acceleration of less than 9.8 m s –2

.

2 Since the gravitational force of attraction between the Earth and the satellite is always

perpendicular to the satellite’s velocity, the speed of the satellite remains constant and therefore

there is no change in its kinetic energy during its circular orbit.

3 The net force acting on the satellite during its orbit " F = mg = 2.3 ! 103 ! 0.22 = 510 N

4 The centripetal force is the gravitational force of attraction between the Earth and the satellite.

5 a a = GM / R2

= (6.67 ! 10 –11

! 1.02 ! 1026

)/(3.55 ! 108)

2

= 5.40 ! 10 –2

m s –2

b a = v2/ R

5.40 ! 10 –2

= v2/3.55 ! 10

8

v = 4.38 ! 103 m s

–1

c T = 2% R/v

= 2%(3.55 ! 108)/(4.38 ! 10

3)

= 5.095 ! 105 s = 5.095 ! 10

5/(24 ! 60 ! 60) = 5.90 days

6 a v = 2% R/T

= 2% ! 1.22 ! 109)/1.38 ! 10

6 = 5550 m s

–1 = 5.55 km s

–1

b a = v2/ R

= (5.55 ! 103)

2/(1.22 ! 10

9) = 2.52 ! 10

–2 m s

–2

7 F =2

2

2

4

T

Rm

R

GMm !

=

Transposing to make M the subject gives:

M =2

324

GT

R!

= 5.64 ! 1026

kg

8 a F = GMm/ R2

= 4%2 R m/T

2

then R3 = GMT

2/4%2

= 6.67 ! 10 –11

! 4.87 ! 1024

! (2.10 ! 107 s)

2/4%2

and R = 1.54 ! 109 m

b v = 2% R/T

= (2% ! 1.54 ! 109)/2.10 ! 10

7

= 460 m s –1

c a = v2/r

= (4.61 ! 102)

2/(1.54 ! 10

9)

= 1.38 ! 10 –4

m s –2





6

9 a i H

A

v

v

=

H

H

A

A

!2

!2

T

R

T

R

=HA

AH

RT

RT = 1.66

iiH

A

a

a

=

2

H

2

A

R

GM

R

GM

=2

A

2

H

R

R= 7.58

b For Atlas: T 2/ R

3 = 0.602

2/(1.37 ! 10

8)

3 = 1.41 ! 10

–25

For Titan: T 2/ R

3 = 1.41 ! 10

–25

T 2 = 1.41 ! 10

–25! (1.2 ! 10

9)

3 = 244

T = 15.6 days

11 The centre of the circular orbit of a satellite must be the centre of the Earth.

12 a g = GM / R2

= (6.67 ! 10

–11! 7.0 ! 10

20)/(385 ! 10

3)

2 = 0.31 N kg

–1

b F = GMm/ R2 = mv

2/ R

then v = & (G M / R)

= & (6.67 ! 10 –11! 7.0 ! 10

20)/(395 ! 10

3)= 340 m s

–1





7

3.4 Energy changes in gravitational fields

1 C is the correct answer. A satellite in a stable circular orbit around the Earth will have a constant

orbital speed and a constant radius.

2 The gravitational field strength is lowest at A and greatest at D.

3 The acceleration is indicated by the gravitational field strength, so it increases as the meteor

travels from A to D.

4 A, B and C are correct. As the meteor falls towards Earth, its gravitational potential energy is

transformed into kinetic energy. If air resistance is ignored, mechanical energy is conserved.

5 a A distance of 100 km above the Earth’s surface represents a distance from the centre of

the Earth =100 ! 103 + 6.4 ! 10

6 = 6.5 ! 10

5 m.

From the graph, F = 9.2 N

b From the graph, F = 5.0 NWhen the distance from the Earth’s centre = 9.0 ! 10

6 m,

then height above Earth’s surface = 9.0 ! 106 – 6.4 ! 10

6 = 2.6 ! 10

6 m.

