Matter and Interaction Chapter 05 Solutions

7/25/2019 Matter and Interaction Chapter 05 Solutions

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1

5.X.1(a) bat, Earth, air (b) string, Earth, air (c) vine, Earth, air (d) snow, Earth, air

5.X.2

F1 +F2 +F3 = 0F

3= F

1 F

2

F3

= 30, 90, 130 N

5.X.3Both dpdt andF

netare0.

5.X.4

(a) dpdt nonzero, directed downward

(b) dpdt nonzero, directed downward

(c) dpdt nonzero, directed upward

(d) dpdt nonzero, directed leftward

5.X.5The best kissing circle is the one with radius R3

, and the net force is directed toward the center.

5.X.6

(a) Arrows representing the force at each point should all point toward Sun and should have correct relative lengths.

(b) zero at A and D, nonzero at all other points

(c) nonzero at all points

(d) zero at A and D, positive at B and D, negative at E and F

(e) changing at all points

5.X.7 Constant speed implies no change in magnitude, so there is no parallel component of net force. The perpendicularcomponent is directed toward the center. Earth and the seat exert forces on the passenger.

5.X.8

|a| |v|2

R

(30 m/s)2

200 m 4.5 m/s2


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2

Compare to9.8 m/s2 for freely falling objects.

5.X.9

dp

dt

= |p||v|R

|p| =dpdt R

c

(2 1010

N)(8 m)

(3 108 ms)

5.3 1018 kg m/s

5.X.10Define the system to be the child.

The parallel component of dpdt is

d |p|dt

p = 0

because|p| is constant.

The speed of the child is

|v| = 2RT

= 2(5 m)

90 s= 0.349 m/s

The magnitude of the momentum is

|p| = m |v|= 14.0 kg m/s

The perpendicular component of dpdt has a magnitude of

|p|

dp

dt

= |p||v|R

= (14 kg

m/s)0.349 m/s

5 m

= 0.977 N

and|p| d pdt points toward the center of the carousel. According to the Momentum Principle, the net force on the child is equto dp/dt which is 0.977 N and points toward the center of the carousel. The only object in the surroundings that exertsforce on the child in the direction of the center of the carousel is the force by the horse on the child.

5.X.11


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3

(a) The only force acting on the satellite is the gravitational force by Earth on the satellite. There is no centripetal forceacting on the satellite. The free-body diagram for the satellite should only show the gravitational force by Earth onthe satellite.

(b) The only force acting on the satellite is the gravitational force by Earth on the satellite. There is no centrifugal forceacting on the satellite. The free-body diagram for the satellite should only show the gravitational force by Earth onthe satellite.

(c) The satellite travels in a circle with a constant speed. Therefore, the net force on the satellite points toward the centerof the circle. The parallel component of the net force on the satellite is zero.

5.X.12

(a) The acceleration of the ball is

a = v

t= 1.43, 8.57, 0 m/s

(b)

p

t = m

v

t= 0.129, 0.771, 0 N

(c)

Fnet

= p

t= 0.129, 0.771, 0 N

5.X.13

Use direction cosines.

(a)

Fnet,x

=F

net

cos x

= (210 N)cos(23)

= 193 N

(b)

Fnet,x =Fnet cos y

= (210 N)cos(90 23)= (210 N)cos(67)

= 82 N

5.P.14


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(a) dpdt is zero. The box is static.

(b) 3 is the correct diagram.

(c) (40 kg)

9.8 Nkg

392 N

(d) Geometrically, its F

T2 cos38 but physically, it must also be 392 N.

(e)F

T2

392 Ncos38 497.4 N(f)

FT2

sin38306.2 N(g) It must be -306.2 N.

5.P.15

(a) dpdt is zero so it has no direction.

(b) Fnet is zero.

(c) rope, floor

(d) (255 N) cos 40195.3 N(e) It must be -195.3 N.

(f) Earth, floor, rope

(g) (255 N) sin 40163.9 N

(h) (30 kg)

9.8 Nkg

294 N

(i) 294 N 163.9 N130 N

5.P.16

Let be the angle between the spring and the vertical. Thensin = 815

and cos =

16115

. The vertical component of tspring tension must null out the balls weight.

M g = ks

s cos

s = Mg

ks

15161

(0.45 kg)

9.8 Nkg

(15)

(110 N/m) (

161)

0.047 m4.7 cmL

o 15 cm 4.7 cm10.3 cm

5.P.17


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(a)

L =

FLo

AY

(60 kg)

9.8 N

kg

(2 m)

(0.35 103 m)2(2 1011 N/m2) 0.015 m1.5 cm

(b) The interatomic bonds are stretched the most at the top of the rod, and progressively less and less from the rods topto the rods bottom.

