PHYSICS 1 FAP0015

Jan2011

Questions & Answers to Physics I Assignments Chapter 1 11. What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is

m? 09.086.2 ±=r Answer: To find the approximate uncertainty in the volume, calculate the volume for the specified

radius, the minimum radius, and the maximum radius. Subtract the extreme volumes. The uncertainty in the volume is then half this variation in volume.

( )

( )( )

( ) ( )

33 1 34 4specified specified3 3

33 1 34 4min min3 3

33 1 34 4max max3 3

1 3 1 3 1 31 1max min2 2

2.86 m 9.80 10 m

2.77 m 8.903 10 m

2.95 m 10.754 10 m

10.754 10 m 8.903 10 m 0.926 10 m

V r

V r

V r

V V V

π π

π π

π π

= = = ×

= = = ×

= = = ×

Δ = − = × − × = ×

The percent uncertainty is 1 3

1 3specified

0.926 10 m100 9.418 9 %

9.80 10 mV

VΔ ×

= × = ≈×

19. Express the following sum with the correct number of significant figures: m. 105.34cm 142.5m 80.1 5 μ×++

Answer: To add values with significant figures, adjust all values so that their units are all the same.

51.80 m 142.5 cm 5.34 10 m 1.80 m 1.425 m 0.534 m 3.759 m 3.76 mμ+ + × = + + = =

When adding, the final result is to be no more accurate than the least accurate number used. In this case, that is the first measurement, which is accurate to the hundredths place.

27. Estimate how long it would take one person to mow a football field using an ordinary home lawn mower. Assume the mower moves with a hkm 1 speed, and has a 0.5 m width.

Answer: The dimension of a football field (FIFA specification) is 64 meters by 100 meters. The mower has a cutting width of 0.5 meters. Thus the distance the mower has to move is

m 0.5m 6400

mower ofWidth field of Area 2

==d = 12800 m = 12.8 km

At a speed of 1 km/hr, then it will take about 12.8 h to mow the field.

51. The diameter of the Moon is 3480 km. What is the volume of the Moon? How many Moons would be needed to create a volume equal to that of Earth?

Answer: The volume of a sphere is found by V = 4/3 πr3.

( )33 6 19 34 4

Moon Moon3 3 1.74 10 m 2.21 10 mV Rπ π= = × = ×

3 33 64EarthEarth 3 Earth3 64

Moon Moon Moon3

6.38 10 m49.3

1.74 10 mRV R

V R Rππ

×= = = =

×

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ . Thus it would take about 49.3 Moons to create a volume equal to that of the Earth.

PHYSICS 1 FAP0015

Jan2011

Chapter 2 13. An airplane travels 3100 km at a speed of 790 km/h and then encounters a tailwind that boosts its speed to

990 km/h for the next 2800 km. What was the total time for the trip? What was the average speed of the plane for this trip?

Answer: The average speed for each segment of the trip is given by d

vt

=Δ

, so d

tv

Δ = for each segment.

For the first segment, 11

1

3100 km3.924 h

790 km hd

tv

Δ = = = .

For the second segment, 22

2

2800 km2.828 h

990 km h

dt

vΔ = = = .

Thus the total time is tot 1 2 3.924 h 2.828 h 6.752 h 6.8 ht t tΔ = Δ + Δ = + = ≈ .

The average speed of the plane for the entire trip is 2tot

tot

3100 km 2800 km873.8 8.7 10 km h

6.752 h

dv

t

+= = = ≈ ×

Δ.

19. A sports car moving at constant speed travels 110 m in 5.0 s. If it then brakes and comes to a stop in 4.0 s, what is its acceleration in m/s2? Express the answer in terms of “g’s,” where 1.0 g = 9.80 m/s2.

Answer: The initial velocity of the car is the average speed of the car before it accelerates.

0

110 m22 m s

5.0 sd

v vt

= = = =Δ

The final speed is v = 0, and the time to stop is 4.0 s. Use Eq. 2-11a to find the acceleration.

( )0

2 202

0 22 m s 1 5.5 m s 5.5 m s 0.56 's

4.0 s 9.80 m s

v v at

v v ga g

t

= + →

− −= = = − = − = −

⎛ ⎞⎜ ⎟⎝ ⎠

37. A ballplayer catches a ball 3.0 s after throwing it vertically upward. With what speed did he throw it, and what height did it reach?

Choose upward to be the positive direction, and take y0 = 0 to be the height from which the ball was thrown. The acceleration is a = −9.81 m/s2. The displacement upon catching the ball is 0, assuming it was caught at the same height from which it was thrown. The starting speed can be found by:

021 2

0 =+= attvy ( )( )0381921

21

2210 2

0 ..atat

v −−=−=−

= = 14.7 m/s ≈ 15 m/s

The height can be calculated from Eq. 2-11c, with a final velocity of v = 0 at the top of the path.

0222 =+= sauv ( )( )8192

7142

22

..

aus

−−

=−

= = 11.0 m

45. A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard 3.2 s later. If the speed of sound is s,m 340 how high is the cliff? Answer: For the falling rock, choose downward to be the positive direction, and y0 =0 to be the height from which the stone is dropped. The initial velocity is v0 = 0 m/s, the acceleration is a = g, the displacement is y = H, and the time of fall is t1. Using Eq. 2-11b, we have H = v0t + ½ at2 = ½ at2. For the sound wave, use the constant speed equation that vs = d/Δt = H/(T− t1) , which can be rearranged to give t1 = T − H/vs , where T = 3.2 s is the total time elapsed from dropping the rock to hearing the sound. Insert this expression for t1 into the equation for H, and solve for H.

2

2 21 12 22

5 2 4

1 0 2

4.239 10 1.092 50.18 0 46.0 m, 2.57 10 ms s s

H g gTH g T H H gT

v v v

H H H−

= − → − + + = →

× − + = → = ×

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

If the larger answer is used in t1 = T− H/vs , a negative time of fall results, and so the physically correct answer is H = 46 m.

