matter and interaction chapter 02 solutions

7/25/2019 Matter and Interaction Chapter 02 Solutions

1/63

1

2.X.1Let pi

=1.5, 1.2, 0.7 kg m/s and pf

=1.6, 0.9, 1.1 kg m/s. The change in momentum is just pf

-pi.

p = pf p

i

= 1.6, 0.9, 1.1 kg m/s 1.5, 1.2, 0.7 kg m/s=

0.1, 0.3, 0.4

kg m/s

2.X.2The net force on a system is always the vector sum, or superposition, of the individual forces acting on the system.

Fnet,sys

= 40, 70, 0 N + 20, 10, 0 N= 60, 60, 0 N

2.X.3This problem is best solved using simple ratio reasoning. Changing the forces magnitude by a factor of two, for thesame duration, must also change the resulting change in momentum. Halving the force must halve the change in momentum.So

|p

|= 1.5 kg m/s.

2.X.4Let the object (whatever it is) be the system. Were given that Fnet,sys

=0.5, 0.2, 0.8 N and t = 2 min = 120 s(dont forget to express tin seconds!).

(a) We have to calculate the impulse as follows:

p = Fnet,sys

t

= (0.5, 0.2, 0.8 N) (120 s)= 60, 24, 96 kg m/s

Note that this is (must be!) the same value we would have gotten had we known both pf

andpi.

(b) Remember thatimpulseandchange in momentumare merely two different words for the same quantity. This quantitycan be calledimpulseif it is calculated from the net force and its duration. It can also be calledchange in momentumif it is calculated from the initial and final momenta. Names aside, it is the same quantity in both cases!

2.X.5Let the system consist of the hockey puck. Assume there are no significant interactions on the system other than fromice and air (we can treat them as one net interaction), and this interaction was constant while it existed. Were given thatp

i=0, 2, 0 kg m/s.

(a) If the puck comes to a stop, its final momentum must be zero. Thereforepf

=0, 0, 0 kg m/s. The impulse, or changein momentum, is merely p = p

f-p

i=0, 0, 0 kg m/s -0, 2, 0 kg m/s =0, 2, 0 kg m/s.

(b) We knowpand tso we can calculate Fnet,sys

using the momentum principle.

p = Fnet,sys

t

Fnet,sys

= p

t

Fnet,sys

= 0, 2, 0 kg m/s

3 s

Fnet,sys

=

0, 2

3, 0

N


2/63

2

2.X.6 This is a straightforward application of the momentum principle. Let the system be the puck, and assume the onsignificant interaction on the system is with the hockey stick.

p = Fnet,sys

t

pf

pi

= Fnet,sys

t

pf

= pi

+Fnet,sys

t

pf

= 10, 0, 5 kg m/s + 0, 0, 2000 N (0.004 s)p

f 10, 0, 13 kg m/s

Note that the pucks x and y components of momentum didnt change because the net force had no components in thodirections.

2.X.7 Were given that vi

=25, 0, 15 m/s, vf

=10, 0, 18 m/s, and mcar

= 1000 kg. Let the system consist of the caand lets assume non-relativistic speeds.

(a)

psys

= pf p

i

psys

mcar

vf m

carv

i

psys

mcar

v

f m

carv

i

p

sys (1000 kg) (10, 0, 18 m/s 25, 0, 15 m/s)

psys

(1000 kg) (15, 0, 3 m/s)p

sys

1.5 104 , 0, 3 103

kg m/s

(b) Remember that impulse and change in momentumrefer to the same physical quantity, so the answer is the same above.

(c)

Fnet,sys

=p

sys

t

Fnet,sys

=

1.5 104 , 0, 3 103

kg m/s

3 s

Fnet,sys

=5 103 , 0, 1 103

N

2.X.8 This is another straightforward application of the momentum principle. Let the system consist of the particle anassume non-relativistic speed.

psys

= Fnet,sys

t

pf p

i= F

net,syst

pf

= pi

+Fnet,sys

t

pf

= 10, 0, 0 kg m/s + (6, 3, 0 N)(0.1 s)p

f 9.4, 0.3, 0 kg m/s


3/63

3

2.X.9 Apply some straightforward reasoning rather than complicated mathematics. For the first hour of the trip, youlltravel50 km. Neglect the duration of the change from 50 km/hto 100 km/h; either you hit the gas pedal hard (not good!) oryou have a car that accelerates quickly (preferable). During the next two hours of the trip, youll travel 200 km. Youll travel

a total of250 km during three hours. By definition, your average velocity will be vavg,x

250 km/h3 h 83.3 km/h. This isntequal to the arithmetic mean of the initial and final velocities because the force that caused the change in velocity wasntconstant for the three hour duration. This, in turn, means that the magnitude of the velocity could not change linearly.

NOTE: text has 50 km/h.

2.X.10

t = 10 s

vi

= 30, 0, 0 m/sv

f= 40, 0, 0 m/s

Sincev is changing at a constant rate,

vavgx

=vix

vfx

2

= 30 m/s + 40 m/s

2= 35 m/s

vavgx

= x

tx = v

avgxt

= (35 m/s)(10 s)

= 350 m

2.X.11

During each time interval, px

= 10 kg m/s.

(a) False, becausepx

= 0

(b) True, becausepx

= 10 kg m/s is constant, i.e. it is the same during each successive 1-second time interval.

(c) False, becausepx

is constant.

(d) False, becausepx

is constant.

2.X.12


4/63

4

Fspring

= ks|s|

= (11 N

m)(0.025 m)

= 0.275 N

This is the magnitude of the force on the spring by the hand. This is also the magnitude of the force on the hand by tspring.

2.X.13

L0

= 0.17 mFspring

= 250 NL = 0.84 m

(a) Fspring

= ks|s|

ks

=

Fspring

|s|

= 250 N

(0.24 m 0.17 m)= 3570

N

m

(b) Fspring

= ks|s|

= (3570 N

m)(0.17 m 0.15 m)

= 71 N

2.X.14

L0

= 0.15 m

(a)

L =0, 0.11, 0 m


5/63

5

L

0.15m

0.11m

Figure 1: A sketch of the situation

(b)

L

= 0.11 m

(c)

L =LL

=

(d)

s = L L0= 0.11 m 0.15 m= 0.04 m

Fon hand by spring

= ks

sL

= (95 Nm

)(0.04 m)< 0, 1, 0>= 0, 3.8, 0 N

Fon spring by hand

= Fon hand by spring

= 0, 3.8, 0 N= 0, 3.8, 0 N

2.X.15


6/63

6

y

x

5kg

Figure 2: A picture of the situation

Draw a sketch of the situation.

Fgrav on object by earth

=mg= (5 kg)(9.8

N

kg)

= 49 NF

grav on object by earth= 0, 49, 0 N

2.X.16

Fgrav on object by earth = mgm =

Fgrav

g

= 1 N

9.8 Nkg= 0.1 kg

2.X.17

L0

= 0.2 m

ks

= 8 N

mm = 0.06 kg

For the third time step oft= 0.1 s


7/63

7

yi

= 0.159 m

viy

= 0.236 m/s

piy

= 0.0141 kg m/s

L = 0.159 m

s =L L

0

= 0.159 m 0.2 m= 0.041 m

Fspring,y

= ks

s

= (8 Nm

)(0.041 m)

= 0.328 N

Fgrav,y

= mg

= (0.06 kg)(9.8 Nkg

)

= 0.588 NFnet,y

= 0.26 N

pfy = piy + Fnet,y t= (0.0141 kg m/s) + (0.26 N)(0.1 s)= 0.0119 kg m/s

vfy

=pfy

m

= 0.0119 kg m/s

0.06 kg

= 0.198 m/syf

= yi

+ vavg,y

t

Usevavg,y

vfy

for small t.

yf

= yi

+ vfy

t

= (0.159 m) + (0.198 m/s)(0.1 s)= 0.139 m


8/63

8

2.X.18

The initial velocity in step 2 was upward and its final velocity was upward. As a result, its y-displacement was upward.

The effect of the downward force was to decrease the y-velocity of the object. As a result, its final y-velocity was less thits initial y-velocity. That is, it slowed down. But because its average velocity was upward during the time interval, it sthad an upward displacement.

2.X.19

(c) 0.5 cm is reasonable if doing the calculation by hand. Since the spring is stretched 5 cm, a small enough time stshould be chosen so that|r| in one time step is significantly less than 5 cm but reasonably large to not require too mancalculations.

2.X.20

(a). For a computer calculation, extremely smallt can be used. However, if it is too small, the computer will take a velong time to complete all necessary calculations and numerical errors will compound.

For a period of 4333 days, 1 day is sufficiently small. A t of0.01 s is too small, as it will require too may calculations fone orbit.

As a general rule, for periodic motion like an orbit, a thousandth of a period (0.001 of 4333 days in this case) is sufficientsmall.

2.X.21

vi

= 0, 0, 10 m/sm = 163 kg

Fnet

=

0, 0,

35

N

t = 60 s

rf

= ?

pf

= pi

+Fnet

t

= mvi

+Fnet

t

= (163 kg)(0, 0, 10 m/s) + (0, 0, 35 N)(60 s)= 0, 0, 1630 kg m/s + 0, 0, 2100 kg m/s= 0, 0, 3730 kg m/s

vf

=p

f

m

= 0, 0, 3730 kg m/s

163 kg

= 0, 0, 22.9 m/s


9/63

9

IfFnet

is constant,

vavg

=v

i+ v

f

2

= 0, 0, 10 m/s + 0, 0, 22.9 m/s

2=

0, 0,

16.5

m/s

rf

= ri

+ vavg

t

= +(0, 0, 16.5 m/s)(60 s)= 0, 0, 90 m

2.X.22

(yf

yi

)is negative because the +y direction is defined to be upward, away from the Earth. Because the ball falls, its final

position is less (or, rather, below) its initial position. Thus, its y-displacement (yf

yi

) is negative.

Fnet y

is negative because the +y direction is defined to be upward and the gravitational force of the Earth on the ball isdownward, in the -y direction. Written as a vector,

Fgrav

=

g is the magnitude of Earths gravitational field near the surface of Earth; therefore, it is always written as g = +9.8 Nkg .Magnitudes are always positive.

2.X.23

(a)

m = 0.02 kgyi

= 4 m

yf

= 0

viy

= 0

t = ?

vfy

= ?