(c) Let the origin be at the very bottom of the vertically hanging rod. Consider a small slice of the rod at the very top.This slice has a thickness L or y. The slice is in static equilibrium, so the force pulling it up must have the samemagnitude as the force pulling it down. The force pulling the slice down is the weight of the rest of the rod, and thisforce has a magnitude V g = Agy where is the materials density. Note that the density is needed to express thisforce as a function ofy . This little slice will stretch an amount L= gYyy. Taking the limit for smaller and smaller

slices gives dL = gYydy and we integrate both sides from y = 0 to y = L. Evaluating the integral gives Lrod = gL2

2Y ,

which is independent of radius (and therefore area). We can express the stretch as L = MgL2AY , which for this rod is

approximately 7.7

107

m.

(d) For a given material, the stretch is independent of radius, and therefore independent of area.

5.P.18

Let the left wire be 1, the right wire be 2, and the vertical wire be 3.

(a)F

1

6098 N, F2

3980 N, F3

7840 N(b) For each wire, the strain will just be the stress divided by Youngs modulus. Wire 1: LL

o

0.049, Wire 2: LLo

0.032,Wire 3: LL

o

0.063

5.P.19

(a) dpdt is zero.

(b) Earth, cable, air

(c) force due to cable pointing to the upper right, force due to air pointing left, force due to Earth pointing down

(d) The vertical component of the tension must null out the packages weight. Therefore,F

cable

9530 N.(e) F

cable=3609, 8820, 0 N

(f) The force due to the air must null out the horizontal component of the tension. Therefore,F

air

3609 N.(g) F

air=3609, 0, 0 N

(h) Yes, the cable will likely break.

5.X.20


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6

(a) distance =2R = 69.1 m

(b)

|v| = distancet

= 69.1 m

12 s

= 5.76 m/s

5.X.21

The speed of the atom is

|v| = 2RT

= 2.24 m/s

and its direction is tangential (i.e. tangent to its path).

5.X.22Assume that Earth travels from A to B in the figure. Thus, it is orbiting counterclockwise around Sun.

The velocity of Earth and the net force on Earth are shown in Figure 1.

Sun

A

B

Figure 1: The velocity of Earth and the net force on Earth at points A and B.

The vector p is toward Sun andpf

is in the direction ofpi

+ p as shown in Figure2.

5.X.23

(a) The momentum is zero, so it has no direction.

(b) dp/dtis downward, toward Earth.

(c) The net force on the ball is downward, toward Earth.

(d) At the turning point, the momentum is zero, so it has no direction.

(e) dp/dtis to the left.

(f) The net force on the cart is to the left.


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Sun

A

B

Figure 2: The change in momentum and the final momentum (after 1 hour) from points A and B.

(g) At this instant, its momentum is zero, so it has no direction.

(h) dp/dt is upward.

(i) The net force on the block is upward.

5.X.24

(a)

|v| = 2RT

= 2(7 m)

12 s= 3.67 m/s

(b) (iii) The direction of the momentum of the atom is tangent to its path.

(c) (i) The direction of the rate of change of the momentum of the atom is inward, toward the center of its circular path.

5.X.25

(a) 1. a

2. h

3. g

4. f

5. e

(b) 1. g

2. f

3. e

4. d

5. c

(c) 1. g


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2. f

3. e

4. d

5. c

5.X.26

(a) c

(b) zero

(c) At location D and also at location F, the perpendicular component of the net force on the comet is nonzero.

(d) zero

(e) changing

(f) e

(g) nonzero

(h) nonzero

(i) positive

(j) changing

(k) h

(l) nonzero

(m) nonzero

(n) negative

(o) changing

5.X.27

Because the speed of the particle is constant, dpdt is perpendicular to the path where the path is curved and is zero whe

the path is straight. For small radius of curvature (of the kissing circle) at a given point, dpdt is large. For a large radius

curvature, dpdt is small. If the path is straight or is an inflection point, then dpdt is zero.

Arrows for dpdt are shown in Figure 3 for given points on the path.

5.X.28

false. At constant speed, the net force on the object is inward toward the center of the kissing circle.

true

true

false. To make an object turn, the net force on the ob ject must be inward toward the center of the kissing circle whiin this case would be toward the left.


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zero

zero

Figure 3: dpdt at given points on the path of a particle.

5.X.29

No. There is no force pushing you to the right as the car turns to the left. According to Newtons first law, your bodycontinues moving in a straight line unless a force acts on your body. As the car turns to the left, your body continues movingin a straight line until hitting the door. Then, the door exerts a force on you to the left, toward the center of the kissing

circle.

5.X.30

(1) If the net force on Moon is zero then it would travel in a straight line at constant speed. The fact that it moves in acircular path shows that the net force on Moon is not zero.

(2) Every force is due to an interaction of two objects. The gravitational force is the force by Earth on Moon. What objectexerts the centrifugal force? There is none, so the force does not exist.

