Phy i assign&answers_2011

PHYSICS 1 FAP0015

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
r = 2.86 ± 0.09 m?
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.

Vspecified = 4 π rspecified = 4 π ( 2.86 m ) = 9.80 × 101 m3
3 3
3 3

Vmin = 4 π rmin = 4 π ( 2.77 m ) = 8.903 × 101 m3
3 3
3 3

Vmax = 3 π rmax = 3 π ( 2.95 m ) = 10.754 × 101 m3
4 3 4 3

ΔV = 1
2 (Vmax − Vmin ) = 12 (10.754 × 101 m3 − 8.903 × 101 m3 ) = 0.926 × 101 m3
ΔV 0.926 × 101 m 3
The percent uncertainty is = × 100 = 9.418 ≈ 9 %
Vspecified 9.80 × 101 m 3

19. Express the following sum with the correct number of significant figures:
1.80 m + 142.5 cm + 5.34 × 10 5 μm.

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

1.80 m + 142.5 cm + 5.34 × 105 μ m = 1.80 m + 1.425 m + 0.534 m = 3.759 m = 3.76 m
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 1 km h 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

Area of field 6400 m 2
d= = = 12800 m = 12.8 km
Width of mower 0.5 m
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.

( )
3
VMoon = 3 π RMoon = 4 π 1.74 × 10 6 m
4 3
3
= 2.21 × 1019 m 3
3
π REarth ⎛ REarth ⎞ ⎛ 6.38 × 106 m ⎞
3 3
4
VEarth
= 4
3
=⎜ ⎟ =⎜ ⎟ = 49.3
VMoon 3
π RMoon ⎝ RMoon ⎠ ⎝ 1.74 × 106 m ⎠
3
.
Thus it would take about 49.3 Moons to create a volume equal to that of the Earth.

Jan2011

PHYSICS 1 FAP0015

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?
d d
Answer: The average speed for each segment of the trip is given by v = , so Δt = for each segment.
Δt v
d1 3100 km
For the first segment, Δt1 = = = 3.924 h .
v1 790 km h
d2 2800 km
For the second segment, Δt 2 = = = 2.828 h .
v2 990 km h
Thus the total time is Δt tot = Δt1 + Δt 2 = 3.924 h + 2.828 h = 6.752 h ≈ 6.8 h .
dtot 3100 km + 2800 km
The average speed of the plane for the entire trip is v = =
= 873.8 ≈ 8.7 ×10 km h .
2

Δttot 6.752 h
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.
d 110 m
v = = = 22 m s = v0
Δt 5.0 s
The final speed is v = 0, and the time to stop is 4.0 s. Use Eq. 2-11a to find the acceleration.
v = v0 + at →

) ⎛ 9.80 m s ⎞
v − v0 0 − 22 m s
a= = = −5.5 m s 2 = −5.5 m s 2 ⎜( 1g
⎟ = −0.56 g 's
⎝ ⎠
2
t 4.0 s
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:
1 2
1 0− at
= − at = − (− 9.81)(3.0) = 14.7 m/s ≈ 15 m/s
1 1
y = v 0 t + at 2 = 0 v0 = 2
2 2 2 2

The height can be calculated from Eq. 2-11c, with a final velocity of v = 0 at the top of the path.
− u 2 − (14.7 )
2
v 2 = u 2 + 2a s = 0 s= = = 11.0 m
2a 2(− 9.81)
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 340 m s, 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
⎛ H⎞ g ⎛ gT ⎞
H = 1 g ⎜T −
2 ⎟ → 2
H2 −⎜ + 1 ⎟ H + 1 gT 2 = 0 →
2
⎝ vs ⎠ 2v s ⎝ vs ⎠
−5
4.239 × 10 H − 1.092 H + 50.18 = 0 → H = 46.0 m, 2.57 × 10 4 m
2

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.

Jan2011

PHYSICS 1 FAP0015

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 ( 45.0 m )
y = y0 + v y 0 t + 1 a y t 2
2
→ 45.0 m = 1
2 ( 9.80 m s ) t
2 2
→ t=
9.80 m s 2
= 3.03 sec

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

Δx = vx t → v x = Δ x t = 24.0 m 3.03 s = 7.92 m s
.

