# Nelson Physics 12 - Solutions Manual ```Section 1.1: Motion and Motion Graphs
Tutorial 1 Practice, page 10
1. (a) Given: !d1 = 1.2
!t1 =
i
!d2 = 1.2
!t2 =
in
Required: va)
Analysis: Calculate the total distance tra)elled, !d = !d1 + !d2 , and the total time
taken, !t = !t1 + !t2 . Then, use the equation va) =
!t1 =
min &quot;
!t1 =
h
!t2 =
h
!d
to calculate the a)erage speed.
!t
1h
/0 min
Solution: !d = !d1 + !d2
= 1.2 km + 1.2 km
!d =
km
!t = !t1 + !t2
=
h+
h
h
!t =
!d
va) =
!t
km
=
h
va) =
km h
Statement: The
r a)erage speed
(b) Given: !d = 1.2 km E !t =
!
Required: va)
min
!
!d
!
Analysis: The a)erage )elocity is the ratio of the displacement and the time taken, va) =
.
!t
1h
!t =
min &quot;
/0 min
!t =
h
!
!d
!
Solution: va) =
!t
1.2 km E
=
h
!
va) = 3.0 km E
Statement: The a)erage )elocity from the house to the farthest position from the house is
3.0
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Chapter 1: Kinematics
1.1-1
r
(c) Given: !d1 = 1.2 km E !t1 =
!
Required: va)
r
!d2 = 1.2 km
!t2 =
h
Analysis: Since the two displacements are equal in magnitude and opposite in direction, the total
Statement:
(d)
A)erage )elocity depends on displacement, which is a )ector quantity. Since the displacements
are not the same, the a)erage )elocities are not the same.
!
2. Given: v =
m s forwar !t = 0.32 s
!
Required: !d
!
r r r
!d
!
Analysis: Rearrange va) =
to sol)e for !d: !d = va) !t.
!t
r r
Solution: !d = va) !t
= ( m s 4forward5 (0.32 s
r
!d =
m 4forward5
Statement: The bus mo)es 8./ m 4forward5 before the dri)er reacts.
3. Given: !d = 200 laps, at
la !t =
h
Required: va) in
va) =
Analysis:
km
Solution: !d = 200 laps &quot;
!d =
va) =
1 lap
!d
.
!t
!d
!t
km
h
va) = 1.20 &quot; 102 km h
km
=
Statement: The a)erage speed of a dri)er who completes 200 laps in /./9 h is 1.20 ! 102 km h .
4. (a) Given: !d1 =
m !t1 = s !d2 =
m !t2 = 21 s
r
Required: va)
Analysis: Calculate the total distance walked, !d = !d1 + !d2 , and the time
taken, !t = !t1 + !t2 . Then, calculate the a)erage speed using va) =
!d
.
!t
Solution: !d = !d1 + !d2
va) =
=
!d =
m+
!t = !t1 + !t2
m
m
=
s + 21 s
!t = 8/ s
=
va) =
!d
!t
m
s
ms
Statement:
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Chapter 1: Kinematics
1.1-2
!
(b) Given: !d1 =
r
Required: va)
m 4E5 !t1 =
!
s !d2 =
m 4 5 !t2 = 21 s
!
!
!
Analysis: Calculate the total displacement, !d = !d1 + !d2 . Then, calculate the a)erage )elocity
!
!d
!
. Use east as positi)e.
using va) =
!t
!
!
!
!
!d
!
Solution: !d = !d1 + !d2
va) =
!t
=
m + (&quot; m
m 4E5
=
=+ m
s
!
!
!d = m 4E5
va) = 1.2 m s 4E5
Statement: The stud
!
!
5. (a) Given: !d1 = /2 km 4S5 !d2 = m 4N5 va) = km h
!
Required: va)
!d
, to determine the
Analysis: Use the total distance, !d = !d1 + !d2 , and a)erage speed, va) =
!t
!
!
!
time taken for the trip, !t . Then, determine the total displacement, !d = !d1 + !d2 . Then
!
!d
!
. Use north as positi)e.
calculate the a)erage )elocity using va) =
!t
!d
Solution: !d = !d1 + !d2
va) =
!t
= /2 km + km
!d
!t =
km
=
va)
!d =
km
km
=
km h
!t =
h (two eGtra digits carrie
!
!
!
!d = !d1 + !d2
= (&quot;62 km) + (+78 km)
= +16 km
!
!d = 16 m [N]
!
!d
!
vav =
!t
16 km [N]
=
2.545 h
!
vav = 6.3 km/h [N]
Statement:
(b) Answers may )ary. Sample answer: The truck turned around at one point during its trip. At
the end, it was only 1/ km away from where it started. As a result, the magnitude of the a)erage
)elocity is quite low compared to the a)erage speed.
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Chapter 1: Kinematics
1.1-3
Tutorial 2 Practice, page 13
1. (a) Velocity is slope on a position–
the )elocity is increasing.
(b)
(a) Initially, the )elocity is large and constant in the east direction. Then, it is lower and constant
in the east direction. Calculate the slope of each section of the position–time graph using data
from the graph.
t (s)
0.0
0.2
0.8
!
0.0
20
30
d (m [E])
!
!
___
r
&quot; 20
!d 20 &quot; 0.0
!d
v (m/h [E])
=
= 100
=
=
&quot; 0.2
!t 0.2 &quot; 0.0
!t
Use these )elocity data to draw a )elocity–time graph.
(b) Initially the )elocity is large and constant in a west direction, then it is zero and finally it is
low and constant in an east direction. Calculate the slope of each section of the position–time
graph using data from the graph.
0.0
1.0
2.0
t (h)
!
0.0
0.0
d ( [ )
!
!
!
___
!
&quot; 0.0
&quot;
!d
!d
!d 0.0 &quot;
v (m/h [W])
=
=
=
=0
=
=&quot;
!t
!t 1.0 &quot;
!t 2.0 &quot; 1.0
&quot; 0.0
Use these )elocity data to draw a )elocity–time graph.
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Chapter 1: Kinematics
1.1-
(c) Initially the )elocity is low and constant in a north direction, and then it is large and constant
in a south direction. Calculate the slope of each section of the displacement–time graph using
data from the graph.
0.0
8.0
t (h)
r
100
&quot;
d (m [/])
!
!
___
r
&quot;
!d (&quot;
!d 100 &quot; &quot;
v (m/h [/])
=
=&quot;
=
= 37.5
!t
!t
&quot; 0.0
&quot;
Use these )elocity data to draw a )elocity–time graph.
Tutorial 3 Practice, page 15
(a) The )elocity–time graph is a straight line showing that the car’s )elocity is changing at a
constant rate. So, the car is mo)ing with constant acceleration.
(b) The car starts from rest at t = 0 s. It mo)es north with increasing speed and constant
(c) Given: )elocity–time graph
!
Required: a
Analysis: Read the coordinates of two points on the graph. Use these points to calculate the
!
! !v
. As in the graph, use north as positi)e.
slope of the line, a =
!t
Solution:
!
! !v
a=
!t
12 m) &quot; 8)
=
&quot;
01 8)
=
=+
8)
1
!
8) 1 4
a=
Statement:
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2
4N5.
Chapter 1: Kinematics
1.1-
(a) Given: )elocity–time graph
!
Required: aa)
Analysis: Read the coordinates of the initial and final points on the graph. Use these points to
!
!v
!
. As in the graph, use forward as positi)e.
calculate the slope of the secant, aa) =
!t
Solution:
!
!v
!
a =
!t
35 8) &quot; 8)
=
&quot;
= + 8) 1
!
a = 8) 1 orward]
2
Statement:
4forward5.
(b) Given: )elocity–time graph
!
Required: a at t = 0 [ a6d a t = 2 [/
Analysis: The instantaneous acceleration at a gi)en time is the slope of the tangent to the
)elocity–time graph at that time. Sketch tangent lines at the required times, read the coordinates
of two points on each tangent line, and then calculate the slopes.
Solution:
For the tangent at 3 s, two points are
r
r !v
a=
!t
52
=
r
r !v
a=
!t
[&quot;
[
s [&quot; s [
=
[ &quot; 3/2 [
[ &quot; 5/2 [
s[
52 [
=
=
[
0/2 [
r
r
5
[ orward]
a=
a = s [ 5 orward]
2
2
Statement:
(c) The slope of the )elocity–time graph increases gradually, so the acceleration–time graph
shou
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Chapter 1: Kinematics
1.1-/
(a) Given: )elocity–time graph
!
Required: aa)
Analysis: Read the coordinates of initial and final points on the graph. Use these points to
!v
. As in the graph, use forward as positi)e.
calculate the slope of the secant, aa) =
!t
Solution:
!v
aa =
!t
s [&quot; s [
=
[&quot; [
&quot; s[
=
[
a = &quot; s [5
2
4backward5.
Statement: The a)erage a
(b) Given: )elocity–time graph
!
Required: a at t = 2 [ t =
6d t = [
Analysis: The instantaneous acceleration at a gi)en time is the slope of the tangent to the
)elocity–time graph at that time. Sketch tangent lines at the required times, read the coordinates
of two points on each tangent line, and then calculate the slopes.
Solution:
For the tangent at 2 s, two points are
e
!
! !v
a=
!t
!
! !v
a=
!t
s[
26 m [ &quot;
=
2[&quot;0[
&quot; s[
=
5[
!
a = &quot; s [5
s [&quot; s [
0 [ &quot;3 [
&quot; s[
=
5[
!
a = &quot; s [5
=
!
! !v
a=
!t
2 m [&quot;
s[
=
[&quot;2 [
&quot; s[
=
5[
!
a = &quot;35 s [ 5
Statement: The instantan
2
4backward5 at / s.
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2
Chapter 1: Kinematics
2
1.1-7
(c) The slope of the )elocity–time graph decreases gradually, so the acceleration–time graph
should be decreasing, possibly linearly. Use the )alues of instantaneous acceleration from
part
to draw the graph.
Section 1.1 Questions, page 16
1. (a) Given: !d1 = 22 m !d5 = 33 s
Required: total distance, !d
Analysis: !d = !d1 + !d2
Solution: !d = !d1 + !d2
= 22 m + 11 m
!d = 00 s
Statement: The total distance tra)elled by the cardinal is 33 m.
(b) Given: !t1 = 2 [ !t5 = 3/2 [
Required: va)
!d
. To determine the total time add the partial times, !t = !t1 + !t2 .
!t
Solution: !t = !t1 + !t2
Analysis: va) =
=
!t =
[ + 3/2 [
[
!d
!t
00 s
=
[
va =
s[
Statement: The a)erage speed of the cardinal
va =
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Chapter 1: Kinematics
1.1-8
!
!
(c) Given: !d1 = 22 m [ ] !d5 = 33 s
!
Required: va)
]
!
!d
!
Analysis: The a)erage )elocity is the ratio of the total displacement to the total time, va) =
.
!t
To determine the total displacement, calculate the )ector sum of the partial displacements,
!
!
!
!d = !d1 + !d2 .
Solution: Determine the magnitude of the displacement.
! 2
! 2
!2
!d = !d1 + !d2
= (22 m)2 + (11 m)2
s5
! =
!d =
s w
ra
[ farr
Determine the angle ! .
!
&quot;d5
ma6 ! = !
&quot;d3
=
33 s
55 s
ma6 ! =
! = ma6 #3
!= &deg;
Determine the a)erage )elocity.
!
!d
!
vav =
!t
24.60 m [E 27&deg; N]
=
4.4 s
!
vav = 5.6 m/s [E 27&deg; N]
Statement: The cardinal’s a)erage )elocity is 5.6 m/s [E 27&deg; N] .
)
2. (a) Given: vav =
Required: va)
Analysis: Con)ert units using 1 km = 1000 m, and 1 h = 3/00 s.
Solution: vav =
=
=
!
8
!
0
0
8) (one extra digit carried)
vav = 4.17 m/s
Statement: The
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Chapter 1: Kinematics
1.1-9
(b) Given: !d =
Required: !t
6 vav = 4.137 m/s
Analysis: Use the relation vav =
Solution: !t =
=
!d
!d
.
6 !t =
!t
vav
!d
vav
4.137 m/s
&quot;
m
1
!t = 353 s
Statement: It takes /9/ s to complete one lap.
3. (a) Given: !d = 16 m6 !t = 0.1 s
Required: a)erage speed, va)
!d
!t
!d
Solution: vav =
!t
13 m
=
0.1 s
vav = 7.3 m/s
Statement:
(b) Given: diameter, = 16 m
Required: time for one lap, !t
Analysis: va) =
Analysis: Calculate the circumference of the pond, ! = &quot;
. Use va) =
time, !t , for one lap.
Solution: ! = &quot;
= &quot; (16 m)
m (t o extra digits carried)
! =
!d
to determine the
!t
!d
!t
!d
!t =
vav
vav =
m
7.315 m/s
!t = 3.3 s
Statement: It takes the skater /./ s to go around the edge of the pond.
=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.1-10
4. (a) Given: !d =
Required: va) ,
Analysis: va) =
!t = 4 min &quot;
=
&quot;
4
m6 !t = 4 min
!d
;
!t
s
1 min
!t = 0.7 &quot;
s
!d
Solution: vav =
!t
&quot; 4m
0.7 &quot;
s
vav =
m/s
Statement:
!
(b) Given: !d =
&deg;
!
Required: va) ,
Analysis: The airplane tra)els in one direction for the whole trip, so the a)erage )elocity is a
)ector equal in magnitude to the a)erage speed.
Solution:
[
&deg; .
r
r
5. Given: vi = m/s6 v = 53
/
6 !t = 4.1 s
!
Required: a
Analysis: Since the rocket starts from rest, its change in )elocity is equal to its final )elocity:
!
! !v
! !
.
!v = v . Its acceleration is the ratio of its change in )elocity to the time inter)al taken: a =
!t
m
1
1 min
! 53
!
!
!
v =
min
1
!
v = 03.37 m/s
(one extra digit carried)
!
! !v
Solution: a =
!t
03.37 m/s
=
4.1 s
!
a=
m/s 0
=
Statement: The rocket accelerates at
Copyright &copy; 2012 Nelson Education Ltd.
m/s 0
.
Chapter 1: Kinematics
1.1-11
!
s6 v = m/s
r
Required: )elocity of the ball before hitting the wall, vi
Analysis: The ball slows down when it hits the wall. Since the acceleration points west, the
!
!v
!
initial )elocity must point east. Use aa) =
to determine the change in )elocity and the initial
!t
)elocity.
!
! !v
Solution: a =
!t
! !
!v = aav !t
!
6. Given: aav =
=(
&quot;
m/s 0
6 !t =
&quot;
m/s 0
)(
&quot;
&quot;
#0
#0
s)
!
!v = 40.744 m/s
(t o extra digits carried)
r r r
!v = v &quot; vi
r
r r
vi = v &quot; !v
= m/s &quot; 40.744 m/s
r
vi = 40.7 m/s
Statement: The ball hits the wall with )elocity 40.7 m/s
.
!
!
7. Given: vi = m/s6 v =
m/s orhar 6 !t =
s
!
Required: a)erage acceleration of the runner, aa)
!
!v
!
Analysis: aa) =
!t
!
!v
!
Solution: aav =
!t
m/s orhar &quot; m/s
=
s
!
0
aav = 0.4 m/s orhar
2
Statement: The runner’s a)erage
4forward5.
(a) Given: position–time graph
!
Required: va) for the entire trip
Analysis: Read the coordinates of initial and final points on the graph. Use these points to
!d
calculate the slope of the secant, va) =
. As in the graph, use east as positi)e.
!t
Solution:
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Chapter 1: Kinematics
1.1-12
vav =
!d
!t
m&quot; m
14 s &quot; s
m
=
14 s
vav = 05 m/s
2
4E5.
Statement: The a)erage )elocity of the car f
(b) Given: position–time graph
!
Required: va) from t
t
Analysis: Use the graph to determine the position at t
t
!d
calculate the slope of the secant, va) =
. As in the graph, use east as positi)e.
!t
Solution:
!d
vav =
!t
m&quot; m
=
14 s &quot; 4 s
m
=
s
vav =
m/s
Statement:
higher than for the trip as a whole because the car is accelerating—the car was mo)ing faster on
a)erage toward the end of the trip.
(c) Given: position–time graph
!
Required: v at t =
s t=
nd at t =
s
Analysis: The instantaneous )elocity at a gi)en time is the slope of the tangent to the position–
time graph at that time. Sketch in tangent lines at the required times, read the coordinates of two
points on each tangent line and then calculate the slopes.
Solution:
For the tangent
For the tangent at 8.0 s, two points are
=
!d
!t
68 m &quot; m
=
s&quot;
s
m
=
v=
v = 17 m/s
Copyright &copy; 2012 Nelson Education Ltd.
v=
=
=
v=
!d
!t
m&quot;
s&quot;
m
m
s
m/s
Chapter 1: Kinematics
1.1-13
For the tangent at 12 s, two points are (10.0 s,
!d
v=
!t
m&quot;
m
=
s&quot;
s
m
=
v=
m/s
Statement:
8.0
(d) The slope of the position–time graph gradually increases throughout the trip showing that the
)elocity is increasing to the east. The )elocity–time graph should show this gradual increase; it
may be linear. Draw a )elocity–time graph using the )alues of instantaneous )elocity. Draw a
smooth line for a reasonable trend.
(a) Given: )elocity–time graph
!
Required: aa)
Analysis: Read the coordinates of the initial and final points on the graph. Use these points to
!v
calculate the slope of the secant, aa) =
. As in the graph, use forward as positi)e.
!t
Solution:
!v
aav =
!t
m/s &quot; m/s
=
s&quot; s
&quot; m/s
=
aav = &quot; m/s 0
Statement: The a)erage acceleration of the car for the entire
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
2
4backward5.
1.1-
(b) Given: )elocity–time graph
!
Required: a at t = s t = 3 s and at t = 5 s
Analysis: The instantaneous acceleration at a gi)en time is the slope of the tangent to the
)elocity–time graph at that time. Sketch in tangent lines at the required times, read the
coordinates of two points on each tangent line and then calculate the slopes.
Solution:
For the tangent at 3 s, two points are
For the tangent at / s, two points are
!
! !v
a=
!t
44 m/s &quot; 50 m/s
=
4 s&quot;2 s
&quot;6 m/s
=
2s
!
a = &quot; m/s 0
!
! !v
a=
!t
04 m/s &quot; m/s
=
7 s&quot; s
&quot;10 m/s
=
0s
!
a = &quot;3 m/s 0
!v
!t
m/s &quot; m/s
=
s&quot; s
&quot; m/s
=
0s
a = &quot;5 m/s 0
2
2
4backwa
4backward5 at
Statement: The
2
4backward5 at 9 s.
(c) The slope of the )elocity–time graph decreases gradually, so the acceleration–time graph
should be decreasing, possibly linearly. Use the )alues of instantaneous acceleration from
part
to draw the graph.
a=
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Chapter 1: Kinematics
1.1-
Section 1.2: Equations of Motion
Tutorial 1 Practice, page 19
!
!
!
1. (a) Given: vi = 15.0 m/s [forward]; vf = 0 m/s; a = 5.0 m/s 2 [backward]
!
Required: !d
! !
! !
1!
! v ! vi
Analysis: Use a = f
to determine the braking time, !t . Then use !d = vi !t + a!t 2 to
&quot;t
2
determine the braking distance. Choose forward to be the positive direction.
Solution:
v ! vi
1
!d = vi !t + a!t 2
a= f
2
&quot;t
1
v ! vi
= (15.0 m/s)(3.0 s) + (&quot;5.0 m/ s 2 )(3.0 s )2
&quot;t = f
2
a
= 45.0 m &quot; 22.5 m
0 m/s ! 15.0 m/s
=
2
!5.0 m/s
!d = 22 m
&quot;t = 3.0 s
Statement: The motorcycle’s braking distance is 22 m [forward].
(b) Answers may vary. Sample answer: In Sample Problem 2 and part (a), I calculated the
braking distance for a motorcyclist slowing down at 5.0 m/s2. When the initial speed is high, it
takes much longer and a much greater distance to stop, so a speeding vehicle has more difficulty
stopping safely.
!
!
2. Given: vi = 0 m/s; !d = 120 m [N]; !t = 15 s
!
Required: a
! !
1!
Analysis: Use !d = vi !t + a!t 2 to determine the acceleration.
2
1
!d = vi !t + a!t 2
2
1
!d = (0 m/s)!t + a!t 2
2
1
!d = a!t 2
2
2!d
a= 2
!t
Choose north to be the positive direction.
2!d
Solution: a = 2
!t
2(120 m)
=
(15 s)2
a = 1.1 m/s 2
Statement: The runner accelerates at 1.1 m/s2 [N].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-1
!
!
!
3. Given: v1 = 22 m/s [E]; !t1 = 12 s; a = 1.2 m/s 2 [W]; vf = 0 m/s
!
Required: total displacement, !d
Analysis: Determine the displacement for the first part of the trip using
!
! !d1
v1 =
!t1
!d1 = v1!t1
Determine the braking time using
! !
! vf ! vi
a=
&quot;t
vf ! vi
&quot;t =
a
! !
1!
Then, use !d2 = vi !t + a!t 2 to determine the displacement for the second part of the trip.
2
!
!
!
Finally, the total displacement is !d = !d1 + !d2 . Use east as the positive direction.
Solution: !d1 = v1!t1
= (22 m/s)(12 s)
!d1 = 264 m (one extra digit carried)
vf &quot; vi
a
0 m/s &quot; 22 m/s
=
&quot;1.2 m/s 2
!t = 18.33 s (two extra digits carried)
!t =
1
!d2 = v1!t + a!t 2
2
1
= (22 m/s)(18.33 s) + (&quot;1.2 m/ s 2 )(18.33 s) 2
2
!d2 = 202 m (one extra digit carried)
!d = !d1 + !d2
= 264 m + 202 m
!d = 470 m
Statement: The total displacement of the bus is 4.7 ! 102 m .
!
!
4. (a) Given: vi = 0 m/s; vf = 9.6 m/s [W]; !t = 4.2 s
!
Required: a
! !
! vf ! vi
Analysis: Calculate the acceleration using a =
. Use west as positive.
&quot;t
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Chapter 1: Kinematics
1.2-2
vf ! vi
&quot;t
9.6 m/s ! 0 m/s
=
4.2 s
= 2.286 m/s 2 (two extra digits carried)
Solution: a =
a = 2.3 m/s 2
Statement: The runner’s acceleration is 2.3 m/s2 [W].
!
!
(b) Given: vi = 0 m/s; !t = 4.2 s; a = 2.286 m/s 2 [W]
!
Required: !d
! !
1!
Analysis: Calculate the displacement using !d = vi !t + a!t 2 .
2
1
Solution: !d = vi !t + a!t 2
2
1
= (0 m/s)(4.2 s) + (2.286 m/ s 2 )(4.2 s) 2
2
= 20.16 m (two extra digits carried)
!d = 20 m
Statement: The displacement of the runner is 2.0 &times; 101 m [W] while accelerating.
!
!
!
(c) Given: v = 9.6 m/s [W]; !d = 100.0 m [W]; !d1 = 20.16 m [W]; !t1 = 4.2 s
Required: total time, !t
!
!
!
!
Analysis: Use !d = !d1 + !d2 to calculate the displacement, !d2 , for the second part of the
race.
!d = !d1 + !d2
!d2 = !d &quot; !d1
!
!d
!
Then use the constant velocity formula, vav =
, to determine the time, !t2 , for the second part
!t
of the race.
!d
v2 = 2
!t2
!t2 =
!d2
v2
Finally, the total time taken is the sum of the two times, !t = !t1 + !t2 .
Solution: !d2 = !d &quot; !d1
= 100.0 m &quot; 20.16 m
!d2 = 79.84 m (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
!t2 =
!d2
v2
79.84 m
9.6 m/s
!t2 = 8.317 s (two extra digits carried)
=
Chapter 1: Kinematics
1.2-3
!t = !t1 + !t2
= 4.2 s + 8.317 s
!t = 13 s
Statement: The total time taken is 13 s.
!
!
5. (a) Given: player 1: d1i = 0 m [right]; v1i = 0 m/s; a = 2.4 m/s 2 [right] ;
!
!
player 2: d2i = 42 m [right]; v2 = 5.4 m/s [left]
Required: time until the players meet, !t
Analysis: Set up an equation for each player that relates position and time. The first player
! !
1!
moves at constant acceleration, !d = vi !t + a!t 2 .
2
The second
player
moves
at
a
constant
velocity:
!
! !d
v=
!t
!d
v2 = 2
!t
!d2 = v2 !t
Compare these equations to solve for either position or time. Use right as positive.
&quot;b &plusmn; b2 &quot; ac
.
2a
player 2:
Solution: player 1:
1
d = v1i t + a t 2
2
1
d1 = (0 m/s) t + (2.4 m/s 2 ) t 2
2
2
d1f = d1i = (1.2 m/s ) t 2
!d2 = v2 !t
d2f &quot; d2i = (&quot;5.4 m/s)!t
d2f &quot; 42 m = (&quot;5.4 m/s)!t
d2f = 42 m &quot; (5.4 m/s)!t
d1f = (1.2 m/s 2 ) t 2
The final positions of the two players are equal.
d1f = d2f
(1.2 m/s 2 )!t 2 = 42 m &quot; (5.4 m/s)!t
(1.2 m/s 2 )!t 2 + (5.4 m/s)!t &quot; 42 m = 0
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-4
This is a quadratic equation in !t .