10/63

10

yf

yi

= viy

t+1

2

Fnet,y

m t2

yf

yi

= 0 +1

2

mgm

t2

yf

yi

= 0 +1

2(g) t2

4 m =

1

2(

9.8 N

kg)t2

t =

2(4 m)

9.8 Nkg

= 0.90 s

(b)

pfy

= piy

+ Fnet,y

t

mvfy

= mviy

+ Fnet,y

t

vfy

= viy

+Fnet,y

m t

vfy

= 0 +

mgm

t

vfy

= (9.8 Nkg

)(0.9 s)

= 8.8 m/s

2.X.24

(a)

ri

= 9, 0, 5 mv

i= 10, 13, 5 m/s

Fnet

= Fgrav

=t = 0.6 s

pf

= pi

+Fnet

t

mvf

= mvi

+Fnet

t

vf

= vi

+F

net

m t

vf

= 10, 13, 5 m/s + < 0, mg, 0>m

(0.6 s)

= 10, 13, 5 m/s + (0, 9.8, 0 Nkg

)(0.6 s)

= 10, 13, 5 m/s + 0, 5.88, 0 m/s= 10, 7.12, 5 m/s


11/63

11

(b) For constantFnet

,

vavg

=v

i+ v

f

2

= 10, 13, 5 /mps + 10, 7.12, 5 m/s

2

= 10, 10.06, 5 m/s

rf

= ri

+ vavg

t

= 9, 0, 5 m + (10, 10.06, 5 m/s)(0.6 s)= 9, 0, 5 m + 6, 6.036, 3 m= 15, 6, 8 m

(c) At maximum height,vfy

= 0, sovf

=10, 0, 5 m/s at max height.

vfy

= viy

+Fnet,y

m t

0 = 13 m/s + (9.8 Nkg

)(t)

t = 13 m/s9.8 Nkg

= 1.33 s

(d) Max height occurs at t = 1.33 s; find yf

.

rf

= ri

+ vavg

t

yf

= yi

+ vavg,y

t

yf

= 0 +

viy

+ vfy

2

t

yf

=

13 m/s + 0

2

(1.33 s)

= 8.65 m


12/63

12

(e) At groundyf

= 0; the ball is initially kicked from the ground atyi

= 0. Use

yf

yi

= viy

t+1

2

Fnet,y

m t2

0 0 = viy

t+1

2

mgm

t2

0 = viy

t

1

2

gt2

1

2gt2 = v

iyt

t =2v

iy

g

= 2(13 m/s)

9.8 Nkg= 2.65 s

Note: this is twice the time at the peak.

(f) Find vfy

when it hits the ground.

vfy

= viy

+

Fnet,y

m

t

= viy

+mgm

t

= viy

gt

= 13 m/s (9.8 Nkg

)(2.65 s)

= 12.97 m/s

Note: it is symmetric. The ball left the ground with a y-velocity of13 m/s, so it returns to the ground with a y-veloci

of13 m/s. Thus vavg,y = 0.r

f= r

i+ v

avgt

= ri

+v

i+ v

f

2 t

= 9, 0, 5 m + 10, 13, 5 m/s + 10, 13, 5 m/s

2

(2.65 s)

= 9, 0, 5 m + (10, 0, 5 m/s)(2.65 s)= 9, 0, 5 m + 26.5, 0, 13.25 m= 35.5, 0, 18.25 m

Note: the y-displacement is 0 since the ball left the ground (y = 0) and returned to the ground (y = 0). Thus, wsimply need the x-displacement and z-displacement. The x-velocity and z-velocity are constant sinceF

net,x= 0 an

Fnet,z

= 0. Thus,

xf

= xi

+ vx

t

= 9 m + (10 m/s)(2.65 s)

= 35.5 m


13/63

13

and

zf

= zi

+ vz

t

= 5 m + (5 m/s)(2.65 s)= 18.25 m

Which is consistent with what was calculated above.

2.X.25

(a) The system was the student on the left.

(b) Fnet,y

= 0 and Fnet,z

= 0.

(c)

Fnet,x

=p

x

t

or

pfx

= pix

+ Fnet,x

t

(d) The initial speed (6 m/s) and displacement of the student (5 cm) were estimated quantities used to calculate t.

(e) The net force on the student was calculated using the Momentum Principle. The force exerted by the right student onthe left student was calculated from net force and superposition. The free-body diagram and the fact that F

net,y= 0

andFnet,z

= 0 were used to show thatFnet,x

was equal to the force on the student on the left due to the other student.

(f) F = 21000 N and Fgrav =mg = (60 kg)(9.8 N

kg )600 N, thus F

Fgrav = 35. The force due to the collision was about 35times the students weight.

2.X.26

The object is moving in the +x direction, but we are not given any information about the change in velocity. Withoutknowing the change in velocity, we cannot know the change in momentum, nor the net force. All of these choices are possible.None of them are false.

2.X.27

F1

= 18, 47, 23 NF

2= 20, 13, 41 N

Fnet

= F1

+F2

= 18 20, 47 13, 23 + 41 N= 2, 34, 18 N


14/63

14

2.X.28

p is in the direction of the net force on the puck. Thus, p is in the direction of arrow (e).

2.X.29

The system is the truck.

pi

= 65000, 0, 0 kg m/sp

f= 26000, 0, 0 kg m/s

t = 4.1 s

Fnet

= p

t

=p

f p

i

t

= 26000, 0, 0 kg m/s 65000, 0, 0 kg m/s

4.1 s

= 39000, 0, 0 kg m/s

4.1 s= 9510, 0, 0 N

2.X.30

Same answer as 2.X.26 but with y instead of x.

2.X.31

Cart B.

pA

= 0 so Fnet on A

= 0.

pB

is to the left, so Fnet on B

is to the left.

pC

is to the right, so Fnet on C

is to the right.

2.X.32

(b),(c),(d),(e),(g) are correct.

In (a), F must be the net force Fnet

.

In (f), v must be a vector v.

In (h), the equation is not correct, nor is it the correct definition. See part (e) for the correct definition of momentum.

In (i), vavg

must be a vectorvavg

.

In (j), Fnet

must be a vector Fnet

.

2.X.33


15/63

15

pi

= 8, 0, 0 kg m/st = 0.13 s

pf

= ?

Fnet

= 7, 5, 0 N

pf

= pi

+Fnet

t

= 8, 0, 0 kg m/s + (7, 5, 0 N)(0.13 s)= 7.09, 0.65, 0 kg m/sp

f

= (7.09 kg m/s)2 + (0.65 kg m/s)2 + (0 kg m/s)2= 7.12 kg m/s

2.X.34

ti

= 16 s

tf

= 16.2 s

m = 4 kg

vi

= 9, 29, 10 m/sv

f= 18, 20, 25 m/s

Fnet

= ?

Fnet

= p

t

= mvt

=m(v

f v

i)

t

= (4 kg)(18, 20, 25 m/s 9, 29, 10 m/s)

(16.2 s 16.0 s)=

(4 kg)(9, 9, 35 m/s)0.2 s

= 180, 180, 700 N

2.X.35Assume motion in the +x direction.

mproton

= 1.7 1027 kgv

i= 0.990c, 0, 0 m/s

vf

= 0.994c, 0, 0 m/s


16/63

16

pi

= mvi

=mv

i1 |vi |

2

c2

= m < 0.99c, 0, 0>

1 (.99c)2

c2

= m < 0.99c, 0, 0>

1 0.992

= (1.7 1027 kg)< (0.99)(3 108 m/s), 0, 0>

1 0.992=

pf

= (1.7 1027 kg)< (0.994)(3 108 m/s), 0, 0>

1 (0.994)2=

4.63 1018, 0, 0

kg m/s

impulse = Fnet

t

= p

= pf p

i

=

4.63 1018, 0, 0 kg m/s 3.58 1018, 0, 0 kg m/s= 1.05 1018 kg m/s

2.X.36

Newtons first law and second law refer to the net force on the system. If you pull the sled in the +x direction at constavelocity, the forces in the x-direction on the sled are the force on the sled by you in the +x direction and the force on thsled by snow (friction) in the -x direction. When added together, these forces sum to zero and the net force on the sledzero, thus satisfying Newtons first and second laws.

Fonsled byfriction Fonsledbyyou

Figure 3: The force vectors cancel


17/63

17

2.P.37 This problem can be confusing. With respect to the shuttle (i.e. in the shuttles frame of reference), the astronaut hasno momentum. With respect to Earth (i.e. in Earths frame of reference), however, she does, and this momentum changesas she and the shuttle orbit Earth.

(a) See Figure4.

(b) See Figure4. Note that the only force on the astronaut is the gravitational force on her due to Earth.

(c) See Figure4.

(d) See Figure4. Note that only the momentums directionhas changed.

(e) Objects in space dont float. The astronautappearsto do so because both she and the shuttle are in a state of free fallaround Earth, subject to the same gravitational field strength. As they attempt to literally fly off on a tangent, they

fall the same distance toward Earths center. Thus, the astronaut never hits the shuttles floor.

p1

p2

F1

F2

p

p2

p1

Figure 4: Astronaut as seen in Earths frame of reference

2.P.38 This is a straightforward application of the momentum principle. Choose the system to consist of the proton; theonly significant interaction with the system is due to the HCl molecule. Further, assume non-relativistic speed so we can useNewtons expression for momentum. Since the net force only has an x component, we should not expect the protons y orzcomponents of momentum, and thus velocity, to change.


18/63

18

psys

= Fnet,sys

t

psys,f

psys,i

= Fnet,sys

t

psys,f

= psys,i

+Fnet,sys

t

vsys,f

msys vsys,i +Fnet,sys tm

sys

vsys,f

(1.673 1027 kg)(3600, 600, 0 m/s) +

1.12 1011 , 0, 0

N (3.4 1014 s)

(1.673 1027 kg)

vsys,f

5.64 1024 , 1.00 1024 , 0

kg m/s

1.673 1027 kgv

sys,f 3370, 600, 0 m/s

Note that the protons position didnt enter into this calculation.

2.P.39 Choose the system to be just the spacecraft and its contents. This choice of system obviously doesnt include thrustexhaust gases. The only significant interaction with the system comes from the thruster exhaust. A popular misconceptionthat air is needed for rockets to work; this is not true. Lets assume non-relativistic speeds so we can use Newtons expressiofor momentum. Lets also treat the spacecraft and its contents as a particle to which we apply the momentum principWe need to apply the momentum principle to see how the systems momentum changes during the thruster burn. Lets alsneglect the loss of mass while firing the engine (due to exhaust).

Notice that the net force is in the+xdirection. As a result, vy does not change. It will remain constant. Onlyvxwill chang

psys

= Fnet,sys

t

psys,f

psys,i

= Fnet,sys

t

psys,f

= psys,i

+Fnet,sys

t

vsys,f

=M

sysv

sys,i+F

net,syst

Msys

vsys,f

=(1.5 105 kg)

0, 20 103 , 0

m/s

+

6 104 , 0, 0

N (3.4 s)

(1.5 105 kg)

vsys,f

=

2.04 105

, 3.00 109

, 0

kg m/s(1.5 105 kg)

vsys,f

=

1.36, 2 104 , 0

m/s

Now we know the systems (spacecrafts) velocity at the end of the burn. Find the position of the spacecraft after the burnSince the net force on the spaceship is constant, use the arithmetic average of the velocity.