PHYSICS 1 FAP0015

Jan2011

Chapter 3 9. An airplane is traveling 735 km/h in a direction 41.5º west of north. (a) Find the components of the

velocity vector in the northerly and westerly directions. (b) How far north and how far west has the plane traveled after 3.00 h?

Answer: (a) vnorth = (735 km/h)(cos 41.5°) = 550 km/h

vwest= (735 km/h)(sin 41.5°) = 487 km/h

(b) Δdnorth = vnorth t= (550 km/h)(3.00 h) = 1650 km Δdwest = vwest t= (487 km/h)(3.00 h) = 1460 km

21. A ball is thrown horizontally from the roof of a building 45.0 m tall and lands 24.0 m from the base. What was the ball’s initial speed? Answer: Choose downward to be the positive y direction. The origin will be at the point where the ball is thrown from the roof of the building. In the vertical direction, v0y = 0, ay = 9.81 m/s2, y0 = 0, and the displacement is 45.0 m. The time of flight is found from applying Eq. 2-11b to the vertical motion.

( ) ( )2 2 21 10 0 2 2 2

2 45.0 m 45.0 m 9.80 m s 3.03 sec

9.80 m sy yy y v t a t t t= + + → = → = =

The horizontal speed (which is the initial speed) is found from the horizontal motion at constant velocity:

24.0 m 3.03 s 7.92 m sx xx v t v x tΔ = → = Δ = =.

27. The pilot of an airplane traveling 180 km/h wants to drop supplies to flood victims isolated on a patch of land 160 m below. The supplies should be dropped how many seconds before the plane is directly overhead? Choose downward to be the positive y direction. The origin is the point where the supplies are dropped. In the vertical direction, vy0 = 0, ay = 9.81 m/s2, y0 = 0, and the final position is y = 160 m. The time of flight is found from applying Eq. 2-11b to the vertical motion.

( )( )

2 2 21 10 0 2 2

2

160 m 0 0 9.80 m s

2 160 m5.71 s

9.80 m s

y yy y v t a t t

t

= + + → = + + →

= =

Note that the speed of the airplane does not enter into this calculation. 47. A swimmer is capable of swimming 0.45 m/s in still water. (a) If she aims her body directly across a 75-m-

wide river whose current is 0.40 m/s how far downstream (from a point opposite her starting point) will she land? (b) How long will it take her to reach the other side? Answer: Call the direction of the flow of the river the x direction, and the direction straight across the river the y direction. Call the location of the swimmer’s starting point the origin.

( ) ( )

( )

swimmer swimmer water rel.rel. shore rel. water shore

0, 0.45 m s 0.40 m s , 0

0.40, 0.45 m s

= + = +

=

v v vr r r

(a) Since the swimmer starts from the origin, the distances covered in the x and y directions will be exactly proportional to the speeds in those directions.

0.40 m s 67 m

75 m 0.45 m sx x

y y

v t vx xx

y v t vΔ Δ

= = → = → Δ =Δ

swimmerrel. water

vr

water rel.shore

vr

swimmerrel. shore

vrθ

PHYSICS 1 FAP0015

Jan2011

Chapter 4 13. An elevator (mass 4850 kg) is to be designed so that the maximum acceleration is 0.0680g. What are the

maximum and minimum forces the motor should exert on the supporting cable? Answer: In both cases, a free-body diagram for the elevator would look like the adjacent diagram. Choose up to be the positive direction. To find the MAXIMUM tension, assume that the acceleration is up. Write Newton’s 2nd law for the elevator.

T F ma F mg= = − →∑

( ) ( ) ( )( )( )2

T 0.0680 4850 kg 1.0680 9.80 m sF ma mg m a g m g g= + = + = + =

4 5.08 10 N= × To find the MINIMUM tension, assume that the acceleration is down. Then Newton’s 2nd law for the

elevator becomes

( ) ( )

( )( ) ( )T T

2 4

0.0680

4850 kg 0.9320 9.80 m s 4.43 10 N

F ma F mg F ma mg m a g m g g= = − → = + = + = − +

= = ×

∑

25. One 3.2-kg paint bucket is hanging by a massless cord from another 3.2-kg paint bucket, also hanging by a

massless cord, as shown in Fig. 4–44. (a) If the buckets are at rest, what is the tension in each cord? (b) If the two buckets are pulled upward with an acceleration of 1.60 m/s2 by the upper cord, calculate the tension in each cord.

Answer: We draw free-body diagrams for each bucket. (a) Since the buckets are at rest, their acceleration is 0. Write Newton’s 2nd law for each bucket, calling

UP the positive direction.

( )( )

( ) ( )

1 T1

2T1

2 T2 T1

2T2 T1

0

3.2 kg 9.8 m s 31 N

0

2 2 3.2 kg 9.8 m s 63 N

F F mg

F mg

F F F mg

F F mg mg

= − = →

= = =

= − − = →

= + = = =

∑

∑

(b) Now repeat the analysis, but with a non-zero acceleration. The free-body diagrams are unchanged.

( ) ( )1 T1

2 2T1

2 T2 T1 T2 T1 T1

3.2 kg 9.80 m s 1.60 m s 36 N

2 73 N

F F mg ma

F mg ma

F F F mg ma F F mg ma F

= − = →

= + = + =

= − − = → = + + = =

∑

∑ 41. A 15.0-kg box is released on a 32º incline and accelerates down the incline at 0.30 m/s2. Find the friction

force impeding its motion. What is the coefficient of kinetic friction? Answer: Start with a free-body diagram. Write Newton’s 2nd law for each

direction.

fr

N

sin

cos 0x x

y y

F mg F ma

F F mg ma

θ

θ

= − =

= − = =

∑∑

Notice that the sum in the y direction is 0, since there is no motion (and hence no acceleration) in the y direction. Solve for the force of friction.