!t =
=
&quot;b &plusmn; b2 &quot; 4ac
2a
&quot;5.4 m/s &plusmn; (–5.4 m/s)2 &quot; 4(1.2 m/s 2 )(–42 m)
2(1.2 m/s 2 )
!t = 4.1 s or &quot; 8.8 s
The second solution is not possible, because the time taken to run is positive.
Statement: The players collide after 4.1 s.
(b) Given: equation for player 1: d1 = (1.2 m/s 2 )!t 2 ; equation for player 2: !d2 = v2 !t ;
!t = 4.1 s
!
!
Required: !d1; !d2
Analysis: Use the time taken by the players and the player’s equations of motion to determine
their displacements.
Solution: player 1:
player 2:
2
2
!d2 = v2 !t
! 1 = (1.2 m/s )!t
= (1.2 m/s 2 )(4.1 s)2
= (&quot;5.4 m/s)(4.1 s)
!d2 = &quot;22 m
! 1 = 20 m
Statement: Player 1 ran 20 m [right] and player 2 ran 22 m [left].
!
!
(c) Given: v1i = 0 m/s; a = 2.4 m/s 2 ; !t = 4.1 s
Required: v1
! !
! vf ! vi
Analysis: Use a =
to solve for the final speed of player 1.
&quot;t
v1 f ! v1i = a&quot;t
v1 f = v1i + a&quot;t
Solution: v1 f = v1i + a!t
= 0 m/s + (2.4 m/s 2 )(4.1 s)
v1 f = 9.8 m/s
Statement: Player 1 is moving at 9.8 m/s when he collides with player 2.
!
!
!
6. (a) Given: v1 = 110 m/s [forward]; v2 = 0 m/s; a = 6.2 m/s 2 [backward]
Required: minimum stopping time, !t
! !
! vf ! vi
Analysis: a =
; Use forward as positive.
&quot;t
v ! vi
&quot;t = f
a
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Chapter 1: Kinematics
1.2-5
vf &quot; vi
a
0 m/s &quot; 110 m/s
=
&quot;6.2 m/s 2
= 17.74 s (two extra digits carried)
Solution: !t =
!t = 18 s
Statement: The airplane needs a minimum of 18 s to stop.
!
!
(b) Given: vi = 110 m/s; !t = 17.74 s; a = 6.2 m/s 2 [backward]
Required: minimum stopping distance, !d
1
Analysis: Use !d = vi !t + a!t 2 to determine the distance travelled while stopping.
2
1
Solution: !d = vi !t + a!t 2
2
1
= (110 m/s)(17.74 s) + (&quot;6.2 m/ s 2 )(17.74 s) 2
2
2
!d = 9.8 # 10 m
Statement: The minimum safe length of the runway is 9.8 ! 102 m .
(c) Answers may vary. Sample answer: The runway should be much longer than the minimum
length because the airplane may land partway along the runway. If the weather is bad, the usual
braking force needed to slow the plane to a stop may be insufficient and more time and distance
may be needed to stop the plane. Also, it is uncomfortable for the passengers to slow a plane
down using the maximum allowed braking force.
Tutorial 2 Practice, page 20
!
!
!
1. Given: vi = 22 m/s [up]; vf = 0 m/s; a = 9.8 m/s 2 [down]
Required: maximum height, !d
Analysis: Use vf2 = vi2 + 2a!d to calculate !d .
vf2 = vi2 + 2a!d
vf2 &quot; vi2
2a
Use up as positive and down as negative.
vf2 &quot; vi2
Solution: !d =
2a
(0 m/s) 2 &quot; (22 m/s) 2
=
2(&quot;9.8 m/ s 2 )
!d =
!d = 25 m
Statement: The maximum height of the apple is 25 m.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-6
!
!
! 1
2. Given: vi = 0 m/s; !d = 10.0 m [down]; a = (9.8 m/s 2 [down])
6
Required: time for an object to fall, !t
1
Analysis: Calculate the acceleration and then use !d = vi !t + a!t 2 to determine !t .
2
&quot;b &plusmn; b2 &quot; ac
.
2a
Use up as positive.
!9.8 m/s 2
Solution: =
6
= !1.633 m/s 2 (two extra digits carried)
1
!d = vi !t + a!t 2
2
1
&quot;10.0 m = (0 m/s)!t + (&quot;1.633 m/s 2 )!t 2
2
2
2
(&quot;0.8165 m/s )!t + (0 m/s)!t + 10.0 m = 0
!t =
=
&quot;b &plusmn; b2 &quot; 4ac
2a
(
)
0 m/s &plusmn; (0 m/s)2 &quot; 4 &quot;0.8165 m/s 2 (10.0 m)
&quot;1.633 m/s
2
!t = 3.5 s
Statement: It would take 3.5 s for an object to fall 10.0 m if g were one-sixth the value of
Earth’s g.
!
!
!
3. (a) Given: vi = 12 m/s [down]; !d = 45 m [down]; a = 9.8 m/s 2 [down]
Required: time to fall, !t
1
Analysis: Use up as positive. Use !d = vi !t + a!t 2 to determine !t .
2
1
a!t 2 + vi !t &quot; !d = 0
2
Use the quadratic formula to determine Δt.
#1 &amp;
&quot;vi &plusmn; vi2 &quot; 4 % a ( (&quot;!d)
\$2 '
!t =
a
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-7
#1 &amp;
&quot;vi &plusmn; vi2 &quot; 4 % a ( (&quot;!d)
\$2 '
Solution: !t =
a
&quot;(&quot;12 m/s) &plusmn; (&quot;12 m/s)2 &quot; 4(&quot;4.9 m/s 2 )(45 m)
&quot;9.8 m/s 2
= 2.044 s or &quot; 4.493 s (two extra digits carried)
=
!t = 2.0 s
Statement: The ball takes 2.0 s to land.
!
!
!
(b) Given: vi = 12 m/s [down]; !d = 45 m [down]; a = 9.8 m/s 2 [down]; !t = 2.044 s
Required: vf
Analysis: Use a =
&quot;v +v %
vf ! vi
or !d = \$ i f ' !t to determine vf . Both solutions are provided.
&quot;t
# 2 &amp;
vf ! vi
&quot;t
vf ! vi = a&quot;t
a=
vf = vi + a&quot;t
&quot;v +v %
!d = \$ i f ' !t
# 2 &amp;
2!d
!t
2!d
vf =
( vi
!t
vi + vf =
Solution: vf = vi + a!t
= (&quot;12 m/s) + (&quot;9.8 m/s 2 )(2.044 s)
= &quot;12 m/s &quot; 20.03 m/s
vf = &quot;32 m/s
2!d
&quot; vi
!t
2(&quot;45 m)
=
&quot; (&quot;12 m/s)
2.044 s
= &quot;44.03 m/s + 12 m/s
vf = &quot;32 m/s
vf =
Statement: The ball is moving at 32 m/s when it lands.
(c) Answers may vary. Sample answer: If the ball were thrown up at 12 m/s, it would rise to a
maximum height and then fall. It would be moving at 12 m/s down as it passed its initial
position. It would then continue down as in the question. The ball would be moving at the same
speed as above, 32 m/s.
!
!
4. (a) Given: !d = 32 m [down]; !t = 1.5 s; a = 9.8 m/s 2 [down]
!
Required: velocity at 32 m, vi
Analysis: The displacement and time taken are known. Calculate average velocity using
! &quot; v! + v! %
!d = \$ i f ' !t . This gives the sum of the velocities.
# 2 &amp;
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Chapter 1: Kinematics
1.2-8
&quot;v +v %
!d = \$ i f ' !t
# 2 &amp;
vi + vf =
2!d
!t
The acceleration is known, and the difference in the (squares of the) velocities can be determined
using vf2 = vi2 + 2a!d . The initial velocity of the ball can then be determined.
vf2 = vi2 + 2a!d
vf2 &quot; vi2 = 2a!d
Although it is not usual, for convenience, use down as the positive direction.
2!d
Solution: vi + vf =
!t
2(32 m)
=
1.5 s
vi + vf = 42.67 m/s (two extra digits carried) (Equation 1)
vf2 ! vi2 = 2a&quot;d
(
)
= 2 9.8 m/s 2 (32 m)
vf2 ! vi2 = 627.2 (m/s)2 (two extra digits carried) (Equation 2)
The left side of Equation 2 can be used to determine the difference of the speeds.
vf2 ! vi2 = 627.2 (m/s)2
(vf ! vi )(vf + vi ) = 627.2 (m/s)2
(vf ! vi )(42.67 m/s) = 627.2 (m/s)2
vf ! vi = 14.70 m/s (two extra digits carried) (Equation 3)
Finally, use equations 1 and 3 to determine v .
f
!
i
= 14.70 m/s
i
!
f
= !14.70 m/s
i
+
f
= 42.67 m/s
(Equation 3)
(Equation 1)
2 i = !14.70 m/s + 42.67 m/s
= 14 m/s
Statement: The ball is moving 14 m/s [downward] when the timing starts at 32 m.
(b) Solutions may vary. Sample answer:
!
!
!
Given: vi = 0 m/s; vf = 14 m/s [downward]; !d2 = 32 m; a = 9.8 m/s 2 [downward]
!
Required: total displacement, !d
Analysis: Calculate the displacement for the first part of the trip using:
i
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-9
vf2 = vi2 + 2a!d
!d1 =
vf2 &quot; vi2
2a
Then calculate the total displacement using !d = !d1 + !d2 . Continue to use down as the
positive direction.
v 2 &quot; vi2
Solution: !d1 = f
! =! 1+! 2
2a
= 10 m + 32 m
(14 m/s)2 &quot; (0 m/s)2
=
! = 42 m
2(9.8 m/ s 2 )
!d1 = 10 m
Statement: The ball’s displacement is 42 m [downward].
!
!
5. (a) Given: !d = 14 m [up]; !t = 1.1 s; a = 9.8 m/s 2 [down]
!
Required: v
Analysis: Use up as positive. Calculate the initial velocity using
1
!d = vi !t + a!t 2
2
1
!d &quot; a!t 2
2
vi =
!t
1
!d &quot; a!t 2
2
Solution: vi =
!t
1
14 m &quot; (&quot;9.8 m/s 2 )(1.1 s)2
2
=
1.1 s
14 m + 5.93 m
=
1.1 s
= 18.12 m/s (two extra digits carried)
vi = 18 m/s
Statement: The initial velocity of the keys was 18 m/s [up].
(b) Solutions may vary. Sample answer:
!
!
!
Given: !d = 14 m [up]; a = 9.8 m/s 2 [downward]; vi = 18.12 m/s
!
Required: vf
Analysis: Use vf2 = vi2 + 2a!d to determine vf . (Any of the equations containing vf could be
used.)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-10
Solution: vf2 = vi2 + 2a!d
= (18.12 m/s)2 + 2(&quot;9.8 m/s 2 )(14 m)
= 328.33 (m/s)2 &quot; 274.4 (m/s)2
= 53.93 (m/s)2
vf = 7.3 m/s
Statement: The velocity of the keys when they were caught was 7.3 m/s [up].
Section 1.2 Questions, page 21
!
!
2
1. (a) Given: vi = 0 m/s; !t = 5.2 s; a = 4.1 m/s [forward]
!
Required: !
! !
1!
Analysis: Calculate the displacement using !d = vi !t + a!t 2 . Use forward as the positive
2
direction.
1
Solution: !d = vi !t + a!t 2
2
1
= (0 m/s)(5.2 s) + (4.1 m/ s 2 )(5.2 s) 2
2
!d = 55 m
Statement: The racehorse’s displacement is 55 m [forward].
!
!
(b) Given: vi = 0 m/s; !t = 5.2 s; a = 4.1 m/s 2 [forward]
!
Required: f
Analysis: Use a =
vf ! vi
&quot;t
vf = vi + a&quot;t
vf ! vi
to determine vf :
&quot;t
a=
(Any of the equations containing vf could be used.)
Solution: vf = vi + a!t
= 0 m/s + (4.1 m/s 2 )(5.2 s)
vf = 21 m/s
Statement: The horse’s final velocity is 21 m/s [forward].
!
!
!
2. (a) Given: vi = 7.72 ! 106 m/s [E]; vf = 2.46 ! 106 m/s [E]; &quot;d = 0.478 m [E]
!
Required:
Analysis: Use east as the positive direction.
Use v 2 = vi2 + 2a!d to calculate the acceleration:
vf2 = vi2 + 2a!d
a=
vf2 &quot; vi2
2!d
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-11
Solution: a =
=
=
vf2 ! vi2
2&quot;d
#
!
m/s 2 !
#
2
m
#
m/s
2
m 2 /s 2
m
a=!
#
m/s 2
Statement: The electron’s acceleration is 5.60 ! 1013 m/s 2 [W].
!
!
!
(b) Given: v =
! 10 m/s [ ]; v =
! 10 m/s [ ]; a =
Required: !t
v ! vi
v &quot; vi
Analysis: Use a = f
to determine !t : !t = f
&quot;t
a
(Any of the equations containing !t could be used.)
v &quot; vi
Solution: !t = f
a
(2.46 # 106 m/s) &quot; (7.72 # 106 m/s)
=
&quot;5.60 # 1013 m/s 2
&quot;5.26 # 106 m/s
=
&quot;5.60 # 1013 m/s 2
! 10 m/s 2 [ ]
!t = 9.39 # 10&quot;8 s
Statement: The acceleration occurs over a time interval of 9.39 ! 10&quot;8 s .
3. Solutions may vary. Sample answer:
!
!
(a) Given: cruiser: d1 = 0 m fiward]; v1i = 0 m/s; a = 3.0 m/s 2 [ffiward] ;
!
!
car: d2i = 0 m [forward]; v2 = 62 km/h [forward]
Required: time until the cruiser catches up with the car, !t
Analysis: The vehicles start from the same position. The displacement and time when the two
vehicles meet are not known. Set up an equation for each vehicle that relates position and time.
! !
1!
The cruiser moves at constant acceleration, !d = vi !t + a!t 2 , and the car at constant velocity,
2
!
! !d
. Compare these equations to solve for time and, later, displacement. Use forward as
v=
!t
positive.
1h
1 min
! 62 km 1000 m
v2 =
!
!
!
1h
60 s
1 km
60 min
!
v2 = 17.22 m/s (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-12
Solution: cruiser:
1
!d1 = vi !t + a!t 2
2
1
= (0 m/s)!t + (3.0 m/s 2 )!t 2
2
2
!d1 = (1.5 m/s )!t 2
The displacements of the two vehicles are equal.
!d1 = !d2
car:
!d2
!t
!d2 = v2 !t
v2 =
!d2 = (17.22 m/s)!t
(1.5 m/s 2 )!t 2 = (17.22 m/s)!t
(1.5 m/s 2 )!t 2 = (17.22 m/s) !t
17.22 m/s
1.5 m/s 2
= 11.48 s (two extra digits carried)
!t =
!t = 11 s
Statement: The cruiser catches up 11 s after starting.
!
m/s [iorward]; !t = 11.48 s
(b) Given: car: v2 =
Required: common displacement of the vehicles, !
Analysis: Substitute the value of !t in the displacement equation of either the cruiser or the car.
Use the equation of the car.
Solution: !d2 = (17.22 m/s)!t
= (17.22 m/s)(11.48 s)
!d2 = 2.0 &quot; 102 m
Statement: The cruiser catches up to the car 2.0 ! 102 m [forward] from the cruiser’s initial
position.
!
(c) Given: cruiser: v1i = 0 m/s; a = 3.0 m/s 2 [forward]; !t = 11.48 s
Required: v1f
! !
! vf ! vi
Analysis: Use a =
to solve for the final speed of the cruiser.
&quot;t
v ! v1i
a= 1
&quot;t
v1f ! v1i = a&quot;t
v1f = v1i + a&quot;t
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-13
Solution: v1i = v1i + a!t
=
2s/ +
2s
=
=
v1 =
s
&quot;
2s / 11.48 s/
1
&quot;
s
1
m
&quot;
m
1f
2f
Statement: The cruiser is moving at 120 km/h when it catches up with the car. This is an unsafe
speed in a school zone.
r
r
(d) Given: cruiser: v1i = 0 m/s; v1f = 72 km/h [forward]; = 3.0 m/s 2 [forward] ; then at constant
!
speed; car: v2 = 17.22 m/s [forward]
Required: time until the cruiser catches up with the car, !t
Analysis: Build a new equation for the displacement of the cruiser and the car.
1h
1 min
72 km 1000 m
v1f =
!
!
!
h
60 s
1 km
60 min
v1f = 20 m/s
Determine how long and how far the cruiser accelerates to reach the speed of 20 m/s. Use
vf = vi + a!t and vf2 = vi2 + 2a!d . Then write an equation for the cruiser’s displacement while
continuing at constant speed. Determine where the car is when the cruiser’s motion changes, and
then write a new equation for the car’s displacement. The rest of the solution will be similar to
part (a).
!
Solution: The cruiser’s final speed is vf = 20 m/s [forward]. It reaches 20 m/s when
vf = vi + a!t
vf &quot; vi
a
20 m/s &quot; 0 m/s
=
3.0 m/s 2
!t = 6.67 s
The cruiser’s displacement is now
vf2 = vi2 + 2a!d
!t =
vf2 &quot; vi2
!d1 =
2a
(20 m/s)2 &quot; (0 m/s)2
=
2(3.0 m/s 2 )
=
400 m 2 / s 2
6.0 m/ s 2
d1 = 66.7 m (one extra digit carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-14
At 6.67 s, the car’s displacement is
!d2 = v2 !t
!d2 = (17.22 m/s)(6.67 s)
d2 = 114.9 m (two extra digits carried)
Treat the motion as if it started at ti = 6.67 s , after which both vehicles move at constant speed.
The later positions of the vehicles are
cruiser:
car:
!d1 = v1!t
!d2 = v2 !t
d1 &quot; 66.7 m = (20 m/s)!t
d2 &quot; 114.9 m = (17.22 m/s)!t
d1 = 66.7 m + (20 m/s)!t
d2 = 114.9 m + (17.22 m/s)!t
The cruiser catches up to the car when the positions are equal.
d1 = d2
66.7 m + (20 m/s)!t = 114.9 m + (17.22 m/s)!t
(20 m/s)!t &quot; (17.22 m/s)!t = 114.9 m &quot; 66.7 m
48.2 m
2.78 m/s
!t = 17.34 s (two extra digits carried)
!t =
d1 = 66.7 m + (20 m/s)!t
= 66.7 m + (20 m/s)(17.34 s)
d1 =
m (t o extra digits carried)
The total time elapsed from the start is
!t = 6.67 s + 17.34 s
!t = 24 s
Statement: Under the second scenario, it takes 24 s for the cruiser to catch the car, by which
time the cruiser and the car have travelled more than 400 m. They might be on the other side of
the school zone and it would be too late to protect the children. Neither of these scenarios is very
reasonable.
4. Answers may vary. Sample answer: I would stand at the top of the cliff with a small stone. My
partner would be at the bottom of the cliff with a stopwatch. I would signal with my hand the
instant I dropped the stone. My partner would time the stone’s fall. Since I would now know the
time interval taken, the acceleration due to gravity, and the initial velocity (0 m/s), I would use
! !
1!
!d = vi !t + a!t 2 to determine the displacement. The magnitude of the displacement would be
2
our estimate of the height of the cliff. The assumption I would have to make is that the height of
the cliff is being measured from sea level.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-15
5. Solutions may vary. Sample answer:
!
(a) Given: !t = 2.4 s; a = 9.8 m/s 2 [down]
!
Required: vi
Analysis: Since the ball is caught at the same height from which it is thrown, its total
! &quot; v! + v! %
displacement is 0 m. From !d = \$ i f ' !t , the final velocity is the opposite of the initial
# 2 &amp;
! !
!
! v ! vi
velocity. Use this fact and a = f
to calculate i .
&quot;t
! !
! v ! vi
a= f
&quot;t
!
!2
vi
!
a=
&quot;t
!
a&quot;t
!
vi = !
2
Use up as the positive direction.
!
a&quot;t
!
Solution: vi = !
2
(!9.8 m/s 2 )(2.4 s)
=!
2
= 11.76 m/s (two extra digits carried)
!
vi = 12 m/s [up]
Statement: The ball’s initial velocity is 12 m/s [up].
!
!
(b) Given: !t = 2.4 s; a = 9.8 m/s 2 [down]; v = 11.76 m/s [up]
Required: maximum height, !d
Analysis: The flight of the ball is symmetric going up and coming down, so the time to reach
maximum height is one-half of the total time interval, 2.4 s: !t = 1.2 s . The speed of the ball at
maximum height is 0 m/s. Use any of the formulas that contain ! to calculate the height. This
&quot;v +v %
solution uses !d = \$ i f ' !t .
# 2 &amp;
&quot;v +v %
Solution: !d = \$ i f ' !t
# 2 &amp;
(11.76 m/s + 0 m/s)(1.2 s)
2
!d = 7.1 m
Statement: The maximum height of the ball is 7.1 m.
=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-16
6. Solutions may vary. Sample answer:
!
!
!
(a) Given: v = 18 m/s [up]; !d =
m [down]; a = 9.8 m/s 2 [down]
Required: time to reach the ground, !t
1
Analysis: Use up as positive. Use !d = vi !t + a!t 2 to calculate !t :
2
1
a!t 2 + vi !t &quot; !d = 0
2
Use the quadratic formula to determine Δt.
#1 &amp;
&quot;vi &plusmn; vi2 &quot; 4 % a ( (&quot;!d)
\$2 '
!t =
a
1
&quot;vi &plusmn; vi2 &quot; 4( a)(&quot;!d)
2
Solution: !t =
a
&quot;18 m/s &plusmn; (18 m/s)2 &quot; 4(&quot;4.9 m/s 2 )(32 m)
=
–9.8 m/s 2
= 4.984 s (or &quot;1.310 s) (two extra digits carried)
!t = 5.0 s
Statement: The ball takes 5.0 s to hit the ground.
!
!
!
(b) Given: vi = 18 m/s [ ; !d = 02 m [down]; a = 9.8 m/s 2 [down]; !t =
s
Required: vf
vf ! vi
to calculate vf = vi + a!t .
&quot;t
Solution: vf = vi + a!t
=
m/s + &quot;9.8 m/s 2
s
= m/s &quot;
m/s
v = &quot; m/s
Analysis: Use a =
Statement: The velocity of the ball when it hits the ground is 3.1 &times; 101 m/s [down].
!
!
!
(c) Given: vi = 18 m/s [up]; vf = 0 m/s ; a = 9.8 m/s 2 [down]
Required: maximum height,
max
vf2 &quot; vi2
Analysis: Use v = v + 2a!d to solve for !d =
.
2a
Then !d = dmax &quot; dv to solve for max .
2
f
2
i
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-17
vf2 &quot; vi2
2a
(0 m/s)2 &quot; (18 m/s)2
=
2(&quot;9.8 m/s 2 )
Solution: !d =
=
324 m 2 / s 2
19.6 m/ s 2
!d = 16.53 m (two extra digits carried)
dmax = di + !d
= 32 m + 16.53 m
dmax = 49 m
Statement: The ball reaches a maximum height of 49 m.
(d) The flight of the ball is not symmetric. The time and distance travelled from my hand to the
maximum are smaller than the time and distance travelled from the maximum to the ground. So,
half the total time does not correspond to the time for either part of the whole flight.
7. Solutions may vary. Sample answer:
!
!
(a) Given: vi = 0 m/s; !t = 5.0 s; a = 39.2 m/s 2 [up]
!
Required: vf
vf ! vi
to determine vf = vi + a!t .
&quot;t
Use up as the positive direction.
Solution: v = vi + a!t
= 0 m/s + (39.2 m/s 2 )(5.0 s)
= 196 m/s (
extra digit carried)
2
v = 2.0 &quot; 10 m/s
Analysis: Use a =
Statement: The rocket is moving at 2.0 ! 102 m/s [up] when the engines stop.
!
!
!
(b) Given: vi = 196 m/s [up]; vf = 0 m/s; a = 9.8 m/s 2 [down]
Required: maximum height of the rocket,
f
1
Analysis: Determine the rocket’s height when the engine shuts off using !d = vi !t + a!t 2 .
2
2
2
Then determine the displacement of the rocket while it slows down using vf = vi + 2a!d .
vf2 = vi2 + 2a!d
vf2 &quot; vi2
2a
Together, these will give us the maximum height. Continue to use up as positive.
!d =
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Chapter 1: Kinematics
1.2-18
1
Solution: When the engine has finished firing, the rocket has risen !d = vi !t + a!t 2 .
2
1
!d = vi !t + a!t 2
2
1
= (0 m/s)(5.0 s) + (39.2 m/ s 2 )(5.0 s) 2
2
!d = 490 m
From when the engines stops to maximum height, the rocket goes up.
vf2 &quot; vi2
!d =
2a
(0 m/s) 2 &quot; (196 m/s) 2
=
2(&quot;9.8 m/ s 2 )
!d = 1960 m
The maximum height is
df = di + !d
= 490 m + 1960 m
= 2450 m
df = 2.4 &quot; 103 m
Statement: The maximum height of the rocket is 1.m ! 2 m .
!
!
!