6 a E k = " mv2

= 0.5 ! 1.0 ! (4.0 ! 103)

2 = 8.0 ! 10

6 J

b The increase in kinetic energy of the rock as it moves between these two points

= corresponding decrease in gravitational energy

= area under the graph from R = 6.5 ! 106 m to R = 9.5 ! 10

6 m

= 2.0 ! 107 J

c ' E k = 2.0 ! 107 J = final E k – 8.0 ! 10

6 J

and the final E k = 2.8 ! 107

Jd E k = " mv

2,

then 2.8 ! 107 J = 0.5 ! 1.0v

2

and v = 7.5 ! 103 m s

–1, i.e. 7.5 km s

–1

7 a v = & (GM / R)

= [(6.67 ! 10 –11

! 6.0 ! 1024

)/(7.0 ! 106 m)]

" = 7.6 ! 10

3 m s

–1 = 7.6 km s

–1

b E k = " mv2

= 0.5 ! 20 ! 103! (7.6 ! 10

3)

2

= 5.7 ! 1011

J

c T = 2% R/v

= (2% ! 7.0 ! 10

6

)/7.6 ! 10

3

= 5.8 ! 10

3 s

8 a v = & (GM / R) = [(6.67 ! 10 –11

! 6.0 ! 1024

)/(9.0 ! 106)]

" = 6.7 km s –1

b E k = " mv2 = 0.5 ! 20 ! 10

3! (6.67 ! 10

3)

2 = 4.4 ! 10

11 J

c T = 2% R/v

= 2% ! 9.0 ! 106/6.67 ! 10

3

= 8.5 ! 103 s





8

9 The increase in gravitational energy per kilogram of the satellite as it moved from its lower orbit

to its higher orbit is equal to the area under the graph between R = 7.0 ! 103 km to R = 9.0 ! 10

3

km

Area = 25 squares ! 1.0 ! 0.5 ! 106 = 1.25 ! 10

7 J kg-1

Increase in gravitational potential energy = area ! mass = 1.25 ! 107 ! 20 000 = 2.5 ! 10

11 J.

10 The initial kinetic energy of the satellite when it was launched = the increase in gravitational

potential energy of the satellite from R = 7.0 ! 103 km to R = 8.0 ! 10

3 km.

From the graph, the increases in gravitational potential energy of a 1.0 kg mass that has moved

between these two points = area under graph = 7.08 ! 103 J

Then for a 240 kg satellite the corresponding area = 240 ! 7.08 ! 103 J)

= 1.7 ! 109 J





9

3.5 Apparent weight and weightlessness1 a Weight F g = mg = 0.15 ! 9.8 = 1.5 N

b Mass = 0.15 kg

2 The toy and Gracie’s hand both move with an acceleration of 9.8 m s –2

down and so the normalforce on the toy is zero.

3 The toy would continue to move with an acceleration of 9.8 m s –2

down and so would move

with the same motion as Gracie. It would stay in the same position relative to her.

4 a Weight F g = mg = 50 ! 9.8 = 490 N

b (F = ma = 490 N down

F g = 490 N down, so the normal force is zero

c Her apparent weight is the same as the normal force i.e. zero.

5 a The weight of the girl is not zero, but she perceives her weight as zero because she feels

zero reaction force from the seat during the plummet. True weightlessness only occurswhen the gravitational field strength is zero. This only happens in deep space.

b Their acceleration must equal g.

6 a Weight F g = mg = 90 ! 9.8 = 880 N down

b (F = ma = 90 ! 35 = 3150 = 3.2 kN up

c (F = F N +F g

F N = (F – F g = 3150) – 880* = 4030 = 4.0 kN

7 a g = GM / R2 = 6.67 ! 10

–11 ! 6.0 ! 10

24/(6.8 ! 10

6)

2 = 8.7 N kg

–1

b F g = mg = 60 ! 8.7 = 520 N

8 a v = & (GM / R) = & (6.67 ! 10 –11

! 6.0 ! 1024

/6.8 ! 106) = 7700 m s

–1 = 7.7 km s

–1

b a = g = 8.7 m s –2

c (F = ma = 60 ! 8.7 = 520 N down

9 This comment is incorrect since the gravitational field strength is not zero. A better comment

would be ‘the astronaut is experiencing apparent weightlessness’.

10 The spacecraft and the astronaut are both moving with an acceleration that is equal to the

gravitational field strength and so the astronaut will experience zero normal force from the

space-craft.