19/63

19

rf

= ri

+ vavg

t

= ri

+

v

i+ v

f

2

t

= m +0, 2

104

, 0 m/s + 1.36, 2 10

4

, 0 m/s2

(3.4 s)

= m

Note that the x-displacement x is very small. The spacecraft barely moved in the x-direction. This is because t and thex-acceleration (F

net,x/m) are very small.

After the burn, the spacecrafts momentum (and thus velocity) is constant (since the net force on the spaceship is zero), andits position after one hour of coasting can be calculated.

rf

= ri

+ vt

rf

m +

1.36, 2 104 , 0

m/s

(3600 s)

rf

1.69 104 , 7.21 107 , 0

m

Notice that the final velocity after the burn has a very small x-component (in comparison to its y-component). As a result,during one hour, it has a much greater y-displacement than x-displacement.

2.P.40The momentum principle is applicable to this problem provided we use Einsteins expression for a particles momentumrather than Newtons expression. If we know the net force on a particle and the particles change in momentum, then themomentum principle gives us the duration of the interaction. Take the system to be the electron. The only significantinteraction with the system is that of the particle accelerator. Assume the particles initial and final momenta are in thesame direction.

psys,i

= 11 |vi |

2

c2

melectron

vi

11 (0.93c)

2

c2

(9.109 1031 kg) 0.93c 11 0.932 (9.109 10

31

kg) 0.93c

6.91 1022 kg m/spsys,f

= 11 |vf|

2

c2

melectron

vf

11 (0.99c)

2

c2

(9.109 1031

kg) 0.99c 11 0.992 (9.109 10

31

kg) 0.99c

1.92 1021 kg m/s

If the initial and final momenta are in the same direction, then the difference in their magnitudes is equal to the magnitudeof their difference.


20/63

20

(a) Now we can apply the momentum principle.

psys

= Fnet,sys

t

t =

psys

F

net,sys

t 1.92 1021

kg

m/s 6.91 1022

kg

m/s2 1012 N

t 6.15 1010 s

(b) Since the net force on the system is assumed constant, we can approximate the particles average velocity as tarithmetic mean of the initial and final velocities.v

avg

0.96c|r|

vavg

t|r| 0.96

3 108 m

s

(6.15 1010 s)

|r| 1.77 101

m18 cm

2.P.41 Choose the system to be just the ball.

(a) Do the following calculations for the early time interval:

vavg,early

rt

pavg,early

mvavg,early

m rt

pavg,early

(2.7

103

kg)0.08, 0.04, 0.02 m

0.02 sp

avg,early 0.011, 0.005, 0.003 kg m/s

(b) Now do the same calculations for the late time interval:

vavg,late

rt

pavg,late

mvavg,late

m rt

pavg,late

(2.7 103 kg) 0.02, 0.36, 0.02 m0.02 s

pavg,late

0.003, 0.049, 0.003 kg m/s

(c) Now use these two results to get the average net force.

Fnet,sys

=p

sys

t =

pavg,late

pavg,early

t

Fnet,sys

0.008, 0.044, 0 kg m/s2 s

Fnet,sys

0.004, 0.022, 0 N


21/63

21

2.P.42

m = 240 kg

ti

= 20.7 s

ri

= 4.3 103, 8.7

102, 0 m

pi

=

4.4 104, 7.6 103, 0 kg m/sF

net=

7 103, 9.2 102, 0 NDefine the system to be the space probe.

(a) At t = 20.9 s,

pf

= pi

+Fnet

t

=

4.4 104, 7.6 103, 0 kg m/s + (7 103, 9.3 102, 0 N)(0.2 s)= 4.4

104,

7.6

103, 0 kg m/s +

1.4

103,

1.86

102, 0 kg m/s

=

4.26 104, 7.79 103, 0 kg m/s(b)

vf

=p

f

m

=

4.26 104, 7.79 103, 0 kg m/s

240 kg

= 177.5, 32.4, 0 m/s

vi

= pim

=

4.4 104, 7.6 103, 0 kg m/s

240 kg

= 183, 31.7, 0 m/s

vavg

=v

f v

i

2

= 183, 31.7, 0 m/s 177.5, 32.4, 0 m/s

2= 180, 32, 0 m/s

rf

= ri

+ vavg

t

=

4.3 103, 8.7 102, 0 m + (180, 32, 0 m/s)(0.2 s)=

4.3 103, 8.7 102, 0 m + 36, 6.4, 0 m

=

4.34 103, 8.64 102, 0 m


22/63

22

2.P.43

m = 5 kgF

net= 29, 15, 40 N

t = 4 sv

f= 114, 74, 112 m/s

vi

= 0

Define the system to be the rock.

Fnet

= p

t

= mv

t

Fnet t = mvF

nett = mv

f mv

i

mvi

= mvf F

nett

vi

= vf

Fnet

m t

= 114, 94, 112 m/s 29, 15, 40 N

5 kg

(4 s)

= 114, 94, 112 m/s 23.2, 12, 32 m/s= 91, 106, 80 m/s

2.X.44

(1) The system is the box. Forces on the system are the gravitational force on the box by Earth and the (contact) foron the box by your hands.

(2) The system is you and the box. Forces on the system are the gravitational force on you and the box by the Earth anthe (contact) force by the floor on you, assuming you are standing on the floor.

(3) The system is you, the box, and Earth. During the short time interval of the lift, there are no significant forces on tsystem.

2.X.45

(a) vx

is negative and increases toward zero, meaning that it becomes less negative as the cart slows down. There is ngraph that demonstrates this motion. Graph (2) is close, but in graph (2), the object reaches zero velocity and thspeeds up as it travels to the right.

(b) vx

is positive and increases. This is graph (6).


23/63

23

(c) vx

is negative and constant. This is graph (3).

(d) vx

is negative, decreases in magnitude until vx

= 0, then increases and is positive. This is graph (2).

(e) vx

is zero and constant. No graphs depict this motion.

(f) vx

is positive and decreases. No graphs depict this motion. Graph (5) is close, but it shows the objects velocitydecreasing to 0, and then the object speeds up in the -x direction.

(g) vx

is positive, decreases to 0, and then becomes more negative (i.e. increases in the -x direction). This is graph (5).

(h) vx

is negative and becomes more negative as the cart speeds up. This is graph (1).

(i) vx

is constant and positive. This is graph (4).

2.X.46

ks

= 40 N

mFon spring

= 2 NL = 18 cm

L0

= ?

Fon spring

= ks|s|

= ks

L L0

L L

0 =

F

on spring

ks

= 2 N40 Nm

= 0.05 m

= 5 cm

L = L0

+ 5 cm

= 18 cm + 5 cm

= 23 cm

2.X.47

Fgrav

= mg= (10 kg)(9.8

N

kg)

= 98 N


24/63

24

On a 20 kg mass, Fgrav

= (20 kg)(9.8 Nkg

)

= 196 N

2.X.48

When doing an iterative calculation, it is desirable to have a small t so that|r| is small. A small enough|r| is (1 109 m, which is 0.1 of the distance from the Sun. (a)1 102 m is too small as it would take too many calculations see the comet move a distance of9 1016 m. On the other hand, (b) and (c) are too large since one would not be able ascertain anything about the orbit of the comet.

2.X.49

(a) 1 second would be 12 the period of the oscillation; thus, no information about the motion during an oscillation could bobtained. All other choices oft are reasonable, with smaller t like 0.01 s being best for a computer calculation.

2.X.50

(a) If we assume a constant net force, then

vavg

=v

i+ v

f

2

Assume motion is in the +x direction.

vf

= 8, 0, 0 m/sv

i= 0, 0, 0 m/s

vavg

= 0, 0, 0 m/s + 8, 0, 0 m/s

2= 4, 0, 0 m/s

(b)

vavg

= r

tr = v

avgt

= (4, 0, 0 m/s)(3 s)= 12, 0, 0 m

|r| = 12 m

(c) Assume direction of motion is the +x direction.

vi

= 18, 0, 0 m/sv

f= 0, 0, 0 m/s

t = 4 s


25/63

25

Assume a constant net force, then

vavg

=v

i+ v

f

2

= 18, 0, 0 m/s + 0, 0, 0 m/s

2= 9, 0, 0 m/s

(d)

r = vavg

t

= (9, 0, 0 m/s)(4 s)= 36, 0, 0 m

|r| = 36 m

2.X.51

(a) t= 3 h :

x = vavg,x

t

= (30mi

h)(3 h)

= 90 mi

t= 1 h :

x = (60mi

h)(1 h)

= 60 mi

xtotal

= 90 mi + 60 mi

= 150 mi

ttotal

= 4 h

vavg,x

=x

total

ttotal

= 150 mi

4 h

= 37.5mi

h

(b) The arithmetic mean velocity is

vavg,x

=vix

+ vfx

2

= 30 mih + 60

mih

2

= 45mi

h


26/63

26

(c) The arithmetic mean is only accurate in the case of constant net force. However the velocity of the car in this caseconstant for 3 h, briefly increases, and is then constant for 1 h. The net force, therefore, is zero for 3 h, briefly non-zerand then zero for 1 h. It is definitely not constant throughout the motion.

2.X.52

ri

= 0.06, 1.03, 0 mt = 0.7 s

vavg

= 17, 4, 6 m/syf

= ?

rf

= ri

+ vavg

t

= 0.06, 1.03, 0 m + (17, 4, 6 m/s)(0.7 s)= 0.06, 1.03, 0 m + 11.9, 2.8, 4.2 m= 12, 3.83, 4.2 m

Thus, yf

= 3.83 m is the height of the ball 0.7 s after it leaves your hand.

2.X.53

(a) px

starts at zero, briefly increases in the +x direction, and then steadily decreases to zero as the cart slows to a stoThe graph that depicts this motion is graph (2).

(b) px

starts at zero and steadily increases until it quickly decreases when it is caught. This is shown in graph (4).

(c) While pushed,px briefly increases. After being released,px decreases to zero (as the cart slows to a stop) and theincreases in the -x direction (i.e. becomes more negative) as it speeds up while traveling in the -x direction. Thisshown in graph (1).

2.X.54

(a) After releasing the cart, the force in the cart (friction) is constant and in the -x direction. This is shown in graph (6The frictional force on the cart becomes zero when the cart stops.

(b) The force by the air on the cart (resulting from the turning fan) is constant in the +x direction. A large force by yohand on the cart is in the -x direction when stopping the cart. This is shown in graph (7).

(c) After the initial push in the +x direction, the force by air on the cart (due to the turning fan) is in the -x directioThis is shown in graph (5).