(c) Given: vi = 0 m/s ; di = 2450 m [up] from part (b); a = 9.8 m/s 2 [down]
Required: time to reach the ground, !t , and total time of the flight, t
Analysis: Use vf = vi + a!t to calculate the time taken from the engine stopping to maximum:
!t =
vf &quot; vi
1
. Then, use !d = vi !t + a!t 2 to determine the time to fall to the ground:
a
2
#1 &amp;
&quot;vi &plusmn; vi2 &quot; 4 % a ( (&quot;!d)
\$2 '
!t =
a
Lastly, add the times needed for the three parts of the flight to determine the total time.
v &quot; vi
Solution: !t = f
a
0 m/s &quot; 196 m/s
=
&quot;9.8 m/s 2
!t = 20 s
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-19
For the rocket to fall to the ground from its maximum height:
#1 &amp;
&quot;vi &plusmn; vi2 &quot; 4 % a ( (&quot;!d)
\$2 '
!t =
a
0 &plusmn; 02 &quot; 4(&quot;4.9)(2450)
=
&quot;9.8
= &plusmn;22.36 s (two extra digits carried)
t =
s + 0/ s + 00.26 s
t = 37 s
Statement: The time for the rocket to fall from rest at the maximum is 22 s. The total time from
being at rest initially is 47 s.
8. Solutions may vary. Sample answer:
Table 2
Acceleration
Reaction
Speed
Braking
(m/s2)
time (s)
(km/h)
distance (m)
(i)
9.5
0.80
60.0
28
(ii)
9.5
0.80
120.0
85
(iii)
9.5
2.0
60.0
48
!
!
km/h [forward]; !t =
s; = 9.5 m/s 2 [backward]
(a) (i) Given: v =
Required: braking distance, d
!d
Analysis: The car moves at constant speed during the reaction time. Use v =
to determine
!t
the distance covered during this time: !d = v!t .
The car then slows to a stop. Determine the distance covered during this time using:
vf2 = vi2 + 2a!d
vf2 &quot; vi2
2a
Then, determine the total of these distances travelled. Use forward as the positive direction.
1h
1 min
! 60.0 km 1000 m
v1 =
!
!
!
1h
60 s
1 km
60 min
!
v1 = 16.67 m/s [forward] (two extra digits carried)
!d =
Solution: !d1 = v1!t
= (16.67 m/s ) (0.80 s)
!d1 = 13.34 m (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-20
!d2 =
=
vf2 &quot; vi2
2a
(0 m/s)2 &quot; (16.67 m/s)2
2(&quot;9.5 m/ s 2 )
!
! 1=!
!
13.34 m = 14.63 m
28 m
2
!d2 = 14.63 m (two extra digits carried)
Statement: The braking distance is 28 m.
r
im/h [forward] = 33.33 m/s [forward]; !t2 = 0.80 s;
(ii) Given: v2 =
!
= 9.5 m/s 2 [backward]
Required: braking distance, d
Analysis: Repeat the steps of part (i).
vf2 &quot; vi2
!d4 =
Solution: !d3 = v2 !t2
2a
= ( 33.33 m/s )
s
(0 )2 &quot; (33.33 m/s)2
=
!d3 =
m w
ra
s arr
2(&quot;9.5 m/ s 2 )
!d4 = 58.47 m (two extra digits carried)
d = !d3 + !d4
= 26.66 m + 58.47 m
d = 85 m
Statement: The braking distance is 85 m.
r
(iii) Given: v3 = 60.0
/ or ar = 16.67 m/s or ar
Required: braking distance, d
Analysis: Repeat the steps of part (i).
Solution: !d5 = v3!t3
r
!t3 = 2.0 s a = 9.5 m/s 2 [backward]
= (16.67 m/s )(2.0 s)
!d5 = 33.34 m (two extra digits carried)
v &quot; vi2
!d6 =
2a
(0 m/s)2 &quot; (16.67 m/s)2
=
2(&quot;9.5 m/ s 2 )
2
f
=!
4
+!
6
=
m+
= 38 m
m
!d6 = 14.63 m (two extra digits carried)
Statement: The braking distance is 48 m.
(b) If a driver uses a cellphone while driving, it is likely that the driver’s reaction time and
braking distance will increase. The table shows that at 60 km/h, increasing the reaction time from
0.80 s to 2.0 s causes the braking distance to increase from 28 m to 48 m (almost double).
Similarly, the driving speed affects the braking distance. For a reaction time of 0.80 s, increasing
the speed from 60 km/h to 120 km/h results in the braking distance increasing from 28 m to 85 m
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-21
Section 1.3: Displacement in Two Dimensions
Tutorial 1 Practice, page 25
!
!
1. Given: !d1 = 1.2 67 4 5 !d2 = 3.1 67 4E &deg; N5
!
!
Required: !d( ; the angle for !d( , !
!
!
!
Analysis: !d( = !d1 + !d2 . Decide on a scale and then draw each vector to scale on a coordinate
axis. Draw the total displace7ent vector.
Solution: An appropriate scale is 1 c7 : 0.3 67. Calculate the lengths of the arrows for the
displace7ent vectors:
r
r
1 c7
1 c7
!d1 = 1.2 67 &quot;
!d2 = 3.1 67 &quot;
0.3 67
0.3 67
=
c7
= 10.3 c7
!
!
Using a ruler and protractor, draw the two vectors, placing the tail of !d2 at the tip of !d1 .
!
!
!
Draw the total displace7ent vector !d( fro7 the tail of !d1 to the tip of !d2 . Measure the
length of the vector, and 7easure the angle the displace7ent vector 7a6es to the horizontal.
(he 7easured length of the total displace7ent vector is
!
0.3 67
!d( =
c7 &quot;
= 2.3 67
1 c7
(he 7easured angle ! is 3)&deg; .
Statement: (he total displace7ent is 2.3 67 4E 3)&deg; N5.
Copyright &copy; 2012 Nelson Education Ltd.
c7. Convert the length to 6ilo7etres.
Chapter 1: Kinematics
1.3-1
!
!
2. Given: !d1 =
7 4E5; !d2 =
74 5
!
!
Required: !d( ; the angle for !d( , !
!
!
!
Analysis: !d( = !d1 + !d2 . (o deter7ine the 7agnitude of the displace7ent, use the
Pythagorean theore7. (o calculate angle ! , use the tangent ratio.
!
Solution: Solve for !d( .
!
! 2
! 2
!d( = !d1 + !d2
=
!
!d( = 122 7
7 2+
7
2
Solve for angle ! .
r
!d2
tan &quot; = r
!d1
=
7
7
=
&quot;=
&deg;
Statement: (he boater’s total displace7ent is 122 7 4E
&deg; S5.
!
!
3. Given: !d1 = 67 4N 32&deg; E5; !d2 = )2 67 4E 21&deg; N5
!
!
Required: !d( ; the angle for !d( , !
!
!
!
Analysis: !d( = !d1 + !d2 . (o deter7ine the 7agnitude of the displace7ent, use the cosine law.
(o calculate the angle ! , use the sine law.
Solution: Ma6e a s6etch of the addition of displace7ent vectors.
! 2 = 21&deg; +
&deg; + 32&deg;
! 2 = 1)3&deg;
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-2
Fro7 the cosine law:
a 2 = b2 + c 2 &quot; 2bccos A
! 2
! 2
! !
! 2
#d( = #d1 + #d2 &quot; 2 #d1 #d2 cos! 2
=
67 2 + )2 67 2 &quot; 2
67 )2 67 cos 1)3&deg;
67 three extra digits carrie
! =
#d( = 1.0 \$ 102 67
Fro7 the sine law:
sin
sin A
=
c
a
sin ! 3 sin ! 2
! = !
#d2
#d(
!
#d2 sin ! 2
!
sin ! 2 =
#d(
=
)2 67 sin 1)3&deg;
67
=
! 3 = sin &quot;1
!3 =
&deg; two extra digits carrie
&quot; = &deg; &quot; 32&deg; &quot;
&deg;
&quot; = ))&deg;
Statement: (he helicopter travels 1.0 ! 102 67 4E ))&deg; N5.
Tutorial 2 Practice, page 26
!
1. (a) Given: !d =
67 4E
&deg; N5
Required: !dx; !dy
Analysis: Draw the displace7ent vector, and then use trigono7etry to deter7ine the
co7ponents. Use east and north as positive.
Solution:
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Chapter 1: Kinematics
1.3-3
!
!d = + !d cos&quot;
=+
!d = +
!
!d = + !d sin &quot;
=+
&deg;
67 cos
!d = +
67
&deg;
67 sin
67
Statement: (he co7ponents of the displace7ent are !d =
67 4E5 and !d =
r
(b) Given: !d =
67 4N )2.0&deg; 5
Required: !dx; !dy
Analysis: Draw the displace7ent vector, and then use trigono7etry to deter7ine the
co7ponents. Use east and north as positive.
Solution:
!
#d = &quot; #d sin !
=&quot;
#d = &quot;
!
!d = + !d cos&quot;
67 sin )2.0&deg;
=+
!d = +
67
67 cos )2.0&deg;
67
Statement: (he co7ponents of the displace7ent are !d =
67 4 5 and !d =
!
(c) Given: !d = 32.3 7 4E
&deg; S5
Required: !dx; !dy
Analysis: Draw the displace7ent vector, and then use trigono7etry to deter7ine the
co7ponents. Use east and north as positive.
Solution:
!
!d = + !d cos&quot;
= + 32.3 7 cos
!d = +
67 4N5 .
67 4N5 .
!
#d = &quot; #d sin !
&deg;
7
= &quot; 32.3 7 sin
#d = &quot;
7
Statement: (he co7ponents of the displace7ent are !d =
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&deg;
7 4E5 and !d =
Chapter 1: Kinematics
7 4S5 .
1.3-)
!
(d) Given: !d =
67 4S 31.2&deg; 5
Required: !dx; !dy
Analysis: Draw the displace7ent vector, and then use trigono7etry to deter7ine the
co7ponents. Use east and north as positive.
Solution:
!
#d x = &quot; #d sin !
=&quot;
#d x = &quot;
!
#d y = &quot; #d cos!
=&quot;
67 sin 31.2&deg;
#d y = &quot;
67
67 cos 31.2&deg;
67
Statement: (he co7ponents of the displace7ent are !d =
67 4 5 and !d =
67 4S5 .
Tutorial 3 Practice, page 28
!
!
1. Given: !d1 =
67 4
&deg; S5; !d2 =
67 4
&deg; S5
!
Required: !d
!
!
!
Analysis: !d = !d1 + !d2 . Deter7ine the x- and y-co7ponents of the given displace7ent
vectors. Add these x- and y-co7ponents to calculate the x- and y-co7ponents of the total
displace7ent. Finally, use the Pythagorean theore7 and tangent ratio to deter7ine the total
displace7ent vector. Use east and north as positive.
Solution: For the first vector,
r
#d1 = &quot; #d1 cos!
=&quot;
67 cos
&deg;
#d1 = &quot;
67 one extra digit carrie
r
#d1 = &quot; #d1 sin !
=&quot;
#d1 = &quot;
67 sin
&deg;
67 two extra digits carrie
For the second vector,
!
#d2 = &quot; #d2 cos!
=&quot;
#d2 = &quot;
67 cos
&deg;
67 two extra digits carrie
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Chapter 1: Kinematics
1.3-
!
#d2 = &quot; #d2 sin !
=&quot;
67 sin
#d2 = &quot;
&deg;
67 two extra digits carrie
#d = #d1 + #d2
=&quot;
67 + &quot;
67
67
#d = &quot;
#d = #d1 + #d2
=&quot;
67 + &quot;
#d = &quot;
67
67
Co7bine the total displace7ent co7ponents to deter7ine the total displace7ent.
r
#d = #d 2 + #d 2
r
#d = &quot;
67 2 + &quot;
67 2
r
#d =
67
\$ #d '
! = tan &quot;1 &amp;
)
% #d (
\$
= tan &quot;1 &amp;
%
67 '
)
67 (
!=
&deg;
Statement: (he total displace7ent of the airplane is 830.0 67 4W 29.09&deg; S5.
!
!
2. Given: !d1 = 120 67 4N 32&deg; W5; !d2 =
67 4W 2)&deg; N5
!
Required: !d
!
!
!
Analysis: !d = !d1 + !d2 . Deter7ine the x- and y-co7ponents of the given displace7ent
vectors. Add these x- and y-co7ponents to calculate the x- and y-co7ponents of the total
displace7ent. Finally, use the Pythagorean theore7 and tangent ratio to deter7ine the total
displace7ent vector. Use east and north as positive.
Solution: For the first vector,
!
#d1 = &quot; #d1 sin !
= &quot; 120 67 sin 32&deg;
#d1 = &quot;
67 two extra digits carrie
!
!d1 = + !d1 cos&quot;
= + 120 67 cos 32&deg;
!d1 = +
67 three extra digits carrie
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Chapter 1: Kinematics
1.3-
For the second vector,
!
#d2 = &quot; #d2 cos!
=&quot;
67 cos 2)&deg;
#d2 = &quot;
67 three extra digits carrie
!
!d2 = + !d2 sin &quot;
=+
67 sin 2)&deg;
!d2 = +
67 two extra digits carrie
#d = #d1 + #d2
=&quot;
67 + &quot;
67
67
#d = &quot;
!d = !d1 + !d2
=+
!d = +
67 +
67
67
Co7bine the total displace7ent co7ponents to deter7ine the total displace7ent.
!
!d = !d 2 + !d 2
=
67 2 +
!
!d =
67
2
67
\$ #d '
! = tan &amp;
)
&amp;% #d )(
&quot;1
\$
= tan &quot;1 &amp;
%
67 '
)
67 (
! = 39&deg;
39&deg; N5.
Statement:
!
!
!
3. Given: !d1 = 12 67 4N5; !d2 = 1) 67 4N 22&deg; E5; !d3 = 11 67 4E5
!
Required: !d
!
!
!
!
Analysis: !d = !d1 + !d2 + !d3 . Deter7ine the x- and y-co7ponents of the given displace7ent
vectors. Add these x- and y-co7ponents to calculate the x- and y-co7ponents of the total
displace7ent. Finally, use the Pythagorean theore7 and tangent ratio to deter7ine the total
displace7ent vector. Use east and north as positive.
Solution: For the first vector,
!d1 = 0 67
!d1 = +12 67
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Chapter 1: Kinematics
1.3-
For the second vector,
!
!d2 = + !d2 sin &quot;
= + 1) 67 sin 22&deg;
!d2 = +
67 two extra digits carrie
!
!d2 = + !d2 cos&quot;
= +(14 km)(cos 22&deg;)
!d2 = +12.98 km (two extra digits carried)
For the third vector,
!d3 = 11 km
!d3 = 0 km
!d = !d + !d + !d
= 0 km + 5.244 km + 11 km
!d = +16.244 km
!d = !d + !d + !d
= + km + 12.98 km + 0 km
!d = +24.98 km
Co7bine the total displace7ent co7ponents to deter7ine the total displace7ent.
!
!d = !d + !d
=
!
!d =
!=
=
16.244 km02 + 24.98 km)2
&quot;
km
!
\$ #d '
&quot;3
&amp; ! )
&amp;% #d )(
&quot;3
\$ 24.98 km '
&amp;% 16.244 km )(
!= &deg;
Statement: (he total displace7ent of the helicopter is 3.0 ! 101 km 4E
&deg; N5.
Section 1.3 Questions, page 29
!
!
1. (a) Given: !d3 = 7.81 km [E 50&deg; N]; !d2 = 5.10 km [W 11&deg; N]
!
!
Required: !d ; the angle for !d , !
!
!
!
Analysis: !d = !d1 + !d2 Decide on a scale and then draw each vector to scale on a coordinate
axis. Draw the total displace7ent vector.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-8
Solution: An appropriate scale is 1 c7 : 1 67. Calculate the lengths of the arrows for the
displace7ent:
! 7.81 km
! 5.10 km
!d3
= 7.81 cm; !d2 =
=
m
1 km m
1 km m
!
!
Using a ruler and protractor, draw the two vectors placing the tail of !d2 at the tip of !d1 . Draw
!
!
!
the total displace7ent vector !d fro7 the tail of !d1 to the tip of !d2 . Measure the length of the
vector, and 7easure the angle the displace7ent vector 7a6es to the horizontal.
(he 7easured length of the total displace7ent vector is
c7, 7easured to the nearest
7illi7etre. Convert to 6ilo7etres.
r
1 km
!d = 7.0 cm &quot;
1 cm
= 7.0 km
(he 7easured angle ! is 0&deg; , 7easured to the nearest degree.
67 4N5.
Statement:
(b) Both an algebraic solution and a co7ponent solution are given.
(i) Algebraic!Method:
!
Required: !d ; the angle for !d , !
!
!
!
Analysis: !d = !d1 + !d2 . (o deter7ine the 7agnitude of the displace7ent, use the cosine law.
(o calculate the angle ! , use the sine law.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-9
Solution: Ma6e a s6etch of the addition of displace7ent vectors.
(he angle ! 2 is
! 2 = 50&deg; + 11&deg;
! 2 = 61&deg;
Fro7 the cosine law,
a 2 = b2 + c 2 &quot; 2bccos A
! 2
! 2
! !
!2
#d = #d1 + #d2 &quot; 2 #d1 #d2 cos! 2
= (7.81 km)2 + (5.10 km)2 &quot; 2(7.81 km)(5.10 km)(cos 61&deg;0
!2
#d = 48.385 m 2
!
#d = 6.956 m (two extra digits carried)
!
#d = 7.0 m
Fro7 the sine law,
si
si A
=
c
a
si ! 3 si ! 2
! =
!
#d2
#d
!
#d2 si ! 2
!
si ! 3 =
#d
=
(5.10 m )(si 61&deg;)
(6.956 m )
= 0.6412
! 3 = si
&quot;1
0.6412
! 3 = 39.88&deg;
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-10
! = 50&deg; + 39.88&deg;
! = 90&deg;
Statement:
(ii) Component
! Method:
Required: !d
!
!
!
Analysis: !d = !d1 + !d2 . Deter7ine the x- and y-co7ponents of the given displace7ent
vectors. Add these x- and y-co7ponents to calculate the x- and y-co7ponents of the total
displace7ent. Finally, use the Pythagorean theore7 and tangent ratio to deter7ine the total
displace7ent vector. Use east and north as positive.
Solution: For the first vector,
!
#d1 = + #d1 cos!
= +(7.81
)(cos 50&deg;)
(two extra digits carried)
#d1 = &quot;5.020
!
!d1 = + !d1 si &quot;
= +(7.81
!d1 = +5.983
)(si 50&deg;)
(two extra digits carried)
For the second vector,
!
#d2 = &quot; #d2 cos!
= &quot;(5.10
)(cos 11&deg;)
(two extra digits carried)
#d2 = &quot;5.006
!
!d2 = + !d2 si &quot;
= +(5.10
!d2 = +0.9732
)(si 11&deg;)
(two extra digits carried)
#d = #d1 + #d2
= +5.020
+ (&quot;5.006
)
#d = +0.014
!d = !d1 + !d2
= +5.983
!d = +6.9562
+ 0.9732
Co7bine the total displace7ent co7ponents to deter7ine the total displace7ent.
!
!d = !d 2 + !d 2
=
(0.014
!
!d = 7.0
) + (6.9562
2
)
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-11
! = ta
&quot;1
= ta
&quot;1
\$ #d '
)
&amp;
&amp;% #d )(
\$ 6.9562
&amp;
% 0.014
'
)
(
! = 90&deg;
Statement:
because care was ta6en to carry extra digits while calculating. In both cases, the final answer was
given to two significant digits, consistent with the given infor7ation. It is so7ewhat surprising
7easure an angle to 1&deg; precision. It is also difficult to draw arrows tip to tail to 1 77 precision.
!
!
!
2. Given: !d1 = 5.0 cm 30.0&deg;
!d2 = 7.5 cm
!d3 = 15.0 cm 10.0&deg;
!
!
Required: !d ; the angle for !d , !
!
!
!
!
Analysis: !d = !d1 + !d2 + !d3 . Decide on a scale and then draw each vector to scale on a
coordinate axis. Draw the total displace7ent vector.
Solution: An appropriate scale is 1 c7 : 2 c7. (he lengths of the arrows for the displace7ent
vectors are
!
1 cm
!d1 = 5.0 cm &quot;
= 2.5 cm
2 cm
!
1 cm
!d2 = 7.5 cm
= 3.75 cm
2 cm
!
1 cm
!d3 = 15.0 cm
= 7.5 cm
2 cm
!
!
Using a ruler and protractor, draw the three vectors placing the tail of !d2 at the tip of !d1 , and
!
!
!
!
the tail of !d3 at the tip of !d2 . Draw the total displace7ent vector !d fro7 the tail of !d1 to
!
the tip of !d3 . Measure the length of the displace7ent vector, and 7easure the angle the
displace7ent vector 7a6es to the horizontal.
(
(
)
)
(he 7easured length of the total displace7ent vector is
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-12
Convert bac6 to the original scale.
!
2 cm
!d = 13.5 cm &quot;
= 27 cm
1 cm
(he 7easured angle ! is 0&deg; , to the nearest degree.
Statement:
c7 4E5.
!
3. Given: !d = 2.50 m
38.0&deg;
Required: !d ; !d
Analysis:
Draw the displace7ent vector, and then use trigono7etry to deter7ine the co7ponents. Use east
and north as positive.
!
!
Solution: #d = &quot; #d si !
!d = + !d cos&quot;
= &quot;(2.50 m)(si 38.0&deg;)
#d = &quot;1.54 m
= +(2.50
)(cos 38.0&deg;)
!d = +1.97 m
Statement: (he co7ponents of the displace7ent are !d = 1.54 m
!
4. Given: !d = 25.0 m 30.0&deg;
and !d = 1.97 m
.
Required: !d ; !d
Analysis: Draw the displace7ent vector, and then use trigono7etry to deter7ine the
co7ponents. Use east and north as positive.
Solution:
!
!d = + !d cos&quot;
= +(25.0 m)(cos 30.0&deg;)
!d = +21.6 m
!
!d = + !d si &quot;
= +(25.0 m)(si 30.0&deg;)
!d = +12.5 m
Statement: (he co7ponents of the displace7ent are !d = 21.6 m
Copyright &copy; 2012 Nelson Education Ltd.
and !d = 12.5 m
Chapter 1: Kinematics
.
1.3-13
!d = 24 m
5. Given: !d = 54 m
!
Required: !d
Analysis: Use the Pythagorean theore7 to deter7ine the length of the displace7ent vector. Use
east and north as positive.
!
Solution: !d = !d 2 + !d 2
= (54 m ) + ( 24 m )
2
2
!
!d = 59 m
Statement:
(b) Required: !
Analysis: Use the tangent ratio to calculate ! .
!
\$ #d '
Solution: ! = ta &quot;1 &amp; ! )
&amp;% #d )(
= ta
&quot;1
\$ 24 m '
&amp;% 54 m )(
! = 24&deg;
Statement: (he vector points 4E 24&deg; N5.
!
!
!
6. Given: !d1 = 15.0
!d2 = 45.0
S !d3 = 32
25&deg;
!
Required: !d
!
!
!
!
Analysis: !d = !d1 + !d2 + !d3 . Deter7ine the x- and y-co7ponents of the given displace7ent
vectors. Add these x- and y-co7ponents to calculate the x- and y-co7ponents of the total
displace7ent. Finally, use the Pythagorean theore7 and tangent ratio to deter7ine the total
displace7ent vector. Use east and north as positive.
Solution: For the first vector,
#d1 = &quot;15.0
!d1 = 0 67
For the second vector,
!d2 = 0
#d2 = &quot;45.0
For the third vector,
!
#d3 = &quot; #d3 si !
= &quot;(32
)(si 25&deg;)
(two extra digits carried)
#d3 = &quot;13.52
!
!d3 = + !d3 cos&quot;
= +(32
!d3 = +29.00
)(cos 25&deg;)
(two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-1)
#d = #d1 + #d2 + #d3
= (&quot;15.0
)+0
+ (&quot;13.52
)
#d = &quot;28.52
#d = #d1 + #d2 + #d3
=0
+ (&quot;45.0
) + 29.00
#d = &quot;16.00
Co7bine the total displace7ent co7ponents to deter7ine the total displace7ent.
!
!d = !d 2 + !d 2
=
( 28.52
!
!d = 33
)2 + (16.00
)2
!
\$ #d '
! = ta &quot;1 &amp; ! )
&amp;% #d )(
= ta
&quot;1
\$ 16.00
&amp;
% 28.52
'
)
(
! = 29&deg;
Statement: (he total displace7ent of the driver is 33 67 4W 29&deg; S5.
!
!
!
7. Given: !d1 = 2.5 m
30.0&deg;
!d2 = 3.6 m S !d3 = 4.9 m 38.0&deg;
!
Required: !d
!
!
!
!
Analysis: !d = !d1 + !d2 + !d3 . Deter7ine the x- and y-co7ponents of the given displace7ent
vectors. Add these x- and y-co7ponents to calculate the x- and y-co7ponents of the total
displace7ent. Finally, use the Pythagorean theore7 and tangent ratio to deter7ine the total
displace7ent vector. Use east and north as positive.
Solution: For the first vector,
!
#d1 = &quot; #d1 cos!
= &quot;(2.5 m)(cos 30.0&deg;)
#d1 = &quot;2.165 m (two extra digits carried)
!