( ) ( )( )fr

2 o 2fr

sin

sin 15.0 kg 9.80 m s sin 32 0.30 m s 73.40 N 73 Nx

x

mg F ma

F mg ma

θ

θ

− = →

= − = − = ≈⎡ ⎤⎣ ⎦ Now solve for the coefficient of kinetic friction. Note that the expression for the normal force comes from the y direction force equation above.

( ) ( )( )fr

fr N 2 o

73.40 Ncos 0.59

cos 15.0 kg 9.80 m s cos 32k k k

FF F mg

mgμ μ θ μ

θ= = → = = =

mgr

TFr

Top (# 2)

mgrT1Fr

T2Fr

Bottom (# 1)

mgr

T1Fr

y

x θ θ

mgr

NFr

frFr

PHYSICS 1 FAP0015

Jan2011

59. A coffee cup on the dashboard of a car slides forward on the dash when the driver decelerates from

45 km/h to rest in 3.5 s or less, but not if he decelerates in a longer time. What is the coefficient of static friction between the cup and the dash? Answer: The free-body diagram for the coffee cup is shown. Assume that the car is moving to the right, and so the acceleration of the car (and cup) will be to the left. The deceleration of the cup is caused by friction between the cup and the dashboard. For the cup to not slide on the dash, and to have the minimum deceleration time means the largest possible static frictional force is acting, so Ffr = μsFN. The normal force on the cup is equal to its weight, since there is no vertical acceleration. The horizontal acceleration of the cup is found from Eq. 2-11a, with a final velocity of zero.

( )0

200

1m s45 km h 12.5 m s

3.6 km h

0 12.5 m s 3.57 m s

3.5 s

v

v vv v at a

t

= =

− −− = → = = = −

⎛ ⎞⎜ ⎟⎝ ⎠

Write Newton’s 2nd law for the horizontal forces, considering to the right to be positive.

( )2

fr N 2

3.57 m s 0.36

9.80 m sx s s s

aF F ma ma F mg

gμ μ μ

−= − = → = − = − → = − = − =∑

mgr

NFr

frFr

PHYSICS 1 FAP0015

Jan2011

Chapter 5 5. Suppose the space shuttle is in orbit 400 km from the Earth’s surface, and circles the Earth about once

every 90 minutes. Find the centripetal acceleration of the space shuttle in its orbit. Express your answer in terms of g, the gravitational acceleration at the Earth’s surface.

Answer: The orbit radius will be the sum of the Earth’s radius plus the 400 km orbit height. The orbital period is about 90 minutes. Find the centripetal acceleration from these data.

r = 6340 km + 400 km = 6780 km = 6.78 × 106 m, T = 90 min (60 s/min) = 5400 s

( )( )

( ) ⎟⎠⎞

⎜⎝⎛=

×== 2

22

62

2

2

m/s 8191m/s 189

s 5400m1078644

.g..

TraR

ππ= 0.937 ≈ 0.9 g’s

Notice how close this is to g, because the shuttle is not very far above the surface of the Earth, relative to the radius of the Earth.

15. How many revolutions per minute would a 15-m-diameter Ferris wheel need to make for the passengers to

feel “weightless” at the topmost point? Answer: The free-body diagram for passengers at the top of a Ferris wheel is as shown. FN is the normal force of the seat pushing up on the passenger. The sum of the forces on the passenger is producing the centripetal motion, and so must be a centripetal force. Call the downward direction positive. Newton’s 2nd law for the passenger is:

2

R NF mg F ma m v r= − = =∑ Since the passenger is to feel “weightless”, they must lose contact with their seat, and so the normal force

will be 0.

mg = mv2/r → v = gr = ( )( )m 57m/s 819 2 .. = 8.6 m/s

( ) ( )( )( )s/min 60m 7.5rev/2 1m/s68 π. = 11 rpm 37. A typical white-dwarf star, which once was an average star like our Sun but is now in the last stage of its

evolution, is the size of our Moon but has the mass of our Sun. What is the surface gravity on this star? Answer: The acceleration due to gravity at any location at or above the surface of a star is given by

gstar = GMstar/r2, where r is the distance from the center of the star to the location in question.

( ) ( )( )

30

11 2 2 7 2sunstar 22 6

Moon

1.99 10 kg6.67 10 N m kg 4.38 10 m s

1.74 10 m

Mg G

R−

×= = × = ×

×

73. A jet pilot takes his aircraft in a vertical loop. (a) If the jet is moving at a speed of 1300 km/h at the lowest point of the loop, determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 6.0 g’s. (b) Calculate the 78-kg pilot’s effective weight (the force with which the seat pushes up on him) at the bottom of the circle, and (c) at the top of the circle (assume the same speed). (a) See the free-body diagram for the pilot in the jet at the bottom of the loop. For aR = v2/r = 6g,

( )

( )

2

22 3

2

1m s1300 km h

3.6 km h6.0 2.2 10 m

6.0 6.0 9.8m sv

v r g rg

= → = = = ×

⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

(b) The net force must be centripetal, to make the pilot go in a circle. Write Newton’s 2nd law for the vertical direction, with up as positive. The normal force is the apparent weight.

2R NF F mg m v r= − =∑

The centripetal acceleration is to be v2/r = 6.0 g,

( )( )2 2 3

N 7 7 78 kg 9.80 m s 5350 N 5.4 10 NF mg m v r mg= + = = = = ×

(c) See the free-body diagram for the pilot at the top of the loop. Notice that the normal force is down, because the pilot is upside down. Write Newton’s 2nd law in the vertical direction, with down as positive.