5.X.31

dp

dt = |p|d p

dt+

d|p|dt

p

Because the speed is constant, d|p|dt p= 0. Thus,

dp

dt = |p|d p

dt

which has a magnitude of

dp

dt = |p||v|

R

= (78 kg m/s)23 m/s

4 m= 449N


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According to the Momentum Principle, the magnitude of the net force is equal to the magnitude of the time rate of chan

of the momentum, soF

net

= 449 N.5.X.32

(a) Because the protons speed is constant, d|p|dt p= 0.

(b)

dp

dt = |p|d p

dtdpdt = |p||v|R

= m |v|2

R since the protons speed is much less than c.

= 1.04 1014 N

The direction of|p| dpdt is toward the center of the circle, which is arrow (h).

5.X.33

(a) d

(b) b

(c) Because the protons speed is constant, d|p|dt p= 0. Thus,

dp

dt

=

|p

|

d p

dtdpdt = |p||v|R

= m |v|2

R since the protons speed is much less than c.

= 8.74 1015 N

5.X.34

(a) d|p|dt p= 0because the childs speed is zero.

(b) The childs speed is

|v| = 2RT

= 3.99 m/s

The magnitude of|p| d pdt is


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|p|dpdt = |p||v|R

= m |v|2

R since the childs speed is much less than c.

= 169 N

The direction of|p| d pdt is toward the center of the carousel.

(c) According to the Momentum Principle, the net force on the child is equal to dpdt which is 169 N, toward the center ofthe carousel.

(d) The object that contributes to this horizontal force on the child is the horse.

5.X.35

The childs speed is

|v| = 2RT

= 1.76 m/s

The tangential component of the net force on the child is zero since the childs speed is constant. Therefore, the net force onthe child is directed toward the center of the circle and has a magnitude

Fnet,

= |p||v|R

= m |v|2

R since the childs speed is much less than c.

= 66.3 N

5.X.36

Convert the period of Earths orbit from years to seconds.

T = (1year) = 3.156 107 s

Earths speed is

|v| = 2RT

= 2(1.5 1011 m)

3.156 107 s= 3.0 104 m/s

Because Earths speed is constant, d|p|dt p= 0.


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The direction of|p| dpdt is toward the center of the circle, which is toward Sun in this case. Its magnitude is

|p|d pdt = |p||v|R

= m |v|2

R since Earths speed is much less than c.

= (6 1024 kg)(3 104 m/s)21.5 1011 m

= 3.6 1022 N

The gravitational force by Sun on Earth is directed toward the center of the circle (toward Sun) and has a magnitude

Fgrav

= GMsunMEarthRorbit

= 3.6 1022 N

Note that we did not use the Momentum Principle, but rather we showed that it is true for Earths orbit. We calculated

from the motion of Earth and found thatdpdt = 3.6 1022 N. Then, we calculated the gravitational force by Sun on Ear

using Newtons Law of Gravitation and showed that it is 3.6 1022 N. These are equal, as is expected from the MomentuPrinciple.

5.X.37

A circular pendulum is analyzed as an example in CH 05 of the textbook. The radius of the pendulum is R= L sin = 0.516 mApplication of the Momentum Principle in the vertical direction shows that

FT

cos = mg

Application of the Momentum Principle in the radial direction shows that

FTsin = m |v|2

R

Solving forFTin the first equation and substituting into the second equation allows one to solve for the speed of the pendulum

mg tan = m |v|2

R

g tan = |v|2

R

|v| =

Rg tan

=

(0.516 m)(9.8 N/kg) tan(28)

= 2.69 m/s


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The period is found using:

|v| = 2RT

T = 1.21 s

5.X.38

(a) Arrow (a)

(b) The net force on Tarzan is upward. The net force is the sum of the force by the vine and the force by Earth on Tarzan.Therefore, the upward force by the vine on Tarzan is greater than the downward gravitational force by Earth on Tarzan,so that when added together, the net result is upward.

5.X.39b.

5.X.40Since the airplanes speed is constant, dpdt for the passenger is directed toward the center of the circular path. In thiscase, it is directed downward.

5.X.41a.

5.X.42

(a) d|p|dt = 0. It is directed in thex direction if speeding up or in the +x direction if slowing down.

(b) dpdt = 0 because the motion of the object is along a straight path.

(c) F

= 0. It is directed in thex direction if speeding up or in the +x direction if slowing down.

(d) F

= 0 because dpdt = 0.

5.X.43

(a) d|p|dt = 0 because the airplanes speed is constant.

(b) dpdt= 0 because the direction of the airplanes momentum is changing. It must be perpendicular to the path; therefore,it could be in the +x,x, +y, ory direction.

(c) F

= 0 because d|p|dt = 0.

(d) F

= 0 because dpdt= 0. It must be perpendicular to the path; therefore, it could be in the+x,x,+y, orydirection.