#d1 = &quot; #d1 si !
= &quot;(2.5 m)(si 30.0&deg;)
#d1 = &quot;1.250 m (two extra digits carried)
For the second vector,
!d2 = 0 m
#d2 = &quot;3.6 m
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-
For the third vector,
!
!d3 = + !d3 cos&quot;
= +(4.9 m)(cos 38.0&deg;)
!d3 = +3.861 m (two extra digits carried)
!
#d3 = &quot; #d3 si !
= &quot;(4.9 m)(si 38.0&deg;)
#d3 = &quot;3.017 m (two extra digits carried)
#d = #d1 + #d2 + #d3
= (&quot;2.165 m) + 0 m + 3.861 m
#d = 1.696 m
#d = #d1 + #d2 + #d3
= &quot;1.250 m + (&quot;3.6 m) + (&quot;3.017 m)
#d = &quot;7.867 m
Co7bine the co7ponents to deter7ine the total displace7ent vector.
!
!d = !d 2 + !d 2
=
(1.696 m )2 + ( 7.867 m )2
!
!d = 8.0 m
!
\$ #d
! = ta &quot;1 &amp; !
&amp;% #d
= ta
&quot;1
'
)
)(
\$ 7.867 m '
&amp;% 1.696 m )(
! = 78&deg;
Statement: (he total displace7ent is 8.0 7 4E 78&deg; S5.
!
!
8. (a) Given: !d1 = 2.70
25.0&deg;
!d2 = 4.80
45.0&deg;
Required: !d !d
!
!
!
Analysis: !d = !d1 + !d2 . Deter7ine the x- and y-co7ponents of the given displace7ent
vectors. Add these x- and y-co7ponents to calculate the x- and y-co7ponents of the total
displace7ent. Use east and north as positive.
Solution: For the first vector,
!
!d1 = + !d1 cos&quot;
= +(2.70
!d1 = +2.4470
)(cos 25.0&deg;)
(two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-
!
!d1 = + !d1 si &quot;
= +(2.70
!d1 = +1.1411
)(si 25.0&deg;)
(two extra digits carried)
For the second vector,
!
!d2 = + !d2 cos&quot;
= +(4.80
)(cos 45.0&deg;)
(two extra digits carried)
!d2 = +3.3941
!
#d2 = &quot; #d2 sik !
= &quot;(4.80 km)(sik 45.0&deg;)
#d2 = &quot;3.3941 km (two extra digits carried)
!d = !d1 + !d2
= +2.4470 km + 3.3941 km
= +5.841 km
!d = +5.84 km
#d = #d1 + #d2
= +1.1411 km + (&quot;3.3941 km)
= &quot;2.2530 km
#d = &quot;2.25 km
Statement: (he co7ponents of the boat’s displace7ent are !d = 5.84 km [E]
and !d = 2.25 km [ .
!
(b) Required: !d
Analysis: Use the Pythagorean theore7 and tangent ratio to deter7ine the total displace7ent
vector.
!
!d = !d 2 + !d 2
=
(5.8411 km )2 + ( 2.2530 km )2
!
!d = 6.26 km
\$ #d '
! = tak &quot;1 &amp;
)
&amp;% #d )(
\$ 2.2530 km '
= tak &quot;1 &amp;
)
% 5.8411 km (
! = 21.1&deg;
Statement:
Copyright &copy; 2012 Nelson Education Ltd.
21.1&deg; S5.
Chapter 1: Kinematics
1.3-
!
!
9. Given: !d1 = 1512.0 km [W 19.30&deg; N]; !d2 = 571.0 km [W 4.35&deg; N];
!
!d3 = 253.1 km [W 39.39&deg; N]
!
Required: !d
!
!
!
!
Analysis: !d = !d1 + !d2 + !d3 . Deter7ine the x- and y-co7ponents of the given displace7ent
vectors. Add these x- and y-co7ponents to calculate the x- and y-co7ponents of the total
displace7ent. Finally, use the Pythagorean theore7 and tangent ratio to deter7ine the total
displace7ent vector. Notice that all of the displace7ents point so7ewhat west and north. Use
west and north as positive.
Solution: For the first vector,
!
!d1 = !d1 cos&quot;
= (1512.0 km)(cos 19.30&deg;)
!d1 = 1427.027 km (two extra digits carried)
!
!d1 = !d1 sik &quot;
= (1512.0 km)(sik 19.30&deg;)
!d1 = 499.738 km (two extra digits carried)
For the second vector,
!
!d2 = !d2 cos&quot;
= (571.0 km)(cos 4.35&deg;)
!d2 = 569.355 km (two extra digits carried)
!
!d2 = !d2 sik &quot;
= (571.0 km)(sik 4.35&deg;)
!d2 = 43.310 km (two extra digits carried)
For the third vector,
!
!d3 = !d3 cos&quot;
= (253.1 km)(cos 39.39&deg;)
!d3 = 195.607 km (two extra digits carried)
!
!d3 = !d3 sin &quot;
=
!d3 =
67 sin 39.39&deg;
67 two extra digits carrie
!d = !d1 + !d2 + !d3
= 1427.027 km + 569.355 km + 195.607 km
!d = 2191.989 km
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-18
!d = !d1 + !d2 + !d3
= 499.738 km + 43.310 km + 160.616 km
!d = 703.664 km
Co7bine the co7ponents to deter7ine the total displace7ent vector.
!
!d = !d 2 + !d 2
= ( 2191.989 km ) + ( 703.664 m )
= 2302.164 km
2
2
!
!d = 2.30 &quot; 103 km
\$ #d '
! = tak &quot;1 &amp;
)
&amp;% #d )(
\$ 703.664 km '
= tak &quot;1 &amp;
)
% 2191.989 km (
! = 17.8&deg;
Statement: (he total displace7ent is 2.30 ! 103 km 4W 17.8&deg; N5.
r
r
10. Given: !d1 = 25 km [N]; !d = 62 km
38&deg; W]
!
Required: !d2
r
r
r
Analysis: !d( = !d1 + !d2 . Deter7ine the x- and y-co7ponents of the given displace7ent
vectors. Subtract these x- and y-co7ponents to calculate the x- and y-co7ponents of the second
displace7ent. Finally, use the Pythagorean theore7 and tangent ratio to deter7ine the second
displace7ent vector. Use east and north as positive.
Solution: For the first vector:
!d1 = 0 67
!d1 = 25 km
For the total vector,
!
#d = &quot; #d sik !
= &quot;(62 km)(sik 38&deg;)
#d = &quot;38.171 km (two extra digits carried)
!
!d = !d cos&quot;
= (62 km)(cos 38&deg;)
!d = 48.857 km (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-19
Subtract the horizontal co7ponents.
#d2 = #d &quot; #d1
= (&quot;38.171 km) + (0 km)
#d2 = &quot;38.171 km
Subtract the vertical co7ponents.
#d2 = #d &quot; #d1
= (48.857 km) + (25 km)
#d2 = 23.857 km
Co7bine the displace7ent co7ponents of the second vector to deter7ine the second
displace7ent.
!
#d2 = #d22 + #d22
=
( &quot;38.171 km )2 + ( 23.857 km )2
!
#d2 = 45 km
!
\$ #d
2
! = tak &quot;1 &amp; !
&amp;% #d2
'
)
)(
\$ 23.857 km '
= tak &quot;1 &amp;
)
% 38.171 km (
! = 32&deg;
32&deg; N5.
Statement:
11. Answers 7ay vary. Sa7ple answer: (he length of the total displace7ent vector is deter7ined
by how the two displace7ent vectors line up.
If the second vector points west, parallel to the first vector, then the total displace7ent is as large
as possible: 450 km [W] + 220 km [W] = 670 km [W]
If the second vector points east, opposite to the first vector, then the total displace7ent is as
s7all as possible: 450 km [W] + 220 km [E] = 230 km [W]
If the second vector points in any other direction, the 7agnitude of the total displace7ent is
between 230 67
not 7atter what order you use when you add two vectors. You can put the tail of the second to
the tip of the first, or you can put the tail of the first to the tip of the second. You get the sa7e
total vector either way.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-20
Section 1.4: Velocity and Acceleration in Two Dimensions
Tutorial 1 Practice, page 32
!
!
!
1. (a) Given: !-1 = 32.0 ;A 87
&deg; 695 !-2 =
;A 8695 !- =
;A 8795 !t =
h
!
Required: !!
!
!
!
Analysis: !- = !-1 + !-2 + !- . DeterAine the x and y coAponents of the displaceAent
vectors. Then deterAine the x and y coAponents of the total displaceAent using
!- x = !-1x + !-2x + !- x and !- y = !-1y + !-2y + !- y . Finally, coAbine these coAponents to
deterAine the total displaceAent vector using the Pythagorean theoreA and the tangent ratio.
Use east and north as positive.
!
Solution: x coAponent of !- :
!- x = !-1x + !-2x + !- x
= &quot; 32.0 ;A cos
=&quot;
&deg; + 0 ;A + &quot;
;A + 0 ;A + &quot;
;A
;A
!- x = &quot;212.4 ;A tHo eItra digits carrie
!
y coAponent of !- :
!- y = !-1y + !-2y + !- y
&deg; + (&quot;
= &quot; 32.0 ;A si
=&quot;
!- y = &quot;
;A + &quot;
;A ) + 0 ;A
;A + 0 ;A
;A one eItra digit carrie
CoAbine the total displaceAent coAponents to deterAine the total displaceAent.
!
!- = !- x2 + !- y2
=
&quot;212.4 ;A 2 + &quot;
=
!
!- =
;A
2
;A tHo eItra digits carrie
;A
\$ &quot;- '
y
! = tan &amp;
)
&amp;% &quot;- x )(
!1
\$
;A '
= tan !1 &amp;
)
% 212.4 ;A (
! = 22
22 69.
Statement:
!
(b) Given: !- = 2/0 ;A 87
695 !t =
h
!
Required: vav
!
!!
to deterAine the average velocity.
Analysis: Use vav =
!t
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-1
!
!!
Solution: vav =
!t
=
;A 87 22 69
h
;A h 87 22 69
!
vav =
Statement:
!
!
(c) Given: !-1 = 32.0 ;A 87 /0.0 695 !-2 =
22 69.
!
;A 8695 !-/ =
;A 8795 !t =
h
Required: vav
Analysis: Calculate the total distance travelled and then use vav =
speed.
Solution: !- = !-1 + !-2 + !-/
!to deterAine the average
!t
!!t
230.0 ;A
=
h
=
;A h
vav = 110 ;A h
vav =
= 32.0 ;A +
;A +
;A
!- = 230.0 ;A tHo eItra digits carrie
Statement: The average speed of the plane is
.
!
!
!
2. (a) Given: !-1 =
;A 8E
N95 !-2 = 20.0 ;A 8695 !-/ =
!
Required: vav
;A 8795 !t = 12 h
!
!
!
!
Analysis: Calculate the total displaceAent using coAponents and !- = !-1 + !-2 + !-/ . Then,
!
!!
calculate the average velocity using vav =
. Use east and north as positive.
!t
!
Solution: x-coAponent of !- :
&quot;- x = &quot;-1x + &quot;-2x + &quot;-/x
=
=
;A cos
;A + 0 ;A + !
+ 0 ;A + !
;A
;A
&quot;- x = 0 ;A
!
y-coAponent of !- :
&quot;- y = &quot;-1y + &quot;-2y + &quot;-/y
=
;A si
+ !20.0 ;A + 0 ;A
= 20.0 ;A + !20.0 ;A + 0 ;A
&quot;- y = 0 ;A
!
nt is !- = 0 ;A .
Statement:
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-2
(b) Given: !-1 =
;A5 !-2 = 20.0 ;A5 !-/ =
;A5 !t = 12 h
Required: vav
Analysis: Calculate the total distance travelled and then use vav =
speed.
Solution: !- = !-1 + !-2 + !-3
= 25.0 km + 20.0 km + 15.0 km
!- = 60.0 km (one extra digit carried)
!to deterAine the average
!t
!!t
60.0 km
=
18 h
vav = 5.0 km/h
vav =
Statement:
(c)
did not Aove
Tutorial 2 Practice, page 34
!
!
1. Given: vi = 20.0 m/s [E]; vf = 20.0 m/s [S]; !t = 12 s
!
Required: aav
Analysis: DraH a vector diagraA of the situation. Calculate the change in velocity using
!
!
!
! ! !
. Use east
coAponents and &quot;v = vf ! v . Then, deterAine the average acceleration using aav =
!
and north as positive.
Solution:
!
x-coAponent of !v :
&quot;v = f !
= 0 m/s ! 20.0 m/s
&quot;
= !20.0 m/s
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-/
!
y-coAponent of !v :
&quot;v = f !
= !20 m/s ! 0 m/s
&quot;
= !20.0 m/s
DeterAine the change in velocity froA its coAponents.
!
&quot;v = &quot;v 2 + &quot; 2
=
!
&quot; =
( !20.0 m/s )2 + ( !20.0 m/s )2
m/s
o extra digits
d)
\$ &quot;v '
! = tan !1 &amp;
)
&amp;% &quot; )(
\$ 20.0 m/s '
= tan !1 &amp;
)
% 20.0 m/s (
! = 45
Calculate the average acceleration.
!
!
!
aav =
!
/ nm 45 as
=
01
!
1
at = 1.4 / nm 45 as
Statement: The average acceleration of the car is 1.4 / 1 nm 45 as .
!
!
2. Given: v =
km/h nm 60.0 s v = 80.0 km/h n 60.0 s !t =
=
!
Required: aav
Analysis: DraH a vector diagraA of the situation. Calculate the change in velocity using
!
!
!
! ! !
coAponents and &quot;v = v ! v . Then, deterAine the average acceleration using aav =
. Use east
!
and north as positive.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-4
Solution:
!
The x-coAponent of !v is
&quot;v =
!
= 80.0 km/h)(cos 60.0 ) ! (!(50.0 km/h)cos 60.0 )
= 40.0 km/h ! (!25.0 km/h)
&quot; = 65.0 km/h
!
The y-coAponent of !v is
!
&quot;v =
&quot;
=(
km/h)(s
=
=
km/h !
km/h (
60.0 ) ! (50.0 km/h)(s
60.0 )
km/h
c
DeterAine the change in velocity froA its coAponents.
!
!v = !v 2 + ! 2
=
!
! =
(65.0 km/h )2 + ( 25.f8 km/h )2
km/h (one extra digit carried)
\$ &quot;v '
! = i !1 &amp;
)
&amp;% &quot; )(
= t
!0
\$ 25.f8 km/h '
&amp;
)
% 65.0 km/h (
!=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-
Calculate the average acceleration.
!
!
!
aav =
!
/ [E
]
=
!
a =
&quot; 102
/
2
[E
]
Statement: The average acceleration of the truc; is 2.80 ! 102 km/h 2 n
s.
!
!
3. (a) Given: !-1 = 800.0 km n
as !-1 =
km n 50 as !t = 18.0 h
Required: !Analysis: Calculate the total distance travelled by adding the individual distances
travelled, !- = !-1 + !-2 .
Solution: !- = !-1 + !-2
= 800.0 km + 400.0 km
&quot; 102 km
!- =
Statement: The total distance travelled by the bird is 1.2 ! 10/ ;A.
!
!
(b) Given: !-1 = 800.0 km nE 7.5 S]; !-2 = 400.0 km [E 51 S]
!
Required: !!
!
!
Analysis: !- = !-1 + !-2 . DeterAine the x- and y-coAponents of the displaceAent vectors.
Then deterAine the x- and y-coAponents of the total displaceAent using !- x = !-1x + !-2x
and !- y = !-1y + !-2y . Finally, coAbine these coAponents to deterAine the total displaceAent
vector using the Pythagorean theoreA and the tangent ratio. Use east and north as positive.
!
Solution: The x-coAponent of !- is
!- x = !-1x + !-2x
=
km
os 7.5 + 400.0 km
os 51
= 793.1 km + 251.7 km
!- x = 1045 km (two extra digits carried)
!
The y-coAponent of !- is:
&quot;- y = &quot;-1y + &quot;-2y
= ! 400.0 km
+ !
km
= !104.4 km + !310.9 km)
&quot;- y = !415.3 km (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-
CoAbine the total displaceAent coAponents to deterAine the total displaceAent.
!
!- = !- x2 + !- y2
=
#1045 km 2 + #
km
2
= 1124 km
!
!- = 1.1&quot; 10 km
\$ &quot;- '
y
! = 8i !1 &amp;
)
&amp;% &quot;- x )(
\$ 415.3 km '
= ta8 !1 &amp;
)
% 1045 km (
! = 22
Statement: The total displaceAent is 1.1! 103 km 8E 22 69.
(c) Given: !t = 18.0 !- = 1200
Required: vav
Analysis: Use vav =
!to deterAine the average speed.
!t
!!t
1200 km
=
18.0 h
vav = 67 km/h
Statement:
Solution: vav =
(d) Given: !t = 18.0
!
Required: vav
r
!- = 0012 km 8 11o a
!
!!
to deterAine the average velocity.
Analysis: Use vav =
!t
r
r
!Solution: vav =
!t
1124 km [E 22 S]
=
18.0 h
r
vav = 61 km/h 8 11 a
Statement: The average velocity of the bird is 61 km/h 8 11 a .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-3
Section 1.4 Questions, page 35
1.
Changes in direction do not affect the average speed. The average velocity is based on the
displaceAent. The displaceAent is affected by changes in direction: it acts li;e the short-cut
because it bac;trac;ed. You do not see the individual displaceAents of its Hal;. The dog’s
sloHly.
!
!
2. (a) Given: !-1 = 25.0 m [E
!-1 =
km v8
a !t =
mi8 = 240 s
Required: !Analysis: Calculate the total distance travelled by adding the individual distances
travelled, !- = !-1 + !-2 .
Solution: !- = !-1 + !-2
= 25.0 m + 75.0 m
!- = 1.0 &quot; 102 m
Statement: The total distance travelled by the loon is 1.0 ! 102 A.
!
!
(b) Given: !-1 = 25.0 m [E 30.0 ]; !-2 = 75.0 km [E 45.0 S]
!
Required: !!
!
!
Analysis: !- = !-1 + !-2 5 DeterAine the x- and y-coAponents of the displaceAent vectors.
Then deterAine the x- and y-coAponents of the total displaceAent using !- x = !-1x + !-2x
and !- y = !-1y + !-2y . Finally, coAbine these coAponents to deterAine the total displaceAent
vector using the Pythagorean theoreA
and the tangent ratio. Use east and north as positive.
!
Solution: The x-coAponent of !- is
!- x = !-1x + !-2x
= (25.0 m)(cos 30.0 ) + (75.0 m)(cos 45.0 )
=
m + 53.0330 m
!- x = 74.684 m (two extra digits carried)
!
The y-coAponent of !- is
&quot;- y = &quot;-1y + &quot;-2y
= (25.0 m)(si
&deg;) + (!(75.0 m)si
= 12.5 m + ( !53.0330 m )
&deg;)
&quot;- y = !40.533 m (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-
CoAbine the total displaceAent coAponents to deterAine the total displaceAent.
r
&quot;- = &quot;- x2 + &quot;- y2
= (74.684 m)2 + (!40.533 m)2
=
m (two extra digits carried)
r
&quot;- = 85.0 m
\$ &quot;- '
y
! = tai &amp;
)
&amp;% &quot;- x )(
!1
\$ 40.533 km '
= tai !1 &amp;
)
% 74.684 km (
! = 28.5&deg;
Statement: The loon’s
(c) Given: !- = 1.0 &quot; 102 m; !t = 4.0 mii = 240 s
Required: vav
Analysis: Use vav =
28.5&deg; 69.
!to deterAine the average speed.
!t
!!t
1.0 &quot; 102 m
=
240 s
vav = 0.42 m/s
Solution: vav =
Statement:
!
(d) Required: vav
!
!!
to deterAine the average velocity.
Analysis: Use vav =
!t
!
!!
Solution: vav =
!t
84.9739 m [E 28.5&deg; S]
=
240 s
!
vav = 0.35 m/s [E 28.5&deg; S]
Statement: The average velocity of the loon is 0.35 m/s [E 28.5&deg; S] .
!
!
3. (a) Given: !-1 = 15.0 km [m 30.0&deg; ]; !-2 = 10.0
[ 75.0&deg; ];
!
i
!-3 = 10.0 km [E 70.0&deg; ]; !t =
!
(b) Required: vav
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-
!
!
!
!
Analysis: Calculate the total displaceAent using coAponents and !- = !-1 + !-2 + !-/ . Then,
!
!!
calculate the average velocity using vav =
. Use east and north as positive.
!t
!
Solution: The x-coAponent of !- is
&quot;- x = &quot;-1x + &quot;-2x + &quot;-3x
= !(15.0 km)(cos 30.0&deg;) + (!(10.0 km)cos 75.0&deg;) + (10.0 km)(cos 70.0&deg;)
= !12.990 km + ( !
km ) + 3.420 km
&quot;- x = !12.158 km (two extra digits carried)
!
The y-coAponent of !- is
!- y = !-1y + !-2y + !-3y
= +(15.0 km)(si8 30.0&deg;) + (10.0 km)(si8 75.0&deg;) + (10.0 km)(si8 70.0&deg;)
= +7.5 km + f.65f km +
km
!- y = +26.556 km (two extra digits carried)
CoAbine the total displaceAent coAponents to deterAine the total displaceAent.
!
&quot;- = &quot;- x2 + &quot;- y2
= (!12.158 km)2 + (26.556 km)2
!
&quot;- = 29.207 km
\$ &quot;- '
y
! = tai !1 &amp;
)
&amp;% &quot;- x )(
\$ 26.556 km '
= tai !1 &amp;
)
% 12.158 km (
! = 65.4&deg;
The average
velocity is
!
!!
vav =
!t
29.207 km [m 65.4&deg; N]
=
0.50 h
!
vav = 58 km/h [W 65.4&deg; N]
Statement:
[m 65.4&deg; N] .
4.
change Aay involve a change in speed or a change in direction. Going around a curve at constant
speed is a situation Hhere there is average acceleration and no change in speed.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-10
5.
aAple ansHers: One Hay to subtract vectors is to use coAponents.
DeterAine the coAponents of each vector. 6ubtract the x-coAponents and then the
y-coAponents. B
second vector. DraH the resulting vector to coAplete the triangle. Then solve the triangle.
!
!
6. Given: !-1 = 150 km [E 12&deg; N]7 !vav =
km/h [N &deg; ]7 !t2 = 2.0 h7 !t =
h
!
Required: !-2
!
!!
Analysis: Use vav =
to calculate the total displaceAent. Then deterAine the x- and
!t
y-coAponents of the first and total displaceAent vectors. 6ubtract these x- and y-coAponents to
!
!
!
deterAine the x- and y-coAponents of the second displaceAent, !- = !-1 + !-2 . Finally, use the
Pythagorean theoreA and tangent ratio to deterAine the second displaceAent vector. Use east
and north as positive.
!
!
Solution:
DeterAine
the
total
displaceAent
.
!
!!
vav =
! !!t
!- = vav !t
= (130 km/h [N 32&deg; E])(3.0
!
!- = 390 km [N 32&deg; E]
!
The x-coAponent of !-2 is
&quot;-2x = &quot;- x ! &quot;-1x
= (390 km)(sii 32&deg;) ! (150 km)(cos12&deg;)
= 206.67 km ! 146.72 km
&quot;-2x = 5f.f5 km (two extra digits carried)
!
The y-coAponent of !-2 is:
&quot;-2y = &quot;- y ! &quot;-1y
= (390 km)(cos 32&deg;) ! (150 km)(si8 12&deg;)
= 330.74 km ! 31.19 km
xtra digits carried)
&quot;-2y = 299.55 km (t
CoAbine the displaceAent coAponents of the second vector to deterAine the second
displaceAent.
!
!-2 = !-2x2 + !-2y2
=
=
(5f.f5 km )2 + ( 2ff.55 km )2
km
!
!-2 = 3.1&quot; 102 km
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-11
\$ &quot;- '
2y
! = tai !1 &amp;
)
&amp;% &quot;-2x )(
\$ 2ff.55 km '
= tai !1 &amp;
)
% 5f.f5 km (
!= &deg;
Statement: The second displaceAent is 3.1! 102 km 8N 11&deg; E9.
!
!
7. Given: vav = 3.5 m/s; ! 1 = 1.8 km [E]; ! 2 =
km [N &deg; 9]
Required: !t
Analysis: The jogger returns to his starting place, so the total displaceAent is 0 A,
!
!
!
!-1 + !-2 + !-3 = 0 m . Use vector subtraction by coAponents to deterAine the Aagnitude of the
third displaceAent vector. Use !- = !-1 + !-2 + !-/ to calculate the total distance the jogger
!. Use east and north as positive.
ran. Finally deterAine the tiAe ta;en using vav =
!t
!
Solution: The x-coAponent of !-3 is
&quot;-1x + &quot;-2x + &quot;-3x = 0 m
&quot;-3x = !&quot;-1x ! &quot;-2x
= !(1.8 km) ! (2.6 km)(sii35&deg;)
= !(1.8 km) ! 1.491 km
&quot;-3x = !3.291 km (two extra digits carried)
!
