www.angelfire.com linux owned2 final

final 01 TAYLOR, JEFFREY W Due: May 10 2006, 1:00 pm 1

Version number encoded for clicker entry:V1:1, V2:1, V3:5, V4:1, V5:5.

Question 1

Part 1 of 1. 10 points.

A heavy liquid with a density 13 g/cm3 ispoured into a U-tube as shown in the left-hand figure below. The left-hand arm of thetube has a cross-sectional area of 9.79 cm2,and the right-hand arm has a cross-sectionalarea of 5.69 cm2. A quantity of 93.1 g of alight liquid with a density 1.1 g/cm3 is thenpoured into the right-hand arm as shown inthe right-hand figure below.

h1

h2

9.79 cm2 5.69 cm2

heavy liquid13 g/cm3

L

9.79 cm2 5.69 cm2

light liquid1.1 g/cm3

If the density of the heavy liquid is13 g/cm3, by what height h1 does the heavyliquid rise in the left arm?1. 0.408594 cm2. 0.42161 cm3. 0.435098 cm4. 0.448565 cm5. 0.462632 cm correct6. 0.477064 cm

Explanation:

Let : m` = 93.1 g ,

A1 = 9.79 cm2 ,

A2 = 5.69 cm2 ,

` = 1.1 g/cm3 , and

h = 13 g/cm3 .

Using the definition of density

` =m`V`

=m`

A2 L

L =m`

A2 `= 14.8746 cm .

After the light liquid has been added to theright side of the tube, a volume A2 h2 of heavyliquid is displaced to the left side, raising theheavy liquid on the left side by a height ofh1 with a displaced volume of A1 h1. Sincethe volume of heavy liquid is not changed, wehave

A1 h1 = A2 h2 .

At the level of the heavy-light liquid interfacein the right side, the absolute pressure is

P = Patm + ` g L ,

and at the same level in the left tube,

P = Patm + h g [h1 + h2] .

Equating these two values, we obtain

` g L = h g [h1 + h2]

= h g

[h1 +

A1A2

h1

]

= h g h1

[1 +

A1A2

].

Solving for h1, we have

h1 =` L

h

[1 +

A1A2

]

=(1.1 g/cm3) (14.8746 cm)

(13 g/cm3)

[1 +

(9.79 cm2)

(5.69 cm2)

]

= 0.462632 cm .

Question 2


A ski jumper travels down a slope andleaves the ski track moving in the horizon-tal direction with a speed of 26 m/s as in the


figure. The landing incline below her falls offwith a slope of = 41 .

The acceleration of gravity is 9.8 m/s2 .

d

41

26 m/s

vf

y

x

Calculate the distance d she travels alongthe incline before landing.1. 154.136 m2. 158.904 m correct3. 163.946 m4. 169.712 m5. 175.259 m6. 182.106 m

Explanation:

It is convenient to select the origin (x =y = 0) at the beginning of the jump. Sincevx0 = 26 m/s and vy0 = 0 m/s in this case,we have

x = vx0 t

y = vy0 t 12

g t2 =1

2g t2 .

The distance d she travels along the inclinebefore landing is related to her x and y coor-dinates by

x = d cos

y = d sin .

Substituting these expressions for x and y intothe two equations above, we obtain

d cos = vx t

d sin =1

2g t2 .

Excluding t from these equations gives

d =2 v2

0sin

g cos2

=(2) (26 m/s)2 sin 41

(9.8 m/s2) cos2 41

= 158.904 m .

Question 3


Determine how long the ski jumper is air-borne.1. 4.46449 s2. 4.61254 s correct3. 4.75773 s4. 4.92699 s5. 5.10204 s6. 5.28332 s

Explanation:

Excluding d rather than t from the systemabove, we obtain

t =2 v0 tan

g

=(2) (26 m/s) tan(41)

9.8 m/s2

= 4.61254 s .

Question 4


What is the magnitude of the relative angle with which the ski jumper hits the slope?1. 15.7047

2. 16.3548

3. 16.9574

4. 17.7627

5. 18.435

6. 19.0932 correct

Explanation:

vy = g t= (9.8 m/s2) (4.61254 s)= 45.2029 m/s , and

vx = 26 m/s , given.


The direction t of the velocity vector (rela-tive to the positive x axis) at impact is

t = arctan

(vyvx

)

= arctan

(45.2029 m/s26 m/s

)= 60.0932 .

Therefore the relative angle of impact onthe slope is

= |t| = (60.0932) (41)= 19.0932 .

Question 5


A runner is jogging at a steady vr =7 km/hr. When the runner is L = 7.5 kmfrom the finish line a bird begins flying fromthe runner to the finish line at vb = 35 km/hr(5 times as fast as the runner). When the birdreaches the finish line, it turns around andflies back to the runner. Even though the birdis a dodo, we will assume that it occupies onlyone point in space, i.e., a zero length bird.

L

vb

vrfinishline

How far does the bird travel?1. 12 km2. 12.5 km correct3. 12.9 km4. 13.3333 km5. 13.76 km6. 14.2286 km

Explanation:

Let, dodo birds fly, anddr be the distance the runner travels.db be the distance the bird travels.vr be the speed of the runner.vb be the speed of the bird.L = dr be the original distance to the finish

line.

L1 be the distance to the finish line afterthe first encounter.

...Li be the distance to the finish line after

the ith encounter.

finishline

L1dr1db1

Since the bird travels 5 times as fast as therunner at the first meeting between the birdand runner,

db1 = 5 dr1 . (1)

The sum of the birds and runners distancesis 5 times L.

db1 + dr1 = 2 L . (2)

Therefore, substituting for db1 from Eq. (1)

dr1 + 5 dr1 = 2 L

dr1 =2

6L =

2

6(7.5 km) = 2.5 km . (3)

Thus the distance the bird flies is

db1 = 5 dr1 =10

6L

=10

6(7.5 km) = 12.5 km , (4)

and the distance for the runner to travel afterthis first encounter is

L1 =4

6L =

4

6(7.5 km) = 5 km .