2 3R N N6 5 3.8 10 NF F mg mv r mg F mg= + = = → = = ×∑

mgr

NFr

mgrNFr

mgrNFr

PHYSICS 1 FAP0015

Jan2011

Chapter 6 13. A spring has k = 88 N/m. Use a graph to determine the work needed to stretch it from x = 3.8 m to

x = 5.8 m where x is the displacement from its unstretched length. Answer: The force exerted to stretch a spring is given by Fstretch =kx (the opposite of the force exerted by the spring, which is given by F = −kx. A graph of Fstretch vs. x will be a straight line of slope k thorough the origin. The stretch from x1 to x2, as shown on the graph, outlines a trapezoidal area. This area represents the work, and is calculated by

( )( ) ( ) ( )

( )( )( )

1 11 2 2 1 1 2 2 12 2

212 88 N m 0.096 m 0.020 m 8.4 10 J .

W kx kx x x k x x x x−

= + − = + −

= = × 25. A 285-kg load is lifted 22.0 m vertically with an acceleration a = 0.160 g by a single cable. Determine

(a) the tension in the cable, (b) the net work done on the load, (c) the work done by the cable on the load, (d) the work done by gravity on the load, and (e) the final speed of the load assuming it started from rest.

Answer: (a) From the free-body diagram for the load being lifted, write Newton’s 2nd law for the vertical direction, with up being positive.

( )( )T

2 3T

0.160

1.16 1.16 285 kg 9.80 m s 3.24 10 N

F F mg ma mg

F mg

= − = = →

= = = ×

∑

(b) The net work done on the load is found from the net force.

( ) ( )( )( )o 2net net

3

cos 0 0.160 0.160 285 kg 9.80 m s 22.0 m

9.83 10 J

W F d mg d= = =

= × (c) The work done by the cable on the load is

( ) ( )( )( )o 2 4cable T cos 0 1.160 1.16 285 kg 9.80 m s 22.0 m 7.13 10 JW F d mg d= = = = ×

(d) The work done by gravity on the load is

( )( )( )o 2 4G cos180 285 kg 9.80 m s 22.0 m 6.14 10 JW mgd mgd= = − = − = − ×

(e) Use the work-energy theory to find the final speed, with an initial speed of 0.

( )

2 21 1net 2 1 2 12 2

32

2 1

2 9.83 10 J20 8.31m s

285 kgnet

W KE KE mv mv

Wv v

m

= − = − →

×= + = + =

29. A 1200-kg car rolling on a horizontal surface has speed v = 65 km/h when it strikes a horizontal coiled

spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring? Answer: Assume that all of the kinetic energy of the car becomes PE of the compressed spring.

( ) ( )

( )

2

22 2 41 1

2 2 22

1m s1200 kg 65km h

3.6 km h 8.1 10 N m

2.2 m

mvmv kx k

x= → = = = ×

⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

kx

x1 x2

kx

F = kx

Stretch distance

Force

TFr

mgr

PHYSICS 1 FAP0015

Jan2011

49. A ski starts from rest and slides down a 22º incline 75 m long. (a) If the coefficient of friction is 0.090, what is the ski’s speed at the base of the incline? (b) If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level? Use energy methods.

Answer: (a) See the free-body diagram for the ski. Write Newton’s 2nd law for forces perpendicular to the direction of motion, noting that there is no acceleration perpendicular to the plane.

N N

fr N

cos cos

cosk k

F F mg F mg

F F mg

θ θ

μ μ θ⊥ = − → = →

= =∑

Now use conservation of energy, including the non-conservative friction force. Subscript 1 represents the ski at the top of the slope, and subscript 2 represents the ski at the bottom of the slope. The location of the ski at the bottom of the incline is the zero location for gravitational PE (y = 0). We have v1 = 0, y1 = d sinθ, and y2 = 0. Write the conservation of energy condition, and solve for the final speed. Note that

fr N cosk kF F mgμ μ θ= =

( ) ( )( )( )

2 21 1NC 2 1 2 1 NC 1 22 2

o 2 2 21 1 1fr 1 1 2 2 22 2 2

o o2

cos180 cos sin

2 sin cos 2 9.80 m s 75 m sin 22 0.090cos 22

20.69 m s 21m s

k

k

W KE PE mv mv mgy mgy W E E

F d mv mgy mv mgy mgd mgd mv

v gd

μ θ θ

θ μ θ 2

= Δ + Δ = − + − → + =

+ + = + → − + = →

= − = −

= ≈

(b) Now, on the level ground, Ff = μk mg, and there is no change in PE. Let us again use conservation of energy, including the non-conservative friction force, to relate position 2 with position 3. Subscript 3 represents the ski at the end of the travel on the level, having traveled a distance d3 on the level. We have v2 = 20.69 m/s, y2 = 0, v3 = 0, and y3 = 0.

( )( )( )

o 2 21 1NC 2 3 3 2 2 3 32 2

222 221

3 2 32

cos180

20.69 m s0 242.7 m 2.4 10 m

2 2 9.80 m s 0.090

f

kk

W E E F d mv mgy mv mgy

vmgd mv d

gμ

μ 2

+ = → + + = + →

− + = → = = = ≈ ×

d

θ mgr

NFrfrF

r

PHYSICS 1 FAP0015

Jan2011

Chapter 7 11. An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and

the remaining nucleus slows to 350 m/s. If the alpha particle has a mass of 4.0 u and the original nucleus has a mass of 222 u, what speed does the alpha particle have when it is emitted? Answer: Consider the motion in one dimension, with the positive direction being the direction of motion of the original nucleus. Let “A” represent the alpha particle, with a mass of 4 u, and “B” represents the new nucleus, with a mass of 218 u. Momentum conservation gives the following.

( )( ) ( )( ) ( )( )

initial final

3

222 u 420 m s 218 u 350 m s4.2 10 m s

4.0 u

A B A A B B

A B B BA

A

p p m m v m v m v

m m v m vv

m

′ ′= → + = + →

′+ − −′ = = = ×

Note that the masses do not have to be converted to kg, since all masses are in the same units, and a ratio of masses is what is significant.