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.
y = y0 + v y 0 t + 1 a y t 2
2
→ 160 m = 0 + 0 + 1
2 ( 9.80 m s ) t 2 2
→

2 (160 m )
t= = 5.71 s
9.80 m s 2
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? r
Answer: Call the direction of the flow of the river the x direction, and the v water rel.
shore
direction straight across the river the y direction. Call the location of the
swimmer’s starting point the origin.
r r r r
v swimmer = v swimmer + v water rel. = ( 0, 0.45 m s ) + ( 0.40 m s , 0 ) v swimmer
r
rel. water
rel. shore rel. water shore v swimmer
θ
= ( 0.40, 0.45 ) m s rel. shore

(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.
Δx vx t vx Δx 0.40 m s
= = → = → Δx = 67 m
Δy vyt vy 75 m 0.45 m s

Jan2011

PHYSICS 1 FAP0015

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. r
∑ F = ma = F T
− mg → FT

FT = ma + mg = m ( a + g ) = m ( 0.0680 g + g ) = ( 4850 kg )(1.0680 ) 9.80 m s 2 ( ) r
mg
= 5.08 × 10 N 4

To find the MINIMUM tension, assume that the acceleration is down. Then Newton’s 2nd law for the
elevator becomes
∑ F = ma = F T
− mg → FT = ma + mg = m ( a + g ) = m ( −0.0680 g + g )

( )
= ( 4850 kg )( 0.9320 ) 9.80 m s 2 = 4.43 × 10 4 N

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. r
r
∑
F1 = FT1 − mg = 0 → FT2 FT1

(
FT1 = mg = ( 3.2 kg ) 9.8 m s 2 = 31 N )
∑F 2
= FT2 − FT1 − mg = 0 → r
FT1
r
mg
r
mg
(
FT2 = FT1 + mg = 2 mg = 2 ( 3.2 kg ) 9.8 m s 2 = 63 N ) Bottom (# 1)
Top (# 2)
(b) Now repeat the analysis, but with a non-zero acceleration. The free-body diagrams are unchanged.
∑F = F
1 T1
− mg = ma →

(
FT1 = mg + ma = ( 3.2 kg ) 9.80 m s 2 + 1.60 m s 2 = 36 N )
∑F 2
= FT2 − FT1 − mg = ma → FT2 = FT1 + mg + ma = 2 FT1 = 73 N

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. r r
∑F x
= mg sin θ − Ffr = max Ffr FN
y
∑F y
= FN − mg cos θ = ma y = 0
x
Notice that the sum in the y direction is 0, since there is no motion θ r θ
(and hence no acceleration) in the y direction. Solve for the force of friction. mg
mg sin θ − Ffr = ma x →
Ffr = mg sin θ − ma x = (15.0 kg ) ⎡ 9.80 m s 2
⎣ ( )( sin 32 ) − 0.30 m s
o 2
⎤ = 73.40 N ≈ 73 N
⎦
Now solve for the coefficient of kinetic friction. Note that the expression for the normal force comes from
the y direction force equation above.
Ffr 73.40 N
Ffr = μ k FN = μ k mg cos θ → μ k = = = 0.59
mg cos θ (15.0 kg ) ( 9.80 m s2 )( cos 32 )
o

Jan2011

PHYSICS 1 FAP0015

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. r
⎛ 1m s ⎞ FN
v0 = ( 45 km h ) ⎜ ⎟ = 12.5 m s r
⎝ 3.6 km h ⎠ Ffr
v − v0 0 − 12.5 m s r
v − v0 = at → a = = = −3.57 m s 2 mg
t 3.5 s
Write Newton’s 2nd law for the horizontal forces, considering to the right to be positive.
a ( −3.57 m s ) = 0.36
2

∑ Fx = − Ffr = ma → ma = − μ s FN = − μ s mg → μ s = − g
=−
9.80 m s 2

Jan2011

PHYSICS 1 FAP0015

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

aR = =
(
4π 2 r 4π 2 6.78 × 10 6 m ) (
= 9.18 m/s 2 )⎛ 9.81 g
⎜
1 ⎞
⎟ = 0.937 ≈ 0.9 g’s
T2 (5400 s )2
⎝ m/s 2
⎠
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 r r
the passenger is: FN mg