5.P.44

The riders speed is


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|v| = 2RT

= 5.98 m/s

The riders speed is constant; therefore, the net force on the rider is always directed toward the center of the Ferris wheand has a magnitude

Fnet,

= |p||v|R

= m |v|2

R since the riders speed is much less than c.

= 200N

At each point in the motion, sketch a free-body diagram. There are two forces acting on the rider: the force by the seat anthe gravitational force by Earth. The gravitational force by Earth always acts downward. The force by the seat on the varieso that when added to the gravitational force, the net force points toward the center of the Ferris wheel and has a constamagnitude.

The gravitational force by Earth on the rider is0, mg, 0=0, 549, 0 N.

In all answers below, define the +y direction to be upward and the +x direction to the right.

(a) dpdt =0, 200, 0 N(b)0, 549, 0 N(c) The net force on the rider is

Fnet

= Fgrav

+Fseat

Solve for the force by the seat on the rider.

Fseat

= Fnet

Fgrav

= 0, 200, 0 N 0, 549, 0 N= 0, 749, 0 N

(d) dpdt =0, 200, 0 N(e)0, 549, 0 N(f)

Fnet

= Fgrav

+Fseat

Fseat

= Fnet

Fgrav

= 0, 200, 0 N 0, 549, 0 N= 0, 349, 0 N

(g) The rider feels heavier at the bottom of the ride becauseF

seat,y

> Fgrav

.


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(h) The rider feels lighter at the top of the ride becauseF

seat,y

< Fgrav

.

5.P.45

(a) a.

(b) Because the speed is constant, d|p|dt p= 0.

(c)

|p|dpdt = |p||v|R

= m |v|2

R since the speed is much less than c.

= 492 N

|p| dpdt is directed toward the center of the circle, which is at this instant in the +x direction.(d) It is in the +x direction.

(e) Fnet =0, 492, 0 N(f)

Fnet

= F1

+F2

F2

= Fnet

+F1

= 492, 0, 0 N 196, 369, 0 N= 296, 369, 0 N

5.P.46

(a) The net force on the car is directed toward the center of the circle and has a magnitude

Fnet,

= |p||v|R

= m |v|2


In the vertical direction (with +y upward), the y-component of the force by the road on the car is equal to thegravitational force by Earth on the car.

Froad,y = mg

The maximum frictional force by the road on the car is parallel to the road and has a magnitude

froad,x

= sFroad,y

froad,x

= smg


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Since the only force on the car that acts toward the center of the circle is the frictional force, then

Fnet,

= froad,x

m |v|2

R = smg

|v| = sgR= 13.6 m/s

This is the maximum possible speed because the maximum possible frictional force (static friction) acts on the car.

(b) (3) and (6) are true.

(c) The maximum speed is the same because it does not depend on mass.

(d) The maximum speed is proportional to the square root of the radius. Thus,|v|= 2(13.6 m/s) = 19.3 m/s

5.P.47

(a)

Fnet,

= |p||v|R

= m |v|2


|v| =

Fnet,

Rm

= 41.3 m/s

(b) 2

(c) Fspring

= ks

ks

=

Fspring

s

= 760 N(9 8.7) m

= 2530 N/m

5.P.48

(a) d|p|dt p= 0


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(b)

|p|dpdt = |p||v|R

= m |v|2


= 8.71 N

(c) 871N

(d) The force by the cord on the child is upward (+y direction) and the gravitational force by Earth on the child is downward(y direction). The net force on the child is in the upward (+y direction), toward the center of the circle.

Fnet

= Fcord

+Fgrav

Fcord

= Fnet

Fgrav

Fcord

= 0, 871, 0 N 0, 255, 0 NFcord = 0, 1130, 0 N

(e) Fspring

= ks

ks

=

Fspring

s

=

1126 N

(4.3

4.22) m

= 1.4 104 N/m

5.P.49

(a) d|p|dt p= 0

|p|

dp

dt

= |p||v|R

= m

|v|

2

R since the speed is much less than c.= 495 N

(b)0, 495, 0 N(c) The force by the cord on the child is upward (+y direction) and the gravitational force by Earth on the child is downward

(y direction). The net force on the child is in the upward (+y direction), toward the center of the circle.