Question 6


After this first encounter, the bird thenturns around and flies from the runner backto the finish line, turns around again and fliesback to the runner. The bird repeats the backand forth trips until the runner reaches thefinish line.


How far does the bird travel from the be-ginning? (i.e., include the distance traveledto the first encounter)1. 36 km2. 37.5 km correct3. 39 km4. 40.5 km5. 42 km6. 43.5 km

Explanation:

Repeating this scenario a second time thedistance for the runner to travel after thesecond encounter is

L2 =4

6L1 =

(4

6

)2L ,

and the third time

L3 =4

6L2 =

(4

6

)3L ,

and the ith time

Li =4

6Li1 =

(4

6

)iL . (5)

Note: The distance the bird travels betweenthe (i 1)th and ith time is [see Eq. (4)]

dbi =10

6L

(4

6

)i(6)

and summing over all terms dbi

db =i=0

dbi =10

6L

[i=0

(4

6

)i](7)

=10

6L

[1 +

4

6+

(4

6

)2+

(4

6

)3(8)

+

(4

6

)4+

(4

6

)5+

(4

6

)6+

]

Or, by factoring1

6from the second term on

db =10

6L

{1 +

4

6

[1 +

4

6+

(4

6

)2(9)

+

(4

6

)3+

(4

6

)4+

(4

6

)5+

]}

By comparing Eq. (8) with (9), and general-izing (` = 4, and k = 6), the infinite series

i=0

(`

k

)i= 1 +

`

k

i=0

(`

k

)i(10)

then solving Eq. (10) fori=0

(`

k

)i

i=0

(`

k

)i=

k

k `k=6, `=4

(11)

i=0

(4

6

)i=

6

6 4 =6

2.

Therefore [from Eq. (7)]

db =10

6L

i=0

(4

6

)i

=10

6L

6

2= 5 L

= 5 (7.5 km) = 37.5 km .

Elegant Alternative Solution: The birdwill travel 5 times as far as the runner in thesame time. Since the bird and jogger travel forthe same length of time, the bird will travel

db = 5 L = 5 (7.5 km) = 37.5 km

Question 7


The two blocks are connected by a lightstring that passes over a frictionless pulleywith a negligible mass. The 4 kg block lieson a rough horizontal surface with a constantcoefficient of kinetic friction 0.3. This blockis connected to a spring with spring constant4 N/m. The second block has a mass of 9 kg.The system is released from rest when thespring is unstretched, and the 9 kg block fallsa distance h before it reaches the lowest point.

The acceleration of gravity is 9.8 m/s2 .Note: When the 9 kg block is at the lowest

point, its velocity is zero.


4 kg

9 kg

4 N/m4 kg

9 kg

h

h

= 0.3

Calculate the mechanical energy removedby friction durning the time when the 9 kgmass falls a distance h .1. 449.467 J correct2. 466.754 J3. 484.042 J4. 512.213 J5. 528.22 J6. 553.19 J

Explanation:

Basic Concepts:

Work-Energy TheoremSpring Potential EnergyFrictional Force according to the Work-Energy TheoremSolution:

W extAB = (KB KA) + (U gB U gA)+ (U spB U spA ) + W disAB .

For the present case, the external workW extAB = 0, A corresponds to the initial stateand B the state where m2 has descended by adistance s. The sum of the kinetic energy ofm1 plus that of m2 at B is given by

K = KB= (U gA U gB) + (U spA U spB )W disAB= m2 g s 1

2k s2 m1 g s . (1)

Based on the Eq. 1 at s = h, KB = 0, wehave

m1 g h = m2 g h 12

k h2 .

In turn,

h =2 g [m2 m1]

k(2)

=2 (9.8 m/s2) [(9 kg) (0.3) (4 kg)]

(4 N/m)= 38.22 m .

We know that Einitial = Efinal +E, whereE is the mechanical energy removed by fric-tion. In order to solve the second part of theproblem we need to calculate the initial and fi-nal energies. Let y1 be the vertical position ofm1 and y2 the vertical position of m2, wheresay y = 0 is the initial vertical position of m2.The total energy of the system is E = U + K,where U = Ugrav1 + Ugrav2 + Uspring. Ini-tially, Ugrav1,in = m1 g y1, Ugrav2,in = 0,Uspringin = 0 (the spring is unstretched) andKin = 0 . In the final situation the masseshave a null velocity, and so we have onceagain Kfin = 0 . The potential energies areUgrav1 = m1 g y1, Ugrav2,fin = m2 g (y1 h)and Uspring =

1

2k h2 .

Finally, letting y1 = 0 m and using Eq. 2for h, we get

E = Einitial Efinal= m2 g (h) 1

2k h2

= m2 g h 12

k h2

=2 g2

k

[m2 (m2 m1) (m2 m1)2

]=

2 m1 g2

k[m2 m1]

=2 (0.3) (4 kg) (9.8 m/s2)2

(4 N/m)

[(9 kg) (0.3) (4 kg)]= 449.467 J .

Question 8


Given: The speed of sound in air is 343 m/s.An open vertical tube has water in it. A

tuning fork vibrates over its mouth. As thewater level is lowered in the tube, the seventhresonance is heard when the water level is143 cm below the top of the tube.


143

cm

What is the wave length of the soundwave?1. 36 cm2. 38 cm3. 40 cm4. 42 cm5. 44 cm correct6. 46 cm

Explanation:

There are seven nodes (N = 7) in the aircolumn.

The number of quarter wavelengths

4in

the length of the pipe is J = 13 . SinceJ = `

4

, then =4 `

J =4 `

13.