∑F R
= mg − FN = ma = m v 2 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 = (9.81 m/s )(7.5 m) = 8.6 m/s
2

(8.6 m/s)(1 rev/2π (7.5 m ))(60 s/min ) = 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.
(1.99 × 10 kg ) =
30

( )
M sun −11
g star = G = 6.67 × 10 4.38 × 10 m s
2 2 7 2
N m kg
(1.74 × 10 m )
2 2
RMoon 6

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 r
⎡ ⎛ 1m s ⎞ ⎤ FN
⎢(1300 km h ) ⎜ 3.6 km h ⎟ ⎥
v r = 6.0 g → r =
2 v2
=⎣
⎝ ⎠ ⎦ = 2.2 × 103 m r
6.0 g (2
6.0 9.8 m s ) mg

(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.
∑F R
= FN − mg = m v 2 r
The centripetal acceleration is to be v2/r = 6.0 g,
(
FN = mg + m v 2 r = 7 mg = 7 ( 78 kg ) 9.80 m s 2 = 5350 N = 5.4 × 103 N )
(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.

∑F R
= FN + mg = m v 2 r = 6mg → FN = 5mg = 3.8 × 103 N
r r
FN mg

Jan2011

PHYSICS 1 FAP0015

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 F = kx
=kx (the opposite of the force exerted by the spring, which is given kx
by F = −kx. A graph of Fstretch vs. x will be a straight line of slope k Force
thorough the origin. The stretch from x1 to x2, as shown on the
graph, outlines a trapezoidal area. This area represents the work, kx
and is calculated by
W = 1
2 ( kx1 + kx2 )( x2 − x1 ) = 12 k ( x1 + x2 )( x2 − x1 ) x1 x2
( 88 N m )( 0.096 m )( 0.020 m ) = 8.4 × 10 J . −2 Stretch distance
= 1
2

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. r
FT
∑F = F T
− mg = ma = 0.160mg →

( )
FT = 1.16mg = 1.16 ( 285 kg ) 9.80 m s 2 = 3.24 × 103 N
(b) The net work done on the load is found from the net force. r
mg
Wnet = Fnet d cos 0 = ( 0.160mg ) d = 0.160 ( 285 kg ) 9.80 m s
o
( 2
) ( 22.0 m )
= 9.83 × 103 J
(c) The work done by the cable on the load is
Wcable = FT d cos 0 o = (1.160 mg ) d = 1.16 ( 285 kg ) 9.80 m s 2 ( ) ( 22.0 m ) = 7.13 × 10 4 J
(d) The work done by gravity on the load is
WG = mgd cos180 o = − mgd = − ( 285 kg ) 9.80 m s 2 ( ) ( 22.0 m ) = −6.14 × 10 4 J
(e) Use the work-energy theory to find the final speed, with an initial speed of 0.
Wnet = KE2 − KE1 = 1 mv2 − 1 mv12 →
2
2
2

v2 =
2Wnet
+v =
2 (
2 9.83 × 103 J ) + 0 = 8.31m s
1
m 285 kg

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
⎡ ⎛ 1m s ⎞ ⎤
(1200 kg ) ⎢( 65 km h ) ⎜ ⎟⎥
⎣ ⎝ 3.6 km h ⎠ ⎦ = 8.1 × 10 4 N m
2
mv
1
mv 2 = 1 kx 2 → k= 2 =
( 2.2 m )
2 2 2
x

Jan2011

PHYSICS 1 FAP0015

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 r d
perpendicular to the direction of motion, noting that there is no acceleration Ffr r
perpendicular to the plane. FN
∑F ⊥
= FN − mg cos θ → FN = mg cos θ → r
mg θ
Ffr = μ k FN = μ k mg cos 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
Ffr = μ k FN = μ k mg cos θ
WNC = ΔKE + ΔPE = 1 mv2 − 1 mv12 + mgy2 − mgy1 → WNC + E1 = E2
2
2
2