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Fnet

= Fcord

+Fgrav

Fcord

= Fnet

Fgrav

Fcord

= 0, 495, 0 N 0, 235, 0 NF

cord=

0, 730, 0

N

(d) Fspring

= ks

ks

=

Fspring

s

=

730 N

(3.1 3.06) m

= 1.8 104

N/m

5.P.50

(a) First, find the speed

Fnet

= |p||v|R

= m |v|2


|v| =

FnetR

m

= 27.8 m/s

Then, calculate the period.

|v| = 2RT

= 1.70 m/s

(b) 27.8 m/s

(c) (4)

5.P.51

The net force on the roller coaster at any instant is


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Fnet

= Ftrack

+Fgrav

Since the roller coaster travels in a circle with constant speed,

Fnet

= |p||v|R

= m |v|2


At the top of the roller coaster, the net force on the roller coaster is toward the center of the circle which is in the downward(y) direction. The gravitational force on the roller coaster is also in the downward (y) direction. At the minimum speedneeded to make it around the loop, F

track= 0 at the top of the loop. Thus,

Fnet

= FgravF

net

= Fgrav

m |v|2

R = mg

|v| =

Rg

A reasonable R is on the order of 10 m which gives a minimum speed of about 10 m/s.

5.P.52

A circular pendulum is analyzed as an example in CH 05 of the textbook. The radius of the pendulum is R = L sin .Application of the Momentum Principle in the vertical direction shows that

FTcos = mg

Application of the Momentum Principle in the radial direction shows that

FTsin = m |v|2

R

Solving forFTin the first equation and substituting into the second equation allows one to solve for the speed of the pendulum.

mg tan = m |v|2

R

g tan = |v|2

R

|v| =

Rg tan


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The speed is|v|= 2RT . Substitute the speed and solve forg in terms of the period T.

2R

T =

Rg tan

g = 42R

T2 tan

To determineg from experimental data, you must measure the period, the radius of the circle, and the angle .

5.P.53

The speed of the engineer is

|v| = 2RT

= 2.93 m/s

Define the system to be the engineer. The only force acting on the engineer is the force by the space station on the engineeThus,

Fnet

= Fspace station

Since the engineer travels in a circle with constant speed,

F

net

= |p||v|R

= m |v|2


Thus,

Fspace station

= Fnet

=

m |v|2R

= 43.0 N

By Newtons third law, the force by the engineer on the space station is also 43.0 N. Thus, the engineer must exert a forcof 43 N on the space station in order to hold on.

5.P.54

First, convert period from days to seconds. T = 1.04day= 8.99 104 s. Define the system to be the NEAR spacecraft. Tnet force on NEAR is equal to the gravitational force by Eros on NEAR. Note that the speed of NEAR is much less than Thus,


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Fnet

= Fgrav

|p||v|

R =

GMm

R2

m |v|2R

= GMm

R2

|v|2 = MGR

The speed of the spacecraft is v = 2RT . Substitute this to get an expression that relates the period and radius of thespacecrafts orbit.

2R

T

2=

M G

R

42R2

T2 =

M G

R

T2 = 42

GMR3

The above expression is known as Keplers third law for circular orbits. Solve for the mass of Eros.

M = 42R3

T2

= 6.65 1012 kg

5.P.55

The force by the seat on the rider is upward (+y direction) and the gravitational force by Earth on the rider is downward(y direction). The net force on the rider is in the upward (+y direction), toward the center of the circle.

Fnet

= Fseat

+Fgrav

Fseat

= Fnet

Fgrav

=

0,

mv2

R , 0

0, mg, 0

=

0,

mv2

R +mg, 0

= 0,m42R

T

2 +mg, 0

Thus,F

seat

= m42RT2 +mg.F

seatupward, in the +y direction, toward the center of the circle. Thus, F

seat=.

5.P.56


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22

(a) The force by the seat on the rider can be upward (+y direction) or downward (y direction) depending on how fathe roller coaster goes over the hill. The gravitational force by Earth on the rider is downward (y direction). At ttop of the hill, the net force on the rider is in the downward ( y direction), toward the center of the circle. To feweightless, the force by the seat on the rider must be zero at the top of the hill.

Fnet

=

0

Fseat

+Fgrav

0,m |v|2

R , 0

= 0, mg, 0

m |v|2R

= mg

|v| =

gR

(b) At the bottom of a hill (or a dip), the net force on the rider is upward. The force by the seat on the rider is alupward and in this case has a magnitude of3mg. The net force on the rider is

Fnet

= Fseat

+Fgrav

0,m |v|2

R , 0

= 0, 3mg, 0 + 0, mg, 0

m |v|2R

= 2mg

|v| =

2gR

5.P.57

(a) T = 6.88 s10revolutions = 0.688s

|v| = 2RT

= 13.7 m/s

(b) Yes, the momentum vector changes because its direction changes. It is not moving in a straight line.

(c) The force by the spring on the ball changes its direction because this force is perpendicular to the balls path.

(d)Fspring

=ks = 300 N

(e)

Fnet

= m |v|2R

= 300 N

Solving for mass gives m = 32.8 kg.


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23

5.P.58

(a) Assume that the parallel component of the net force on the car is zero; therefore, its speed might be constant or thecar be transitioning from speeding up to slowing down as it goes over the hill. The net force on the car at the top ofthe hill is thus directed toward the center of the hill, in they direction. The forces that act on the car are: (1) thegravitational force by Earth on the car and (2) the force by the road on the car.