The number of quarter wavelengths is oddwhich is indicative of a node at one end andan anti-node at the other end of the tube.

=4 `

2N 1 , where N = 7

=4 `

2 (7) 1=

4 (143 cm)

2 (7) 1

= 44 cm .

Question 9


What is the frequency of the sound wave;i.e., the tuning fork?1. 635.185 s1

2. 659.615 s1

3. 686 s1

4. 714.583 s1

5. 745.652 s1

6. 779.545 s1 correct

Explanation:

The frequency is

f =v

=(343 m/s)

(44 cm) (0.01 m/cm)

= 779.545 Hz .

Question 10


The water continues to leak out the bottomof the tube.

When the open vertical tube next resonateswith the tuning fork, what is its length.1. 139.5 cm2. 144 cm3. 148.75 cm4. 154 cm5. 159.75 cm6. 165 cm correct

Explanation:

The next resonance will occur when theopen vertical tube has a length of one-half wavelength greater than its initial wavelength.

` = ` +

2

= (143 cm) +(44 cm)

2= 165 cm .


Question 11


You are given f1(x), a transverse wave thatmoves on a string that ends and is FIXED inplace at x = 5 m. As the problem begins, thewave is moving to the right at v = 1 m/s.

v

0 1 2 3 4 5-3

-2

-1

0

1

2

3

Am

plitu

de

(cen

tim

eter

)

Distance (meter)

What is the shape of the wave on the stringafter 5 s?

1.

0 1 2 3 4 5-3

-2

-1

0

1

2

3

Distance (meter)

2.

0 1 2 3 4 5-3

-2

-1

0

1

2

3

Distance (meter)

3.

0 1 2 3 4 5-3

-2

-1

0

1

2

3

Distance (meter)

4.

0 1 2 3 4 5-3

-2

-1

0

1

2

3

Distance (meter)

correct

5.

0 1 2 3 4 5-3

-2

-1

0

1

2

3

Distance (meter)

6.

0 1 2 3 4 5-3

-2

-1

0

1

2

3

Distance (meter)

Explanation:

Consider the image of the wave reflected


about the FIXED point x = 5 m in the fol-lowing diagram. The image will be movingto the left at v = 1 m/s (in the oppositedirection from the real wave).

The initial wave (real) on the string is rep-resented with a dashed line and its reflectedwave (imaginary) is represented with a dottedline.

v v

0 1 2 3 4 5 6 7 8 9 10-3

-2

-1

0

1

2

3

Am

plitu

de

(cen

tim

eter

)

Distance (meter)

Initial time, t = 0 s

After 5 s the positions of the two wavesare have both moved 5 meters in oppositedirections. The resultant sum of the twowaves is the light gray line.

0 1 2 3 4 5 6 7 8 9 10-3

-2

-1

0

1

2

3

Am

plitu

de

(cen

tim

eter

)

Distance (meter)

Superposition, at t = 5 s

0 1 2 3 4 5-3

-2

-1

0

1

2

3

Am

plitu

de

(cen

tim

eter

)

Distance (meter)

Resultant, at t = 5 s

Question 12


Assume: When the disk lands on the sur-face it does not bounce.

The disk has mass 8 kg and outer radius50 cm with a radial mass distribution (whichmay not be uniform) so that its moment of

inertia is5

9m R2 .

The disk is rotating at angular speed7 rad/s around its axis when it touches thesurface, as shown in the figure below. Thedisk is carefully lowered onto a horizontal sur-face and released at time t0 with zero initiallinear velocity along the surface. The coef-ficient of friction between the disk and thesurface is 0.01 .

The kinetic friction force between the sur-face and the disk slows down the rotation ofthe disk and at the same time gives it a hor-izontal acceleration. Eventually, the diskslinear motion catches up with its rotation,and the disk begins to roll (at time trolling)without slipping on the surface.


I=5

9m R2

50 cm , radius8 kg

7 rad/s

= 0.01

How long t = trolling t0 does it take forthe ball to roll without slipping?


1. 12.7551 s correct2. 13.3333 s3. 14.0625 s4. 14.5005 s5. 15.3061 s6. 16.3265 s

Explanation:

Let : r = 50 cm = 0.5 m ,

0 = 7 rad/s ,

m = 8 kg , and

= 0.01 .

From the perspective of the surface, let thespeed of the center of the disk be vsurface.Using the frictional force f , we can determinethe acceleration

f = m g , andFsurface = m a , or

m a = m g , so

a = g , and

= g

R. (1)

Since

surface = t , we have

= g

Rt . (2)

After pure rolling begins at trolling there is nolonger any frictional force and consequentlyno acceleration. From the perspective of thecenter of the disk, let the tangential velocityof the rim of the disk be vdisk and the angularvelocity be ; the angular acceleration is

= I , so

=

I

= m g R5

9m R2

=9

5

g

R(3)

=9

5

(0.01) (9.8 m/s2)

(0.5 m)

= 0.3528 rad/s2 .

The time dependence of is

= 0 t= 0 9

5

g

Rt . (4)

When the disk reaches pure rolling, the veloc-ity from the perspective of the surface will bethe same as the velocity from the perspectiveof the center of the disk; that is, there will beno slipping. Setting the velocity disk fromEq. 4 equal to surface from Eq. 2 gives

disk = surface

0 95

g

Rt =

g

Rt , or

14

5 g t = R 0 , so

t =5

14

R 0 g

. (5)

=5

14

(0.5 m) (7 rad/s)

(0.01) (9.8 m/s2)

= 12.7551 s .