2.X.55

(a) The cart starts at the origin (x = 0) and slows down due to friction until it stops. Then its x-position remains constaat some positive value of x. This is depicted in graph (9).


27/63

27

(b) The slope of x vs. t increases in the +x direction until the cart is caught and then x is constant. This is shown in graph(12).

(c) The slope of x vs. t is initially positive, then decreases, then goes to zero (whenvx

= 0), then becomes negative andincreases (becomes steeper) until the cart is caught. Presumably the cart returns to where it started. This is depictedin graph (10).

2.X.56

(a) The net force is the gravitational force on the ball by Earth, and is directed downward toward Earth.

(b) Only the y-component of the momentum of the ball changes.

(c) The x-component of the momentum of the ball is constant becauseFnet,x

= 0.

(d) The y-component of the momentum of the ball changes at a constant rate becauseFnet,y

=mg and is constant.(e) The z-component of the momentum of the ball is constant becauseF

net,z= 0.

(f) vavg,y

=

viy

+vfy

2 is accurate because F

net,yis constant.

2.P.57

vi

= 20 m/s

i= 50

(a) Sketch a picture showing the trajectory of the ball, like the one shown in Figure5.

vi

x

x

y

Figure 5: A sketch of the trajectory of the ball

Use direction cosines to writevi

as a vector.

vix

=v

i

cos x

= (20 m/s)cos59

= 10.3 m/s


28/63

28

viy

=v

i

cos y

= (20 m/s)cos(90 59)= (20 m/s)cos31

= 17.1 m/s

Define i to be when the ball is kicked and f when it hits the ground.

yi

= 0

yf

= 0

Use

(yf

yi

) = viy

t+1

2

Fnet,y

m t2

0 0 = viy

t+1

2

mgm

t2

viyt =

1

2

gt2

t =2v

iy

g

= 2(17.1 m/s)

9.8 Nkg= 3.9 s

(b) Fnet,x

= 0 therefore vx

is constant.

vx

= x

t

x = vx t= (10.3 m/s)(3.49 s)

x = 35.9 m

(c) In this case, the motion is symmetric. Thus, the time when the ball is at its peak is 12 the time it takes to leave tground and return to the ground. Thus, t

peak= 12 (3.49 s) = 1.75 s. The y-velocity of the ball at its peak is zero.

rf

= ri

+ vavg

t

yf

= yi

+ vavg,y

t

yf

= yi

+vfy

+ viy

2t

= 0 +

0 + 17.1 m/s

2

(1.75 s)

= 15 m

2.P.58


29/63

29

ri

= 9, 0, 6 mv

i= 11, 16, 6 m/s

(a) vf aftert= 0.5 s? According to the momentum principle, with the system defined to be the ball.

pf

= pi

+Fnet

t

mvf

= mvi

+Fnet

t

vf

= vi

+F

net

m t

vf

= vi

+

m

t

vf

= vi+< 0, g, 0> t

= 11, 16, 6 m/s+< 0, 9.8 Nkg

, 0> (0.5 s)

= 11, 16, 6 m/s + 0, 4.9, 0 m/s= 11, 11.1, 6 m/s

(b) vavg

= v

i+v

f

2 gives the most accurate result for the average velocity becauseF

netis constant. The arithmetic mean is

exact in this case.

(c)

vavg

=v

i+ v

f

2=

11, 16, 6 m/s + 11, 11.1, 6 m/s2

= 11, 13.6, 6 m/s

Note thatvx

andvz

are constant since Fnet,x

= 0 and Fnet,z

= 0.

(d)

rf

= ri

+ vavg

t

=

9, 0,

6

m + (

11, 13.6,

6

m/s)(0.5 s)

= 9, 0, 6 m + 5.5, 6.8, 3 m= 3.5, 6.8, 9 m

(e) vy

= 0 at its highest point because this is a "turning point" in the y-direction. When an object changes direction, vy

must be zero at the instant it changes directions. In terms of calculus, y is a maximum at the balls highest point.vy

= dydt must be zero since the derivative of a function at its maximum is zero.


30/63

30

(f)

pfy

= piy

+ Fnet,y

t

mvfy

= mviy

+ Fnet,y

t

vfy

= viy

+

mgm

t

0 = 16 m/s + (9.8 Nkg

)t

(g)

t = 16 m/s

9.8 Nkg= 1.63 s

(h)

vavg,y

=viy

+ vfy

2

= 16 m/s + 0

2= 8 m/s

yf

= yi

+ vavg,y

t

= 0 + (8 m/s)(1.63 s)

= 13 m

2.P.59

m = 1.5 kg

vi

= 5, 8, 0 m/sr

i=

(a) Sketch a picture of the motion as shown in Figure6.

At the peak, vfy

= 0. Thus

vpeak

= 5, 0, 0 m/s


31/63

31

vi

x

y

vf

Figure 6: A sketch of the motion

(b)

pf

= pi

+Fnet

t

mvf = mvi +Fnet t

vf

= vi

+F

net

m t

vfy

= viy

+Fnet,y

m t

vfy

= viy

+

mgm

t

vfy

= viy

gt

0 = 8 m/s (9.8 Nkg

)t

t = 8 m/s

9.8 Nkg= 0.816 s

(c) SinceFnet

is constant,

vavg

=v

i+ v

f

2

= 5, 8, 0 m/s + 5, 0, 0 m/s

2= 5, 4, 0 m/s

(d)

rf

= ri

+ vavg

t

= +(5, 4, 0 m/s)(0.816 s)= 4.1, 3.3, 0 m


32/63

32

(e) The motion is symmetric in this case. The total time will be 2(tpeak

) = 2(0.816 s) = 1.63 s. Alternatively, use the fa

thatvfy

=viy

=8 m/s

vfy

= viy

gt

8 m/s = 8 m/s (9.8 Nkg

)t

t = 8 m/s + 8 m/s9.8 Nkg= 1.63 s

Which agrees with what was obtained by calculating 2 times the time interval to reach the peak.

(f)

rf

= ri

+ vavg

t

Note: vavg,y

= 0, so one only needs to consider the x direction.

xf

= xi

+ vavg,x

t

= 0 + (5 m/s)(1.63 s)

= 8.2 m

(g) Because of symmetry,

vfy

= viy

= (8 m/s)= 8 m/s

(h) Sketch a picture of the vector, as shown in Figure7

vi

vix

viy

x

Figure 7: A sketch of the vector and components


33/63

33

Use direction cosines

cos x

=vixv

i

v

i

= (5 m/s)2 + (8 m/s)2 + (0 m/s)2= 9.43 m/s

cos x =

5 m/s

9.43 m/s

x

= 58

(i) Sketch a picture ofvf

when the ball hits the ground.

vf

vfx

vfy

x


vfx

= 5 m/s

vfy

=

8 m/s

Due to symmetry,x

is the same as what was calculated for vi

when the ball was released, except that now it is belowthe x-axis. Thus,

x

=58

2.P.60

Case (1):

Assume the ball is released from a height of 6 m from the floor at an angle of45. Find the initial speed needed to make the

free-throw.Sketch a picture with a coordinate system, like the one shown in Figure 9.

vix

=v

i

cos x

=v

i

cos45=

vi

2


34/63

34

yi

=2myf=3m

x i=0x

f=4.3m

y

x


vi

vix

viy

x

Figure 10: Initial velocity vector and components

viy

=vi

2

= vix

Since = 45. Write

rf

= ri

+ vavg

t

In component form.

xf

= xi

+ vavg,x

t

and

yf

= yi

+ vavg,y

t


35/63

35

The x-velocity is constant since Fnet,x

= 0, so

xf

= xi

+ vix

t

4.3 m = 0 +

vi

2

t

t = (4.3 m)

2

vi

This equation has 2 unknowns. Thus, analyze the y-direction.

yf

= yi

+ vavg,y

t

= yi

+

vfy

+ viy

2

Substitute

vfy

= viy

+Fnet,y

m t

= viy

+

mgm

t

= viy

gt

yf

= yi

+

(v

iy gt) + v

iy

2

t

yf

= yi

+ viy

t 12

gt2

3 m = 2 m +vi 2t 1

2

(9.8 N

kg

)t2

1 =

vi

2t 4.9t2

This equation also has 2 unknowns. However, we now have 2 equations with 2 unknowns. They are:

t=(4.3)

2v

i

and

1 =v

i

2t 4.9t2

Use substitution to solve, or use the solve() function on a TI calculator. The result is

t = 0.821 svi

= 7.41 m/s


36/63

36

Check that the result is correct by using the initial speed and time interval to calculate the final position of the basketbaand see if it matches the hoop.

vix

=v

i

cos 45= (7.41 m/s)cos45

= 5.24 m/s

Since = 45,

viy

= 5.24 m/s

vi

= 5.24, 5.24, 0 m/s

vf

= vi

+

F

net

m

t

= vi

+< 0, mg, 0>

m t

= vi+< 0, g, 0> t

= 5.24, 5.24, 0 m/s + (0, 9.8, 0 Nkg

)(0.821 s)

= 5.24, 5.24, 0 m/s + 0, 8.05, 0 m/s= 5.24, 2.81, 0 m/s

Note vfy

is negative as it should be for a ball traveling downward into the basket.

vavg

=v

i+ v

f

2

= 5.24, 5.24, 0 m/s + 5.24, 2.81, 0 m/s

2= 5.24, 1.22, 0 m/s

rf = ri + vavg t= 0, 2, 0 m + (5.24, 1.22, 0 m/s)(0.821 s)= 0, 2, 0 m + 4.3, 1.0, 0 m= 4.3, 3, 0 m

Score! Thats the position of the basket.

Case (2):

Choose a different angle, say = 60. Then,

vix

=v

i

cos60

=vi

2

viy

=v

i

cos (90 60)=

vi

cos30=

3

2

vi


37/63

37

Follow the same procedure. You will get

t= 8.6v

i

and

1 =

3

2vi

t 4.9t2

Solve for tandv

i

and they aret = 1.15 sv

i

= 7.50 m/sCheck to see if they give the correct final position of the ball.

vix

= (7.50 m/s)cos60

= 3.75 m/s

viy

= (7.50 m/s)cos30

= 6.5 m/s

vi

= 3.75, 6.5, 0 m/sv

f= v

i+< 0, g, 0> t

=

3.75, 6.5, 0

m/s + (

0,

9.8, 0

N

kg

)(1.15 s)

= 3.75, 4.77, 0 m/s

vavg

=v

i+ v

f

2

= 3.75, 6.5, 0 m/s + 3.75, 4.77, 0 m/s

2= 3.75, 0.865, 0 m/s

rf = ri + vavg t= 0, 2, 0 m + (3.75, 0.865, 0 m/s)(1.15 s)= 4.3, 3.0, 0 m/s

Score! Note that it was a high-arching shot, thus viy

was greater and t was greater than for the 45 shot. But both shotsmake it into the hoop.