(b) The net force on the car is in they direction, toward the center of the circular path. Thus, the downward force byEarth is greater than the upward force by the road. A diagram is shown in Figure 4.

Figure 4: Forces on a car traveling over a hill.

(c)

Fnet

= Froad

+Fgrav

Froad

= Fnet

Fgrav

=

0,

m |v|2R

, 0

0, mg, 0

Froad,y

= mg m |v|2

R

Note that it is less than mg as expected.

(d) If the cars speed is great enough such that m|v|2

R =mg, then the car will momentarily leave the road and the force bythe road on the car will be zero.

5.P.59

The net force on the ball is equal to the force by the spring on the ball which isF

spring

= ks

. The distance the spring

stretches is s = R L. Because the speed of the ball is constant and is much less than the speed of light,F

net

= m|v|2

R .

Therefore,

Fnet

= Fspring

m |v|2

R = k

s(R L)

|v| =

ks

R(R L)m


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24

For circular motion,|v|= 2RT . Substitute for|v| and solve for T.

2R

T =

k

sR(R L)

m

T = 42mR

ks

(R

L)

5.P.60

Define the system to be Earth. The net force on Earth is the gravitational force by Sun on Earth. Use this to solve for tspeed of Earth.

Fnet

= Fgrav

|p||v|

R =

GMm

R2

m

|v

|2

R =

GMm

R2

|v|2 = GMR

Earth is in a nearly circular orbit, thus its speed is v = 2RT . Substitute this to get an expression that relates the period anradius for Earths orbit.

2R

T

2=

GM

R

42R2

T2 =

GM

R

T2 = 42

GMR3

The above expression is known as Keplers third law for circular orbits.

From the solution above, Earths orbital speed as determined from Newtons law of gravitation and The Momentum Principis

|v| =

GM

R

=

6.6742 10

11

N m2

kg2

(2 1030 kg)1.5 1011 m

= 3.0 104 m/s

Note that the radius of Earths orbit is defined as 1 A.U. (Astronomical Unit) which is 1.5 1011 m.

5.P.61


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25

(a) The

dpdt = |p||v|R

= m |v|2

R since the persons speed is much less than c.

= 1.58 N

(b) The direction of the rate of change of momentum of the person is toward the center of the circle. In this case, it appearsthat the angle with respect to the +x axis (to the right) is about 120. Thus, the direction of dp/dt is given bydirection cosines: cos(120), cos(30), 0=0.5, 0.866, 0.

(c)F

net

= 1.58 N. A sketch of the net force on the person is shown in Figure5.

Fnet

Figure 5: The net force on a person riding a Ferris Wheel.

5.P.62The net force on the charged particle is equal to the magnetic force on the particle. Since the particles speed is constant,

the net force on the particle is toward the center of the circle and its magnitude isF

net,

=|p||v|r . wherer is the radius ofits path.

Fnet

= FmagneticF

net

= Fmagnetic

|p||v|

r = q|v| B

|p|

= qBr

5.P.63

(a) For a star orbiting a central body of mass M at the center of the galaxy, the net force on the star is equal to thegravitational force by the central body on the star. The net force is also equal to the time rate of change of themomentum of the star. If it orbits with constant speed in a circular orbit of radius r , then


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Fnet

= FgravF

net

= Fgrav

|p||v|

r =

GMm

r2

|v| =

GMr

Thus, the speed of the star should be proportional to r1/2. Stars further from the center of the galaxy should havesmaller orbital speed.

(b) Rubin found that stars speeds were independent of distancer for stars further out in the galaxy. As a result, they agoing faster than expected. This means that there is more mass in the galaxy that is not found at the center of tgalaxy. It must be spread throughout the disc of the galaxy, even where there is little luminous matter.

5.P.64

(a) Convert the units for the period of the satellite to seconds. T = 24 h = 86400 s.

Use Keplers third law for a circular orbit that was derived in Problem 5.P.60 from Newtons law of gravitation anThe Momentum Principle for an object moving at constant speed along a circular path.

T2 = 42

GMR3

R = 4.24 107 m

Its useful to calculate the altitude h of the satellite, which is the height of the satellite above Earths surface. Taltitude is h = R

RE= 4.24

107 m

6.4

106 m = 3.6

107 m.

(b) The distance from one person to the satellite is h. A radio wave travels from person 1 to the satellite back to personfor a total distance of2h. Then person 2 responds and a radio wave travels from person 2 to the satellite and back tperson 1 for a distance of2h. The total distance traveled by the radio waves is d = 4h.

d = 4h

= 1.44 108 mt =

d

c= 0.48 s

This is the minimum delta time simply due to the travel time for light (in this case, the light has a radio wavelengtand is not visible).