23. A 0.450-kg ice puck, moving east with a speed of 3.00 m/s has a head-on collision with a 0.900-kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed and direction of each object after the collision? Answer: Let A represent the 0.450-kg puck, and let B represent the 0.900-kg puck. The initial direction of puck A is the positive direction. We have vA = 3.00 m/s, and vB = 0. Use Eq. 7-7 to obtain a relationship between the velocities.

( )A B A B B A A v v v v v v v′ ′ ′ ′− = − − → = + Substitute this relationship into the momentum conservation equation for the collision.

( )( )( )

( ) ( )

( )

A A B B A A B B A A A A B A A

A BA A

A B

B A A

0.450 kg3.00 m s 1.00 m s 1.00 m s west

1.350 kg

3.00 m s 1.00 m s 2.00 m s east

m v m v m v m v m v m v m v v

m mv v

m m

v v v

′ ′ ′ ′+ = + → = + + →

− −′ = = = − =+

′ ′= + = − =

35. A 920-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8 m before stopping. The police officer, knowing that the coefficient of kinetic friction between tires and road is 0.80, calculates the speed of the sports car at impact. What was that speed? Use conservation of momentum in one dimension. Call the direction of the sports car’s velocity the positive x direction. Let A represent the sports car, and B represent the SUV. We have vB = 0 and vA

′ = vB

′. Solve for vA.

( ) A B

initial final A A A B A A AA

0 m m

p p m v m m v v vm+′ ′= → + = + → =

The kinetic energy that the cars have immediately after the collision is lost due to work done by friction. The work done by friction can also be calculated using the definition of work. We assume the cars are on a level surface, so that the normal force is equal to the weight. The distance the cars slide forward is Δx. Equate the two expressions for the work done by friction, solve for vA

′, and use that to find vA.

( ) ( )

( )( ) ( )

21afterfr final initial A B A2collision

ofr fr A B

21A B A A B A2

0

cos180

2

k

k k

W KE KE m m v

W F x m m g x

m m v m m g x v g x

μ

μ μ

′= − = − +

= Δ = − + Δ

′ ′− + = − + Δ → = Δ

( )( )( )2A B A BA A

A A

920 kg 2300 kg2 2 0.80 9.8 m s 2.8 m

920 kg

23.191m s 23m s

k

m m m mv v g x

m mμ

+ + +′= = Δ =

= ≈

47. The distance between a carbon atom (mC = 12 u) and an oxygen atom (mO = 16 u) in the CO molecule is 1.13 × 10−10 m. How far from the carbon atom is the center of mass of the molecule? Answer: Choose the carbon atom as the origin of coordinates.

( )( ) ( )( )1011C C O O

CMC O

12 u 0 16 u 1.13 10 m6.5 10 m

12 u 16 um x m x

xm m

−

−+ ×+

= = = ×+ + from the C atom.

PHYSICS 1 FAP0015

Jan2011

Chapter 8 15. A centrifuge accelerates uniformly from rest to 15,000 rpm in 220 s. Through how many revolutions did it

turn in this time? Answer: The angular displacement can be found from the following uniform angular acceleration

relationship.

( ) ( )( )( ) 41 12 2 0 15000 rev min 220 s 1min 60s 2.8 10 revo tθ ω ω= + = + = ×

25. Two blocks, each of mass m, are attached to the ends of a massless rod which pivots as shown in

Fig. 8–40. Initially the rod is held in the horizontal position and then released. Calculate the magnitude and direction of the net torque on this system.

Answer: There is a counterclockwise torque due to the force of gravity on the left block, and a clockwise torque due to the force of gravity on the right block. Call clockwise the positive direction.

( )2 1 2 1 , clockwisemgL mgL mg L Lτ = − = −∑

29. A small 650-gram ball on the end of a thin, light rod is rotated in a horizontal circle of radius 1.2 m.

Calculate (a) the moment of inertia of the ball about the center of the circle, and (b) the torque needed to keep the ball rotating at constant angular velocity if air resistance exerts a force of 0.020 N on the ball. Ignore the rod’s moment of inertia and air resistance.

Answer: (a) The small ball can be treated as a particle for calculating its moment of inertia. I = MR2 = (0.650 kg)(1.2 m)2 = 0.94 kgm2.

(b) To keep a constant angular velocity, the net torque must be zero, and so the torque needed is the same magnitude as the torque caused by friction.

Στ = τapplied − τfr = 0 → τapplied = τfr = Ffr r = (0.020 N)(1.2 m) = 2.4 × 10−2 mN 49. Two masses, m1 = 18.0 kg and m2 = 26.5 kg are connected by a rope that hangs over a pulley. The pulley is

a uniform cylinder of radius 0.260 m and mass 7.50 kg. Initially, m1 is on the ground and m2 rests 3.00 m above the ground. If the system is now released, use conservation of energy to determine the speed of m2 just before it strikes the ground. Assume the pulley is frictionless. Answer: The only force doing work in this system is gravity, so mechanical energy will be conserved. The initial state of the system is the configuration with m1 on the ground and all objects at rest. The final state of the system has m2 just reaching the ground, and all objects in motion. Call the zero level of gravitational potential energy to be the ground level. Both masses will have the same speed since they are connected by the rope. Assuming that the rope does not slip on the pulley, the angular speed of the pulley is related to the speed of the masses by ω = v/R. All objects have an initial speed of 0.

i fE E=

2 2 2 2 2 21 1 1 1 1 11 2 1 1 2 2 1 2 1 1 2 22 2 2 2 2 2i i i i i f f f f fm v m v I m gy m gy m v m v I m gy m gyω ω+ + + + = + + + +

( )