The y-coAponent of !-3 is
&quot;-1y + &quot;-2y + &quot;-3y = 0 m
&quot;-3y = !&quot;-1y ! &quot;-2y
= !(0 km) ! (2.6 km)(cos 35&deg;)
= !(0 km) ! 2.130 km
&quot;-3y = !2.130 km (two extra digits carried)
CoAbine these displaceAent coAponents to deterAine the Aagnitude of the third displaceAent.
!
&quot;- = &quot;- x2 + &quot;- y2
=
( !3.291 km )2 + ( 2.130 km )2
!
&quot;- = 3.920 km
The total distance run is
!- = !-1 + !-2 + !-3
= (1.8 km) + (2.6 km) + (3.920 km)
!- = 8.320 km
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-12
The tiAe ta;en is
!!t =
vav
8.320 km 1000 m
&quot;
3.5 m/s
1 km
= 2377 s
=
!t = 2.4 &quot; 103 s
Statement: The jogger’s run ta;es 2.4 ! 103 s .
r
r
8. Given: v1 = 6.4 m/s [E 30.0&deg; S]; v2 = 8.5 m/s [E 30.0&deg; N]7 !t = 3.8 s
!
Required: aav
! ! !
Analysis: Calculate the change in velocity using coAponents and &quot;v = vf ! vi . Then deterAine
!
!
!
. Use east and north as positive.
the average acceleration froA aav =
!
!
Solution: The x-coAponent of !v is
&quot;v = f ! i
= (8.5 m/s)(cos 30.0&deg;) ! (6.4 m/s)(cos 30.0&deg;)
= 7.361 m/s ! 5.543 m/s
&quot; = 1.818 m/s (two extra digits carried)
!
The y-coAponent of !v is
&quot;v = f ! i
= (8.5 m/s)(sii 30.0&deg;) ! (!(6.4 m/s)(sii 30.0&deg;))
= 4.25 m/s ! (!3.2 m/s)
&quot;
= 7.45 m/s (o8e extra digit carried)
DeterAine the change in velocity froA its coAponents.
!
!v = !v 2 + ! 2
=
(1.818 m/s )2 + ( 7.45 m/s )2
!
! = 7.669 m/s
\$ &quot;v '
! = tai !1 &amp;
)
&amp;% &quot; )(
\$ 7.45 m/s '
= tai !1 &amp;
)
% 1.818 m/s (
! = 76&deg;
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-1/
Calculate the average acceleration.
!
!
!
aav =
!
7.669 m/s 76&deg;
=
s
!
2
76&deg;
a = 2.1 m/s
Statement: The average acceleration of the bird is 2.0 m/s 2 [E 76&deg;
!
!
9. Given: vi = 50.0 m/s [8 v = 45.1 m/s [S]; !t = 45.0 s
!
Required: aav
.
! ! !
Analysis: DraH a vector diagraA shoHing the change in velocity using &quot;v = vf ! vi . 6olve the
!
!
!
triangle using trigonoAetry. Then deterAine the average acceleration froA aav =
. Use east
!
and north as positive.
Solution:
!
Use the Pythagorean theoreA to deterAine !v .
r
!v =
r2
r
!vi + !vf
2
= (50.0 m/s)2 + (35.0 m/s)2
= 61.033 m/s (two extra digits carried)
!
\$ &quot;vf '
!1
! = tai &amp; ! )
% &quot;vi (
\$ 35.0 m/s '
= tai !1 &amp;
)
% 50.0 m/s (
! = 35.0&deg;
Calculate the average acceleration.
r
r
!
aav =
!
61.033 m/s [E 35.0&deg; S]
=
45.0 s
r
aav = 1.36 m/s 2 [E 35.0&deg; S]
Statement: The average acceleration of the helicopter is 1.36 m/s 2 [E 35.0&deg; S] .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-14
r
r
10. Given: vi = 8.2 m/s [E 25&deg; S]; vf = 8.2 m/s [E 25&deg; N]7 !t = 3.2 ms = 3.2 &quot; 10#3 s
!
Required: aav
Analysis: DraH a vector diagraA of the situation. Calculate the change in velocity using
!
!
!
! ! !
coAponents and &quot;v = v ! v/ . Then, deterAine the average acceleration froA aav =
. Use east
!
and north as positive.
Solution: CoAponents for the initial velocity vector:
CoAponents for the final velocity vector:
!
The x-coAponent of !v is
&quot;v = S ! /
= E.2 ] s Nfs 2;&deg; ! E.2 ] s Nfs 2;&deg;
=
] s!
]s
&quot; =0 ]s
!
The y-coAponent of !v is
&quot;v = S ! /
= E.2 ] s s/ 2;&deg; ! ! E.2 ] s s/ 2;&deg;
&quot;
=
] s! !
=
]s
]s
m/ s N
/
DeterAine the change in velocity froA its coAponents.
r
!v = !v 2 + ! 2
=
(0 m/s )2 + ( 6.941 m/s )2
r
! = 6.941 m/s (two extra digits carried)
!
!
6ince !v = 0 m/s , !v points north.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-
Calculate the average acceleration.
!
!
!
aav =
!
6.941 m/s [
=
3.2 &quot; 10#3 s
!
a = 2.2 &quot; 103 ] s 2
Statement: The average acceleration of the pool ball is 2.2 ! 103 m/s 2 [5 .
!
!
11. Given: vi = 6.4 m/s [8 45&deg; 5[ a = 2.2 ] s 2 [ ! = 8.0 s
!
Required: vS
Analysis: Calculate the change in velocity froA the acceleration and tiAe interval:
!
!
!
aav =
!
! !
! = aav !
! ! !
!
Using coAponents and &quot;v = vS ! v/ , deterAine vS . Use east and north as positive.
Solution: The change in velocity is
! !
!v = aav !
= ( 2.2 m/s 2 [S]) (4.0 s)
!
! = 8.8 m/s [S]
!
The x-coAponent of vS is
vS =
/
=!
+&quot;
] s Nfs 3;&deg; + 0 ] s
=!
=!
] s+0 ] s
]s
S
!
The y-coAponent of vS is
vS =
=
=
S
/
+&quot;
] s s/i 3;&deg; + ( !E.E ] s )
m/s + (!8.8 m/s)
= !5.129 m/s
DeterAine the final velocity froA its coAponents.
2
2
r
vS = vS + S
r
S
= (#5.246 m/s)2 + (#5.129 m/s)2
= 7.3 m/s
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-
#v &amp;
S
! = tai &quot;1 %
(
%\$ S ('
#
m/s &amp;
= tai &quot;1 %
(
\$ 5.246 m/s '
! = 44&deg;
44&deg; 69.
Statement:
!
!
2
2
12. Given: vS = 3.6 &quot; 10 km/h [5[; a8v = 5.0 m/s ; ! = f.2 s
!
Required: vS
Analysis: Convert the final velocity to Aetres per second.
1h
1000 m
!
vS = 3.6 ! 102 km /h !
!
3600 s 1 km
!
2
vS = 1.0 ! 10 ]/s [5[
Calculate the change in velocity froA the acceleration and tiAe interval:
!
!
!
aav =
!
! !
! = aav !
! ! !
!
Using coAponents and &quot;v = vS ! v/ , deterAine vi . Use east and north as positive.
Solution: The change in velocity is
! !
!v = aav !
= (5.0 m/s 2 ) (f.2 s)
!
! = 46 m/s 
!
The x-coAponent of vi is
vi =
S
!&quot;
= 0 ]/s ! (!46 m/s)
i
= 46 m/s
!
The y-coAponent of vi is
vi =
i
S
!&quot;
= 100 ]/s ! 0 m/s
= 100 m/s
DeterAine the initial velocity froA its coAponents.
2
2
r
vi = vi + i
= (46 m/s)2 + (100 m/s)2
r
i
= 110.1 m/s
= 1.1! 102 m/s
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-13
#v &amp;
i
! = tai &quot;1 %
(
%\$ i ('
# 100 m/s &amp;
= tai &quot;1 %
(
\$ 46 m/s '
! = 65&deg;
Statement: The initial velocity of the boat is 1.1! 102 m/s 8E 65&deg; N9.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-
Section 1.5: Projectile Motion
Mini Investigation: Analyzing the Range of a Projectile, page 38
A. There appears to be no relationship between the horizontal component of velocity and the
maximum height of a projectile.
B. The maximum height of a projectile is greatest when the launch angle is largest.
C. A projectile has maximum range when launched at an intermediate angle, around 45&deg;.
D. The range of a projectile is the same whether launched at an angle or at the complement of
that angle. (Two angles are complementary if they add up to 90&deg;; for example, 25&deg; and 65&deg; are
complementary angles, as are 11&deg; and 79&deg;.)
Tutorial 1 Practice, page 40
1. (a) Given: viy = 0 m s; !d y = 76.5 cm = 0.765 m; g =
m s2
Required: !t
Analysis: Set the table top as di = 0. Therefore ! dy = –0.765 m. In the vertical direction, I know
the displacement, initial velocity, and acceleration. Use down as positive, so the displacement
1
will be negative. Use !d y = v1y !t &quot; g!t 2 to determine !t using the quadratic formula.
2
1
g!t 2 &quot; v1y !t + !d y = 0
2
#1 &amp;
v1y &plusmn; v1y2 &quot; 4 % g ( (!d y )
\$2 '
!t =
g
#1 &amp;
v1y &plusmn; v1y2 &quot; 4 % g ( (!d y )
\$2 '
Solution: !t =
g
!t =
#1
0 &plusmn; 0 &quot; 4% (
\$2
&amp;
m s 2 )( (&quot;0.765 m)
'
m s2
s (two extra digits carried)
!t = &plusmn;
!t = 0.40 s
Statement: The marble hits the floor after 0.40 s.
(b) Given: vx =
m s; !t =
s
Required: !d x
Analysis: Since I know the time of flight of the marble and its horizontal velocity. I can
determine its horizontal range using !d x = x !t .
Solution: !d x =
=(
!
x
x
!t
m s )(
s)
= 0.76 m
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Chapter 1: Kinematics
1.5-1
Statement: The range of the marble is 0.76 m, or 76 cm.
m s2
(c) Given: vx = 1.9/ m s; i = 0 m s; !t = 0./951 s; =
!
Required: vf
Analysis: The horizontal component of the velocity is constant throughout. The vertical
component changes with constant acceleration: !v y = &quot;g!t . Determine the vertical component
of the final velocity and then construct the final velocity vector.
Solution: Determine the vertical component of the final velocity.
!v y = &quot;g!t
vf y = viy &quot; g!t
=0 m s&quot;(
vf y = &quot;
m s 2 )(0./951 s)
m s (two extra digits carried)
!
Since vfx = x = 1.9/ m s , I can combine the components to determine the magnitude of vf .
!
vf = (vfx )2 + ( f )2
= (1.9/ m s)2 + (&quot;/.372 m s)2
!
f
= 4./ m s
!
The angle below the horizontal axis of vf is
#v &amp;
fy
! = tan %
(
%\$ vf x ('
&quot;1
# /.372 m s &amp;
= tan &quot;1 %
(
\$ 1.9/ m s '
! = 64&deg;
&deg; below the horizontal].
Statement: The final velo
2. Given: viy = 0 m s; !d y = &quot;0.3/ m; !d x = 13.4 m; v = 9.3 m s 2
Required: vx
Analysis: I know the horizontal displacement and want to determine the constant horizontal
speed. The appropriate formula is !d x = x !t , but I do not know the time taken. Looking at the
vertical motion, use !d y = v1y !t &quot;
1
g!t 2 to determine !t using the quadratic formula.
2
1
g!t 2 &quot; v1y !t + !d y = 0
2
#1 &amp;
v1y &plusmn; v1y2 &quot; 4 % g ( (!d y )
\$2 '
!t =
g
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Chapter 1: Kinematics
1.5-2
Solution: Using the vertical motion,
#1 &amp;
v1y &plusmn; v1y2 &quot; 4 % g ( (!d y )
\$2 '
!t =
g
!t =
#1
&amp;
0 &plusmn; 0 &quot; 4 % (9.3 m s 2 )( (&quot;0.3/ m)
\$2
'
9.3 m s 2
!t = &plusmn;0.4116 s (two extra digits carried)
Using the horizontal motion,
!d x = x !t
! x
!t
13.4 m
=
0.4116 s
= 45 m s
x
Statement: The initial horizont
r
3. (a) Given: vi = 12 m s 42&deg; above the horizontal]; !d y = &quot;9.5 m; g = 9.3 m s 2
x
=
Required: !t
Analysis: First, determine the components of the initial velocity. Then, use the vertical motion
#1 &amp;
v1y &plusmn; v1y2 &quot; 4 % g ( (!d y )
\$2 '
1
and !d y = v1y !t &quot; g!t 2 to solve for the time taken: !t =
g
2
Solution: The components of the initial velocity are
vix = (12 m s)(cos 42&deg;)
ix
= 3.913 m s (two extra digits carried)
i
= (12 m s)(sin 42&deg;)
i
= 3.0/0 m s (two extra digits carried)
Using the vertical motion,
#1 &amp;
v1y &plusmn; v1y2 &quot; 4 % g ( (!d y )
\$2 '
!t =
g
=
3.0/0 m s &plusmn; (&quot;3.0/0 m s)2 &quot; 4(4.9 m s 2 )(&quot;9.5 m)
2(4.9 m s 2 )
!t = 2.4/5 s or &quot; 0.796 s (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-/
The time of flight cannot be a negative value.
!t = 2.4 s
Statement: The rock’s time of flight is 2.4 s.
(b) Given: vix = 3.913 m s; !t = 2.4/5 s; v = 9.3 m s 2
Required: width of the moat, !d x
Analysis: Since I know the time of flight of the rock and its horizontal velocity, I can determine
its horizontal range (the width of the moat) using !d x = x !t .
Solution: !d x =
x
!t
= (3.913 m s)(2.4/5 s)
! x = 22 m
Statement: The width of the moat is 22 m.
(c) Given: vix = 3.913 m s; i = 3.0/0 m s; !t = 2.4/5 s; v = 9.3 m s 2
!
Required: vf
Analysis: The horizontal component of the velocity is constant throughout. The vertical
component changes with constant acceleration: !v y = &quot;g!t . Determine the vertical component
of the final velocity and then construct the final velocity vector.
Solution: Determine the vertical component of the final velocity.
!v y = &quot;g!t
vf y = viy &quot; g!t
= 3.0/0 m s &quot; (9.3 m s 2 )(2.4/5 s)
vf y = &quot;15.3/ m s (two extra digits carried)
r
Since vfx = x = 3.913 m s , I can combine the components to determine the magnitude of vf .
!
vf = (vfx )2 + ( f )2
= (3.913 m s)2 + (&quot;15.3/ m s)2
= 13.17 m s
!
f
= 13 m s
r
The angle below the horizontal axis of vf is
#v &amp;
fy
! = tan &quot;1 %
(
%\$ vf x ('
# 15.3/ m s &amp;
= tan &quot;1 %
(
\$ 3.913 m s '
! = 61&deg;
Statement: The final ve
61&deg; below the horizontal].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-4
!
4. (a) Given: vi = 4./ m s 42&deg; below the horizontal]; !d y = –/.9 m + 1.4 m = &quot;2.5 m;
g = 9.3 m s 2
Required: !t
Analysis: First determine the components of the initial velocity. Then use the vertical motion
#1 &amp;
v1y &plusmn; v1y2 &quot; 4 % g ( (!d y )
\$2 '
1
and !d y = v1y !t &quot; g!t 2 to solve for the time taken: !t =
g
2
Solution: The components of the initial velocity are
= &quot;(4./ m s)(sin 42&deg;)
vix = (4./ m s)(cos 42&deg;)
i
= /.196 m s
= &quot;2.377 m s
ix
i
Using the vertical motion,
&quot;2.377 m s &plusmn; (2.377 m s)2 &quot; 4(4.9 m s 2 )(–2.5 m)
9.3 m s 2
!t = 0.4737 s or &quot; 1.066 s (two extra digits carried)
The time the ball is in the air cannot be a negative value.
!t = 0.43 s
Statement: The baseball’s time of flight is 0.43 s.
(b) Given: vix = /.196 m s; !t = 0.4737 s
!t =
Required: !d x
Analysis: Since I know the time of flight of the baseball and its horizontal velocity, I can
determine its horizontal range using !d x = x !t .
Solution: !d x =
x
!t
= (/.196 m s)(0.4737 s)
! x = 1.5 m
Statement: The horizontal distance from the window is 1.5 m.
(c) Given: vix = /.196 m s; i = &quot;2.377 m s; !t = 0.4737 s; v = 9.3 m s 2
!
Required: vf
Analysis: The horizontal component of the velocity is constant throughout. The vertical
component changes with constant acceleration: !v y = &quot;g!t . Determine the vertical component
of the final velocity and then construct the final velocity vector.
Solution: Determine the vertical component of the final velocity.
!v y = &quot;g!t
vf y = viy &quot; g!t
= &quot;2.377 m s &quot; (9.3 m s 2 )(0.4737 s)
vf y = &quot;7.563 m s
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-5
r
Since vfx = x = /.196 m s , I can combine the components to determine the magnitude of vf .
!
vf = (vfx )2 + ( f )2
!
f
= (/.196 m s)2 + (&quot;7.563 m s)2
= 3.2 m s
Statement:
Tutorial 2 Practice, page 42
r
1. (a) Given: vi = 2.2 ! 102 m s 45&deg; above the horizontal]; diy = df y ; g = 9.3 m s 2
Required: !t
Analysis: Since the projectile (the marble) lands at the same height from which it was launched,
2v sin !
the time taken is given by !t = i
.
g
2v sin !
Solution: !t = i
g
2(220 m s)(sin 45&deg;)
=
9.3 m s 2
!t = /2 s
Statement: The time of flight is /2 s.
!
s2
(b) Given: vi = 2.2 ! 102 m s 45&deg; above the horizontal] ; g
Required: !d x
Analysis: Use the range formula, !d x =
Solution: !d x =
2
i
2
i
sin 2!
v
sin 2!
v
(220 m s)2 (sin 90&deg;)
9.3 m s 2
= 4.9 &quot; 10/ m
=
!
x
Statement: The horizontal range of the projectile is 4.9 ! 10/ m , or 4.9 km.
!
(c) Given: vi = 2.2 ! 102 m s 45&deg; above the horizontal] ; g
s2
Required: maximum height, d y
Analysis: The maximum height occurs when v y = 0 m s . Use the acceleration formula
vf2y = viy2 # 2g!d y to determine d y .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-6
Solution: Since viy = vi sin ! and vf y = 0 m s , d y can be calculated.
vf2y = viy2 # 2g!d y
!d y =
vf2y
2g
(vi sin # )2
!d y =
2g
((220 m s)sin 45&deg;)2
=
2(9.3 m s 2 )
= 12/5 m
!d y = 1.2 \$ 10/ m
Statement: The projectile’s maximum height is 1.2 &quot; 10/ m above the ground.
!
(d) Given: vi = 2.2 ! 102 m s 45&deg; above the horizontal]; diy = df y
r
Required: velocity on impact, vf
Analysis: The projectile’s flight is symmetric. Its final velocity is the same as the initial velocity
except that the direction of its vertical component is reversed.
[45&ordm; below the
Statement: The velocity of the marble when it hits the floor is 2.2 &quot; 102
horizontal].
r
2. (a) Given: vi = 14.5 m s [/5.0&deg; above horizontal]; diy = df y ; g = 9.3 m s 2
Required: maximum height, d y
Analysis: The maximum height occurs when v y = 0 m s . Use the acceleration formula
vf2y = viy2 # 2g!d y to determine d y .
vf2y = viy2 # 2g!d y
!d y =
vf2y
2g
Solution: Since viy = vi sin ! and vf y = 0 m s , d y can be calculated.
(vi sin ! )2
!d y =
2g
((14.5 m s)sin /5&deg;)2
=
2(9.3 m s 2 )
!d y = /.5 m
Statement: The projectile’s maximum height is /.5 m above the ground.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-7
!
(b) Given: vi = 14.5 m s [/5.0&deg; above horizontal] ; g
2
Required: horizontal range, !d x
Analysis: Use the range formula !d x =
Solution: !d x =
2
i
2
i
sin 2!
v
.
sin 2!
v
(14.5 m s)2 (sin 70.0&deg;)
9.3 m s 2
= 2.0 &quot; 101 m
=
!
x
Statement: The horizontal range of the projectile is 2.0 &quot; 101 m.
!
2
(c) Given: vi = 14.5 m s [/5.0&deg; above horizontal] ; g
Required: time to maximum height, !t
Analysis: By symmetry, the time to maximum height is half of the total time !t =
Solution: !t =
2vi sin !
.
g
vi sin !
g
(14.5 m s)(sin /5.0&deg;)
9.3 m s 2
!t = 0.35 s
Statement: The time for the projectile to reach its maximum height is 0.35 s.
3. Solutions may vary. Sample answer:
2v sin !
. So !t is proportional to vi . When vi doubles,
(a) The time of flight is given by !t = i
g
!t doubles.
2
sin 2!
. So !d x is proportional to vi2 . When vi doubles, !d x
(b) The range is given by !d x = i
v
increases by a factor of four.
(vi sin ! )2
(c) The maximum height is given by !d y =
. So !d y is proportional to vi2 . When vi
2g
doubles, !d y increases by a factor of four.
=
Section 1.5 Questions, page 43
1. Given: viy = 0 m s; !d y = #1.5 m; !d x = 3./ m; v = 9.3 m s 2
Required: rock’s initial speed, vx
Analysis: I know the horizontal displacement and want to determine the constant horizontal
speed. The appropriate formula is !d x = x !t , but I do not know the time taken.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-3
!d x =
x
!t
! x
!t
Looking at the vertical motion, determine !t from
1
!d y = v1y !t # g!t 2
.
2
x
=
#1 &amp;
v1y &plusmn; v1y2 # 4 % g ( (!d y )
\$2 '
!t =
g
Solution: Using the vertical motion,
#1 &amp;
v1y &plusmn; v1y2 &quot; 4 % g ( (!d y )
\$2 '
!t =
g
0 m s &plusmn; (0 m s)2 &quot; 4(4.9 m s 2 )(&quot;1.5 m)
9.3 m s 2
!t = &plusmn;0.55// s (two extra digits carried)
Using the horizontal motion,
!
vx = x
!t
3./ m
=
0.55// s
= 15 m s
x
Statement: The initi
r
2. (a) Given: vi = 1.1! 10/ m s [45&deg; above the horizontal]; diy = df y ; g = 9.3 m s 2
=
Required: !t
Analysis: Since the projectile lands at the same height from which it was launched, the time
2v sin !
taken is given by !t = i
.
g
2v sin !
Solution: !t = i
g
=
2(1.1&quot; 10/ m s )(sin 45&deg;)
9.3 m s 2
= 153.7 s
!t = 1.6 &quot; 102 s
Statement: The object is in the air for 1.6 ! 102 s .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-9
!
(b) Given: vi = 1.1 ! 10/ m s [45&deg; above the horizontal] ; g
s2
Required: !d x
Analysis: Use the range formula !d x =
2
i
Solution: !d x =
=
=
2
i
sin 2!
v
.
sin 2!
v
(1.1&quot; 10/ m s)2 (sin 90&deg;)
9.3 m s 2
(1.1&quot; 10/ m s )(1.1&quot; 10/ m s )(sin 90&deg;)
9.3 m s 2
= 1.2 &quot; 105 m
!
x
= 1.2 &quot; 102 km
Statement: The horizontal range of the projectile is 1.2 &quot; 102 km.
r
(c) Given: vi = 1.1! 10/ m s [45&deg; above the horizontal]; g = 9.3 m s 2
Required: maximum height, d y
Analysis: Use the formula !d y =
Solution: !d y =
=
(vi sin ! )2
2g
(vi sin ! )2
2g
((1100 m s)sin 45&deg;)2
2(9.3 m s 2 )
= /.1&quot; 104 m
!d y = /1 km
Statement: The projectile’s maximum height is /1 km above the ground.
r
3. (a) Given: vi = 6.0 m s [/2&deg; below the horizontal]; !t = /.4 s; g = 9.3 m s 2
Required: !d y
Analysis: First determine the components of the initial velocity. Then use the vertical motion
1
and !d y = v1y !t # g!t 2 to solve for the vertical displacement.
2
Solution: The components of the initial velocity are
vix = (6.0 m s)(cos /2&deg;)
= 5.033 m s (two extra digits carried)
ix
i
i
= #(6.0 m s)(sin /2&deg;)
= #/.130 m s (two extra digits carried)
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Chapter 1: Kinematics
1.5-10
Using the vertical motion,
1
!d y = v1y !t # g!t 2
2
1
= (#/.130 m s )(/.4 s ) # (9.3 m s 2 )(/.4 s )2
2
!d y = –67 m
Statement: The ball fell 67 m, so the window was 67 m above the ground.
2
(b) Given: vix = 5.033 m s; i = #/.130 m s; !t = /.4 s ; g
r
Required: vf
Analysis: The horizontal component of the velocity is constant throughout. The vertical
component changes with constant acceleration:
!v y = #g!t
vf y = viy # g!t
Determine the vertical component of the final velocity and then construct the final velocity
vector.
Solution: Determine the vertical component of the final velocity.
vf y = viy # g!t
= #/.130 m s # (9.3 m s 2 )(/.4 s )
vf y = #/6.50 m s (two extra digits carried)
r
Since vfx = vx = 5.033 m s , I can combine the components to determine the magnitude of vf .