(c) Use Keplers third law with the radius of the orbit approximately equal to Earths radius (RREarth= 6.4 106 m

T2 = 42

GMR3

T = 5073 s = 85 minute= 1.4hour


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27

(d) Its speed is

|v| = 2RT

= 7.9 103 m/s

(e) The orbital radius in this case is approximately the radius of Moon, 1.74 106

m. The mass of Moon is 7.35 1022

kg. Use Keplers third law to find the orbital period.

T2 = 42

GMR3

T = 6500 s

The time interval that the astronauts lose radio contact is half the period, 1/2T= 3250 s.

5.P.65

A picture of the situation is shown in Figure6. The vectors from the center of the circle to the stars are their position vectors.

a a

a

1

2

v

v

v

Figure 6: Three stars in a particular three-body orbit.

The stars are equidistant from another and form an equilateral triangle of side lengtha. The stars move in uniform circularmotion with speed v (that is much less than the speed of light) and radius r from the center of the circle. Therefore, the netforce on each star has a magnitude

Fnet

= m |v|2r


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28

and is directed toward the center of the circle. Select star 2 to be the system". Apply the Momentum Principle to star We will need to calculate the gravitational force of each of the other stars on star 2; therefore, we need to know the positioof the stars at this instant. Define the +x direction to the right and the +y direction upward. Use direction cosines to gthe unit vector for each stars position (i.e. r =< cos x, cos y, 0 >). Note that because it is an equilateral triangle, tangles between the positions of the stars are 120.

r1

= r =r

r2

= r =r r

3= r =r

The vectors from star 2 to each of the other stars are:

r32

= r3

r2

= r

=

3r

r12

= r1

r2

= r r = r

=

3r

Calculate the gravitational force by star 1 on star 2 and the gravitational force by star 2 on star 3. The net force on staris the sum of these forces.

F32 = Gmmr

32

r32=

Gmm

3r2

F12

= Gmmr

12

r12=

Gmm

3r2

The net force on star 2 is

Fnet

= F32

+F12

= Gmm

3r2 +

Gmm

3r2

= Gmm

3r2


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29

Its magnitude is:

Fnet

= Gmm3r2

(1.5 )2 + (0.866 )2 + (0 )2

= Gmm

3r2

(1.5 )

2+ (0.866 )

2+ (0 )

2

= Gmm3r2

3

= Gmm

3r2

According to the Momentum Principle, the net force is equal to dp/dt which has a magnitude ofv 2/r. Thus,

m |v|2r

= Gmm

3r2

m |v|2r

= Gmm

3r2

|v|2 = Gm3r

The speed of a star in uniform circular motion is v = 2r/T. Substitute this for the speed and solve for the period of thestar.

42r2

T2 =

Gm3r

T2 =

342

GMr3

Note that this looks like Keplers law for a star orbiting a central object of mass m except for the constant

3. The additionalgravitational force due to the second star in the three-body problem increases the period compared to a star orbiting a centralbody in the two-body problem.

The period of the star will be

T =

3

42r3

GM

1/2

5.P.66

(a) The period of the star can be found from the figure. From 1995 to 2004 the star travels approximately 90 degrees, or1/4 of a revolution around the circle. Therefore, the period of the star is T= 4(9 y) = 36 y. Convert this to seconds,T = 1.136 109 s.The speed of the star is


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30

|v| = 2rT

= 2(2.9 1014)

1.136 109 s= 1.60 106 m/s

In terms of the speed of light, the stars speed is (1.60 106 m/s)/(3 108 m/s) = 0.005c(b) The stars speed is less than one-hundredth the speed of light. As a result, it is considered non-relativistic, andpm(c) Use Keplers third law.

T2 = 42

GMr3

M = 1.11 1037 kg

Divide this by the mass of Sun to get the mass in units of solar masses. (1.11 1037

kg)/(2 1030

kg) = 5.6 millisolar masses.

5.P.67

At the North Pole (NP), the net force on the person standing on the scale is zero. The upward force by the scale on tperson is equal in magnitude to the gravitational force on the person. The calibrated scale reads 10 kg. The force by tscale on the person at the NP is

F

scale, N.P.

=F

grav

= mgNP

= m GMEarth

RNP

= (10 kg)

6.6742 1011 N m2

kg2

(6 1024 kg)

(6.357 106 m)2= (10 kg)(9.9 N/kg)

= 99.0 N

At the equator, the person is in uniform circular motion with a period T = 24 h = 86400 s. The net force on the persontoward the center of Earth and has a magnitude mv2/R. Call this the positive radial" direction. Then, the net force on tperson is the sum of the gravitational force (which poiints toward Earth) and the force by the scale on the person (whicpoints away from Earth).

Fnet

= Fgrav

+Fscale

m|v|2R

= mGM

R2 F

scale

F

scale

= m GMR2

m |v|2

R


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31

Substitutev = 2R/T.