38/63

38

2.P.61

Typically such throws are at a small angle, less than 45. Probably30 is a good estimate. If the ball is caught at the samheight that it is thrown, then y

f= 0 and y

i= 0.

To find the time interval that it is in the air, use

yf

yi

= viy

t+1

2

Fnet,y

m t2

0 0 = viy

t+12

mgm

t2

0 = viy

t 12

gt2

viyt =

1

2gt2

t =2v

iy

g

Use

viy

=

v

i cos

y

=vi

cos90 30=

vi

cos60

t=2v

i

cos60g

There are two unknowns, so consider the x-direction. vx

is constant because Fnet,x

= 0. Thus,

xf

= xi

+ vavg,x

t

xf

= xi

+v

i

cos 30t

30 = 0 + vi cos30tvi

= 30cos 30t

Substitutet= 2|v

i| cos60g . v

i

= 30cos30

2|v

i| cos 60g

v

i

2 = 30(g)2cos30 cos60

vi 2

= 30(9.8)

232 1

2v

i

2 = 30(9.8)3

2vi

=

30(9.8)

32

= 18.4 m/s


39/63

39

t =2v

i

cos60g

= 2(18.4 m/s) 12

9.8 Nkg= 1.88 s

Double check that our calculations are correct using

xf

= xi

+ vx

t

= 0 + |v| cos30t= (18.4 m/s)cos30(1.88 s)

= 30 m

exactly as expected.

Because the motion is symmetric (yi

= 0 and yf

= 0), the time interval to the peak is 12 the total time which is 1

2 (1.88 s) =

0.94 s. Also, vfy

= 0 at the peak.

The height after t= 0.94 s is

yf

= yi

+ vavg,y

t

= yi

+

viy

+ vfy

2

t

yf

= 0 +

vi

cos60+ 02

(0.94 s)

= (18.4 m/s)cos60

2 (0.94 s)

= 4.3 m

This is the peak height above where the ball was released.

2.P.62

First, sketch a picture of the situation.

6 frames of video corresponds to 5 time intervals, with each time interval being 130 s = 0.033 s. The first frame is at t = 0.

The average velocity of the flower pot is

vavg,y

= y

t

=yf

yi

t

Wheret= 5( 130 s) = 530 s =

16 s

vavg,y

= 1.87 m

16 s

= 11.2 s


40/63

40

Windowy=0

yf=0.852.2m=1.87m


This is the only information that can be ascertained from the given information.

Suppose that the first frame occurs when the flower pot is released from rest. Then,

viy

= 0

vfy

= viy

+

Fnet,y

m

t

= viy

+mgm

t

= viy gt= 0 (9.8 N

kg)(

1

6s)

= 1.63 m/s

Then,

vavg,y

=viy

+ vfy

2

=

0

1.63 m/s

2= 0.815 m/s

Which is much smaller than the measured value of11.2 m/s.

There are two possible conclusions: (1) the flower pot fell from a much greater height than the fourth floor or (2) The flowpot was thrown downward with an initial y-velocity much greater than 0.


41/63

41

We can calculate the initial y-velocity at the top frame of the video.

vfy

= viy

gt

= viy

(9.8 Nkg

)(1

6s)

= viy

1.63 m/s

vfy

viy

= 1.63 m/sv

y = 1.63 m/s

Sincevavg,y

=11.2 m/s then,

vfy

= 11.2 m/s 1.632

m/s

=

12 m/s

viy

= 11.2 m/s +1.632

m/s

= 10.4 m/s

Thus, the flower pot either fell from a greater height or was thrown downward with an initial y-velocity no greater than10.4 m/sdownward.

2.P.63

There are two regions to consider.

(1) in between the plates where Fnet

= and is constant.

(2) after leaving the plates where Fnet

= 0 and v is constant.

Calculate y in region (1) and y in region (2). Add them together for the total y. Define yi

= 0 to be the verticalposition of the electron before going through the plates.

In region (1),

pfy

= piy

+ Fnet,y t

vfy

= viy

+Fnet,y

m t

= viy

+F

mt

= 0 +F

mt


42/63

42

vavg,y

=viy

+ vfy

2yf

yi

= vavg,y

t

y =

viy

+ vfy

2

t

= 0 + Fmt

2

t

= 1

2

F

mt2

To know t, use the x-motion. Since Fnet,x

= 0, vx

is constant and is v0

.

x = vx

t

d = v0

t

t = d

v0

Substitute to get y for region (1).

y = 1

2

F

mt2

= 1

2

F

m

d

v0

2

Now for region (2),

The velocity is constant and is

v =

wheret= dv0

from region (1). Thus,

v =

y = vavg,y

t

= F

m

d

v0

t

Wheret is the time interval in region (2).To get t in region (2), use the x-motion. Note that v

x = v

0throughout the motion sinceF

net,x= 0.

x = vavg,x

t

L = v0

t

t = L

v0


43/63

43

Substitute into the expression for y.

y =

F

m

d

v0

L

v0

y = F

m

dL

v20

The total y-displacement across regions (1) and (2) is

ytotal

= 1

2

F

m

d

v0

2+

F

m

dL

v20

= F

mv20

(1

2d2 +dL)

Notice that the dimensions on each term are the same.

force

mass

distance2

speed2

So that gives us some confidence. Now, check that the units result in meters.

N

kg

m2

m2

s2

=kg

m

s2m2

kgm2s2

= m

So the units do give meters which is correct.

2.P.64

Just after release, yi

=h and viy

= 0.

Just before hitting the floor, yf

= 0.

To find t, use

yf

yi

= viy

t+1

2

Fnet,y

m t2

0 h = 0 +12

mgm

t2

h = 12

gt2

t =

2h

g

Forvfy

, use the momentum principle applied to the system of the ball.

pfy = piy + Fnet,y t

vfy

= viy

+Fnet,y

m t

= 0 +mgm

t

= g

2h

g


44/63

44

Note that vfy

is negative as expected.

The results for t and vfy

do not depend on the mass m. As a result, if you drop two balls with the same mass m, th

will hit the floor at the same time and with the same velocity, as long as air resistance is negligible since we assumed thaF

net=F

grav in this problem.

2.X.65

(a), (c), and (e) are correct.

(b) is incorrect because0.1 m/s is 10 cm/s which is much too small.

(d) is incorrect becausetcan be calculated by estimating the distance compressed during the collision.

2.P.66

m = 2200 kg

viy

= 40 m/sy = 0.06 mvfy

= 0

Define the system to be the safe. Draw a free-body diagram like the one shown in Figure 12.

Fground

Fgrav

Figure 12: A free-body diagram of the safe

Assume thatFground

is constant as the safe is slowing down due to the collision.

Apply the momentum principle.

Fnet

= p

tp

f=

pi

= mvi

= (2200 kg)(0, 40, 0 m/s)= 0, 88000, 0 kg m/s


45/63

45

To calculatet, assume constant net force. Then,

vavg

=v

i+ v

f

2

= 0, 40, 0 m/s+< 0, 0, 0>

2= 0, 20, 0 m/s

vavg,y

= y

t

t = y

vavg,y

= 0.06 m

20 m/s= 0.003 s

This is such a small time interval that the approximation of constant net force is reasonable. Any error from this approximationwould be small.

The principle of superposition gives

Fnet

= Fground

+Fgrav

= Fground

+< 0, mg, 0>

= Fground

+< 0, (2200 kg)(9.8 Nkg

), 0>

= Fground

+ 0, 221560, 0 N

Substitute into the momentum principle.

Fnet

= p

t

Fground

+ 0, 21560, 0 N = pf pit

Fground

+ 0, 21560, 0 N = 0, 88000, 0 kg m/s0.003 s

Fground

+ 0, 21560, 0 N = 0, 2.93 107, 0 NF

ground=

0, 2.93 107, 0 N 0, 2.156 104, 0 N

= 0, 2.93 107, 0 N

Note that Fgrav


46/63

46

vi


m = 2500 kgvix

= 24 m/s

vfx

= 0

x = 0.72 m

(a) Assuming constantFnet

during the collision, then,

vavg,x

=vix

+ vfx

2

= 24 m/s + 02

= 12 m/s

(b)

x = vavg,x

t

t = x

vavg,x

= 0.72 m

12 m/s

= 0.06 s

(c) Define the system to be the truck. The only significant force on the truck during the collision is the force by the wa


47/63

47

Apply the momentum principle.

Fnet

= p

t

Fby wall

= p

t

=p

f p

i

t

=mvi

t

= (2500 kg)(24, 0, 0 m/s)

0.06 s

= 1.0 106, 0, 0 NFby wall on truck

= 1.0 106 Nweight = mg

= (2500 kg)(9.8 N

kg)

= 2.45 104

N

Fby wall on truck

Fgrav on truck

= 1.0 106 N

2.45 106 N= 41

The wall exerts a force on the truck that is more than 40 times the weight of the truck.

(d)

vavg =

vi

+ vf

2

was valid.

2.P.68

(a) Image (B) is correct because the kick is brief and after the kick,v is constant so the block will move with a constantvelocity.

(b) Only the y-component ofp changes due to the kick because the kick is in the +y direction.

(c) Fnet,x

is constant so vfx

= vix

= 2.5 m/s

Use a picture ofvf after the kick (shown in Figure14) to find vfz .

tan22 =vfz

vfx

vfz

= vfx

tan22

= (2.5 m/s)tan22

= 1.0 m/s


48/63

48

x

vfz

vf

2.5m /s


vf

= 2.5, 0, 1.0 m/s

|v| =

(2.5 m/s)2

+ (0 m/s)2

+ (1.0 m/s)2

= 2.69 m/s

v = v

|v|=

2.5, 0, 1.0 m/s2.69 m/s

=

p is the same as v, so

p =

(d)

pfx

= mvfx

= (0.7 kg)(2.5 m/s)

= 1.75 kg m/s

(e)

|p| = m |v|= (0.7 kg)(2.69 m/s)

= 1.88 kg m/s

(f)

pfz

= mvfz

= (0.7 kg)(1.0 m/s)

= 0.7 kg m/s


49/63

49

(g)

Fnet

= p

t

Choose the system to be the block of ice. The only component of the net force that is non-zero is the z-component.

Fnet,z

= p

z

t

=pfz

piz

t

= 0.7 kg m/s 0

0.003 s= 233 NF

net

= 233 N

This is called the average net force because we assumed that it is constant.