!
vf = (vfx )2 + ( f )2
= (5.033 m s)2 + (#/6.50 m s)2
!
f
= /7 m s
r
The angle below the horizontal axis of vf is
#v &amp;
fy
! = tan &quot;1 %
(
%\$ vfx ('
# /6.50 m s &amp;
= tan &quot;1 %
(
\$ 5.033 m s '
! = 32&deg;
32&deg; below the horizontal].
Statement: The final ve
4. (a) Given: ball’s initial direction, ! = 5/&deg; above the horizontal; !d x = 25 m; !t = 2.1 s
r
Required: initial velocity, vi
Analysis: Draw a diagram of the situation. The horizontal component of velocity is constant.
Determine it from the time interval and the horizontal distance using !d x = x !t . Then use the
cosine of the initial angle to determine the initial speed.
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Chapter 1: Kinematics
1.5-11
Solution:
r
The x-component of vi is
!d x =
x
!t
! x
!t
25 m
=
2.1 s
= 11.91 m s (two extra digits carried)
x
Determine the initial speed.
vix
cos! =
x
=
i
ix
i
=
i
= 2.0 &quot; 101 m s
cos!
11.91 m s
=
cos 5/&deg;
= 19.79 m s (two extra digits carried)
[5/&deg; above the horizontal].
Statement: The initial velocity of the soccer ball is 2.0 &quot; 101
r
(b) Given: vi = 19.79 m s [5/&deg; above the horizontal]; !d y = 7.2 m; g = 9.3 m s 2
Required: horizontal range, !d x
Analysis: I know the horizontal component of the initial velocity and want the horizontal
distance. I can use !d x = x !t once I know the time of flight. For this I need to look at the
vertical motion, viy = vi sin ! .
#1 &amp;
v1y &plusmn; v1y2 # 4 % g ( (!d y )
\$2 '
1
.
Use !d y = v1y !t # g!t 2 to determine !t =
g
2
Then solve the horizontal motion.
Solution: The vertical component of the initial velocity is
viy = vi sin !
= (19.79 m s)( sin 5/&deg;)
viy = 15.30 m s (two extra digits carried)
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Chapter 1: Kinematics
1.5-12
#1 &amp;
viy &plusmn; viy2 # 4 % g ( (!d y )
\$2 '
!t =
g
15.30 m s &plusmn; (–15.30 m s)2 &quot; 4(4.9 m s 2 )(7.2 m)
9.3 m s 2
!t = 2.675 s or 0.5495 s (two extra digits carried)
The ball lands on the building at about 2.7 s. The other time corresponds to the ball moving
through a height of 7.2 m on the way up.
The horizontal displacement of the ball is
!d x = x !t
= (11.91 m s )(2.675 s )
! x = /2 m
Statement: The ball’s horizontal range is /2 m.
2
(c) Given: !d x = 25 m; x = 11.91 m s; i = 15.30 m s ; g
=
Required: ball’s clearance above wall based on !d y
Analysis: I have calculated the components of the initial velocity above. Use the x-component
and !d x = x !t to determine the time for the ball to reach the building. Use the y-component and
1
g!t 2 to determine the ball’s height at that time. Then calculate how much above
2
7.2 m the ball is.
Solution: Calculate the time for the ball to reach the wall.
!d x = x !t
!d y = v1y !t #
!t =
!
x
x
25 m
11.91 m s
!t = 2.101 s (two extra digits carried)
The vertical displacement of the ball at this time is
1
!d y = v1y !t # g!t 2
2
= (15.30 m s)(2.101 s) # (4.9 m s 2 )(2.101 s)2
=
= //.19 m # 21.6/ m
!d y = 11.56 m
Determine the distance by which the ball clears the wall.
! = 11.56 m # 7.2 m
! = 4.4 m
Statement: The ball clears the wall by 4.4 m.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-1/
r
5. (a) Given: vi = 26 m s [52&deg; above horizontal]; vf y = 0 m s; g = /.7 m s 2 ; g = 9.3 m s 2
Required: maximum height, df y , based on !d y
Analysis: Determine the vertical component of the initial velocity using viy = vi sin ! .
Then use vf2y = viy2 # 2g!d y .
Solution: The vertical component of the initial velocity is
viy = vi sin !
= (26 m s)(sin 52&deg;)
viy = 20.49 m s (two extra digits carried)
The vertical displacement is
vf2y = viy2 # 2g!d y
(0 m s)2 = (20.49 m s)2 # 2(/.7 m s 2 )!d y
!d y =
(20.49 m s ) 2
2(/.7 m s 2 )
!d y = 57 m
Statement: The rock rises to a maximum height of 57 m.
!
(b) Given: vi = 26 m s [52&deg; above horizontal]; !d y = 12 m; g = /.7 m s 2
Required: time of flight, !t
#1 &amp;
v1y &plusmn; v1y2 # 4 % g ( (!d y )
\$2 '
1
Analysis: Use !d y = v1y !t # g!t 2 to determine !t =
.
g
2
20.433 m s &plusmn; (#20.433 m s)2 # 4(1.35 m s 2 )(12 m)
Solution: !t =
2(1.35 m s 2 )
= 10.45 s or 0.621 s (two extra digits carried)
!t = 1.0 &quot; 101 s
Statement: The rock strikes the hill after 1.0 ! 101 s .
!
(c) Given: vi = 26 m s [52&deg; above horizontal]; ! = 10.45 s
Required: horizontal range, !d x
Analysis: Calculate the horizontal component of the initial velocity using vix =
Using !d x =
i
cos! .
!t , determine the horizontal displacement.
r
Solution: The horizontal component of vi is
vix =
i
x
cos!
= (26 m s)(cos 52&ordm; )
ix
= 16.01 m s (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-14
The horizontal distance travelled is
!d x = x !t
(
)
= 16.01 m s (10.45 s )
!
x
= 1.7 &quot; 102 m
Statement: The rock’s range is 1.7 ! 102 m .
r
6. Given: vi = 16 m s [65&deg; above horizontal]; angle of inclination of hill ! = /0&deg;; g = 9.3 m s 2
Required: distance up the hill where rock lands, l
Analysis: Draw a diagram of the situation. I know the initial velocity of the rock. From
1
!d x = x !t and !d y = v1y !t # g!t 2 , I can determine where the rock is at any later time. The
2
&quot;d y
rock will hit the hill when its displacement components are in the correct ratio: tan ! =
.
&quot;d x
Solution:
Determine the x- and y-components of the initial velocity.
vix = i cos!
= (16 m s)( cos 65&deg;)
ix
= 6.762 m s (two extra digits carried)
viy = vi cos!
= (16 m s)( sin 65&deg;)
viy = 14.50 m s (two extra digits carried)
Write equations for the x- and y-components of the displacement after time !t .
1
!d x = x !t
!d y = v1y !t # g!t 2
2
! x = (6.762 m s)!t
!d y = (14.50 m s)!t # (4.9 m s 2 )!t 2
Use tan ! =
&quot;d y
&quot;d x
to build an equation for the time taken to reach the hill, !t .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-15
tan ! =
!d y
!d x
(14.50 m/s)!t # (4.9 m/s 2 )!t 2
tan30&deg; =
(6.762 m/s)!t
(0.5774)(6.762 m/s)!t = (14.50 m/s)!t # (4.9 m/s 2 )!t 2
(4.9 m/s 2 )!t 2 + (3.904 m/s)!t # (14.50 m/s)!t = 0
(4.9 m/s 2 )!t 2 # (10.60 m/s)!t = 0
[(4.9 m/s 2 )!t # (10.60 m/s)]!t = 0
One solution is !t = 0 s, when the rock is thrown from the bottom of the hill. The other solution
is as follows:
(4.9 m/s 2 )!t # (10.60 m/s) = 0
10.60 m/s
4.9 m/s 2
!t = 2.163 s (two extra digits carried)
The x- and y-components of the displacement at this time are
!d x = (6.762 m/s)!t
!t =
= (6.762 m/s)(2.163 s)
!
x
= 14.63 m (two extra digits carried)
!d y = (14.50 m/s)!t # (4.9 m/s 2 )!t 2
= (14.50 m/s )(2.163 s ) # (4.9 m/ s 2 )(2.163 s )2
= 31.37 m # 22.92 m
!d y =
The distance up the hill is
l = (14.63 m)2 + (9.23
)0
l = /6
Statement: The rock lands 17 m up the hill.
7. Solutions may vary. Sample answer:
r
Given: vi = 45 m s [/5&deg; above horizontal]; dix = 0 m;
i
= 12 m; v = 9.3 m s 2
Required: whether the snowball lands on a 25 m high, /5 m wide building, 150 m away
Analysis: One approach is to determine the horizontal displacement when the snowball is 25 m
up. If the snowball is between 150 m and 135 m horizontally from its launch position, then it is
over the building—and lands on the building.
Determine the components of the initial velocity, calculate the time to reach d y = 25 m , and then
determine d x at that time.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-16
Solution: The x- and y-components of the initial velocity are
vix = i cos!
= (23
ix
s)( cos
=
&deg;)
s (two extra digits carried)
viy = vi cos!
= (23
viy = 03.9/
s)( sim
&deg;)
s (two extra digits carried)
The time to reach a height of /5 m starting at 12 m is
1
!d y = v1y !t # g!t 2
2
13 m = (03.9/
s 0 )!t 0 # (03.9/
(
s)!t +
/
s)!t # (
0
s 0 )!t 0
=
Solve the quadratic equation for !t .
(25.31 m s) &plusmn; (25.31 m s)2 # 4(4.9 m s 2 )(1/ m)
!t =
9.3 m s 2
!t = 0.564 s or 4.70/ s (two extra digits carried)
The first solution represents the snowball passing through the height of 25 m on its way up.
The second solution is the one where the snowball may be over the building. Determine the
horizontal displacement when !t = 4.703 s .
!d x = (
s)!t
=(
s)(
s)
! x=
The horizontal displacement is indeed between 150 m and 135 m.
Statement: Yes, the snowball lands on top of building 2.
8. Solutions may vary. A written explanation and an algebraic solution are presented.
Written Explanation: All objects fall with the same acceleration regardless of mass. When a
projectile is fired at a target, its path is formed by the action of gravity together with its given
velocity. If gravity were to stop acting, the projectile would follow a straight line directly from
the launcher to the target. When gravity is acting, both the projectile and falling target fall at the
same speed, even though the projectile is travelling faster horizontally than the target. So at some
point along the path of the dropped target, the projectile will hit it because both the projectile and
the target are falling at the same speed. Given enough distance, they will both hit the floor at the
same time. If the projectile is fired with a greater velocity, the target will not fall as far before it
is hit by the projectile. In this case the projectile will follow a straighter path to the target. If the
projectile is fired more slowly, it will follow a more curved path and hit the target farther down
toward the ground. As long as the projectile launcher is aimed directly at the target and the
projectile has enough velocity to reach the target before it hits the ground, the projectile will hit
the target as it falls.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-17
Algebraic Solution:
Given:
ecti e s oti
diy = dix tam !
=
!
i
4/s
at ! a
i omta
!
i
=
target
!
i
=
s
2
Required: Show that the projectile hits the target.
Analysis: Write the equations for the position of the projectile after time !t . Also do this for the
target. Then compare the equations to determine the time when the x-components of the two
objects coincide. Check whether the y-components coincide at the same time.
Solution:
projectile:
target:
!d x = i !
!d x = i !
!
=
i
cos! &quot;
=
i
cos! &quot;
!
=
(Equation 1)
1
&quot;2
2
1
= 1 sin ! &quot; #
&quot;
2
=0 m
(Equation 3)
i
1
!2
2
1
2
(Equation 2)
! = (0 m/s)! #
!2
2
1
= i #
!2
(Equation 4)
2
Compare the horizontal positions of the projectile and the target using Equations 1 and /.
vi cos! !t = dix
&quot;
=
1
&quot; #
!t =
!
=
1
! #
ix
cos
Compare the vertical positions of the projectile and the target using Equations 2 and /, and also
the given fact diy = dix tan ! .
i
1
1
&quot;2=
2
sin
!&quot; =
1
sin ! &quot; #
&quot; =
=
=
&quot; =
#
i
1
&quot;
2
2
i
i
sin !
1
tan !
sin !
1
i
sin !
cos!
sin !
1
i
i
cos!
i
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-13
diy
!t =
1
=
=
! =
sin !
tan !
sin !
1
i
sin !
cos!
sin !
1
i
i
cos!
i
The x-positions of the projectile and target are equal at the same time that the y-positions are
equal. In other words, the projectile hits the target.
Statement: The equations of motion describe motion at constant velocity, except for the
1
# ! 2 term in the y-direction. Since this term is the same for both the projectile and target,
2
they will be equally affected by gravity. The projectile was aimed directly at the target, so it will
remain headed for the target as gravity pulls them both downward. So the projectile will hit the
target.
9. Solutions may vary. Sample answer:
!
Given: football: vi = 18 m/s [39&deg; above horizontal]; dix = 0 m; i = 22 m ;
r
player: v = 6.0 m/s [horizontally]; dix = 12 m; i = 0 m; v = 9.8 m/s 2
Required: whether the player can catch the ball
Analysis: Draw a diagram for the situation. There is enough information to determine where and
1
when the football would hit the ground. Use !d y = v1y !t # g!t 2 to
2
#1 &amp;
v1y &plusmn; v1y2 # 4 % g ( (!d y )
\$2 '
determine !t =
.
g
Check whether or not the player has enough time to get to the correct position while running at
constant speed.
Solution:
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-19
Determine the vertical speed.
v1y = v1 sin !
= 18 m/s sin
v1y =
&deg;
m/s tpo extra digits carried)
Determine when the football hits the ground.
#1 &amp;
v1y &plusmn; v1y2 # 4 % g ( (!d y )
\$2 '
!t =
g
(11.33 m/s) &plusmn; (11.33 m/s)2 &quot; 4(4.9 m/s 2 )(&quot;22 m)
=
9.8 m/s 2
!t = 3.570 s or &quot; /.039 s (two extra digits carried)
Time must be a positive quantity, so the time when the football hits the ground is /.570 s.
!d x = ix !t
!
x
=
i
cos! !t
= (18 m/s)(cos 39&deg;)(3.570 s)
! x = 64 m
The football will hit the ground about 64 m from the cliff, about /.6 s after being thrown.
At this time the player can make it to the position
!d x = x !t
= (6.0 m/s)(3.570 s)
!
x
= 21.42 m
x
= 12 m + 21.42 m
= 32 m
The player is about /2 m short of catching the football.
Statement: No, the player cannot run far enough to catch the ball.
x
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-20
Section 1.6: Relative Motion
Tutorial 1 Practice, page 47–48
!
!
1. (a) Given: v =
8)s 37or?ard4/ v = 1.1 8)s 37or?ard4
!
Required: v (
!
!
!
Analysis: Use v , = v ( + v(,
!
!
!
Solution: v , = v ( + v(,
= 1.1 8)s 37or?ard4 +
8)s 37or?ard4
!
8)s 37or?ard4
v,=
Statement:
!
!
(b) Given: v(, =
8)s 37or?ard4/ vT( = 1.1 8)s 3 c:?ard4
!
Required: vT(
!
!
!
Analysis: Use vT, = vT( + v(,
!
!
!
Solution: vT, = vT( + v(,
= 1.1 8)s 3 c:?ard4 +
=
8)s 37or?ard4 +
8)s 37or?ard4
8)s 37or?ard4
!
vT, =
8)s 37or?ard4
Statement: The velocity o7 the teenagers ?ith respect
!
!
2. Given: v5 =
:8)h 3N4/ v = :8)h 3E &deg; N4
!
Required: v5
Component Method:
= v5
Analysis: Use v5 x = v5 x + v x and v5
positive.
Solution: xv5 x = v5
x
+ v
= 0 :8)h +
v5
x
:8)
cos
&deg;
0 :8)h =
:8)h
:8)h t?o eMtra digits carrie
v5
yv5
+ v
= v5
+ v
=
:8)h +
=
=
:8)h +
:8)h
:8)h t?o eMtra digits carrie
:8)
sin
Copyright &copy; 2012 Nelson Education Ltd.
&deg;
Chapter 1: Kinematics
1.6-1
!
No? use these co8ponents to deter8ine v5 .
!
v5 =
v5
2
x
=
=
!
v5 =
#
! 2 = tan &quot;1 %
\$
2
+ v5
:8)h 2 +
:8)h
2
!2 =
:8)h t?o eMtra digits carrie
:8)h &amp;
(
:8)h '
&deg;
:8)h
Statement:
3E &deg; N4.
Geometry Method:
Analysis: I
Dra? a diagra8 o7 the situation. Use the cosine and sine la?s to deter8ine the speed and
direction o7 the plane.
sin
sin
sin
2
=
=
= 2 + 2 &quot; 2 cos
Solution:
!2 =
The cosine la? gives
v52 = v52 + v 2 &quot; 2v5 v
v
2
5
=
:8)h 2 +
=
:8)h
v5 =
v5 =
&deg;=
&deg;=
&deg;.
cos! 2
:8)h 2 &quot; 2
:8)h
:8)h cos
&deg;
:8)h t?o eMtra digits carrie
:8)h
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-2
The sine la? gives
sin !
sin ! 2
=
v
v5
:8)
sin ! =
! =
!=
sin
&deg;
:8)h
&deg;
&deg; &quot;!
=
&quot;
!= &deg;
Statement: The velocity o7 the
!
!
3. Given: v =
:8)h 3(4/ v
!
Required: v
&deg;N4.
=
:8)h 3E4
!
Analysis: The directions o7 the helicopter and the ?ind 7or8 a right angle ?ith v as the
hypotenuse o7 a right-angled triangle. Use the 5ythagorean theore8 and the tangent ratio to
!
deter8ine v .
!
Solution: Deter8ine the 8agnitude o7 v .
v
= v2 + v2
=
v
=
:8)h 2 +
:8)h
2
:8)h
!
Deter8ine the direction o7 v
!
#v &amp;
&quot;1
! = tan % ! (
\$v '
#
= tan &quot;1 %
\$
:8)h &amp;
(
:8)h '
!= &deg;
Statement: The velocity o7 the helicopter ?ith respect to the ground is
!
!
4. Given: ! =
:8 3(4/ ! =
h/ v =
:8)h 3E4
!
Required: v5
Analysis:
&deg;(4.
!
v5 is also south.
!
Deter8ine the 8agnitude o7 v5 using the given displace8ent and ti8e. Then dra? the vector
!
!
!
the 5ythagorean
addition diagra8 7or v5 = v5 + v . (ince this is a right!
theore8 and the tangent ratio to deter8ine v5 .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-
Solution: The ground speed is
!
v5 =
!
:8
=
h
v5 =
:8)h
!
Deter8ine the 8agnitude o7 v5 .
v5 =
v5
=
2
+ v
2
:8)h 2 +
:8)h
2
v5 = 160 :8)h
!
Deter8ine the direction o7 v5 .
!
#v &amp;
&quot;1
! = tan % ! (
\$ v5 '
#
= tan &quot;1 %
\$
:8)h &amp;
(
:8)h '
!= &deg;
Statement: The plane should hea
!
!
5. (a) Given: vFE =
8)s 3N4/ vCF =
!
Required: vCE
!
!
!
Analysis: Use vCE = vCF + vFE
!
!
!
Solution: vCE = vCF + vFE
= +
8)s + +
&deg; ,4 ?ith an airspeed o7 160 :8)h.
8)s 3N4
8)s
=+
8)s
!
8)s 3N4
vCE =
Statement: ,he
8)s 3N4.
Copyright &copy; 2012 Nelson Education Ltd.
he velocity o7 the child ?ith respect to Earth is
Chapter 1: Kinematics
1.6-
!
!
(b) Given: vFE =
8)s 3N4/ vCF =
!
Required: vCE
!
!
!
Analysis: Use vCE = vCF + vFE
!
!
!
Solution: vCE = vCF + vFE
= +
8)s + &quot;
8)s 3(4
8)s
= +1.0 8)s
!
vCE = 1.0 8)s 3N4
he velocity o7 the child ?ith respect to Earth
Statement:
is 1.0 8)s 3N4.
!
!
(c) Given: vFE =
8)s 3N4/ vCF =
8)s 3E4
!
Required: vCE
!
!
!
Analysis: Use vCE = vCF + vFE . This is a rightand the tangent ratio.
!
Solution: Deter8ine the 8agnitude o7 vCE .
vCE =
vCF
=
2
+ vFE
2
8)s 2 +
vCE =
8)s
2
8)s
!
Deter8ine the direction o7 vCE .
!
# vFE &amp;
&quot;1
! = tan % ! (
\$ vCF '
#
= tan &quot;1 %
\$
8)s &amp;
(
8)s '
!= &deg;
Statement:
&deg;
&deg; E4.
!
!
2
6. (a) Given: v5 =
! 10 :8)h 3N &deg; ,4/ v = 62 :8)h 3(4
!
Required: v5
Analysis: (ince
= v5 + v
solution. Use v5 x = v5 x + v x and v5
positive.
Solution: xv5 x = v5
= &quot;
v5
x
=&quot;
x
+ v
x
:8)
sin
&deg; = 0 :8)h
:8)h t?o eMtra digits carrie
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-
yv5
= v5
+ v
=
:8)
&deg; = &quot;62 :8)
cos
=
:8)h t?o eMtra digits carrie
!
No? use these co8ponents to deter8ine v5 .
v5
!
v5 =
2
v5
x
=
2
+ v5
:8)h 2 +
=
!
v5 =
:8)h
2
#
! = tan &quot;1 %
\$
!=
:8)h &amp;
(
:8)h '
&deg;
:8)h t?o eMtra digits carrie
! 102 :8)h
Statement: The velocity o7 plane ?ith respect to the ground is
!
(b) Given: v5 =
:8)h 3N
,4/ ! = 1.2 h
!
Required: !
Analysis:
r r
Solution: ! = v5 !
r
=
! =
:8)h 3N
&deg; ,4 1.2
&quot; 10 :8)h 3N
&deg; ,4
2
Statement: The plane’s displace8ent a7ter 1.2 h is
!
!
7. (a) Given: v5, =
8)s 3N4/ v,E =
8)s 3E4
!
Required: v5E
! 102 :8)h
&deg; ,4.
r
r r
r
!
vav =
/ ! = vav ! .
!
! 102 :8)h 3N
&deg; ,4 .
r
Analysis: The directions o7 the current and the s?i88er 7or8 a right angle ?ith v5E as the
hypotenuse o7 a right-angled triangle. Use the 5ythagorean theore8 and the tangent ratio to
!
deter8ine v5E .
!
Solution: Deter8ine the 8agnitude o7 v5E .
v5E =
v5,
=
2
+ v,E
8)s 2 +
v5E =
2
8)s
2
8)s
!
Deter8ine the direction o7 v5E .
!
# v,E &amp;
&quot;1
! = tan % ! (
\$ v5, '
#
= tan &quot;1 %
\$
!=
8)s &amp;
(
8)s '
&deg;
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-6
Statement:
!
(b) Given: ! =
Required:
&deg; E4.
8 3N4
!
r
r
!
!
!
!
Analysis: The co8ponent o7 v5E pointing north is v5, . Use vav =
.
/! =
!
vav
Solution: ! =
!
v5,
8
8)s
! = 1.2 &quot; 102 s
Statement: It ta:es the s?i88er 1.2 ! 102 s to cross the river.
!
(c) Given: vWE = 0.40 m/s [E]; ! = 120 s
!
Required:
r
r r
r
!
!
!
/ ! = vav ! .
Analysis: The co8ponent o7 v5E pointing east is vWE . Use vav =
!
Solution: ! = vWE !
=
= (0.40 m/s )(120 s)
! = 48 m
Statement:
r
(d) Given: vW =
m/s v
!
Required: direction o7 v5,
= 0.40 m/s
r
vW
E
!
Analysis: The vectors 7or8 a right-angled triangle ?ith v5, as the hypotenuse. is a right-angled
r
!
triangle. Use the sine ratio to deter8ine the direction o7 v5, ?ith respect to vWE [E]
sin ! =
opposite side
hypoteuse
# opposite side &amp;
! = sin &quot;1 %
\$ hypoteuse ('
!
Solution: Deter8ine the direction o7 v5, .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-
!
# vWE &amp;
! = sin % ! (
\$ vWW '
&quot;1
# 0.40 m/s &amp;
= sin &quot;1 %
(
\$ 0.70 m/s '
! = 35&deg;
Statement:
!
8. Given: v
E
!
=v
1E
!
4v
=
m/s P
&deg; ,4 to land directly north o7 her starting point.
!
nimW4 v = 2.9 m/s [downstream]
1
Required: v
!
=v
!
+v
!
and w = v
1
!
v
Analysis: I
(?itching to a si8pler notati
!
v= v
!
=v
2
1
1
!
and v
21
!
=v
2
!
+v
1
.
. I ?ill re?rite the relative velocity
I can solve 7or the required speed.
Solution: The relative velocity equations are
!
!
!
vC 1 = vC W + vW1
&quot;
m/s = &quot;v + w
!
!
!
vC 1 = vC W + vW1
2
1
t on
2
+2.9 m/s = v + w
1 t on
&quot;;.2 m/s + +2.9 m/s = &quot;v + w + v + w
m/s = 2w
m/s
w=
Statement:
!