Question 13


Once the disk rolls without slipping, whatis its angular speed?1. 2.22222 rad/s2. 2.30769 rad/s3. 2.4 rad/s4. 2.5 rad/s correct5. 2.625 rad/s6. 2.72727 rad/s

Explanation:

Using Eqs. 1 and 5, we have

rot = t

= g

Rt

= g

T

(5

14

0 R

g

)

=5

140 , (6)

=5

14(7 rad/s)

= 2.5 rad/s .


or using Eqs. 4 and 5, we have

rot = 0 95

g

Rt

= 0 (

9

5

g

R

) (5

14

R

g

)0

=5

140 . (6)

Question 14


How far s does the ball slide until it beginsto roll without slipping?1. 6.94444 m2. 7.18006 m3. 7.41817 m4. 7.68469 m5. 7.97194 m correct6. 8.22857 m

Explanation:

Starting at rest, the ball slides at a constantacceleration. Using Eq. 5, we have

s =1

2a t2

=1

2 g

[5

14

R 0 g

]2

=25

392

R2 20

g(7)

=25

392

(0.5 m)2 (7 rad/s)2

(0.01) (9.8 m/s2)

= 7.97194 m .

Question 15


Through what angle does the disk ro-tate while sliding before it begins to roll with-out slipping?1. 8.20643 rev2. 8.50254 rev3. 8.94874 rev4. 9.23223 rev5. 9.64268 rev correct6. 9.94718 rev

Explanation:

The ball spins at a constant deceleration.Using Eq. 3, we have

= 0 t 12

t2

= 0

(5

14

R 0 g

)

12

(9

5

g

R

) (5

14

R 0 g

)2

=

[5

14 45

392

]R 2

0

g

=95

392

R 20

g(8)

=95

392

(0.5 m) (7 rad/s)2

(0.01) (9.8 m/s2)

= 60.5867 rad

= 3471.36 .

The number of revolutions is

rev =

360

=(60.5867 rad)

360

= 9.64268 rev .

Note: When comparing Part 3 to Part 4,the ball spins more than it slides due to thefact it is slipping on the surface.

R > s95

392

R2 20

g>

25

392

R2 20

g.

Question 16


What is the ratio of the final kinetic en-ergy (after pure rolling occures) to the initialkinetic energy?

1.KfK0

=2

5

2.KfK0

=3

7

3.KfK0

=5

14correct


4.KfK0

=7

15

5.KfK0

=1

4

6.KfK0

=9

19

Explanation:

Using Eq. 6, disk = rolling =5

140 ,

K0 =1

2I 20 , and

Kf =1

2I 2rolling +

1

2m v2rolling

=1

2I 2rolling +

1

2m R2 2rolling

=1

2I 2rolling

[1 +

m R2

I

]

=1

2I

[5

140

]2 [1 +

9

5

]

=1

2I 20

[5

14

]2 [14

5

]

=5

14K0 , so

KfK0

=5

14.

Question 17


A simple U-tube that is open at both ends ispartially filled with heavy liquid. The densityof heavy liquid is 1000 kg/m3 . A liquid ofdensity 611 kg/m3 is then poured into onearm of the tube, forming a column 8.8 cm inheight, as shown in the following diagram.

h

8.8 cm light liquid611 kg/m3

heavy liquid1000 kg/m3

What is the difference h in the heights ofthe two liquid surfaces?1. 2.9155 cm2. 3.008 cm3. 3.1059 cm4. 3.204 cm5. 3.315 cm6. 3.4232 cm correct

Explanation:

Let : ` = 8.8 cm ,

` = 611 kg/m3 , and

h = 1000 kg/m3 .

Basic Concepts: gauge pressure, varia-tion of pressure with depth

Because the liquid in the U-tube is static,the pressure exerted by the heavy liquid col-umn of height ` h in the left branch of thetube must balance the pressure exerted bythe liquid of height h poured into the rightbranch. Therefore,

P0 + (` h) h g = P0 + ` ` g .Solving for h,

h = `

[1 `

h

]

= (8.8 cm)

[1 (611 kg/m

3)

(1000 kg/m3)

]

= 3.4232 cm .


Question 18


The cylindrical disk has mass 7 kg andouter radius 8 cm with a radial mass distribu-tion (which may not be uniform) so that its

moment of inertia is3

5m r2 .

The disk rolls (perpendicularly to the axis)without slipping in a cylindrical trough, seefigure below.


40cm

8 cm

Determine (for small displacements fromequilibrium) the period of harmonic oscilla-tion which the disk undergoes.1. 1.30074 s2. 1.3464 s3. 1.39055 s4. 1.43616 s correct5. 1.48096 s6. 1.53294 s

Explanation:

Basic Concepts: Let r be the radius ofthe disk and R be the radius of the cylin-drical trough. The disk is rolling withoutslipping. Choose the point of contact as ouraxis. Around this point, the rotational inertiaof the disk is, using parallel axis theorem,

I =3

5m r2 + m r2 =

8

3m r2 . (1)

Solution: Let the angle of rotation aroundthis instantaneous contact point be and theangle the center of the disk makes from thecenter of the trough to the vertical be .

R

Rr

r

S

S

S is the com-mon arc length

S = r ( + )S = R

Note: The dotted curve is ahypocycloid denoting the path of thecontact point at equilibrium as thedisk rolls back and forth in the cylin-drical trough.

The arc length along the disk (which rollsback and forth) must equal the arc lengthalong the cylindrical trough (both arc lengthsare labeled S in the figure).

Therefore, and are related by

r ( + ) = R , or

r = (R r) , sor

d

dt= (R r) d

dt,

from the rolling without slipping condition.Now the torque equation around the point ofcontact is

: m g r sin = I d2

dt2

Id2

dt2+ m g r = 0 ,

since sin . Substituting in place of ,we have

IR r

r

d2

dt2+ m g r = 0

d2

dt2+

m g r2

I (R r) = 0 .Substituting I from Eq. 1, we have

d2

dt2+

5

8

m g r2

(R r) m r2 = 0

d2

dt2+

5

8

g

(R r) = 0 . (2)


Equation 2 is the differential equation for sim-ple harmonic motion. The coefficient of is2. Therefore

=

5

8

g

(R r) , and

T = 2 pi

8

5

(R r)g

= 2 pi

8

5

(0.4 m) (0.08 m)(9.8 m/s2)

= 1.43616 s .