2.P.69

Draw a sketch of the situation.

vi

vf=0

Figure 15: A sketch of the collision

m = 0.057 kg

vi

= 50, 0, 0 m/sv

f= 0 when instantaneously stopped during the collision

x = 0.02 m during the collision when slowing down


50/63


51/63

51

15

16 # d e f i ne c a rt ' s i n i t i a l momentum ( and i n i t i a l v e l o c i t y ) i n SI u n it s 17 # r e ad t h e v a r i a b l e name f rom r i g h t t o l e f t : momentum o f t h e o b j e c t named " c a r t " 18 c a r t . p = c a r t .m v e c t o r ( 1 / 3 , 0 , 0 )19

20 # d e f in e n et f o rc e on c a rt 21 F ne t = v e c t o r ( 0 , 0 , 0 )22

23 # t i m e s t e p24 d e l t a t = 0. 0 125

26 # i n i t i a l t ime 27 t = 028

29 # i n i t i a l i z e graph d i sp l ay 30 g ra ph 1 = g d i s p l a y ( x =4 50 , y =5 0, t i t l e = ' x v s . t ' , x t i t l e= 't ( s ) ' ,31 y t i t l e = ' x(m) ' , xm in= 0. , xmax= 6. 0, ymi n= 0. , ymax= 2. 0)32 # i n i t i a l i z e t he f un ct io n t o be p l o t t e d 33 xvt = gcurve ( co l o r=co l o r . green )34

35 # pa us e and w a it f o r u se r t o c l i c k mouse t o s t a r t a ni ma ti on 36 s c e n e . mous e . g e t c l i c k ( )37

38 # l o op t o a ni ma te t h e c a r t 39 while t < 6 . 0 1:40 # l i m i t an im at io n r a t e 41 r a t e ( 1 0 0 )42 # a p p ly momentum p r i n c i p l e t o u p d at e c a r t ' s momentum43 c a r t . p = c a r t . p + F ne t d e l t a t44 # u pd at e c a r t ' s p o s i t i o n , a ss u mi ng New ton ' s e x p r e s s i o n f o r momentum 45 c a r t . p o s = c a r t . p o s + ( c a r t . p / c a r t .m) d e l t a t46 i f c a r t . x > 2 :47 break

48 # u p da t e e l a p s e d t im e 49 t = t + d e lt a t50 # u p d at e g r ap h 51 xvt . pl o t ( pos=(t , car t . x) )

(b) Here is the program for this part.

1 from __future__ import d i v i s i o n2 from v i s u a l import 3 from vi su al . graph import 4

5 # make a t a b l e 6 # Note t h a t t he t a b l e i s a b i t l o ng e r t ha n 2 m t o a ccommodate t he c a rt 7

tra ck = box( pos= vecto r (1 , 0 . 0 5 , 0 ) , s i z e = ( 2 . 1 , 0 . 0 5 , 0 . 1 0 ) , c o l o r =c o l o r . w h it e )8

9 # make a c a r t 10 c a r t = bo x ( p o s=v e c t o r ( 0 , 0 , 0 ) , s i z e = ( 0 . 1 , 0 . 0 4 , 0 . 0 6 ) , c o l o r =c o l o r . r e d )11

12 # d e f i ne c a rt ' s mass i n S I u n i t s 13 # r ea d t h e v a r i a b l e name from r i g h t t o l e f t : mass o f t h e o b j e c t named " c a r t " 14 c a r t . m = 0 . 515


52/63

52

16 # d e f i ne c a rt ' s i n i t i a l momentum ( and i n i t i a l v e l o c i t y ) i n SI u n it s 17 # r e ad t h e v a r i a b l e name f rom r i g h t t o l e f t : momentum o f t h e o b j e c t named " c a r t " 18 c a r t . p = c a r t .m v e c t o r ( 0 . 3 3 3 3 , 0 , 0 )19

20 # d e f in e n et f o rc e on c a rt 21 F ne t = v e c t o r ( 0 . 2 , 0 , 0 )22

23

# t i m e s t e p24 d e l t a t = 0. 0 125

26 # i n i t i a l t ime 27 t = 028

29 # i n i t i a l i z e graph d i sp l ay 30 g ra ph 1 = g d i s p l a y ( x =4 50 , y =5 0, t i t l e = ' x v s . t ' , x t i t l e= 't ( s ) ',31 y t i t l e = ' x(m) ' , xmi n= 0. , xmax= 6. 0, ymi n= 0. , ymax= 2. 0)32 # i n i t i a l i z e t he f un ct io n t o be p l o t t e d 33 xvt = gcurve ( co l o r=co l o r . green )34

35 # pa us e and w a it f o r u se r t o c l i c k mouse t o s t a r t a ni ma ti on

36 s c e n e . mo us e . g e t c l i c k ( )37

38 # l o op t o a ni ma te t h e c a r t 39 while t < 6 . 0 1:40 # l i m i t an im at io n r a t e 41 r a t e ( 1 0 0 )42 # a p p l y momentum p r i n c i p l e t o u p da t e c a r t ' s momentum43 c a r t . p = c a r t . p + F ne t d e l t a t44 # u p da t e c a r t ' s p o s i t i o n , a ss u mi ng Ne wt on ' s e x p r e s s i o n f o r momentum 45 c a r t . p o s = c a r t . p o s + ( c a r t . p / c a r t .m) d e l t a t46 # s ee i f c ar t runs out o f t r ac k 47 i f c a r t . x > 2 :48 break


(c) Here is the program for this part.


5 # make a t a b l e 6 # Note t h a t t he t a b l e i s a b i t l o ng e r t ha n 2 m t o a ccommodate t he c a rt 7

trac k = box( pos= vecto r (1 , 0 . 0 5 , 0 ) , s i z e = ( 2 . 1 , 0 . 0 5 , 0 . 1 0 ) , c o l o r = c o l o r . w h it e )8

9 # make a c a r t 10 c a r t = bo x ( p o s=v e c t o r ( 0 , 0 , 0 ) , s i z e = ( 0 . 1 , 0 . 0 4 , 0 . 0 6 ) , c o l o r =c o l o r . r e d )11

12 # d e f i ne c a rt ' s mass i n SI u n i t s 13 # r ea d t h e v a r i a b l e name from r i g h t t o l e f t : mass o f t h e o b j e c t named " c a r t " 14 c a r t . m = 0 . 515


53/63

53


20 # d e f in e n et f o rc e on c a rt 21 Fnet =0.25 c a r t . p / 3 . 022

23

# t i m e s t e p24 d e l t a t = 0. 0 125

26 # i n i t i a l t ime 27 t = 028

29 # i n i t i a l i z e g ra ph d i s p la y and a l lo w g ra ph d i s p la y t o a u t os i z e 30 g ra ph 1 = g d i s p l a y ( x =4 50 , y =5 0, t i t l e = ' x v s . t ' , x t i t l e= 't ( s ) ' , y t i t l e = ' x(m) ')31

32 # i n i t i a l i z e t he f un ct io n t o be p l o t t e d 33 xvt = gcurve ( co l o r=co l o r . green )34

35 # pa us e and w a it f o r u se r t o c l i c k mouse t o s t a r t a ni ma ti on

36 s c e n e . mous e . g e t c l i c k ( )37

38 # l o op t o a ni ma te t h e c a r t 39 while t < 2 4 . 01 :40 # l i m i t an im at io n r a t e 41 r a t e ( 1 0 0 )42 # a p p ly momentum p r i n c i p l e t o u p d at e c a r t ' s momentum43 c a r t . p = c a r t . p + F ne t d e l t a t44 # u pd at e c a r t ' s p o s i t i o n , a ss u mi ng New ton ' s e x p r e s s i o n f o r momentum 45 c a r t . p o s = c a r t . p o s + ( c a r t . p / c a r t .m) d e l t a t46 # s ee i f c ar t runs o f f e i t he r end of t r ac k 47 i f c a r t . x > 2 :48 break

49 i f c a r t . x < 0 :50 break


(d) Here is the program for this part.


5

# make a t a b l e 6 # Note t h a t t he t a b l e i s a b i t l o ng e r t ha n 2 m t o a ccommodate t he c a rt 7 tra ck = box( pos= vecto r (1 , 0 . 0 5 , 0 ) , s i z e = ( 2 . 1 , 0 . 0 5 , 0 . 1 0 ) , c o l o r =c o l o r . w h it e )8

9 # make a c ar t , b ut p ut c a rt on t he r i g h t end o f t he t r ac k 10 c a r t = bo x ( p o s=v e c t o r ( 2 , 0 , 0 ) , s i z e = ( 0 . 1 , 0 . 0 4 , 0 . 0 6 ) , c o l o r =c o l o r . r e d )11

12 # d e f i ne c a rt ' s mass i n S I u n i t s 13 # r ea d t h e v a r i a b l e name from r i g h t t o l e f t : mass o f t h e o b j e c t named " c a r t "


54/63

54

14 c a r t . m = 0 . 515


20 # d e f in e n et f o rc e on c a rt 21

# THINK ABOUT WHY THIS LINE DOESN'T CHANGE FROM THAT IN PART ( c ) !

22 Fnet =0.25 c a r t . p / 3 . 023

24 # t i m e s t e p25 d e l t a t = 0. 0 126

27 # i n i t i a l t ime 28 t = 029

30 # i n i t i a l i z e g ra ph d i s p la y and a l lo w g ra ph d i s p la y t o a u t os i z e 31 g ra ph 1 = g d i s p l a y ( x =4 50 , y =5 0, t i t l e = ' x v s . t ' , x t i t l e= 't ( s ) ', y t i t l e = ' x(m) ')32

33 # i n i t i a l i z e t he f un ct io n t o be p l o t t e d

34 xvt = gcurve ( co l o r=co l o r . green )35

36 # pa us e and w a it f o r u se r t o c l i c k mouse t o s t a r t a ni ma ti on 37 s c e n e . mo us e . g e t c l i c k ( )38

39 # l o op t o a ni ma te t h e c a r t 40 while t < 2 4 . 01 :41 # l i m i t an im at io n r a t e 42 r a t e ( 1 0 0 )43 # a p p l y momentum p r i n c i p l e t o u p da t e c a r t ' s momentum44 c a r t . p = c a r t . p + F ne t d e l t a t45 # u p da t e c a r t ' s p o s i t i o n , a ss u mi ng Ne wt on ' s e x p r e s s i o n f o r momentum 46 c a r t . p o s = c a r t . p o s + ( c a r t . p / c a r t .m) d e l t a t47 # s ee i f c ar t runs o f f e i t he r end of t r ac k 48 i f c a r t . x > 2 :49 break

50 i f c a r t . x < 0 :51 break


2.P.71Start your program simply. Create a table, a block, and a spring such that the spring is attached between the top of ttable and the bottom of the block. It helps to place the position of the spring aty = 0 and to move the table downwarddistance equal to half its height. The code below creates a spring that is 0.1 m long that is attached between the tabletand the bottom of the block.