(b) Given: vC 1 = 2.9 m/s [downstream]E vW1 = w = 0.85 m/s
2
!
!
Required: v = vC W = vC W
;
2
Analysis: (ubstitute the speed o7 the ?
Solution:
+2.9 m/s = v + w
quation 2.
v = +2.9 m/s &quot; w
= +2.9 m/s &quot; 0.85 m/s
v=
m/s
Statement: The canoeists paddle at 2.0 8)s ?ith respect to the ?ater.
!
!
!
9. (a) Given: v =
em/d ]Ea9 v;C = em/d ][a9 !d =
em ]Ea
Required: !
!
!
!
Analysis: Use v5 = v5A + vA
!
!
!
!
velocity o7 the plane. Then use va =
to deter8ine the ti8e ! =
.
!
va
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-
Solution: The ground velocity is
!
!
!
v =v +v
= +
=+
!
vC=
! =
=
em/ + &quot;
em/d
em/d ]Ea
!
em/
n
n
!
v
750
/
! =
Statement:
!
!
(b) Given: v =
em/d ]Ea9 v;C =
Required: !
!
!
!
Analysis: Use v5 = v5A + vA
!
em/d ]Ea9 !d =
em ]Ea
!
!
!
velocity o7 the plane. Then use va =
to deter8ine the ti8e ! .
!
Solution: The ground velocity is
!
!
!
v =v +v
= +
em/ + +
em/
=+
em/d
!
em/d ]Ea
n
vWC =
The required ti8e ! is
!
va =
!
!
! =
v
=
or
rn
750
/
! = ;.;
Statement: The 7light ti8e is 1.1 h ?hen there is a tail ?ind. This ti8e is shorter than ?hen
there is an opposing ?ind because the plane 8oves 8ore quic:ly ?ith respect to the ground.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-9
!
!
em/d9 v;C = em/d ] a9 !d =
em ]Ea
!
Required: direction o7 v5A !
Analysis: The pilot needs to head so8e?hat ?est o7 north to co8pensate 7or the ?ind that heads
!
!
!
v5 = v5A + vA . (ince the vector addition triangle is a rightthe sine ratio to deter8ine the direction o7 the
air velocity and the 8agnitude o7 the ground velocity. Then deter8ine the 7light ti8e using
r
r
!
!
.
/! =
vav =
!
vav
=
(c) Given: v
Solution:
!
Deter8ine the direction o7 v5A .
!
#v &amp;
&quot;;
! =sn % ! (
\$ v '
#
= s n &quot;; %
\$
&amp;
(
/ '
/
!= &deg;
!
Deter8ine the 8agnitude o7 v5 .
2
v
v
= v
=
2
v
+ v
2
&quot; v
2
2
=
em/d &quot;
=
em/d
em/d
wn
or s
Copyright &copy; 2012 Nelson Education Ltd.
rn
Chapter 1: Kinematics
1.6-10
! =
=
!
v
750
/
! = ;.2
Statement:
&deg; ,4. The ne? 7light ti8e is 1.2 h.
Section 1.6 Questions, page 49
!
1. (a) Given: v W = ;.2 m/sE vW1 =
!
m/s E !d; =
!
[W]E !d2 =

Required: ! = ! ; + ! 2
Analysis: Loo: 7irst at the upstrea8 8otion. Deter8ine the speed !o7 the person ?ith respect to
!
!
!
!
!
Earth using v 1 = v W + vW1
rearrange the equation va =
to deter8ine the ti8e
!
!
required/ ! =
. Repeat this procedure 7or th
va
! =!;+! 2
Solution:
d
r
r
r
v 1 = v W + vW1
!
!
!
v 1 = v W + vW1
= &quot;;.2 m/s + +
=&quot;
!
!;=
vW
m/s
m/s
m
m/s
!;=
s two e ra dp ts arrpe
The total ti8e 7or the s?i8 is
! =!;+! 2
=
=
= +;.2 m/s + +
=+
!
!2=
vW
=
!2=
m/s
m/s
m
m/s
s
wn
or s
rn
s + 588.2 s
= 2017 s &quot;
1 min
60 s
! = 34 min
Statement:
(b)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-11
(c) Given: v = ;.2 m/sE !d =
Required: !
!
!
/ ! =
Analysis: va =
!
va
Solution: ! =
=
!
v
m
;.2 m/s
= 1667 s &quot;
1 min
.
60 s
9
s 9 i
! = 28 min
Statement:
7or 7ully by s?i88ing ?ith the current 7or the sa8e distance. The trip ta:es longer ?hen there is
a current because the current slo?s the s?i88er ?hen s?i88ing against it.
!
!
2. (a) Given: v =
m/s [W]E v =
m/s [ ]
!
Required: vCE
!
!
!
Analysis: v5 = v5A + vA . This is a rightthe tangent ratio.
!
Solution: Deter8ine the 8agnitude o7 v5 .
2
v
2
v
= v
2
+ v
=
m/s 2 +
2
m/s
2
= 43 600 m 2 s 2
v
=
m s . o e ra dp ts arrpe
v
=
m/s
!
Deter8ine the direction o7 v5 .
!
#v &amp;
&quot;;
! = tan % ! (
\$ v '
# 60 m s &amp;
= tan &quot;; %
(
ms'
\$
! = 20&deg;
Statement: The ground velocity o7 the plane is 200 8)s 3, 20&deg; N4.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-12
!
!
!
(b) Given: v =
m/s [W]E v =
m/s E !d =
[W]
Required: !
!
!
!
Analysis:
v5 = v5A + vA to deter8ine the ground velocity. Then rearrange the
!
!
!
!
to calculate the 7light ti8e/ ! =
.
equation va =
!
va
Solution: The ground velocity is
r
r
r
v =v +v
r
v
=
m/s [W] +
=
m/s [W] +
=
m/s [W] one e ra dp t arrpe
! =
m/s 
m/s [W]
!
v
300
1000 m
&quot;
140 m/s
1
1 min
= 2143 s &quot;
60 s
! = 36 min
Statement:
8in.
!
!
3. (a) Given: v =
m/s [ ]E v 1 =
!
Required: v 1
!
!
!
Analysis: v 1 = v + v 1
r
r
r
Solution: v 1 = v + v 1
=
r
v
1
=
m/s [ ] +
=
!
Statement:
!
(b) Given: v =
!
Required: v 1
!
!
Analysis: v 1 = v
!
!
Solution: v 1 = v
!
v
Statement:
1
;
m/s [ ]
m/s [ ]
!
m/s [ ]E v
!
+v
!
+v
m/s [ ]
.0 ! 101 8)s 3N4.
1
=
m/s [ ]
1
1
=
=
m/s [ ] +
m/s [ ] +
=
m/s [ ]
m/s [ ]
m/s [ ]
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-
!
(c) Given: v
!
Required: v
!
m/s [ ]E v
=
1
!
Analysis: The vectors v
=
1
!
=v
!
+v
m/s [W]
!
7or8 a right-angled triangle ?ith v
!
Use the 5ythagorean theore8 and the tangent ratios to deter8ine v 1 .
Solution: v
1
=
=
v
1
2
v
1
+ v
1
as the hypotenuse.
2
1
m/s 2 +
=
1
m/s
2
m/s
!
Deter8ine the direction o7 v 1 .
!
# v 1&amp;
&quot;;
! = tan % ! (
\$v '
#
= tan &quot;; %
\$
m/s &amp;
(
m/s '
!= &deg;
Statement:
!
(d) Given: v
!
Required: v
1
xv
x
=
!
m/s [ ]E v
&deg; ,4.
1
=
m/s [
&deg; W]
!
Analysis: Dra? the vector addition diagra8 7or the situation. Use co8ponents to deter8ine v 1 .
Use east and north as positive.
Solution:
v
1 x
1 x
= v
+ v
1 x
=
m/s + &quot;
m/s spn
=
m/s + &quot;
m/s
=&quot;
&deg;
m/s two e ra dp ts arrpe
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-
yv
1
= v
=
+ v
m/s +
1
m/s
os
&deg;
=
=
m/s +
m/s
m/s two e ra dp ts arrpe
v 1
!
Deter8ine the 8agnitude o7 v 1 .
v
1
2
v
=
12.04 m/s 2 + 75.38 m/s)2
1 x
+ v
2
=
1
vHE = 76 m/s
!
Deter8ine the direction o7 v 1 .
# v!
&amp;
&quot;;
! = tan % ! 1 x (
%\$ v 1 ('
# 12.04 m/s &amp;
= tan &quot;; %
(
\$ 75.38 m/s '
!= &deg;
Statement: The velocity o7 the
!
m/s 4 !d = 88 m
4 !d =
4. (a) Given: v W =
!
Required: vW1
Analysis: (:etch the river and the displace8ent vectors.
Copyright &copy; 2012 Nelson Education Ltd.
&deg; ,4.
m
4
Chapter 1: Kinematics
s
1.6-
!
!
!
Also dra? v 1 = v W + vW1 .
!
!
!
Vectors v5, and vW1 are perpendicular co8ponents o7 v5E .
I
I can
deter8ine the ti8e !
0 8 at
!d
!d
8)s. Use va =
to deter8ine the crossing ti8e and the va =
to deter8ine the speed
!v
!v
o7 the current. Use south and ?est as the positive directions.
Solution:
Dri7ting do?n
!d
!
va =
va =
!
!v
m
!d
vW1 =
!v =
s
W
vW1 =
m/s
m
=
m/s
!v =
s
Statement:
m 1
(b) Given: !d = m [W]E !d =
!
Required: v5E
Analysis: T
-angled triangles.
!
the velocity triangle to calculate v5E .
Use the displace8ent triangle to deter8ine the angle !
Solution:
!
\$ &quot;d '
!;
! = tan &amp; ! )
&amp;% &quot;d )(
\$
= tan !; &amp;
%
!=
m '
m )(
&deg;
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-16
v
2
1
= v
2
W
+ vW1
2
= 0.65 m/s)2 + (
m/s
2
v 1 = 0.78 m/s
&deg; ,4.
!
!
= 0.65 m/s 4 vW1 = 0.44 m/s [W]; v 1
[ ]
W
!
Required: direction o7 v5,
!
!
!
Analysis: Use v 1 = v W + vW1 to dra? the relative velocities. This is a right-angled triangle. Use
!
the sine ratio to deter8ine the direction o7 v5, .
Statement:
(c) Given: v
r
r
# vW1 &amp;
vW1
&quot;;
spn ! = r E ! = spn % r (
vW
\$vW'
!
# vW1 &amp;
&quot;;
Solution: ! = spn % ! (
\$vW'
# 0.44 m/s &amp;
= spn &quot;; %
(
\$ 0.65 m/s '
! = 43&deg;
Statement:
!
5. Given: ! =
&quot;
!
Required: v5A
[
&deg; E4.
!
&deg; 1]E ! =
v
=
/ [ ]
!
!
!
Analysis: Dra? the relative velocities using v5 = v5A + vA . (olve the triangle using the cosine
2
sin
=
2
=
+
sin
2
=
&quot;2
cos
sin
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-
Solution:
r
v
!
=
!
r
&quot;
=
=
2
v
2
v
&deg; 1]
[
&deg; 1]
/ [
two e ra dp ts arrpe
= v
2
+ v
2
&quot;2 v
v
os! 2
= v
2
+ v
2
&quot;2 v
v
os! 2
=
2
v
=
v
= 370
+.
.
2
9
&quot; 2.
.
.
43&deg;
s 9 i
spn !
spn ! 2
=
v
v
spn ! =
v spn ! 2
v
.sin 43&deg;
=
!3 =
&deg;
! = 180 &quot; ! 3
= 180 &quot;
&deg;
!= &deg;
Statement: The plane 8ust
Copyright &copy; 2012 Nelson Education Ltd.
! 102
Chapter 1: Kinematics
&deg; E4.
1.6-
!
6. (a) Given: !d =
!
=
/ [ 36&deg; E]E
!
Required: direction o7 v5A
Analysis:
!
!
!
?ind direction. (o I can dra? v5 = v5A + vA .
[ ]E
=
v
/
can be deter8ined
spn 36&deg;
v
um/ sr 36&deg;)
(230 um/
!=
&deg; . o extra digits carried)
! = 6.2&deg;
Statement: The heading o7 the plane should be 3N 6.2&deg; ,4.
!
!
(b) Given: !d =
P
=
P 16&deg; A
=
7G
7
Required: !
Analysis: Use the vector triangle and the sine la?
The calculate ! using v
Solution:
! + &quot; + 16&deg; =
!=
!=
!=
4
I :no? the
angle !
The direction o
using the sine la?.
spn ! spn 36&deg;
=
Solution:
v
v
spn ! =
= 130
G
=
&quot; = 6./64&deg;
v5 .
!d
!d
.
( !t =
!t
vG
!.
&deg;
&deg; # &quot; # 16&deg;
&deg; # 6./64&deg; # 16&deg;
&deg; th
xtra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-19
sin ! sin 16&deg;
=
v7
vG
v
G
v
G
!t =
=
v 7 sin !
sin 16&deg;
210 km
sin
&deg;)
=
sin 16&deg;
= 262.6 km
tho extra digits carried)
=
!d
vG
220 km
262.6 km
!t =
Statement:
!
7. (a) Given: v 7 =
!
( v7G =
ms
ms A
Required: v5
!
!
!
Analysis: Use v5 = v5A + vA to deter8ine v5 .
!
!
!
Solution: v G = v 7 + v7G
= 230 m s
+ 30.0 m s A
= 230 m s
+
!
2
v G = 2.0 ! /0 m s
ms
)
Statement: The ground speed o7 the plane on its ?ay ?est is 2.0 ! /02 m s .
!
!
(b) Given: v 7 = 230 m s A ( v7G = 30.0 m s A
Required: v5
!
!
!
Analysis: v5 = v5A + vA
!
!
!
Solution: vPG = vPA + vAG
= 250 m/s [E] + 50.0 m/s [E]
!
vPG = 3.0 ! 102 m/s [E]
Statement: The ground speed o7 the plane on its ?ay east is 3.0 ! 102 m/s .
!
!
8. (a) Given: vPS = 2.0 m/s [up]; vSW = 3.2 m/s [E]
!
Required: v5,
!
!
!
!
Analysis: T
-angled ?ith v5, as the
v W = v S + vSW
hypotenuse. Use the 5ythagorean theore8 and the tangent ratio to deter8ine the 8agnitude and
!
direction o7 v5, .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-20
Solution: v
2
W
v
v
W
W
= v
S
=
v
=
2.0 m/s 2 + 3.2 m/s
2
+ vSW
2
S
2
+ vSW
2
2
= 3.P m/s
#v &amp;
! = tan &quot;1 % PS (
\$ vSW '
# 2.0 m/s &amp;
= tan &quot;1 %
(
\$ 3.2 m/s '
! = 32&deg;
Statement:
&deg; up4 relative to the ?ater.
!
!
(b) Given: vPS = 2.0 m/s [E 3P&deg; up]; vSW = 3.2 m/s [E]
!
Required: v5,
!
!
!
Analysis: (:etch
vPW = vPS + vSW . Use co8ponents to deter8ine
!
v5, ?ith east and up as positive.
Solution:
x(vPW ) x = (vPS ) x + (vSW ) x
= +(2.0 m/s)cos38&deg; = (=3.2 m/s)
= 1.576 m/s + 3.2 m/s
(vPW ) x = 4.776 m/s (two extra digits carried)
y(vPW ) y = (vPS ) y + (vSW ) y
= (2.0 m/s)sin 3P&deg; = 0 m/s
= 1.231 m/s + 0 m/s
(vPW ) y = 1.231 m/s (two extra digits carried)
!
No? use these co8ponents to deter8ine v5, .
!
vPW =
(vPW )
2
+ (vPW ) y
2
= (4.776 m/s)2 + (1.231 m/s)2
!
vPG = 4.9 m/s
Copyright &copy; 2012 Nelson Education Ltd.
# 1.231 m/s &amp;
! = tan &quot;1 %
(
\$ 4.776 m/s '
! = 14&deg;
Chapter 1: Kinematics
1.6-21
Statement:
!
9. (a) Given: v67 =
!
Required: v)6
Analysis:
!
/ P7W2 v
&deg; up4 relative to the ?ater.
&deg; GW
3 P
rain.
!
!
!
Also dra? v)7 = v)6 + v67 .
Convert :ilo8etres per hour to 8etres per second.
60
!
m
!
s
=
m/s ( o extra digits carried)
(olve 7or v)6 using the sine ratio.
Solution: sin70&deg; =
v67
v)6
v67
s 70.0&deg;
16.667 m/s
=
sin70.0&deg;
= 17.737 m/s (two extra digits carried)
v)6 =
v)6 = 17.7 m/s
Statement:
!
(b) Given: v)6 3 17.737 m/s [down 70.0&deg; W]
!
Required: v
Analysis:
&deg; ,4 ?ith respect to the car.
v
Solution: cos70&deg; =
v/[
v
v/[ = v d
=
using the cosine ratio.
70.0&deg;
m1o d
70.0&deg;
v/[ = C.07 m1o
Statement:
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-22
!
10. (a) Given: vPG
!
s 30.0&deg; W]R v
!
Required: direction o7 v5A
=
velocity triangle is 6
Calculate t
relative-velocity triangle using the sine la?.
Solution:
sin ! 2 sin 60.0&deg;
=
vPA
vAG
=
=
1
&deg;
I
!
!
!
!
I can dra? v5 = v5A + vA . The angle opposite v5A in the relative
[ 30.0&deg; W] is equivalent to [W 60.0&deg; ]
Analysis:
sin ! 2 =
1 sW]R v
! 2 and ! in the scalene
vAG sin 60.0&deg;
vPA
1
(48
! =
o nC0.0&deg;
1
&deg;
! 3 = 180&deg; &quot; 60.0&deg; &quot; ! 2
= 180&deg; &quot; 60.0&deg; &quot;
!3 =
&deg;
!=
=
&deg;
&deg; &quot; !3
&deg;&quot;
&deg;
!= &deg;
The heading o7 the plane should be 3, 69&deg; N ?hich is equivalent to 3N 21&deg; ,4.
Statement: The heading o7 the plane should be 3N 21&deg; ,4.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-
!
(b) Given: vPG
sW C0.0&deg; ]R v
1 R !3 =
=
Required: v5
Analysis: Use the relative-velocity triangle
Solution:
and the sine la? to deter8ine v5 .
sin ! 3 sin 60.0&deg;
=
vPA
vPG
vPA sin ! 3
sin 60.0&deg;
1 on
(260
=
o n C0.0&deg;
v =
1
Statement: The ground speed
vPG =
Copyright &copy; 2012 Nelson Education Ltd.
&deg;
Chapter 1: Kinematics
1.6-
Section 2.1: Forces and Free-Body Diagrams
Tutorial 1 Practice, page 65
1. (a)
(b)
(c)
(d)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.1-1
2. While the ball is in the air (from just after it leaves your hand, until just before it makes
contact
with the object that it will hit), it is only acted upon by one force—the force of gravity,
!
Fg . Therefore, for (a), (b), and (c), the FBD of the force acting on the ball is shown below.
3.
Tutorial 2 Practice,
page 68
!
!
1. (a) Given: Fa = 25 N [forward 40.0&deg; up] ; Fg = 4.2 N [down]
!
Required: !F
!
⎛ ΣF ⎞
Analysis: !F = (!Fx )2 + (!Fy )2 ; θ = tan −1 ⎜ y ⎟ . Choose forward and up as positive.
⎝ ΣFx ⎠
Solution:! For!the x-component of the force,
Fax F cos!
!
Fax
(25 N)cos 40.0=
19.15 N (two extra digits carried)
!
!
!
!Fx = Fax + Fgx
= 19.15 N + 0.0 &quot;
!
!Fx = 19.15 N (two extra digits carried)
For the y-component
of the force,
!
!
Fay = F sin !
= (25 N)sin 40.0&deg;
!
Fay = 16.07 N (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.1-2
!
!Fy
!
!Fy
!
!
Fay + (&quot; Fgy )
16.07 &quot; 4.2 N
11.87 N (two extra digits carried)
!
Construct !F :
!
!F = (!Fx )2 + (!Fy )2
= (19.15 N)2 + (11.87 N)2
= 22.53 N
!
!F = 23 N
!
\$
'
#
F
! = tan &quot;1 &amp; !y )
% #Fx (
\$ 11.87 N '
= tan &quot;1 &amp;
% 19.15 N )(
! = 32&deg;
Statement: The net force acting on the soccer ball is 23 N [32&deg; above the horizontal].
!
!
(b) Given: F1 = 15 N [N 35&deg; E] ; F2 = 25 N [N 54&deg; W]
!
Required: !F
!
\$
'
!
#
F
Analysis: !F = (!Fx )2 + (!Fy )2 ; ! = tan &quot;1 &amp; !y ) . Choose east and north as positive.
% #Fx (
Solution:! For !the x-component of the force,
F1x = F cos!
= (15 N)cos35&deg;
!
F1x = 12.29 N (two extra digits carried)
!
!
F2x = F cos!
= (25 N)cos54&deg;
!
F2 x = 14.70 N (two extra digits carried)
!
!
!
!Fx = F1x + F2x
= 12.29 N + 14.70 N
!
!Fx = 26.99 N (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.1-3
For the y-component
of the force,
!
!
F1y = F sin !
= (15 N)sin35&deg;
!
F1y = 8.604 N (two extra digits carried)
!
!
F2y = F sin !
= (&quot;25 N)sin54&deg;
!
F2 y = &quot;20.22 N (two extra digits carried)
!
!
!
!Fy = F1y + F2y
= 8.604 &quot; 20.22
!
!Fy = &quot;11.62 N (two extra digits carried)
!
Construct !F :
!
!F = (!Fx )2 + (!Fy )2
= (26.99 N)2 + (&quot;11.62 N)2
= 29.39 N
!
!F = 29 N
!&quot;
\$
#
Fy'
! = tan &quot;1 &amp; !&quot; )
% #Fx (
\$ &quot;11.62 N '
= tan &quot;1 &amp;
% 26.99 N )(
! = 23&deg;
Statement: The net force acting on the sled is 29 N [N 23&deg; W].
!
!
!
(c) Given: Fg = 4.4 ! 102 N [down]; F1 = 4.3! 102 N [up 35&deg; left]; F2 = 2.8 ! 102 N [up]
!
Required: !F
!
\$
'
!
#
F
Analysis: !F = (!Fx )2 + (!Fy )2 ; ! = tan &quot;1 &amp; !y ) . Choose right and up as positive.
% #Fx (
Solution: !For the
of the force,
! x-component
!
!
!Fx = Fgx + F1x + F2x
= (0 N) + (&quot;430 N)sin35&deg; + (0 N)
!
!Fx = &quot;246.6 N (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.1-4
For the y-component
!
!
! of the
! force,
!Fy = Fgy + F1y + F2y
= (&quot;440 N) + (430 N)cos35&deg; + (280 N)
= &quot;440 N + 352.2 N + 280 N
!
!Fy = 192.2 N (two extra digits carried)
!
Construct !F :
!
!F = (!Fx )2 + (!Fy )2
= (246.6 N)2 + (192.2 N)2
= 310 N
!
!F = 3.1&quot; 102 N
!
\$
'
#
F
! = tan &quot;1 &amp; !y )
% #Fx (
\$ 192.24 N '
= tan &quot;1 &amp;
% 246.64 N )(
! = 38&deg;
Statement: The net force acting on the performer is 3.1&times;102 N [up 38&deg; left] or [left 52&deg; up].
!
!
!
2. (a) Given: F1 = 1.2 ! 104 N [E 12&deg; N] ; F2 = 1.2 ! 104 N [E 12&deg; S] ; !F = 0 N
!
Required: Ff
! ! ! !
Analysis: !F = F1 = F2 = Ff . Choose east and north as positive.
! ! ! !
Solution: !F = F1 + F2 + Ff
! ! !
0 = F1 + F2 + Ff
!
! !
Ff = &quot; F1 &quot; F2
!
!
!
Ffx = ! F1x ! F2x
= !(1.2 &quot; 104 N)cos12&deg; ! (1.2 &quot; 104 N)cos12&deg;
!
Ffx = !2.3&quot; 104 N
Statement: The force of friction on the rock is 2.3 &times;104 N [W].
(b) Answers may vary. Sample answer: I notice that the forces of the tractors are equal in
magnitude but act in directions symmetric about the east. So the net force of the tractors has to
be due east, making the force of friction due west. Another way to say this is that the north
component of one tractor force cancels the south component of the other.
2
3. Given: m1! = 15.0
! kg;
! m2 = 7.0 kg; m3 = 13.0 kg; g = 9.8 m/s
Required: FT1; FT2 ; FT3
Analysis: Draw an FBD for each mass. Choose up as positive.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.1-5
Solution: The FBDs are shown below.
Mass 1 ( m1 = 15.0 kg )
!
!
!
!
!Fy = FT1y = FT2y = Fgy
!
!
0 N = FT1 &quot; FT2 &quot; (15.0 kg)g
!
!
FT1 &quot; FT2 = (15.0 kg)g (Equation 1)
Mass 2 ( m2 = 7.0 kg )
!
!
!
!
!Fy = FT2y + FT3y + Fgy
!
!
0 N = FT2 &quot; FT3 &quot; (7.0 kg)g
!
!
FT2 &quot; FT3 = (7.0 kg)g (Equation 2)
Mass 3 ( m3 = 13.0 kg )
!