Alternative Solution: The kinetic en-ergy,

K =1

2m

(d x

dt

)2+

1

2I

(d

dt

)2

=1

2m (R r)2

(d

dt

)2

+3

10m r2

(d

dt

)2

=1

2m (R r)2

(d

dt

)2

+3

10m (R r)2

(d

dt

)2

=4

5m (R r)2

(d

dt

)2,

since Idisk =3

5m r2, v = r , and =

d

dt.

The potential energy is

U = m g h

= m g (R r) (1 cos ) 1

2m g (R r) 2 ,

since 1cos 12

2. Energy is conserved,

therefore E = K+U = constant, and we have

E =4

5m (R r)2

(d

dt

)2

+1

2m g (R r) 2

= constant , so

d E

dt=

8

5m (R r)2 d

dt

d2

dt2

+m g (R r) d dt

= 0 . Therefore,

d2

dt2+

5

8

g

(R r) = 0 . (3)

Equations 2 and 3 are the same equation forsimple harmonic motion.

Question 19


The figure below shows a complex wavepattern on a string moving towards a rigidhook at the wall on the right. After sometime, the wave is reflected from the wall.

v

Select the wave pattern for the reflected wave.

1.

2.

v

correct

3.

v

4.

v

5.

v

Explanation:

Consider the wave pattern image reflectedabout the rigid hook on the wall.


v

v

After the time it takes for the wave to bereflected from the wall, this image is the wavepattern traveling to the left along the string.

Note: Reflection about a point (hook) is thesame as reflection about the y-axis (wall) fol-lowed by reflection about the x-axis (string).The leading part of the wave must remain infront and the wave is flipped over.

This is the first wave pattern of four possi-ble wave patterns presented by this question.

Question 20


Consider a uniform ladder leaning againsta smooth wall and resting on a smooth floorat point P . There is a rope stretched horizon-tally, with one end tied to the bottom of theladder essentially at P and the other end tothe wall. The top of the ladder is at a heightis h up the wall and the base of the ladder isat a distance b from the wall.

The weight of the ladder is W1 . Jill, with a

weight W2, is one-fourth the way

(d =

`

4

)up

the ladder. The force which the wall exerts onthe ladder is F .

P

W1

W2

`

d

T

h

b

F

Note: Figure is not to scale.

The torque equation about P is given by

1.h

2W2 + h W1 = F b

2.b

2W2 + b W1 = F h

3.b

4W2 +

b

2W1 = F h correct

4. (W1 + W2)h

2= F b

5. (W1 + W2)b

2= F h

6.h

4W2 +

h

2W1 = F b

Explanation:

Pivot

F

T

Nf

W2

W1

Fx : T F = 0 , (1)Fy : Nf W2 W1 = 0 , and (2)

P: W2 d cos + W1

`

2cos (3)

F ` sin = 0 ,where d is the distance of the person from thebottom of the ladder. Therefore

2 F ` sin = 2 W2 d cos + W1 ` cos .

Since sin =h

`and cos =

b

`, the torque

equation about P is given by

b

4W2 +

b

2W1 = F h . (4)


Question 21


A flexible chain weighing 44.2 N hangs be-tween two hooks located at the same height.At each hook, the tangent to the chain makesan angle = 35.8 with the horizontal.

Find the magnitude of the force each hookexerts on the chain.1. 34.343 N2. 35.492 N3. 36.6252 N4. 37.7805 N correct5. 38.9907 N6. 40.3285 N

Explanation:

Basic Concept: In equilibrium,~F = 0

Solution: In equilibrium

~F = 0. By

symmetry each hook supports half the weightof the chain, so

Ty = Te sin =W

2

Therefore

Te =W

2 sin

=44.2 N

2 sin 35.8

= 37.7805 N

Question 22


Hint: For this part, make a free-body dia-gram for half the chain.

Find the tension in the chain at its mid-point.1. 29.7217 N2. 30.6424 N correct3. 31.5973 N4. 32.7227 N

5. 33.8495 N6. 35.108 N

Explanation:

At the midpoint of the chain, there is only ahorizontal component of the tension. Sincethe chain is in equilibrium, the tension at themidpoint must equal the horizontal compo-nent of the force of a hook.

Te cos = Tm

Therefore

Tm = Te cos

= (37.7805 N) cos(35.8)

= 30.6424 N

Question 23


A string of linear mass density =1.94 g/m is stretched by the weight of anadjustable mass m as shown on the picturebelow.

vibrator

= 1.94 g/mm18

2.28 m

Note: The wave pattern showabove is for illustrative pur-poses only.

Near the end of the string a vibrator is at-tached at a constant but unknown frequency;the length of the string which vibrates isL = 2.28 m. For some values of the massmn, the string resonates with the vibratorsfrequency and develops visible standing waveswith n 1 nodes (not counting the nodes aseach end) and n anti-nodes between the vi-brator and the pulley.

A lab student who performed this experi-ment recorded two consecutive resonances formn = 2.25 kg and for mn1 = 2.56 kg. Unfor-tunately, he forgot to record the actual node


numbers for the resonances but only wrotethat there were no resonances for any massesbetween 2.25 kg and 2.56 kg.

How many antinodes did the resonant wavehave for m = 2.25 kg ?1. 132. 143. 154. 16 correct5. 176. 18

Explanation:

A resonant standing wave with (n 1)nodes between two fixed ends of the string which are also nodes occupies length ofn half-wavelengths, hence the resonance oc-curs whenever

L =n n

2= n = 2 L

n.