22 2 21 1 1 1

2 1 2 12 2 2 2 2f

f f

vm gh m v m v MR m gh

R= + + +

⎛ ⎞⎜ ⎟⎝ ⎠

( )( )

( )( )( )( )( )

2

2 1

1 11 2 2 2

2 26.5 kg 18.0 kg 9.80 m s 3.00 m23.22 m s

26.5 kg 18.0 kg 7.50 kgf

m m ghv

m m M

−−= = =

+ + + +

m1

m2

h

R M

PHYSICS 1 FAP0015

Jan2011

Chapter 9 9. A 75-kg adult sits at one end of a 9.0-m-long board. His 25-kg child sits on the other end. (a) Where should

the pivot be placed so that the board is balanced, ignoring the board’s mass? (b) Find the pivot point if the board is uniform and has a mass of 15 kg. Answer: The pivot should be placed so that the net torque on the board is zero. We calculate torques about the pivot point, with counterclockwise torques as positive. The upward force PF at the pivot point is shown, but it exerts no torque about the pivot point. The mass of the board is mB, and the CG is at the middle of the board.

(a) Ignore the force mB g. ( )

( )( )

( )

0

25 kg9.0 m 2.25m 2.3 m from adult

25 kg 75 kg

Mgx mg L x

mx L

m M

τ = − − = →

= = = ≈+ +

∑

(b) Include the force mB g.

( ) ( )B 2 0Mgx mg L x m g L xτ = − − − − =∑ ( )

( )( )

( )( )B

B

2 25 kg 7.5kg9.0 m 2.54 m 2.5 m from adult

75kg 25kg 15kgm m

x LM m m

+ += = = ≈

+ + + + 20. A shop sign weighing 245 N is supported by a uniform 155-N beam as shown in Fig.

9–54. Find the tension in the guy wire and the horizontal and vertical forces exerted by the hinge on the beam. Answer: The beam is in equilibrium. Use the conditions of equilibrium to calculate the tension in the wire and the forces at the hinge. Calculate torques about the hinge, and take counterclockwise torques to be positive.

( )( )( ) ( )( )

( )( )

T 2 1 1 2 1

1121 1 2 12

T o2

2

sin 2 0

155 N 1.70 m 245 N 1.70 msin 1.35 m sin 35.0

708.0 N 7.08 10 N

F l m g l m gl

m gl m glF

l

τ θ

θ

= − − = →

++= =

= ≈ ×

∑

( )

( ) ( )

o 2H T H T

H T 1 2

oH 1 2 T

cos 0 cos 708 N cos35.0 579.99 N 5.80 10 N

sin 0

sin 155 N 245 N 708 N sin 35.0 6.092 N 6 N down

x x x

y y

y

F F F F F

F F F m g m g

F m g m g F

θ θ

θ

θ

= − = → = = = ≈ ×

= + − − = →

= + − = + − = − ≈ −

∑∑

43. A 15-cm-long tendon was found to stretch 3.7 mm by a force of 13.4 N. The tendon was approximately

round with an average diameter of 8.5 mm. Calculate the Young’s modulus of this tendon. Answer: The Young’s Modulus is the stress divided by the strain.

( ) ( )( ) ( )

2312 6 2

3 2

13.4 N 8.5 10 mStressYoung's Modulus 9.6 10 N m

Strain 3.7 10 m 15 10 mo

F AL L

π −

− −

× ×= = = = ×

Δ × ×

⎡ ⎤⎣ ⎦

51. (a) What is the minimum cross-sectional area required of a vertical steel cable from which is suspended a 320-kg

chandelier? Assume a safety factor of 7.0 (b) If the cable is 7.5 m long, how much does it elongate? Answer: (a) The area can be found from the ultimate tensile strength of the material.

Tensile StrengthSafety Factor

FA

=,

( ) ( )2 5 26 2

Safety Factor 7.0320 kg 9.8m s 4.4 10 m

Tensile Strength 500 10 N mA F −= = = ×

×⎛ ⎞⎜ ⎟⎝ ⎠

(b) The change in length can be found from the stress-strain relationship, equation (9-5).

( )( )( )( )( )

230

5 2 9 20

7.5 m 320 kg 9.8 m s 2.7 10 m

4.4 10 m 200 10 N mL FF L

E LA L AE

−

−

Δ= → Δ = = = ×

× ×

L/2 - x

mgrBm grMgr

PFr

x L x−L

1m gr

TFr

θHFr

1 2l

2 l

1 l

2m gr

PHYSICS 1 FAP0015

Jan2011

Chapter 10 19. An open-tube mercury manometer is used to measure the pressure in an oxygen tank. When the

atmospheric pressure is 1040 mbar, what is the absolute pressure (in Pa) in the tank if the height of the mercury in the open tube is (a) 28.0 cm higher, (b) 4.2 cm lower, than the mercury in the tube connected to the tank?

Answer: The pressure in the tank is atmospheric pressure plus the pressure difference due to the column of mercury, as given in Eq. 10-3c.

(a) P = P0 + ρgh = 1.04 bar + ρHggh = (1.04 bar)(1.00 × 105 N/m2/bar) + (13.6 × 103 kg/m3)(9.81 m/s2)(0.280 m) = 1.41 × 105 N/m2

(b) P = (1.04 bar)(1.00 × 105 N/m2/bar) + (13.6 × 103 kg/m3)(9.81 m/s2)(−0.042 m) = 9.84 × 104 N/m2

25. A spherical balloon has a radius of 7.35 m and is filled with helium. How large a cargo can it lift,

assuming that the skin and structure of the balloon have a mass of 930 kg? Neglect the buoyant force on the cargo volume itself.

Answer: The buoyant force of the balloon must equal the weight of the balloon plus the weight of the helium in the balloon plus the weight of the load. For calculating the weight of the helium, we assume it is at 0oC and 1 atm pressure. The buoyant force is the weight of the air displaced by the volume of the balloon.