Fscale

= m GMR2

m 42R

T2

= (10 kg)6.6742 1011 N m2

kg2

(6 1024 kg)

(6.378 106

m)2

(10 kg)

42(6.378 106 m)

(86400 s)2

= 98.4 N 0.337 N= 98.1 N

The slightly larger radius at the equator results in a slightly smaller gravitational force of 98.4 N. The rotational motionfurther reduces the scale force by 0.337 N to give a net force of 98.1 N.

What will the scale read? It was calibrated at the NP to give the mass in kg. Thus, we must divide by g at the NP.

scale reading = 98.1 N

9.9 N/kg

= 9.91 kg

5.P.68

It is easiest to use the dot product ofFnet

p. The dot product is defined as:

Fnet

p =F

net

|p| cos

where is the angle between the net force and momentum vectors. VPython has a function dot(A,B) which calculates the

dot product of two vectors A and B. The parallel component of the net force isF

net

cos . Thus, it can be calculated using

Fnet,

=F

net

cos p=

Fnet

p

|p| p

The net force can be written as Fnet

=F

+F

. Use this to solve for the perpendicular component of the net force.

F

= Fnet

F

Use these equations to calculate the parallel and perpendicular components of the net force. Draw arrows for the net forceand its components. Multiply the arrow by a scale factor to scale its length so that it can be seen. Use the distance fromEarth to the spacecraft to scale the length of the arrow.

An example program is shown below for a spacecraft orbiting Earth in an elliptical orbit.


32/33

32

1 from __future__ import d i v i s i o n2 from v i s u a l import 3

4 RE = 6. 4 e6 #r a d i u s o f E ar th 5

6 s p a c e c r a f t = s p h e r e ( p o s =(10RE , 0 , 0 ) , c o l o r =c o l o r . c ya n , r a d i u s = 0. 25RE)7 E a rt h = s p h e r e ( c o l o r =c o l o r . b l u e , r a d i u s =RE)8

9 m=1.5e4 #mass o f s p a c e c r a f t 10 ME = 6e 2411 G = 6 . 6 7 e1112

13 v=1.2sq rt (GME/mag( spa ce cr af t . pos ) ) v e c t o r ( 0 , 1 , 0 ) # i n i t i a l v e l o c i ty o f s p ac e cr a ft 14 p=mv #i n i t i a l momentum o f s p a c e c r a f t 15

16 t=017 dt= 0. 13600 #t i m e s t e p18

19 rmag=mag( s p ac ec ra ft . pos ) ; #d i s t a nc e o f s p a c e c r a f t from E ar th 20

21 t r a i l =c u r ve ( c o l o r =s p a c e c r a f t . c o l o r )22 F p er p ar r ow = a r ro w ( p o s= s p a c e c r a f t . p os , a x i s = v e c t o r ( 0 , 0 , 0 ) , c o l o r =c o l o r . g r e e n )23 F p ar a rr o w = a r ro w ( p o s= s p a c e c r a f t . p os , a x i s = v e c t o r ( 0 , 0 , 0 ) , c o l o r =c o l o r . r e d )24 F n et a rr o w = a r ro w ( p o s= s p a c e c r a f t . p os , a x i s = v e c t o r ( 0 , 0 , 0 ) , c o l o r =c o l o r . y e l l o w )25

26 s c a l e = 10RE/100027

28 while rmag>RE: #s t o p i f rmag < RE 29 r a t e ( 1 0 0 )30 r= s p a c e c r a f t . p o s31 rmag=mag( r )32 runit=r/rmag33

34 Fgrav=GmME/rmag 2r u n i t35 Fnet=Fgrav36

37 p = p + F ne tdt38 v = p/m39 s p a c e c r a f t . p os = s p a c e c r a f t . p os + vdt40

41 #c a l c u l a t e t h e p a r a l l e l c omponent o f t he n et f o r c e

42 pmag = mag(p )43 phat = p/pmag44 F p a r a l l e l = d o t ( F ne t , p ) / pmagphat45

46 #c a l c u l a t e t h e p e r p e n di c u l ar component o f t h e ne t f o r c e

47 Fperp = Fnet F p a r a l l e l48

49 #u pd at e a rr ow s f o r th e f o r c es

50 Fperparrow . pos = spa ce cr af t . pos51 Fperparrow . ax i s = Fperp s c a l e52 Fpararrow . pos = spa ce cr af t . pos53 F pa ra rr ow . a x i s = F p a r a l l e l s c a l e54 F n et a rr o w . p o s = s p a c e c r a f t . p o s


33/33

33

Fnetarrow . ax i s = Fnet s c a l e

t r a i l . a pp en d ( p o s= s p a c e c r a f t . p o s )

t = t + dt

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Matter and Interaction Chapter 05 Solutions

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