The wavelength depends on the oscillatorsfrequency f and the speed

v =

F

=

m g

of the transverse waves on the string,

f = v .

Consequently, the resonance happens when

f2 L

n= f n = vn =

mn g

,

i. e., for the string-stretching mass

mn =1

n24 L2 f2

g.

Now consider the ratio of masses for twoconsecutive resonances

mn1mn

=n2

(n 1)2 ,

which does not depend on anything but n.Consequently,

n2

(n 1)2 =2.56 kg

2.25 kg= 1.13778 ,

n

n 1 =

1.13778 = 1.06667,

n 1 = 11.06667 1 = 15 .

Therefore, for the mass m16 = 2.25 kg , theresonant wave has n = 16 antinodes andn1 = 15 nodes (not counting the two nodesat each end), as seen in the figure below.

vibrator

= 1.94 g/m2.25 kg

2.28 m

Question 24


What is the frequency of the oscillator?1. 374.075 Hz correct2. 388.512 Hz3. 401.554 Hz4. 416.671 Hz5. 431.679 Hz6. 448.881 Hz

Explanation:

In the first part of the problem, we derived

f 2 Ln

= f n = vn =

mn g

.

Now that we know n, we use this formula toevaluate the frequency

f =n

2 L

mn g

= 374.075 Hz .

Question 25


Needing help, the secretary of theUnited States Department of Agricultureasked your teacher, If a chicken-and-a-half can lay an egg-and-a-half in a day-and-a-half, how many days will it take


two chickens to lay twenty-four eggs?

Please help your teacher select the correctanswer to the secretarys question.

1. Two chickens will lay twenty-four eggs intwenty-two days.

2. Two chickens will lay twenty-four eggs inten days.

3. Two chickens will lay twenty-four eggs ineighteen days. correct

4. Two chickens will lay twenty-four eggs intwenty-one days.

5. Two chickens will lay twenty-four eggs intwelve days.

6. Two chickens will lay twenty-four eggs intwenty days.

Explanation:

Basic Concept: The information givenin the question is the rate of egg productionin one instance and you must make this ratecompatible with another instance. The rateof egg production is constant. The number ofeggs per chicken per day is a constant.Solution: Since it takes a chicken-and-a-halfa day-and-a-half to lay an egg-and-a-half, itwill take one chicken one-and-a-half days tolay one egg. Therefore, to lay twenty-foureggs it will take two chickens eighteen days.

Alternative Method: Unit analysis isbasic to every physics problem and is centralto this problem. The rate of egg production isthe number of eggs produced per chicken perday. In the given instance the rate is

rate ={3/2 eggs}

{3/2 chickens} {3/2 days}=

2

3

eggs

chickens days . (1)

In the requested instance, the number ofchickens is (2 chickens) and the number ofeggs is (24 eggs). The number of days N isto be determined. Therefore in the requestedinstance, the rate is

rate ={24 eggs}

{2 chickens} {N} . (2)

The rate is constant, so equating the ratesEqs. (1) and (2), we have

2 eggs

3 chicken days =24 eggs

2 chickens NSolving for the number of days N , we have

N = 24 eggs2 chickens

3 chickens days2 eggs

= 18 days .

The correct answer: Two chickens will laytwenty-four eggs in eighteen days.

Note: The early chicken catches the worm.

Question 26


Assume: h = 12 m, L = 6 m, and = 60,and that the cross-sectional area at A is verylarge compared with that at B.Assume: y = 0 at B.

Measure the height from the top of theangled tube. The figure below shows a watertank with a valve at the bottom.


Valve

A

B12 m

6m

60

hmax

Figure: Not drawn to scale.

If this valve is opened, what is the max-imum height hmax attained by the waterstream coming out of the spigot on the rightside of the figure?


1. 4.5051 m2. 4.64933 m3. 4.79733 m4. 4.94634 m5. 5.10289 m correct6. 5.26413 m

Explanation:

Let us first compute the speed with whichthe water leaves the tank at B by applyingBernoullis equation between A and B. Sincethe cross-sectional area at A is much largerthan the one at B, we can neglect the speedof the water at A compared to that at B. Thepressure at both points is going to be equal tothe atmospheric pressure Patm. The equationis then

Patm + w g (h L sin ) = Patm + 12

w v2

B ,

and so

vB =

2 g (h L sin ) .Now the problem reduces to that of projec-tile motion, for which the maximum height isgiven by

hmax =v2B sin

2

2 g

= [h L sin ] sin2 = [(12 m) (6 m) sin(60)] sin2(60)= 5.10289 m .

Question 27


The system shown below is released fromrest and moves 94.7 cm in 1.51074 s.


3 kg

M

= 0

What is the value of the mass M? Assumeall surfaces are frictionless.1. 0.277538 kg correct2. 0.286754 kg3. 0.296683 kg4. 0.306103 kg5. 0.317439 kg6. 0.327494 kg

Explanation:

Given : m = 3 kg ,

y0 = 0 m ,

y = 94.7 cm = 0.947 m ,

t0 = 0 s ,

t = 1.51074 s , and

= 0 .

m M

a

T

N m g

a

T

M g

From kinematics, we have

y y0 = v0 + 12

a t2 =1

2a t2

a =2 y

t2

The only external force acting on the sys-tem is the weight M g suspended from therope, so from Newtons second law we have

M g = (M + m) a

M =m a

g a (3)

=m

2 y

t2

g 2 yt2

t2

t2

=2 m y

g t2 2 y=

2 (3 kg) (0.947 m)

(1.51074 s)2 (9.8 m/s2) 2 (0.947 m)= 0.277538 kg .