Ffr d cos180o + 1 mv12 + mgy1 = 1 mv2 + mgy2 → − μ k mgd cos θ + mgd sin θ = 1 mv2 →
2 2
2
2
2

(
v2 = 2 gd ( sin θ − μ k cos θ ) = 2 9.80 m s 2 ) ( 75 m ) ( sin 22 o
− 0.090 cos 22o )
= 20.69 m s ≈ 21m s

(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.
WNC + E2 = E3 → Ff d 3 cos180o + 1 mv2 + mgy2 = 1 mv32 + mgy3 →
2
2
2

( 20.69 m s )
2 2
v2
− μ k mgd 3 + 1 mv2 = 0 → d 3 =
2
= = 242.7 m ≈ 2.4 × 10 2 m
2 g μk 2 ( 9.80 m s 2 ) ( 0.090 )
2

Jan2011

PHYSICS 1 FAP0015

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.
pinitial = pfinal → ( mA + mB ) v = mAv′ + mB vB →
A
′
( mA + mB ) v − mB vB ( 222 u )( 420 m s ) − ( 218 u )( 350 m s )
′
v′ =
A
= = 4.2 × 103 m s
mA 4.0 u
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.
vA − vB = − ( vA − vB ) → vB = vA + vA
′ ′ ′ ′
Substitute this relationship into the momentum conservation equation for the collision.
mA vA + mB vB = mA vA + mB vB → mA vA = mA vA + mB ( vA + vA ) →
′ ′ ′ ′
( mA − mB ) −0.450 kg
′
vA = vA = ( 3.00 m s ) = −1.00 m s = 1.00 m s ( west )
( mA + mB ) 1.350 kg
vB = vA + vA = 3.00 m s − 1.00 m s = 2.00 m s ( east )
′ ′
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.
mA + mB
pinitial = pfinal → mA vA + 0 = ( mA + mB ) vA → vA =
′ ′
vA
mA
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.
Wfr = ( KEfinal − KEinitial )after = 0 − 1 ( mA + mB ) vA2
2
′
collision

Wfr = Ffr Δx cos180o = − μ k ( mA + mB ) g Δx
− 1 ( mA + mB ) vA2 = − μ k ( mA + mB ) g Δx → vA =
2
′ ′ 2 μ k g Δx
mA + mB mA + mB 920 kg + 2300 kg
vA =
mA
′
vA =
mA
2 μ k g Δx =
920 kg
(
2 ( 0.80 ) 9.8 m s 2 ) ( 2.8 m )
= 23.191m s ≈ 23 m s
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.
mC xC + mO xO (12 u )( 0 ) + (16 u ) (1.13 × 10−10 m )
xCM = = = 6.5 × 10 −11 m
mC + mO 12 u + 16 u
from the C atom.

Jan2011

PHYSICS 1 FAP0015

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.
θ= 1
2 (ωo + ω ) t = 12 ( 0 + 15000 rev min )( 220 s )(1min 60 s ) = 2.8 × 10 4 rev

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.
∑τ = mgL 2
− mgL1 = mg ( L2 − L1 ) , clockwise

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 M
reaching the ground, and all objects in motion. Call the zero level of gravitational R
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 m2
ω = v/R. All objects have an initial speed of 0.
Ei = E f h
m1

1
2
m1vi2 + 1 m2 vi2 + 1 I ωi2 + m1 gy1i + m2 gy2 i = 1 m1v 2 + 1 m2 v 2 + 1 I ω 2 + m1 gy1 f + m2 gy2 f
2 2 2 f 2 f 2 f

⎛ v2 ⎞
m2 gh = 1 m1v 2 + 1 m2 v 2 + 1
2 f 2 f 2 ( 1
2 )
MR 2 ⎜
f
⎟ + m1 gh
R2 ⎠
⎝

vf =
2 ( m2 − m1 ) gh
=
(
2 ( 26.5 kg − 18.0 kg ) 9.80 m s 2 ) ( 3.00 m ) = 3.22 m s
( m1 + m2 + 1
2
M) ( 26.5 kg + 18.0 kg + ( ) 7.50 kg )
1
2