10

Chapter review

1 F = GMm/ R2

then R2 = GMm/ F

= (6.67 ! 10 –11

! 5.69 ! 1026

! 1.05 ! 1021

)/(2.79 ! 1020

)and R = 3.78 ! 10

8 m

2 At a height of 2 Earth radii above the Earth’s surface, a person is a distance of 3 Earth radii

from the centre of the Earth.

Then F = 900/32 = 900/9 = 100 N so D.

3 a (F = F N +F g

F N = (F – F g = 2400) – 780* = 3200N ) so D

b The normal force is about 4 times greater than usual, so the astronaut will feel about 4

times heavier, so B.

c Weight F g = mg = 80 ! 9.8 = 780 N, so C.

d During orbit, the astronaut travels with an acceleration of 8.2 m s –2 and so is in free-fall.

F N = 0, so A.

e Weight F g = mg = 80 ! 8.2 = 660 N, so D.

4 M /(0.8 R)2 = m/(0.2 R)

2

and M /m = 1/42 = 1 : 16 = 0.063

5 a g = GM / R2

= (6.67 ! 10 –11

! 1.02 ! 1026

)/(2.48 ! 107)

2

= 11.1 N kg –1

b C, it will accelerate at a rate given by the gravitational field strength, i.e. at 11.1 m s –2

6 The orbital speed of the Moon around the Earth is given by:

v = & (GM / R)

= [(6.67 ! 10 –11

! 5.98 ! 1024

)/(3.84 ! 108)]"

= 1.02 ! 103 m s

–1

The period of the Moon around the Earth T = 2% R/v

= 2% ! 3.84 ! 108/(1.02 ! 10

3)

= 2.367 ! 106 s

= 2.367 ! 106 / (60 ! 60 ! 24) = 27.4 days

7 a v = & (GM / R)

= [(6.67 ! 10

–11

!

1.90 ! 10

27

)/1.10 ! 10

10

]

"

= 3.39 ! 10

3

m s

–1

b a = v2/ R

= (3.39 ! 103)

2/(1.10 ! 10

10)

= 1.05 ! 10 –3

m s –2

towards Jupiter

c T = 2% R/v

= 2% ! 1.10 ! 1010

/3.39 ! 103

= 2.048 ! 107 s = 2.048 ! 10

7 / (60 ! 60 ! 24) = 235 days

8 a C, a satellite is said to be in a geosynchronous orbit if its orbital period is the same as that

of the Earth’s rotation (i.e. 24 hours or 86 400 s).

b This satellite will always be directly above the same point on the Earth’s surface, and can

therefore transmit radio signals to any point that can see it.

c R3 = GMT 2/4%2,





11

then R3 = (6.67 ! 10

–11 ! 6.0 ! 10

24 ! 86400)

2/4%2

and R = 4.2 ! 107 m

9 a R3 = GMT

2/4%2

= (6.67 ! 10 –11

! 3.30 ! 1023

! (5.07 ! 106)

2/4%2

and R = 2.43 ! 108

mb v = 2% R/T

= 2% ! 2.43 ! 108/(5.07 ! 10

6) = 301 m s

–1

c a = v2/r

= (301)2/(2.429 ! 10

8) = 3.73 ! 10

–4 m s

–2 towards Mercury

10 a The increase in the kinetic energy of the rock

= area under the graph from R = 2.5 ! 106 m to R = 3.0 ! 10

6 m

= 5.6 squares ! 10 ! 0.5 ! 106

=2.8 ! 107 J

b The initial E k = " mv2

= 0.5 ! 20 ! (1.0 ! 103)

2 = 1.0 ! 10

7 J

The final E k = 2.8 ! 107 J + 1.0 ! 10

7 J = 3.8 ! 10

7 J

c 3.8 ! 107 J = 0.5 ! 20v

2

and v = 1.9 ! 103 m s

–1

d At r = 2.5 ! 106 m, F g = 70 N = mg

g = 70/20 = 3.5 N kg –1

11 a (T 1/T 2)2 = ( R1/ R2)

3

and T 1/T 2 = ( R1/2 R)3/2

= 1/ 8 so D

b v1/v2 = (2%R 1/T1)/(2%R 2/T2) = (T 2/T 1)( R1/ R2) = 8 /2 = 2 , so B.

c a1/a2 = (v12/ R1)/(v2

2/ R2) = 2 ! 2 = 4, so D.