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


55/63

55

t a b l e = b o x ( p o s = ( 0 , 0 . 00 2 , 0 ) , s i z e =( 1 , 0 . 0 0 4 , 1 ) , c o l o r =c o l o r . w h it e ) b l o c k=b ox ( p os = ( 0 , 0 . 1 2 , 0 ) , s i z e = ( 0 . 2 , 0 . 0 4 , 0 . 2 ) , c o l o r =c o l o r . r e d ) s p r i n g =h e l i x ( p o s = ( 0 , 0 , 0 ) , a x i s = ( 0 , 0 . 1 , 0 ) , r a d i u s = 0 . 1 , c o l o r =c o l o r . y e l l o w )

Now, define constants and initial conditions. L is the length of the spring. Update the axis of the spring and the position ofthe block after defining L. This way, if you change L, the length of the spring and the position of the block will be correct.See the program below. Change L to other values like 0.25 m, for example, to see if the length of the spring and position ofthe block are adjusted accordingly. By testing your code as you write it, you will be able to find any mistakes in syntax orerrors in logic.

from __future__ import d i v i s i o nfrom v i s u a l import


m = 0 . 0 6g = 9. 8k=8L0 = 0 . 2

L = 0 . 1 s p r i n g . a x i s = v e c t o r ( 0 , L , 0 ) b l o c k . p o s . y = L + b l o c k . s i z e . y / 2

v=vec to r (0 ,0 ,0 )p=mv

dt=0.01

In a while loop, calculate the net force on the block and update its momentum and position. The net force on the block isthe sum of the gravitational force by Earth on the block and the force by the spring on the block. Calculate the force by thespring on the block using the procedure shown in Chapter 2. After updating the position of the block, be sure to update theaxis (i.e. length) of the spring. The code below produces a working simulation of the spring-block system.



m = 0 . 0 6g = 9. 8k=8L0 = 0 . 2L = 0 . 1

s p r i n g . a x i s = v e c t o r ( 0 , L , 0 ) b l o c k . p o s . y = L + b l o c k . s i z e . y / 2


dt=0.01

while 1 :


56/63

56

22 r a t e ( 1 0 0 )23 Fgrav= vector(0,mg , 0 )24

25 L = s p r i n g . a x i s . y26 s = a bs ( L ) L027 L ve c = v e c t o r ( 0 , L , 0 )28 Lhat = Lvec/mag( Lvec)29

Fspri ng=k

s

Lhat30

31 Fnet=Fgrav + Fspr ing32

33 p=p+Fnet dt34 v=p/m35 b l o c k . p o s = b l o c k . p o s + vdt36

37 s p r i n g . a x i s = v e c t o r ( 0 , b l o c k . p o s . y b l o c k . s i z e . y / 2 , 0 )

Now, we are ready to add the graph. Be sure to label the axes with the quantities being plotted and their units. Also, ada variable t to keep track of the clock reading since it will be plotted on the horizontal axis of the graph.

1 from __future__ import d i v i s i o n2 from v i s u a l import 3 from v i s u a l . g r ap h import 4

5 t a b l e = b o x ( p o s = ( 0 , 0 . 00 2 , 0 ) , s i z e =( 1 , 0 . 0 0 4 , 1 ) , c o l o r = c o l o r . w h it e )6 b l o c k=b ox ( p os = ( 0 , 0 . 1 2 , 0 ) , s i z e = ( 0 . 2 , 0 . 0 4 , 0 . 2 ) , c o l o r = c o l o r . r e d )7 s p r i n g =h e l i x ( p o s = ( 0 , 0 , 0 ) , a x i s = ( 0 , 0 . 1 , 0 ) , r a d i u s = 0 . 1 , c o l o r =c o l o r . y e l l o w )8

9 yGraph = g d i s p l a y ( x =0 , y =4 00 , w id th = 40 0 , h e i g h t = 30 0 , t i t l e = ' y v s . t ' , x t i t l e= 't ( s ) ' ,y t i t l e = ' x(m) ')

10 y P l o t = g c u r v e ( c o l o r =b l o c k . c o l o r )11

12 m = 0 . 0 613 g = 9 . 814 k=815 L0 = 0 . 216 L = 0 . 117 s p r i n g . a x i s = v e c t o r ( 0 , L , 0 )18 b l o c k . p o s . y = L + b l o c k . s i z e . y / 219

20 v=vec to r (0 ,0 ,0 )21 p=mv22

23 dt=0.0124 t=025

26

while 1 :27 r a t e ( 1 0 0 )28 Fgrav= vector(0,mg , 0 )29

30 L = s p r i n g . a x i s . y31 s = a bs ( L ) L032 L ve c = v e c t o r ( 0 , L , 0 )33 Lhat = Lvec/mag( Lvec)34 Fspri ng=k s Lhat


57/63

57

Fnet=Fgrav + Fsp ring

p=p+Fnet dt v=p/m b l o c k . p o s = b l o c k . p o s + vdt

s p r i n g . a x i s = v e c t o r ( 0 , b l o c k . p o s . y b l o c k . s i z e . y / 2 , 0 ) yPl ot . pl o t ( pos=(t , bl ock . pos . y) )

t = t+dt

(a) Take the previous program and use a condition in the while statement to stop the simulation at t= 1 s. Adjust thetime step todt = 0.1and print data to compare the simulation results to the example from Chapter 2 of the textbook.


5 t a b l e = b o x ( p o s = ( 0 , 0 . 00 2 , 0 ) , s i z e = ( 1 , 0 . 0 0 4 , 1 ) , c o l o r =c o l o r . w h i te )6 b l o c k=b ox ( p o s = ( 0 , 0 . 1 2 , 0 ) , s i z e = ( 0 . 2 , 0 . 0 4 , 0 . 2 ) , c o l o r =c o l o r . r e d )7 s p r i n g =h e l i x ( p o s = ( 0 , 0 , 0 ) , a x i s = ( 0 , 0 . 1 , 0 ) , r a d i u s = 0 . 1 , c o l o r =c o l o r . y e l l o w )8

9 yGraph = g d i s p l a y ( x =0 , y =4 00 , w id th = 40 0 , h e i g h t = 30 0 , t i t l e = ' y v s . t ' , x t i t l e= 't ( s ) ' ,y t i t l e = ' x(m) ')

10 y P l o t = g c u r v e ( c o l o r =b l o c k . c o l o r )11

12 m = 0 . 0 613 g = 9. 814 k=815 L0 = 0 . 216 L = 0 . 117 s p r i n g . a x i s = v e c t o r ( 0 , L , 0 )18 b l o c k . p o s . y = L + b l o c k . s i z e . y / 219

20 v=vect or (0 ,0 ,0 )21 p=mv22

23 dt=0.124 t=025

26 while t< 1:27 r a t e ( 1 0 0 )28 Fgrav= vector(0,mg , 0 )29

30 L = s p r i n g . a x i s . y31 s = a bs ( L ) L032 L ve c = v e c t o r ( 0 , L , 0 )33 Lhat = Lvec/mag( Lvec)34 Fspri ng=k s Lhat35

36 Fnet=Fgrav + Fsp ring37


58/63

58

38 p=p+Fnet dt39 v=p/m40 b l o c k . p o s = b l o c k . p o s + v dt41

42 s p r i n g . a x i s = v e c t o r ( 0 , b l o c k . p o s . y b l o c k . s i z e . y / 2 , 0 )43

44 yPl ot . pl o t ( pos=(t , bl ock . pos . y) )45

46 print " t i me : " , t , " yp os : " , s p r i n g . a x i s . y

The resulting graph and data are identical to what was calculated in the corresponding example in Chapter 2 of thtextbook, with long straight lines between data points on the graph due to the relatively large time interval.

(b) Take the previous program and changedt to 0.01 s. Comment out or delete theprint statement because it will slodown the simulation. The resulting graph is now a smooth curve. (Technically, there are still straight lines betwedata points in the graph, but they are very short straight lines so the curve appears smooth.)

(c) Though dt = 0.001 should result in a more accurate calculation, the resulting graph appears to be just as smooth afor dt= 0.01. One way to check the difference is to print the y-axis of the spring at t= 1 s and compare the resultFor dt= 0.01, the y-axis of the spring at t= 1 is 0.111 m. For dt= 0.001, it is 0.112 m. The difference is only 0.0m (thats 1 mm) during a 1-second time interval. This is a very small difference, so dt = 0.01 is sufficiently small

produce fairly accurate results.

(d) Visual inspection of the graph gives a period of little more than 0.5 s. ChangeL to another value such as 0.05 m, anrun the simulation. The period is the same as before. Evidently, the period of oscillation is independent of the amouthe that the spring is initially compressed or stretched.

2.P.72

Start your program by creating a tabletop, wall, block, and spring. Place the wall at the left side of the tabletop. One enof the spring should be attached to the surface of the wall. The other end of the spring should be attached to the end of thblock. The spring should be 0.5 m long which is its unstretched length. An example is shown below.

1 from __future__ import d i v i s i o n2 from v i s u a l import 3 from v i s u a l . g r ap h import 4

5 w a l l = b ox ( p os = ( 0 . 5 1 , 0 . 1 , 0 ) , s i z e = ( 0 . 0 2 , 0 . 2 , 0 . 2 ) , c o l o r = ( 1 , 0 . 5 , 0 ) )6 f l o o r = box ( p o s = (0 , 0 . 01 , 0 ) , s i z e = ( 1 , 0 . 0 2 , 0 . 2 ) , c o l o r = c o l o r . w h it e )7 b l o c k = b ox ( p os = ( 0 . 1 , 0 . 0 5 , 0 ) , s i z e = ( 0 . 2 , 0 . 1 , 0 . 1 ) , c o l o r = c o l o r . r e d )8 s p r i n g = h e l i x ( p o s = ( 0 . 5 , 0 . 0 5 , 0 ) , a x i s = ( 0 . 5 , 0 , 0 ) , c o l o r =c o l o r . y e l l o w , r a d i u s = 0 . 04 )

Now, define constants. Define the axis of the spring and the position of the block in terms of the length of the spring L shown below.