Jan2011

PHYSICS 1 FAP0015

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 L
board is zero. We calculate torques about the pivot point, with
x L−x
counterclockwise torques as positive. The upward force F P at the r r r r
pivot point is shown, but it exerts no torque about the pivot point. The Mg mBg mg
mass of the board is mB, and the CG is at the middle of the board.
FP
(a) Ignore the force mB g.
∑
τ = Mgx − mg ( L − x ) = 0 → L/2 - x

m ( 25 kg )
x= L= ( 9.0 m ) = 2.25 m ≈ 2.3 m from adult
m+M ( 25 kg + 75 kg )
(b) Include the force mB g.
∑τ = Mgx − mg ( L − x ) − m g ( L 2 − x ) = 0
B

( m + mB 2 ) ( 25 kg + 7.5 kg )
x= L= ( 9.0 m ) = 2.54 m ≈ 2.5 m from adult
( M + m + mB ) ( 75 kg + 25 kg + 15 kg )
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.
r
∑τ = ( F T
sin θ ) l2 − m1 g l1 2 − m2 gl1 = 0 → r
FH
FT

1
m1 gl1 + m2 gl1 1
(155 N )(1.70 m ) + ( 245 N )(1.70 m ) θ
FT = 2
= 2

(1.35 m ) ( sin 35.0o )
r r
l2 sin θ l1 2 m1g m2 g

= 708.0 N ≈ 7.08 × 102 N l2
l1

∑F x
= FH x − FT cos θ = 0 → FH x = FT cos θ = ( 708 N ) cos 35.0o = 579.99 N ≈ 5.80 × 102 N

∑F y
= FH y + FT sin θ − m1 g − m2 g = 0 →

FH y = m1 g + m2 g − FT sin θ = 155 N + 245 N − ( 708 N ) sin 35.0o = −6.092 N ≈ −6 N ( down )
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.
(13.4 N ) ⎡π ( 12 × 8.5 × 10−3 m ) ⎤
2

Young's Modulus =
Stress
=
F A
= ⎣ ⎦ = 9.6 × 106 N m 2
Strain ΔL Lo ( 3.7 × 10 m ) (15 × 10 m )
−3 −2

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 Strength F A = F ⎛ Safety Factor ⎞ = ( 320 kg ) 9.8 m s2
= ⎜ Tensile Strength ⎟
7.0
500 ×106 N m2
( )
= 4.4 ×10−5 m2
Safety Factor A, ⎝ ⎠
(b) The change in length can be found from the stress-strain relationship, equation (9-5).
F ΔL L0 F ( 7.5 m )( 320 kg ) ( 9.8 m s2 )
=E → ΔL = = = 2.7 × 10−3 m
A L0 AE ( 4.4 ×10 −5
m 2
)( 200 ×10 9
N m 2
)

Jan2011

PHYSICS 1 FAP0015

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.
Pinside + 1 ρ vinside + ρ gyinside = Poutside + 1 ρ voutside + ρ gyoutside →
2
2
2
2

Fair
Pinside − Poutside = 1 ρ air voutside =
2
2
→
Aroof

Fair = 1 ρ air voutside Aroof =
2
2 1
2 (1.29 kg m3 ) ( 35 m s ) ( 240 m ) = 1.9 × 10 N
2 2 5

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.
mwood mwood
( mwood + mCu ) g = Vwood ρwater g = ρwater g → mwood + mCu = ρwater →
ρwood ρwood
⎛ ρwater ⎞ ⎛ 1000 kg m3 ⎞
mCu = mwood ⎜ − 1⎟ = ( 0.50 kg ) ⎜ − 1⎟ = 0.33kg
⎝ ρwood ⎠ ⎝ 600 kg m ⎠
3

Jan2011

PHYSICS 1 FAP0015

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?

k 12 N/m
Answer: (a) v = (A2 − x2) = [(0.15 m)2 − (0.10 m)2] = 0.77 m/s.
m 0.25 kg
(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
v 0.750 m/s
A= = = 0.1875 m = 0.188 m.
ω 4.00 rad/s
(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?
v 345 m/s
Answer: λmin = f = = 0.017 m = 1.7 cm.
20 × 103 Hz
345 m/s
λmax = 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
λ1
Answer: (a) L = 2 , λ1 = 2L = 2(3.0 m) = 6.0 m.
L 3.0 m
(b) L = 1.5λ3, λ3 = 1.5 = 1.5 = 2.0 m.