12 Transposing GMm/ R2 = 4%2

Rm/T 2 to make M the subject gives: M = 4%2

R3m/GT

2.

Substituting into this using R = 1.5 ! 1011

m and T = 365.25 ! 24 ! 60 ! 60 gives M =

2.0!1030

kg.

13 From graph, when distance = 300 km = 3.0 ! 105 m, g = 9.0 N kg

–1

14 D is correct. Area units are: N kg –1

! m = N m kg –1

= J kg –1

15 C is the correct answer. The area under the graph gives the energy change per kilogram. In this

example, the satellite is falling closer to Earth, so it is losing gravitational potential energy.

16 This can be determined by finding the area under the graph between R = 2 ! 105 m and

R = 6 ! 105 m.

Counting squares give area = 35 ! 1.0 ! 1.0 ! 105 = 3.5 ! 10

6 J kg-1

Energy = mass ! area = 1000 ! 3.5 ! 106 = 3.5 ! 10

9 J

17 No, air resistance would mean that a significant amount of kinetic energy would be transformed

into heat energy.

18 D, the mass of the moons would need to be known before the forces can be compared.

19 Charon: T 2/ R

3 = 6.4

2/1960

3 = 5.0 ! 10

–9

Nix: T

2

/ R

3

= 5.0!

10

–9





12

T 2/4400

3 = 5.0 ! 10

–9

T = & (5.0 ! 10 –9

! 44003) = 21.5 days

20 F =2

2

2

4

T

Rm

R

GMm !

=

Transposing to make M the subject gives:

M =2

324

GT

R!

= 1.3 ! 1022

kg





13

Exam-style questions (Motion in one and twodimensions)

1 'U g = area under graph = 8.5 squares = 8.5 ! 1.0 ! 104 ! 0.4 ! 106 = 4.3 ! 1010 J

2 At 600 km: E k = " mv2 = 0.5 ! 10 000 ! 1500

2 = 1.1 ! 10

10 J

So E k at 100 km = 1 ! 1010

+ 4.3 ! 1010

= 5.3 ! 1010

J

" mv2 = 5.3 ! 10

10

0.5 ! 10 000v2 = 5.3 ! 10

10

v = 3.3 ! 103 m s

–1

3 a r = 6400 + 3600 = 10 000 km = 10 ! 106 m

F g = 4.0 ! 104 N (from graph)

b r = 6.0 ! 105 + 6.4 ! 10

6 = 7.0 ! 10

6 m

F g = 8.1!

10

4

N (from graph)

4 At 600 km, F g = 8.1 ! 104 N, so a = F g/m = 8.1 m s

–2

At 100 km, F g = 9.2 ! 104 N, so a = F g/m = 9.2 m s

–2

The acceleration increases from 8.1 m s –2

to 9.2 m s –2

5 Distance = area = 3350 m = 3.4 km

6 a C

b This section has the largest acceleration.

7 From 10–40 s and 60–80 s

8 Never

9 unbalanced, balanced

10 v = 2%r /T = 2 ! % ! 5.00/2.50 = 12.6 m s –1

11 a = v2/r = 31.6 m s

–2

12 F N = " F = ma = 60 ! 31.6 = 1.89 ! 103 N

13 T = 10.0/6 = 1.67

f = 1/T = 1/1.67 = 0.600 Hz

142

2

2

4

T

Rm

R

GMm !

=

R = (GMT 2/4%2

)1/3

= (1.25 ! 1023

)1/3

= 5.0 ! 107 m

15 v = 2%r /T = 2 ! % ! 5.0 ! 107/(0.31 ! 86 400) = 1.17 ! 104 m s

–1=11.7 km s

–1

16 Thalassa: T 2/ R

3 = 0.31

2/(5.0 ! 10

7)

3 = 7.69 ! 10

–25

Proteus: T 2/ R3 = 7.69 ! 10 –25





14

1.12/ R

3 = 7.69 ! 10

–25

R = & 1.57 ! 1034

= 1.16 ! 108 m

17 At maximum height: v = 0, a = –9.8 m s –2

, t = 1.0 s, x = ?