Fbuoyant = ρair Vballoon g = mHe g + mballoon g + mcargo g

→ mcargo = ρair Vballoon − mHe − mballoon = ρair Vballoon − ρHe Vballoon − mballoon = (ρair − ρHe )Vballoon − mballoon

= (1.29 kg/m3 − 0.179 kg/m3) 4/3 π (7.35 m)3 − 930 kg = 920 kg = 9.0 × 103 N 43. If wind blows at 35 m/s over a house, what is the net force on the roof if its area is 240 m2 and is flat?

Answer: We assume that there is no appreciable height difference between the two sides of the roof. Then the net force on the roof due to the air is the difference in pressure on the two sides of the roof, times the area of the roof. The difference in pressure can be found from Bernoulli’s equation.

2 21 1inside inside inside outside outside outside2 2 P v gy P v gyρ ρ ρ ρ+ + = + + →

( )( ) ( )

2 air1inside outside air outside2

roof

22 3 2 51 1air air outside roof2 2

1.29 kg m 35 m s 240 m 1.9 10 N

FP P v

A

F v A

ρ

ρ

− = = →

= = = ×

77. A copper (Cu) weight is placed on top of a 0.50-kg block of wood (density = 0.60 × 103 kg/m3) floating in

water, as shown in Fig. 10–57. What is the mass of the copper if the top of the wood block is exactly at the water’s surface? Answer: The buoyant force on the block of wood must be equal to the combined weight of the wood and copper.

( )

( )

wood woodwood Cu wood water water wood Cu water

wood wood

3water

Cu wood 3wood

1000kg m1 0.50kg 1 0.33kg

600kg m

m mm m g V g g m m

m m

ρ ρ ρρ ρ

ρρ

+ = = → + = →

= − = − =⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

PHYSICS 1 FAP0015

Jan2011

Chapter 11 19. In Exercise 18 [A mass–spring system is in SHM in the horizontal direction. The mass is 0.25 kg, the

spring constant is 12 N/m and the amplitude is 15 cm.], (a) what is the speed of the mass if it is at x = 10 cm? (b) What is the magnitude of the force exerted by the spring on the mass?

Answer: (a) v = k m (A

2 − x2) = 12 N/m 0.25 kg [(0.15 m)2 − (0.10 m)2] = 0.77 m/s.

(b) |Fs| = kx = (12 N/m)(0.10 m) = 1.2 N. 45. The velocity of a vertically oscillating mass–spring system is given by v = (0.750 m/s) sin(4t). Determine

(a) the amplitude and (b) maximum acceleration of this oscillator. Answer: (a) Since v = ωA sin ωt = (0.750 m/s) sin (4t), ω = 4.00 rad/s and

A = v ω =

0.750 m/s 4.00 rad/s = 0.1875 m = 0.188 m.

(b) Since a = ω2 A cos ωt , amax = ω2 A = (4.00 rad/s)2 (0.1875 m) = 3.00 m/s2. 71. The range of sound frequencies audible to the human ear extends from about 20 Hz to 20 kHz. If the speed

of sound in air is 345 m s, what are the limits of this audible range, expressed in wavelengths?

Answer: λmin = v f =

345 m/s 20 × 103 Hz = 0.017 m = 1.7 cm.

λmax = 345 m/s 20 Hz =17 m.

96. A standing wave is formed in a stretched string that is 3.0 m long. What are the wavelengths of (a) the first

harmonic and (b) the third harmonic? (a) 6.0 m (b) 2.0 m

Answer: (a) L = λ1 2 , λ1 = 2L = 2(3.0 m) = 6.0 m.

(b) L = 1.5λ3, λ3 = L

1.5 = 3.0 m

1.5 = 2.0 m.

PHYSICS 1 FAP0015

Jan2011

Chapter 12 41. The intensity levels of two people holding a conversation are 60 dB and 70 dB, respectively. What is the

intensity of the combined sounds?

Answer: (a) β = 10 log I Io ,

I Io = 10β/10, so I = 10β/10 Io.

so I1 = 106.0 (10−12 W/m2) = 10−6 W/m2 and I2 = 107.0 (10−12 W/m2) = 10−5 W/m2. The intensity for the combined sound is then I = I1 + I2 = 10−5 W/m2 + 10−6 W/m2 = 1.1 × 10−5 W/m2.

45. A compact speaker puts out 100 W of sound power. (a) Neglecting losses to the air, at what distance

would the sound intensity be at the pain threshold? (b) Neglecting losses to the air, at what distance would the sound intensity be that of normal speech? Does your answer seem reasonable? Explain.

Answer: (a) I = P

4π R2 , R = P

4π I = 100 W

4π (1.00 W/m2) = 2.82 m.

(b) Normal speech is at 60 dB or I = 10−6 W/m2.

R = 100 W

4π (10−6 W/m2) = 2.82 × 103 m.

This is unreasonable . In reality, the distance is less due to absorption by air. 53. A 1000-Hz tone from a loudspeaker has an intensity level of 100 dB at a distance of 2.5 m. If the speaker

is assumed to be a point source, how far from the speaker will the sound have intensity levels (a) of 60 dB and (b) barely high enough to be heard?

Answer: (a) From Exercise 14.50, I2 I1 = 10Δβ/10 = 10−4.0 = 10−4.

Also I2 I1 =

R12

R22 , R2 =

I1 I2 R1 = 102 (2.5 m) = 2.5 × 102 m.

(b) The threshold of hearing is at 0 dB. I2 I1 = 10−10.0. R2 = 1010.0 R1 = 105.0 (2.5 m) = 2.5 × 105 m.

This number is a bit unrealistic, because we ignored loss of sound during propagation. 72. The frequency of an ambulance siren is 700 Hz. What are the frequencies heard by a stationary pedestrian

as the ambulance approaches and moves away from her at a speed of 90.0 km/h. (Assume that the air temperature is 20°C.) 755 Hz approaching, 652 Hz moving away

Answer: 90.0 km/h = 25 m/s.