Question 28


The system is in equilibrium and the pulleysare frictionless and massless.


5 kg

7 kg

9 kg

1

2

3

T

Find the force T .1. 205.8 N2. 215.6 N3. 225.4 N4. 235.2 N5. 245 N correct6. 254.8 N

Explanation:

Let : m1 = 5 kg ,

m2 = 7 kg , and

m3 = 9 kg .

m1

m2

m3

1

2

3

T

T4

T3 T3 T3

T2

T1T1

The mass m1 defines the tension T1:

T1 = m1 g .

At pulley 3, T1 acts down on either side of thepulley and T2 acts up, so

T2 = 2 T1 = 2 m1 g .

At the mass m2, T3 acts up, and m2 g and T2act down, so

T3 = m2 g + T2 = m2 g + 2 m1 g .

At pulley 2, T3 acts up on either side of thepulley and T4 acts down, so

T4 = 2 T3

= 2 m2 g + 4 m1 g .

At the mass m3,

T4 = T + m3 g

T = T4 m3 g= (2 m2 + 4 m1 m3) g= [2 (7 kg) + 4 (5 kg) (9 kg)] (9.8 m/s2)= 245 N .

Question 29


An object of mass m is moving with speedv0 to the right on a horizontal frictionlesssurface, as shown, when it explodes into two

pieces. Subsequently, one piece of mass10

13m

moves with a speed v10/13 =v05

to the left.

m v0

before

10

13m

v05

3

13m v3/13

after

What is the speed ~v3/13 of the other pieceof the object?

1. ~v3/13 = 5 v0. correct

2. ~v3/13 =39

10v0.

3. ~v3/13 =13

3v0.


4. ~v3/13 =13

10v0.

5. ~v3/13 =23

3v0.

6. ~v3/13 = 3 v0.Explanation:

The horizontal component of the momen-tum is conserved, so

0 + m v0 =10

13m v10/13 +

3

13m v3/13

0 + m v0 =10

13m

(v0

5

)+

3

13m v3/13

m v0 = 1065

m v0 +3

13m v3/13

3

13v3/13 =

(65

65+

10

65

)v0

3

13v3/13 =

75

65v0

v3/13 =75

65

13

3v0

~v3/13 = 5 v0 .

Question 30


A police car is traveling at a speed, vc, tothe right. A truck is traveling at a speed, vt,to the left. The frequency of the siren on thepolice car is fc.

The speed of sound in air is va. Let vt bethe speed of the observer in the truck, andvc be the speed of the source, the police car.

Police

vc vt

Truck

What is the frequency, ft, heard by anobserver in the moving truck?

1. ft =va vtva vc fc

2. ft =va vtva + vc

fc

3. ft =va + vtva vc fc correct

4. ft =va + vtva + vc

fc

Explanation:

Basic Concepts: The Doppler shifted fre-quency, f , heard in the truck is

f =va v0va vs f , (1)

where va is the speed of sound in air, vo is thespeel of the observer, and vs is the speed ofthe source,

The upper sign is used when the relativevelocities are toward one-another, and viceversa.

Solution: The relative velocity of the ob-server is towards the source so the upper signis used in the numerator ( +), and therelative velocity of the source is towards theobserver so the upper sign is used in the de-nominator ( ). Therefore Eq. 2 be-comes

ft =va + vtva vc fc .

This is version three of four versions.

Question 31


A police car is traveling at a speed, vc, tothe right. A truck is traveling at a speed, vt,to the right. A wind is blowing in the oppositedirection as that of the truck with a speed, vw,to the left. The frequency of the siren on thepolice car is fc.

The speed of sound in air is va.

Police

vc vt

Truck

vwwind

What is the frequency, ft, heard by anobserver in the moving truck?

1. ft =va vt vwva + vc + vw

fc


2. ft =va vt + vwva vc vw fc

3. ft =va + vt + vwva + vc + vw

fc

4. ft =va vt vwva vc vw fc correct

5. ft =va + vt vwva + vc + vw

fc

6. ft =va + vt + vwva vc vw fc

Explanation:

The problem must be worked in the frameof reference relative to the air. vt + vw isthe relative velocity of the truck (observer),vo. vc + vw is the relative velocity of the car(source), vs, therefore

ft =va (vt + vw)va (vc + vw) fc . (2)

The relative velocity of the observer is awayfrom the source so the lower sign is used inthe numerator ( ), and the relativevelocity of the source is towards the observerso the upper sign is used in the denominator( ). Therefore Eq. 2 becomes

ft =va (vt + vw)va (vc + vw) fc , so

=va vt vwva vc vw fc .

This is version four of eight versions.

Question 32


The pulley is massless and frictionless. Amassless inextensible string is attached tothese masses: 3 kg, 2 kg, and 6 kg.


1.8 m

21.1 cm

2 kg

3 kg

6 kg

T2

T1

T3

What is the tension T1 in the string be-tween the block with mass 3 kg and the blockwith mass 2 kg (on the left-hand side of thepulley)?1. 22.8667 N2. 24.1231 N3. 24.9455 N4. 26.1333 N5. 27.1385 N6. 32.0727 N correct

Explanation:

Let : R = 21.1 cm ,

m1 = 3 kg ,

m2 = 2 kg ,

m3 = 6 kg ,

h = 1.8 m ,

v = R ,

I =1

2M R2 , and

Kdisk =1

2I 2 =

1

4M v2 .

Consider the free body diagrams

3 kg 2 kg 6 kg

T1

T2

T3

m1g

T1

m2g

m3g

a a


Basic Concept : For each mass in thesystem

~Fnet = m~a .

Solution : Since the string changes direc-tion around the pulley, the forces due to thetensions T2 and T3 are in the same direction(up). The acceleration of the system will bedown to the right (m3 > m1 + m2), and eachmass in the system accelerates at the samerate (the string does not stretch). Let this ac-celeration rate be a and the tension over thepulley be T T2 = T3.