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

from v i s u a l . g r ap h import

4

5 w a l l = b ox ( p os = ( 0 . 5 1 , 0 . 1 , 0 ) , s i z e = ( 0 . 0 2 , 0 . 2 , 0 . 2 ) , c o l o r = ( 1 , 0 . 5 , 0 ) )6 f l o o r = box ( p o s = (0 , 0 . 01 , 0 ) , s i z e = ( 1 , 0 . 0 2 , 0 . 2 ) , c o l o r = c o l o r . w h it e )7 b l o c k = b ox ( p os = ( 0 . 1 , 0 . 0 5 , 0 ) , s i z e = ( 0 . 2 , 0 . 1 , 0 . 1 ) , c o l o r = c o l o r . r e d )8 s p r i n g = h e l i x ( p o s = ( 0 . 5 , 0 . 0 5 , 0 ) , a x i s = ( 0 . 5 , 0 , 0 ) , c o l o r =c o l o r . y e l l o w , r a d i u s = 0 . 04 )9

10 m = 0 . 2 511 k=1.3


59/63

59

L0 = 0 . 5L = 0 . 7

s p r i n g . a x i s = v e c t o r ( L , 0 , 0 ) b l o c k . p o s . x = s p r i n g . p o s . x + L + b l o c k . s i z e . x / 2


dt=0.01 t=0

Create a while loop and use it to calculate the net force on the block and update the blocks momentum and position. Besure to also update the axis of the spring. The program below results in a working simulation of the block-spring system.

from __future__ import d i v i s i o nfrom v i s u a l import from v i s u a l . g r ap h import

w a l l = b ox ( p os = ( 0 . 5 1 , 0 . 1 , 0 ) , s i z e = ( 0 . 0 2 , 0 . 2 , 0 . 2 ) , c o l o r = ( 1 , 0 . 5 , 0 ) ) f l o o r = box ( p os = ( 0 , 0 . 01 , 0 ) , s i z e = ( 1 , 0 . 0 2 , 0 . 2 ) , c o l o r = c o l o r . w h it e ) b l o c k = b ox ( p os = ( 0 . 1 , 0 . 0 5 , 0 ) , s i z e = ( 0 . 2 , 0 . 1 , 0 . 1 ) , c o l o r = c o l o r . r e d )

s p r i n g = h e l i x ( p o s = ( 0 . 5 , 0 . 0 5 , 0 ) , a x i s = ( 0 . 5 , 0 , 0 ) , c o l o r =c o l o r . y e l l o w , r a d i u s = 0 .0 4 )m = 0 . 2 5k=1.3L0 = 0 . 5L = 0 . 7

s p r i n g . a x i s = v e c t o r ( L , 0 , 0 ) b l o c k . p o s . x = s p r i n g . p o s . x + L + b l o c k . s i z e . x / 2


dt=0.01

t=0

while 1 : r a t e ( 1 0 0 0 ) L ve c = s p r i n g . a x i s L = mag(Lvec) Lhat = Lvec/L s=L L0 F s p r in g =k s Lhat

F ne t = F s p r i n g p = p + F ne t dt v = p/m b l o c k . p o s = b l o c k . p o s + vdt

s p r i n g . a x i s = s p r i n g . a x i s + vdt

t = t+dt

Now, add a graph and plot the blocks x-position vs. time as shown in the program below.

from __future__ import d i v i s i o n


60/63

60

2 from v i s u a l import 3 from v i s u a l . g r ap h import 4

5 w a l l = b ox ( p os = ( 0 . 5 1 , 0 . 1 , 0 ) , s i z e = ( 0 . 0 2 , 0 . 2 , 0 . 2 ) , c o l o r = ( 1 , 0 . 5 , 0 ) )6 f l o o r = box ( p o s = (0 , 0 . 01 , 0 ) , s i z e = ( 1 , 0 . 0 2 , 0 . 2 ) , c o l o r = c o l o r . w h it e )7 b l o c k = b ox ( p os = ( 0 . 1 , 0 . 0 5 , 0 ) , s i z e = ( 0 . 2 , 0 . 1 , 0 . 1 ) , c o l o r = c o l o r . r e d )8 s p r i n g = h e l i x ( p o s = ( 0 . 5 , 0 . 0 5 , 0 ) , a x i s = ( 0 . 5 , 0 , 0 ) , c o l o r =c o l o r . y e l l o w , r a d i u s = 0 . 04 )9

10 x Gr aph = g d i s p l a y ( x = 0, y =4 00 , w i dt h = 40 0 , h e i g h t = 30 0 , t i t l e = ' x v s . t ' , x t i t l e= 't ( s ) ' , y t i t l= ' x(m) ')

11 x P l o t = g c u r v e ( c o l o r =b l o c k . c o l o r )12

13 m = 0 . 2 514 k=1.315 L0 = 0 . 516 L = 0 . 717 s p r i n g . a x i s = v e c t o r ( L , 0 , 0 )18 b l o c k . p o s . x = s p r i n g . p o s . x + L + b l o c k . s i z e . x / 219

20 v=vec to r (0 ,0 ,0 )

21 p=mv22

23 dt=0.0124 t=025

26 while 1 :27 r a t e ( 1 0 0 0 )28 L ve c = s p r i n g . a x i s29 L = mag( Lvec)30 Lhat = Lvec/L31 s=L L032 F s p r in g =k s Lhat33

34 F ne t = F s p r i n g35 p = p + F ne t dt36 v = p/m37 b l o c k . p o s = b l o c k . p o s + vdt38

39 s p r i n g . a x i s = s p r i n g . a x i s + vdt40

41 t = t+dt42

43 xPl ot . pl o t ( pos=(t , bl ock . pos . x) )

(a) The oscillation is symmetric. Thus, the period should not depend on whether the spring is initially compressed (0.1

from equilibrium, for example) or whether it is initially stretched the same amount.(b) To stretch the spring 14 cm, change the initial value ofL to 0.64 m. (Thus, s= 0.64 m 0.5 m = 0.14 m at t = 0

The system will oscillate with a period of approximately 2.8 s.

(c) To compress the spring 14 cm, change the initial value ofL to 0.36 m. (0.5 m 0.14 m = 0.36 m) The system oscillatwith the same period of approximately 2.8 s.

(d) Change the initial value ofL to 0.5 m. (0.5 m 0.5 m = 0, so the spring is not stretched nor compressed.) The systedoes not oscillate at all. In order to make the system oscillate, the initial velocity can be given a non-zero value in th


61/63

61

x-direction. For example, use v=vector(-0.4,0,0). Though the spring is unstretched at t= 0, the block is moving,so the system will oscillate.

(e) Change the initial value ofLto 0.4 m and the initial value ofv to v=vector(-0.4,0,0)m/s. The system still oscillates;however, the graph is not at a minimum at t = 0, so the spring gets further compressed. Its maximum compressionwill be greater than the initial amount.

2.P.73

A program that simulates the motion of the spring-mass system is shown below. Note that the initial value of L can be usedto change the initial length of the spring. The axis of the spring and and the initial position of the mass are calculated afterthe constant L. These are the initial conditions of the mass-spring system and can be adjusted.

Its important to calculate the net force on the mass, which in this case is the sum of the gravitational force on the mass byEarth and the force on the mass by the spring. See the example below.


s u p p o r t = b ox ( p o s = ( 0 , 0 . 5 5 , 0 ) , s i z e = ( 1 , 0 . 1 , 0 . 5 ) , c o l o r = ( 1 , 0 . 5 , 0 ) ) s p r i n g = h e l i x ( p o s = ( 0 , 0 . 5 , 0 ) , a x i s = ( 0 ,

0 . 2 , 0 ) , r a d i u s = 0 .1 , c o l o r =c o l o r . y e l l o w )

m as s = s p h e r e ( p o s = ( 0 , 0 . 6 , 0 ) , r a d i u s = 0 . 1 , c o l o r =c o l o r . r e d )

t r a i l =curve ( co l o r=m ass . co l o r )

m = 0 . 0 2g = 9. 8k = 0 . 9L0 = 0 . 2L = 0 . 3

L ha t = v e c t o r ( 0 , 1,0) s p r i n g . a x i s = LLhat

m as s . p o s = s p r i n g . p o s + s p r i n g . a x i s + m as s . r a d i u s Lhat


d t = 0 . 0 1 t = 0

while 1 : r a t e ( 1 0 0 )

F gr av = v e c t o r ( 0 , mg , 0 )

L ve c = s p r i n g . a x i s Lmag=mag( Lv ec ) Lhat = Lvec/Lmag s = mag(Lvec) L0 F s p r in g =k s Lhat

F ne t = F gr av + F s p r i n g p = p + F ne t dt v = p/m


62/63

62

40 mass . pos = mass . pos + v dt41

42 s p r i n g . a x i s = s p r i n g . a x i s + vdt43

44 t r a i l . append( pos=mass . pos )45

46 t = t +d t

(a) Make a prediction. The mass-spring system should oscillate up and downward with sinusoidal motion. Running tprogram shows the validity of this prediction, as the mass oscillates sinusoidally in the vertical direction.

(b) Make a prediction. The force by the spring results in oscillatory motion along the axis of the spring. But the gravitationforce, with the mass displaced a certain angle from vertical, causes the mass to oscillate back and forth like a penduluabout the vertical. Thus, its reasonable to predict that the resulting motion will be a combination of the two.

In your program, the spring should have a length L, but be stretched in both the x and y directions. Define an ang(from vertical) and calculate the unit vector. Then multiply the unit vector by L to get the springs initial axis. Bsure to write the angle theta in radians. Use direction cosines to get the unit vector for the axis of the spring bason the angle from vertical.

If you run the program below, you will notice that the motion is indeed a superposition of oscillations, one oscillatio

back and forth along the axis of the spring and another oscillation about the vertical like a pendulum.1 from __future__ import d i v i s i o n2 from v i s u a l import 3 from vi su al . graph import 4

5 s u p p o r t = b ox ( p o s = ( 0 , 0 . 5 5 , 0 ) , s i z e = ( 1 , 0 . 1 , 0 . 5 ) , c o l o r = ( 1 , 0 . 5 , 0 ) )6 s p r i n g = h e l i x ( p o s = ( 0 , 0 . 5 , 0 ) , a x i s = (0 , 0 . 2 , 0 ) , r a d i u s = 0. 1 , c o l o r = c o l o r . y e l l o w )7 m as s = s p h e r e ( p o s = ( 0 , 0 . 6 , 0 ) , r a d i u s = 0 . 1 , c o l o r =c o l o r . r e d )8

9 t r a i l =curve ( co l o r=m ass . co l o r )10

11 m = 0 . 0 212 g = 9. 813 k = 0. 914 L0 = 0 . 215 L = 0 . 316

17 t h e ta = 45 p i / 1 8 018 Lhat = vec tor ( cos ( pi /2 thet a ) , cos ( the ta ) ,0)19 s p r i n g . a x i s = LLhat20 m as s . p o s = s p r i n g . p o s + s p r i n g . a x i s + m as s . r a d i u s Lhat21

22 v=vect or (0 ,0 ,0 )23 p=mv24

25

d t = 0 . 0 126 t = 027

28 while 1 :29 r a t e ( 1 0 0 )30

31 F gr av = v e c t o r ( 0 ,mg , 0 )32

33 L ve c = s p r i n g . a x i s

7/25/2019 Ma

matter and interaction chapter 02 solutions

Documents