Approaching: fo = v

v − vs fs =

343 m/s 343 m/s − 25 m/s × (700 Hz) = 755 Hz.

Moving away: fo = v

v + vs fs = 343 m/s

343 m/s + 25 m/s × (700 Hz) = 652 Hz.

PHYSICS 1 FAP0015

Jan2011

Chapter 13 6. In an alcohol-in-glass thermometer, the alcohol column has length 11.82 cm at 0.0°C and length 22.85 cm

at 100.0°C. What is the temperature if the column has length (a) 16.70 cm, and (b) 20.50 cm? Answer: Assume that the temperature and the length are linearly related. The change in temperature

per unit length change is as follows.

cm11.82 - cm 8522

C0.0C0100.

-.LT °°

=ΔΔ

= 9.066 °C/cm

Then the temperature corresponding to length L is T(L) = 0.0°C + (L − 11.82 cm)(9.066°C/cm). (a) T(16.70 cm) = 0.0°C + (16.70 cm− 11.82 cm)(9.066°C/cm) = 44.2°C (b) T(20.50 cm) = 0.0°C + (20.50 cm − 11.82 cm)(9.066°C/cm) = 78.7°C 7. A concrete highway is built of slabs 12 m long (20°C). How wide should the expansion cracks between the

slabs be (at 20°C) to prevent buckling if the range of temperature is −30°C to +50°C? Answer: When the concrete cools in the winter, it will contract, and there will be no danger of buckling.

Thus the low temperature in the winter is not a factor in the design of the highway. But when the concrete warms in the summer, it will expand. A crack must be left between the slabs equal to the increase in length of the concrete as it heats from 20oC to 50oC.

( )( )( )6 o o o 3

0 12 10 C 12 m 50 C 20 C 4.3 10 mL L Tα − −Δ = Δ = × − = ×

13. An ordinary glass is filled to the brim with 350.0 mL of water at 100.0°C. If the temperature decreased to

20.0°C, how much water could be added to the glass? Answer: The amount of water that can be added to the container is the final volume of the container

minus the final volume of the water. Also note that the original volumes of the water and the container are the same. We assume that the density of water is constant over the temperature change involved.

( ) ( ) ( )( )( ) ( )

2 22added 0 0 container H O container H O 0container H O

6 o 6 o o 27 10 C 210 10 C 350.0 mL 80.0C 5.12 mL

V V V V V V V V Tβ β

− −

= + Δ − + Δ = Δ − Δ = − Δ

= × − × − =

PHYSICS 1 FAP0015

Jan2011

Chapter 14 15. How long does it take a 750-W coffeepot to bring to a boil 0.75 L of water initially at 8.0°C? Assume that

the part of the pot which is heated with the water is made of 360 g of aluminum, and that no water boils away.

Answer: The heat must warm both the water and the pot to 100oC. The heat is also the power times the time.

Q = Pt = (mAlcAl + mwatercwater)ΔTwater

( ) ( )( ) ( )( )[ ]( )

W750C92CJ/kg 4186kg 750CJ/kg 900kg 360waterwaterwaterAlAl °°+°

=Δ+

=..

PTcmcm

t = 425 s = 7 min

25. A cube of ice is taken from the freezer at Cº5.8− and placed in a 95-g aluminum calorimeter filled with

310 g of water at room temperature of 20.0°C. The final situation is observed to be all water at 17.0°C. What was the mass of the ice cube?

Answer: The heat lost by the aluminum and 310 g of liquid water must be equal to the heat gained by

the ice in warming in the solid state, melting, and warming in the liquid state.

( ) ( ) ( ) ( )( ) ( )( ) ( )( )( )

( )( ) ( )( )

2 2 2 2Al Al Al eq H O H O H O eq ice ice melt ice fusion H O eq melt

o o o o3

ice o o 5 o o

0.095 kg 900 J kg C 3.0C 0.31 kg 4186J kg C 3.0C9.90 10 kg

2100J kg C 8.5C 3.3 10 J kg 4186 J kg C 17 C

i i im c T T m c T T m c T T L c T T

m −

− + − = − + + −

+= = ×

+ × +

⎡ ⎤⎣ ⎦

⎡ ⎤⎣ ⎦ 35. (a) How much power is radiated by a tungsten sphere (emissivity e = 0.35) of radius 22 cm at a temperature

of 25°C? (b) If the sphere is enclosed in a room whose walls are kept at −5°C, what is the net flow rate of energy out of the sphere?

Answer: (a) The power radiated is given by Eq. 14-5. The temperature of the tungsten is 273K + 25K = 298K.

( )( ) ( ) ( )2 44 8 2 40.35 5.67 10 W m K 4 0.22 m 298 K 95 W

Qe AT

tσ π−Δ

= = × =Δ

(b) The net flow rate of energy is given by Eq. 14-6. The temperature of the surroundings is 268 K.

( ) ( ) ( ) ( ) ( ) ( )2 4 44 4 8 2 41 1 0.35 5.67 10 W m K 4 0.22 m 298 K 268 K

33 W

Qe A T T

tσ π−Δ

= − = × −Δ

=

⎡ ⎤⎣ ⎦

41. A 100-W lightbulb generates 95 W of heat, which is dissipated through a glass bulb that has a radius of

3.0 cm and is 1.0 mm thick. What is the difference in temperature between the inner and outer surfaces of the glass?

Answer: This is an example of heat conduction, and the temperature difference can be calculated by Eq.

14-4.

( )( )( ) ( )

3o1 2

2o 2

95 W 1.0 10 m 10C

0.84 J s m C 4 3.0 10 m

T TQ PlP kA T

t l kA π

−

−

×−= = → Δ = = =

×