Jan2011

PHYSICS 1 FAP0015

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?
I I β/10 β/10
Answer: (a) β = 10 log I , Io = 10 , so I = 10 Io.
o
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.
P P 100 W
Answer: (a) I = , R= = = 2.82 m.
4π R2 4π I 4π (1.00 W/m2)
(b) Normal speech is at 60 dB or I = 10−6 W/m2.
100 W
R= = 2.82 × 103 m.
4π (10−6 W/m2)
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?
I2
Answer: (a) From Exercise 14.50, I = 10Δβ/10 = 10−4.0 = 10−4.
1
I2 R12 I1 2 2
Also I = R 2 , R2 = I2 R1 = 10 (2.5 m) = 2.5 × 10 m.
1 2
I2
(b) The threshold of hearing is at 0 dB. = 10−10.0. R2 = 1010.0 R1 = 105.0 (2.5 m) = 2.5 × 105 m.
I1
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.
v 343 m/s
Approaching: fo = f = × (700 Hz) = 755 Hz.
v − vs s 343 m/s − 25 m/s
v 343 m/s
Moving away: fo = v + v fs = × (700 Hz) = 652 Hz.
s 343 m/s + 25 m/s

Jan2011

PHYSICS 1 FAP0015

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.
ΔT 100.0°C-0.0°C
= = 9.066 °C/cm
ΔL 22.85 cm - 11.82 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.
( ) ( )
ΔL = α L0 ΔT = 12 × 10−6 Co (12 m ) 50o C − 20o C = 4.3 × 10−3 m

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 2
( )
Vadded = (V0 + ΔV )container − (V0 + ΔV ) H O = ΔVcontainer − ΔVH O = β container − β H O V0 ΔT
2

(
= 27 × 10 −6
C − 210 × 10
o −6
C ) ( 350.0 mL ) ( −80.0 C ) = 5.12 mL
o o

Jan2011

PHYSICS 1 FAP0015

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
(mAlcAl + mwatercwater )ΔTwater [(0.36 kg)(900 J/kg°C) + (0.75 kg)(4186 J/kg°C)](92°C) = 425 s = 7 min
t= =
P 750 W

25. A cube of ice is taken from the freezer at −8.5º C 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
)
mAl cAl Ti Al − Teq + mH O cH O Ti H O − Teq = mice ⎡cice (Tmelt − Ti ice ) + Lfusion + cH O (Teq − Tmelt ) ⎤
⎦ 2

( 0.095 kg ) ( 900 J )( 3.0 C ) + ( 0.31 kg ) ( 4186 J kg C )( 3.0 C ) = 9.90 ×10
kg C o o o o
−3
mice = kg
⎡( 2100 J kg C )( 8.5C ) + 3.3 × 10 J kg + ( 4186 J kg C )(17 C ) ⎤
⎣
o o 5
⎦
o o

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.
ΔQ
Δt
( )
= eσ AT 4 = ( 0.35 ) 5.67 × 10−8 W m2 K 4 4π ( 0.22 m ) ( 298 K ) = 95 W
2 4

(b) The net flow rate of energy is given by Eq. 14-6. The temperature of the surroundings is 268 K.
ΔQ
Δt
( ) ( 2
)
= eσ A T14 − T14 = ( 0.35) 5.67 × 10−8 W m 2 K 4 4π ( 0.22 m ) ⎡( 298 K ) − ( 268 K ) ⎤
4 4

⎣ ⎦
= 33 W

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.
Q T1 − T2 Pl ( 95 W ) (1.0 ×10−3 m )
= P = kA → ΔT = = = 10 Co
( 0.84 J s m C ) 4π ( 3.0 ×10 m ) −2 2
t l kA o

Jan2011

Phy i assign&answers_2011

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