x = vt = " at 2 = 0 =0.5 ! –9.8 ! 1.0

2 = 4.9 m

18 9.8 m s –2

down

19 H: v = x/t = 8.0/2.0 = 4.0 m s –1

V: v = 0, a = -9.8 m s –2, t = 1.0 s, u = ?

v = u + at

0 = u – 9.8 ! 1.0

u = 9.8 m s –1

Use Pythagoras to find actual speed at launch: v = & (4.02 + 9.8

2) = & 112 = 11 m s

–1

20 D, if air resistance is ignored, the only force acting is the gravitational force.

21 ' p = pf – pi = (0.20 ! 30 south) – (0.20 ! 20 north) = 6.0 kg m s –1

south – 4.0 kg m s –1

north =10 kg m s

–1 south

22 Impulse = ' p = 10 N s south

23 a " F 't = ' p

" F = 1.0 ! 103 N south

b Using Newton’s third law, the average force that the ball exerts on the bat is 1.0 ! 103 N

north.

24 C and E are correct. The area under a force–time graph represents impulse or change in

momentum.

25 At 300 km, g = 3.0 N kg –1

F g = mg = 20 ! 3.0 = 60 N

26 Area = 9 squares = 9 ! 1.0 ! 2.0 ! 105 = 1.8 ! 10

6 J kg-1

' E k = area ! mass = 1.8 ! 106 ! 20 = 3.6 ! 10

7 J

27 Determine the energy associated with each grid square by multiplying each area by the mass of

20 kg. Calculate altitude at which total area starting from zero height is equal to 40 MJ.

28 W (by prospector) = F P ! d = 100 ! 20 = 2.0 ! 103 N m

29 W (on trolley) = " F ! d = 70 ! 10 = 1.4 ! 103 N m

30 ' E k =1.4 ! 103 J

31 ' E k =1.4 ! 103 J

" mv2 = 1.4 ! 10

3

v2 = 14

v = 3.7 m s –1

32 W (friction) = F f ! d = 30 ! 20 = 600 J

33 P = W (by prospector)/time = 2.0 ! 103/10 = 200 W





15

34 U s = " kx2 = 0.5 ! 150 ! (0.075)

2 = 0.42 J

35 vISS/vO= & (GM / RISS) / & (GM / RO) = & ( RO/ RISS) = & (4.24 ! 107/6.78 ! 10

6) = 2.5

36 F =2

2

24

T Rm

RGMm !

=

Transposing to make T the subject gives:

T = & (4%2 R

3/GM )

T ISS/T O = & ( RISS3/ RO

3) = & (4.1 ! 10

–3) = 0.064

37 aISS/aO = (G M / RISS2)/(GM / RO

2) = RO

2/ RISS

2 = (4.24 ! 10

7)

2/(6.78 ! 10

6)

2 = 39.1

38

39 k = gradient + 20 N m –1

40 U s = " kx2 = 0.5 ! 20 ! 15

2 = 2.3 ! 10

3 J

41 C, the force is becoming smaller so acceleration will decrease.

42 E k = " mv2 = 2250

0.5 ! 60 ! v2 = 2250

v = & 75 = 8.7 m s –1

43 " pi = " pf

(6.0 ! 120) + 0 = (120 + 45) ! vc

vc = 4.4 m s –1

44 ' p = pf – pi = (120 ! 4.4) – (120 ! 6.0) = – 200 kg m s –1

The ruckman loses 200 kg m s –1

of momentum.

45 In this collision, the momentum gain of the bag is equal to the momentum loss of the ruckman,

i.e. 200 kg m s –1

.

46 inelastic, momentum, kinetic energy

47 a a = v2/r = 6.0

2/2.0 = 18 m s

–2 up

b (F = ma = 55!

18 = 990 N )





16

F g = mg = 55 ! 9.8 = 539 N *

( F = F N + F g

F N = (F – F g = 990 N ) – 539 N * = 1529 N ) = 1.5 ! 103 N up

48 The apparent weight is given by the normal force of 1350 N. This is almost three times larger

than the weight force and so the skater would feel much heavier than usual.

49

50 a Cb One force acts on the floor and the other force acts on the ball. These forces are equal in

magnitude as described in Newton’s third law.

Chapter 3 Gravity and Satellites

Documents

Transcript of Chapter 3 Gravity and Satellites