In free-body diagram for the lower left-handmass m1 the acceleration is up and

T1 m1 g = m1 a . (1)In free-body diagram for the upper left-handmass m2 the acceleration is up and

T T1 m2 g = m2 a . (2)In free-body diagram for the right-hand massm3 the acceleration is down and

T + m3 g = m3 a . (3)Adding Eqs. (1), (2), and (3), we have

(m3 m1 m2) g = (m1 + m2 + m3) a . (4)Therefore

a =m3 m1 m2m1 + m2 + m3

g (5)

=6 kg 3 kg 2 kg3 kg + 2 kg + 6 kg

g

=1 kg

11 kg(9.8 m/s2)

=1

11(9.8 m/s2)

= 0.890909 m/s2 .

The tension in the string between block m1and block m2 (on the left-hand side of thepulley) can be determined from Eq. (1).

T1 = m1

(1

11+ 1

)g (6)

= (3 kg)

(12

11

)(9.8 m/s2)

=

(36

11kg

)(9.8 m/s2)

= 32.0727 N .

Question 33


What is the magnitude of the accelerationof the block 3 kg?1. 0.890909 m/s2 correct2. 1.08889 m/s2

3. 1.4 m/s2

4. 1.63333 m/s2

5. 1.96 m/s2

6. 2.26154 m/s2

Explanation:

The acceleration is the same for every mass,since the string is inextensible. See Part 1, Eq.(5).

Question 34


All angles are measured in a counter-clockwise direction from the positive x-axis.

A hiker makes four straight-line walks (A,B, C, and D) in random directions andlengths starting at position (41 km, 41 km) ,listed below and shown below in the plot.

A 17 km at 46

B 33 km at 351

C 13 km at 306

D 19 km at 193

A

B

C

D

Figure: Drawn to scale.

How far from the starting point is the hikerafter these four legs of the hike?1. 31.322 km2. 32.3045 km3. 33.3462 km


4. 34.4094 km correct5. 35.4878 km6. 36.5875 km

Explanation:

A

B

C

e

DE

Scale: 10 km =

Note: ~E = 34.4094 km and e = 167.027 .

ax = (17 km) cos 46 = 11.8092 km ,

ay = (17 km) sin 46 = 12.2288 km ,

bx = (33 km) cos 351 = 32.5937 km ,

by = (33 km) sin 351 = 5.16225 km ,

cx = (13 km) cos 306 = 7.64123 km ,

cy = (13 km) sin 306 = 10.5172 km ,

dx = (19 km) cos 193 = 18.513 km ,

dy = (19 km) sin 193 = 4.2741 km ,

x0 = 41 km , starting point

y0 = 41 km , starting point

ax = (11.8092 km) + (41 km)

= 52.8092 km ,

ay = (12.2288 km) + (41 km)

= 53.2288 km ,

bx = (32.5937 km) + (52.8092 km)

= 85.4029 km ,

by = (5.16225 km) + (53.2288 km)= 48.0665 km ,

cx = (7.64123 km) + (85.4029 km)

= 93.0441 km ,

cy = (10.5172 km) + (48.0665 km)= 37.5493 km ,

dx = (18.513 km) + (93.0441 km)= 74.5311 km ,

dy = (4.2741 km) + (37.5493 km)= 33.2752 km ,

Therefore

E ={[(74.5311 km) (41 km)]2

+ [(33.2752 km) (41 km)]2}1/2= 34.4094 km ,

E = arctan

{[(33.2752 km) (41 km)][(74.5311 km) (41 km)]

}= 167.027 .

Question 35


A particle of mass m moves along the xaxis. Its position varies with time accordingto x = (6 m/s3) t3 + (5 m/s2) t2 .

What is the work done by the force fromt = 0 to t = t1 ?

1. W = 18m t21

[(16 m/s6) t21

+(8 m/s5) t1 + (1 m/s4)]

2. W = 2 m t21

[(144 m/s6) t21

+(120 m/s5) t1 + (25 m/s4)]

3. W = 2 m t21

[(81 m/s6) t21

+(90 m/s5) t1 + (25 m/s4)]correct

4. W =9

2m t21

[(25 m/s6) t21

+(40 m/s5) t1 + (16 m/s4)]

5. W =9

2m t21

[(81 m/s6) t21

+(36 m/s5) t1 + (4 m/s4)]

6. W = 2 m t21

[(36 m/s6) t21

+(60 m/s5) t1 + (25 m/s4)]

Explanation:

Let : x = a t3 + b t2 ,

a = 6 m/s3 , and

b = 5 m/s2 ,

Since the force is time dependent

W xf

xi

~F d~x


=

xfxi

m a dx

= m

xfxi

d v

dtdx

= m

xfxi

d v

dx

d x

dtdx

= m

vfvi

v dv

=1

2m v2f

1

2m v2i .

The velocity of the particle is

v =d x

dt

=d

dt

[a t3 + b t2

]=

[3 a t2 + 2 b t

]=

[3 (6 m/s3) t2 + 2 (5 m/s2) t

]= (18 m/s3) t2 + (10 m/s2) t .

Therefore work done on the particle is thechange in kinetic energy. For this case,

W = K = Kf Ki=

1

2m (v21 v20)

=1

2m v21

=1

2m

[3 a t21 + 2 b t1

]2=

1

2m

[9 a2 t41 + 6 a b t

3

1 + 4 b2 t21

]=

1

2m

[9 (6 m/s3)2 t41

+6 (6 m/s3) (5 m/s2) t31+4 (5 m/s2)2 t21

]= 2 m t21

[(81 m/s6) t21

+(90 m/s5) t1 + (25 m/s4)].

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