!
!
!Fy = F3y + Fgy
!
0 N = FT3 &quot; (13.0 kg)g
!
FT3 = (13.0 kg)g (Equation 3)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.1-6
Solve for the tensions in the wires, working from equation (3) to equation (2) to equation (1):
!
!
!
!F = (!Fx )2 + (!Fy )2
= (0.3392 N)2 + (1.2880 N)2
!
!F = 1.3 N
!
FT3 = (13.0 kg) (Equation 1)
!
FT3 = 1.3! 102 N
!
!
FT2 &quot; FT3 = (7.0 kg)
(Equation 2)
!
!
FT2 = FT3 = (7.0 kg)
= (13.0 kg) = (7.0 kg)
!
F2 = 2.0 ! 102 N
!
!
FT1 &quot; FT2 = (15.0 kg) (Equation 3)
!
!
FT1 = FT2 = (15.0 kg)
= (13.0 kg) = (7.0 kg) = (15.0 kg)
!
FT1 = 3.4 ! 102 N
Statement: The tension in the top wire is 3.4 &times;102 N , in the middle wire it is 2.0 ! 102 N, and in
the bottom !wire it is 1.3 ! 102 N.
!
4. Given: Fair = 0.40 N [32&deg; above the horizontal]; Fg = 1.5 N [down]
!
Required: !F
!
\$ #Fy '
!
&quot;1
2
2
Analysis: !F = (!Fx ) + (!Fy ) ; ! = tan &amp; ! ) ; use forward and up as positive.
% #Fx (
Solution: !For the
of the force,
! x-component
!
!Fx = Fairx + Fgx
= (&quot;0.40 N)cos32&deg; + (0 N)
!
!Fx = &quot;0.3392 N (two extra digits carried)
For the y-component
!
!
!of the force,
!Fy = Fairy + Fgy
= (0.40 N)sin32&deg; + (&quot;1.5 N)
!
!Fy = &quot;1.288 N (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.1-7
!
Construct !F :
!
\$ #Fy '
! = tan &amp; ! )
% #Fx (
&quot;1
\$ 1.288 N '
(two extra digits carried)
= tan &quot;1 &amp;
% 0.3392 N )(
! = 75&deg;
Statement: The magnitude of the net force on the ball is 1.3 N [75&deg; below the horizontal].
5. (a) Since the ball is at rest, the net force on it is 0 N.
(b) If I suddenly remove my hand, the only force acting on the ball is gravity. The net force is
16 N [down].!
!
(c) Given: Fg = 16 N [down]; Fa = 12 N [right]
!
Required: !F
!
\$
'
!
#
F
Analysis: !F = (!Fx )2 + (!Fy )2 ; ! = tan &quot;1 &amp; !y ) . Choose forward and up as positive.
% #Fx (
Solution: ! For !the x-component
of the force,
!
!Fx = Fax + Fgx
= 12 N + 0 N
!
!Fx = 12 N
For the y-component
!
!
! of the force,
!Fy = Fay = Fgy
= (0 N) = (&quot;16 N)
!
!Fy = &quot;16 N
!
Construct !F :
!
!F = (!Fx )2 + (!Fy )2
= (12 N)2 + (16 N)2
!
!F = 20 N
!=
=
!
\$
'
#
F
y
&quot;1
!
&amp;
)
% #Fx (
&quot;1
\$ N'
&amp;% 12 N )(
! = 53&deg;
Statement: The net force on the basketball is 20 N [right 53&deg; down].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.1-8
!
!
(d) Given: Fg = 16 N [down]; Fa = 26 N [up 45&deg; right]
!
Required: !F
!
\$ #Fy '
!
&quot;1
2
2
Analysis: !F = (!Fx ) + (!Fy ) ; ! = tan &amp; ! ) . Choose forward and up as positive.
% #Fx (
Solution: ! For !the x-component
of the force,
!
!Fx = Fax + Fgx
= (26 N)cos 45&deg; + 0 N
!
!Fx = 18.38 N (two extra digits carried)
For the y-component
!
!
! of the force,
!Fy = Fay + Fgy
= (26 N)sin45&deg; + (&quot;16 N)
!
!Fy = &quot;2.385 N (two extra digits carried)
!
Construct !F :
!
!F = .!F )2 + (!Fy )2
= (18.38 N)2 + (2.385 N)2
!
!F = 19 N
!
\$ #Fy '
! = tan &amp; ! )
% #Fx (
&quot;1
\$ 2.385 N '
= tan &quot;1 &amp;
% 18.38 N )(
! = 7.4&deg;
Statement: The net force on the basketball is 19 N [right 7.4&deg; up].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.1-9
Section 2.1 Questions, page 69
1. Table 1 Common Forces
Name
force of
gravity
Symbol
!
Fg
normal force
string tension
friction
!
FN
!
FT
!
Ff
Contact/
non-contact
non-contact
contact
contact
contact
static friction
!
FS
contact
kinetic friction
!
FK
contact
air resistance
!
Fair
contact
!
F
contact
applied force
(push or pull)
Direction
down
perpendicular to
surface
away from object
opposite to direction of
motion or tendency to
motion
along the surface,
opposite to sum of the
other forces
along the surface,
opposite to direction of
motion
opposite to direction of
motion
any direction
Example in daily life
A peanut butter sandwich falls to
the floor because it is pulled by
gravity.
A tea cup sits on the surface of a
table, held up by the normal force.
A child uses tension in a leash to
pull her dog.
Friction in the car brake pads
causes the car to slow down.
Static friction between a sled and
the snow has to be overcome before
the sled will slide.
Kinetic friction between my skate
blades and the ice causes me to
slow down.
A falling sheet of paper is subject to
air resistance as well as the force of
gravity.
My friends help me push my car
out of the ditch.
2. A pulley is a device that changes the direction of string tension but does not change its
magnitude. This means that the tension in the cord is 22 N throughout. Therefore, the student’s
statement is not valid.
3. Ropes can only pull and never push because the rope just sags when you push on it and,
therefore, you cannot exert a force.
4. (a) The forces acting on the textbook are the force of gravity, the normal force, the applied
force, and the force of kinetic friction.
(b) The FBD of the textbook is shown below.
!
!
!
5. Given: FA = 2.3 N [S 35&deg; W] ; FB = 3.6 N [N 14&deg; W] ; FC = 4.2 N [S 34&deg; E]
!
!
!
(a) Required: FA + FB + FC
!
!
!
Analysis: F + F + F . Choose north and east as positive.
!
!
!
!
Solution: !Fx = FAx + FBx + FCx
= (&quot;2.3 N)sin35&deg; + (&quot;3.6 N)sin14&deg; + (4.2 N)sin 24&deg;
= &quot;1.319 N &quot; 0.8709 N + 1.708 N
!
!Fx = &quot;0.4819 N (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.1-10
!
!
!
!
!Fy = FAy + FBy + FCy
= (&quot;2.3 N)cos35&deg; + (3.6 N)cos14&deg; + (&quot;4.2 N)cos 24&deg;
= &quot;1.884 N + 3.493 N &quot; 3.836 N
!
!Fy = &quot;2.227 N (two extra digits carried)
!
Construct !F :
!
!F = !Fx 2 + !Fy 2
=
!
!F = 2.3 N
N 2+
N
2
!
\$
'
#
F
! = tan &quot;1 &amp; !y )
% #Fx (
\$ 2.227 N '
= tan &quot;1 &amp;
% 0.4819 N )(
! = 78&deg;
!
!
!
Statement: FA + FB + FC = 2.3 N [W 78&deg; S] or [S 12&deg; W]
!
!
(b) Required: F ! FC
!
!
Analysis: F ! FC . Choose north and east as positive.
Solution:
For the x-component of the force,
!
!
!
!
FB ! FC F = BF ! CF
(
)
= (!3.6 N)sin14&deg; ! (4.2 N)sin 24&deg;
(
!
B
!
!
= !0.8709 N ! 1.708 N
)
C F
= !2.579 N (two extra digits carried)
For the y-component of the force,
!
!
!
!
FB ! FC y = FBy ! FCy
(
)
= (3.6 N)cos14&deg; ! (!4.2 N)cos 24&deg;
(
!
!
FB ! FC
)
= +3.493 N + 3.836 N
y
= 7.329 N (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.1-11
!
Construct !F :
!
!
Fa ! FC =
!
(F
a
=
!
! Fn
!
) +(F
2
x
a
2.579 N 2 +
!
!
Fa ! FC =
!
! Fn
)
2
y
N
2
N
!
!
# FB &quot; FC
!
! = tan &quot;1 % !
%\$ FB &quot; FC
(
(
)
)
&amp;
(
(
x'
y
# 7.329 N &amp;
= tan &quot;1 %
\$ 2.579 N ('
! = 71&deg;
!
!
Statement: F ! F = 7.8 N
71&deg; Na nt N 19&deg; a
!
!
!
6. Given: FA = 33 N [E 22&deg; N] ; FB = 42 N [S 15&deg; E] ; !F = 0 N
!
Required: FC
!
!
!
Analysis: FA + FB + FC = 0 N . Choose north and east as positive.
Solution:! For the
of the force,
! x-component
!
FCx = ! FAx ! FBx
= !(33 N)cos 22&deg; ! (42 N)sin15&deg;
= !30.60 N ! 10.87 N
!
FCx = !41.47 N (two extra digits carried)
For the y-component
of
!
!
! the force,
FCy = ! FAy ! FBy
= !(33 N)sin 22&deg; ! (!42 N)cos15&deg;
= !12.36 N + 40.57 N
!
FCy = 28.21 N (two extra digits carried)
!
Construct F :
!
!
!
FC = ( FCx )2 = ( FCy )2
= (41.47 N)2 = (28.21 N)2 (two extra digits carried)
!
FC = 50 N
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.1-12
!
# Fny &amp;
! = tan % ! (
\$ Fnx '
&quot;1
# 28.21 N &amp;
= tan &quot;1 %
\$ 41.47 N ('
! = 34&deg;
!
!
!
Statement: The force needed so that FA + FB + FC = 0 N is 50 N [W 34&deg; N].
!
!
!
!
7. (a) Given: F1 = 15 N [N 24&deg; E] ; direction of F2 is [S] ; direction of F3 is [W] ; !F = 0 N
! !
Required: F ; F
!
! !
Analysis: F1 = ! F2 ! F3 . Choose north and east as positive.
Solution: For the !x-components
!
! of the force:
F1x = ! F2x ! F3x
&quot;!
(15 N)sin 24&deg; = !(0 N) ! F 3x
!
F3x = !6.101 N (two extra digits carried)
!
F3 = 6.1 N [W]
For the y-components
! of the
! force:
!
F1y = ! F2y ! F3y
&quot;!
(15 N)cos 24&deg; = ! F 2y ! (0 N)
!
F2y = !13.70 N (two extra digits carried)
!
F2 = 14 N [S]
Statement: The second child pulls with a force of 14 N [S] and the third child with a force of
6.1 N [W].
(b) Since the net
! force was zero when the second child lets go, the new net force has the same
magnitude as F2 = 14 N but the opposite direction.
Therefore, the
! net force is 14 N [N].!
(c) Given: F1 = 15 N [N 24&deg; E] ; !F = 0 N
! ! !
Analysis: !F = F1 + F3 . Choose north and east as positive.
! ! !
Solution: !F = F1 + F3
!
0 = 15N + F3
!
F3 = 0 &quot; 15 N
!
F3 = &quot;15 N
Statement: The third child must exert a force of 15 N [S 24&deg; W] to cancel the force of the first
child on her own.
!
!
8. Given: !F = 180 N [E] ; F1 = 120 N [E 14&deg; ]
!
Required: F
! ! !
Analysis: !F = F1 + F2 . Choose north and east as positive.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.1-13
Solution:! For the
of the force:
! x-components
!
F2x = !Fx &quot; F1x
= (180 N) &quot; (120 N)cos14&deg;
!
F2x = 63.56 N (one extra digit carried)
For the y-components
!
! ! of the force:
F2y = !Fy &quot; F1y
= (0 N) &quot; (120 N)sin14&deg;
!
F2y = &quot;29.03 N (one extra digit carried)
!
F = (F x ) + (F y )
= (
N) + (
=
N(
!
F = 50 N
N)
extra digit carried)
!
#
&amp;
F
! = tan &quot;1 % ! y (
\$ Fx'
#
= tan &quot;1 %
\$
N&amp;
N ('
!= &deg;
Statement: The second student exerts a force of 70 N [E 25&deg; N].
!
!
!
9. Given: Fa = 55 N [forward 28&deg; up] ; Fg = 120 N ; !F = 0 N
(a) The FBD of the sled is shown below.
!
!
!
(b) Given: Fa = 55 N [forward 28&deg; up] ; Fg = 120 N ; !F = 0 N
!
Required: FN
!
!
!
!
Analysis: !Fy = Fay + F y + F y . Choose forward and up as positive.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.1-14
!
!
!
!
Solution: !Fy = Fay + Fgy + FNy
!
0 N = (55 N)sin 28&deg; + (&quot;120 N) + FN
!
FN = &quot;25.82 N + 120N
= 94.18 N
!
FN = 94 N
Statement: The normal force acting on the sled is 94 N [up]. The magnitude of the normal force
is less than the magnitude of the force of gravity because the applied force has an upward
component. In effect, the applied force lifts some of the weight from the surface, reducing the
normal force.!
!
(c) Given: Fa = 55 N [forward 28&deg; up] ; !F = 0 N
!
Required: F
!
!
!
Analysis: !Fx = Fax = FSx . Choose forward and up as positive.
!
!
!
Solution: !Fx = Fax + FSx
!
0 N = (55 N)cos 28&deg; &quot; FS
!
FS = 48.56 N
!
FS = 49 N
Statement: The force of static friction acting on the sled is 49 N [backward].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.1-15
Section 2.2: Newton’s Laws of Motion
Tutorial 1 Practice, pages 72–73
!
!
1. (a) Given: m = 1.2 &times;102 kg ; F1 = 1.5 ! 102 N [N] ; F2 = 2.2 ! 102 N [W]
!
Required: a
!
⎛ ΣF ⎞ !
!
Analysis: !F = (!Fx )2 + (!Fy )2 ; θ = tan −1 ⎜ y ⎟ ; !F = ma . Choose east and north as
⎝ ΣFx ⎠
positive.
Solution: !For the
of the force:
! x-components
!
!Fx = F1x + F2x
= (0 N) + (&quot;220 N)
!
!Fx = &quot;220 N
For the y-components
!
!
! of the force:
!Fy = F1y + F2 y
= (150 N) + (0 N)
!
!Fy = 150 N
!
Construct !F :
!
!F = (!Fx )2 + (!Fy )2
= (220 N)2 + (150 N)2
!
!F = 266.3 N (two extra digits carried)
!
\$ #Fy '
! = tan &amp; ! )
% #Fx (
&quot;1
\$ 150 N '
= tan &quot;1 &amp;
% 220 N )(
! = 34&deg;
!
Calculate !a :
!
!F = ma
!
! !F
a=
m
266.3 N [W 34&deg; N]
=
120 kg
!
a = 2.2 m/s 2 [W 34&deg; N]
Statement: The acceleration of the mass is 2.2 m/s2 [W 34&deg; N] or 2.2 m/s2 [N 56&deg; W].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.2-1
!
!
(b) Given: m = 26 kg ; F1 = 38 N [N 24&deg; E] ; F2 = 52 N [N 02&deg; N]
!
Required: a
!
⎛ ΣF ⎞ !
!
Analysis: !F = (!Fx )2 + (!Fy )2 ; θ = tan −1 ⎜ y ⎟ ; !F = ma . Choose east and north as
⎝ ΣFx ⎠
positive.
Solution: !For the
of the force:
! x-components
!
!Fx = F1x + F2 x
= (38 N )sin 24&deg; + (52 N)sin36&deg;
= 15.46 N + 30.57 N
!
!Fx = 46.03 N (two extra digits carried)
For the y-components
!
!
! of the force:
!Fy = F1y + F2 y
= (38 N)cos 24&deg; + (52 N)cos36&deg;
= 34.72 N + 42.07
!
!Fy = 76.79 N (two extra digits carried)
!
Construct !F :
!
!F = (!Fx )2 + (!Fy )2
= (46.03 N)2 + (76.79 N)2
!
!F = 89.53 N (two extra digits carried)
!
\$ #Fy '
! = tan &amp; ! )
% #Fx (
&quot;1
\$ 76.79 N '
= tan &quot;1 &amp;
% 46.03 N )(
! = 59&deg;
!
Calculate !a :
!
!F = ma
!
! !F
a=
m
89.53 N [E 59&deg; N]
=
26 kg
!
a = 3.4 m/s 2 [E 59&deg; N]
Statement: The acceleration of the mass is 3.4 m/s2 [E 59&deg; N] or 3.4 m/s2 [N 31&deg; E].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.2-2
!
!
!
2. Given: m = 65 kg ; F1 = 2.2 ! 102 N [E 42&deg; N] ; Ff = 1.9 ! 102 N [W] ; a = 2.0 m/s 2 [E]
!
Required: F2
!
! ! ! !
!
Analysis: !F = ma ; !F = F1 + F2 + Ff . Choose east and north as positive.
Solution:
!
Calculate !the net force !F :
!
!F = ma
= (65 kg)(2.0 m/s 2 )
!
!F = 130 N [E]
!
Find F2 in terms of the other forces:
! ! ! !
!F = F1 + F2 + Ff
!
! ! !
F2 = !F &quot; F1 &quot; Ff
!
Calculate the components of F2 .
For the x-components
!
! ! of the
! force:
F2 x = !Fx &quot; F1x &quot; Ffx
= (130N) &quot; (220 N)cos 42&deg; &quot; (&quot;190 N)
= 130 N &quot; 163.5 N + 190 N
!
F2 x = 156.5 N (two extra digits carried)
For the y-components
!
!
! of the
! force:
F2 y = !Fy &quot; F1y &quot; Ffy
= (0N) &quot; (&quot;220 N)sin 42&deg; &quot; (0 N)
!
F2 y = 147.2 N (two extra digits carried)
!
Construct F2 :
!
F2 = (F2 x )2 + (F2 y )2
= (156.5 N)2 + (147.2 N)2
= 214.9 N
!
F2 = 2.1! 102 N
!
#
&amp;
F
! = tan &quot;1 % !2 y (
\$ F2 x '
# 147.2 N &amp;
= tan &quot;1 %
\$ 156.5 N ('
! = 43&deg;
Statement: The second student applies a force of 2.1 ! 102 N [E 43&deg; N] to the trunk.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.2-3
!
!
3. Given: m = 1.5 &times;102 kg ; Fg = 1.47 ! 103 n [down] ; F1 = 1.8 ! 103 N [up 30.0&deg; left] ;
!
F2 = 1.8 ! 103 N [up 30.0&deg; right]
!
Required: a!
!
Analysis: !F = ma . Choose right and up as positive.
Solution: ! For !the x-components
of the force:
!
!
!Fx = Fgx + F1x + F2x
= (0 N) + (&quot;1800 N)sin30.0&deg; + (1800 N)sin30.0&deg;
!
!Fx = 0 N
For the y-components
!&quot;
!&quot;
!&quot;of the
!&quot; force:
! F y = F gy + F 1y + F 2y
= (&quot;1470 N) + (1800 N)cos30.0&deg; + (1800 N)cos30.0&deg;
!&quot;
! F y = 1648 N (two extra digits carried)
!
!
Since the x-component of !F is zero, !F = 1648 N [up].
!
Calculate !a :
!
!F = ma
!
! !F
a=
m
1648 N [up]
=
150 kg
!
a = 11 m/s 2 [up]
Statement: The acceleration of the beam is 11 m/s2 [up].
Tutorial 2 Practice, page 74
!
(a) The rocket engine pushes down on the burning fuel ( FR on F ) while the burning fuel pushes up
!
on the rocket ( FF on R ).
!
(b) The plane pushes back on the air passing through the jets ( FP on A ) while the air passing
!
through the jets pushes the plane forward ( FA on P ).
!
(c) The runner pushes down on the ground ( FR on G ) while the ground pushes up on the runner’s
!
foot ( FG on R ). This last force takes the form of the normal force of the ground on the runner.
!
2. Given: m = 56 kg ; Δt1 = 0.75 s ; vi = 0 m/s ; vf = 75 cm/s [W] = 0.75 m/s [W]
!
Required: a
!
! !
Analysis: a =
. Choose east as positive.
!
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.2-4
! !v
Solution: a =
!t
(&quot;0.75 m/s) &quot; (0 m/s)
=
0.75 s
= &quot;1.0 m/s 2
!
a = 1.0 m/s 2 [W]
Statement: The magnitude of the (constant) acceleration is 1.0 m/s2 [W].
!
(b) Given: m = 56 kg ; a = 1.0 m/s 2 [W]
!
Required: FS on W
!
!
Analysis: FS on W = ma
Solution:! Force of the swimmer on the wall:
!
FS on W = a
!
= (56 kg)(1.0 m/s 2 )
= 56 N [E]
Statement: !The force exerted by the swimmer on the wall is 56 N [E].
(c) Given: FS on W = 56 N [E]
!
Required: FW on S
!
!
!
Analysis: FW on S = a . The force of the swimmer on the wall, FS on W , is the action−reaction
!
partner to the force of the wall on the swimmer, FW on S . It is the force of the wall on the swimmer
that causes the swimmer’s acceleration.
Solution:! The force
! of the wall on the swimmer:
FW on S = ! FS on W
S on W
= !56 N [E]
!
FW on S = 56 N [W]
Statement: The force exerted by the wall on the swimmer is 56 N [W].
!
(d) Given: a = 1.0 m/s 2 [W] ; Δt2 = 1.50 s − 0.75 s = 0.75 s
!
!
!
Required: ! = ! 1 + ! 2
1
2
Analysis: Δd1 = vi Δt + a ( Δt ) ; Δd 2 = vf Δt2 .
2
Solution: The first distance covered is Δd1 :
1r
2
Δd1 = vi Δt + a ( Δt )
2
1
2
Δd1 = (0 m/s)(0.75 s) + (1.0 m/s 2 ) ( 0.75 s )
2
Δd1 = 0.2813 m (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.2-5
The second distance covered is Δd 2 :
Δd 2 = vf Δt2
= (0.75 m/s)(0.75 s)
Δd 2 = 0.5625 m (two extra digits carried)
The total distance covered is:
!
!
!
! =! 1+! 2
!
= 0.2813 m + 0.5625 m
! = 0.84 m
Statement: Both parts of the swimmer’s motion were away from the wall. The total
displacement is 84 cm [W].
!
!
3. (a) Given: mboy = 32.5 kg ; mmattress = 2.50 kg ; !Fboy = 0 N ; !F 1 [[ = . N
!
Required: FW on M
!
!
!
Analysis: FB on M + FW on M + F = 0 N
Solution: Equation
for
!
! upward
! force of the water on the mattress:
FB on M + FW on M + Fg = 0 N
!
!
!
! FB on M + FW on M ! Fg = 0 N
!
!
!
FW on M = FB on M + Fg
= (32.5 kg)(9.8 m/s 2 ) + (2.50 kg)(9.8 m/s 2 )
= 318.5 N + 24.5 N
!
FW on M = 3.4 &quot; 102 N
2
Statement: The upward force of the water on the
! mattress is 3.4
! ! 10 N.
(b) Given: mboy = 32.5 kg ; mmattress = 2.50 kg ; !Fboy = 0 N ; !Fmattress = 0 N
!
Required: FB on M
!
!
Analysis: F on M = ! FM on . Choose up as positive.
Solution:
!
!
!
Since F on M = ! FM on ; solve by determining the upward force of the mattress on the boy, FM on B .
!
!
FM on B + Fg = 0 N
!
!
FM on B = Fg
= (32.5 kg)(9.8 m/s 2 )
= 318.5 N (two extra digits carried)
!
FM on B = 3.2 ! 102 N
!
Statement: The force that the boy exerts on the mattress, FB on M , is 3.2 ! 102 N.
!
!
(c) Since F on M = ! FM on , the upward force of the mattress on the boy is 3.2 ! 102 N.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.2-6
!
!
4. Given: mP = 0.20 kg ; aP = 25 m/s 2 [forward] ; aL = 0.25 m/s 2 [backward]
Required: mL
!
!
Analysis: FP on = ! F on W . Choose forward as positive.
!
!
Solution: FP on L = ! FL on P
!
!
mL aL = !mP aP
!
!mP aP
mL = !
aL
mL =
–(0.20 kg)(25 m/s 2 )
–0.25 m/s 2
mL = 20 kg
Statement: The mass of the launcher is 20 kg.
Section 2.2 Questions, page 76
1. According to Newton’s first law of motion, when the snowboarder suddenly encounters the
rough patch on the hill, her body will continue in a forward motion, but her snowboard may stick
2. As you are sitting in the bus tossing the ball vertically, both you and the ball are acted upon by
the forward motion of the bus. By Newton’s first law of motion, both you and the ball continue
in a forward direction. From your point of view, the ball remains directly in front of you and will
not hit you in the face unless a horizontal force is exerted.
!
3. (a) Given: vi = 4.2 m/s [E]; vf = 0 m/s ; m = 41 kg ; Ff = 25 N [W]
!
Required: a!
!
Analysis: !F! = a
!
Solution: !F = a
!
! !
a=
! 25 N [W]
a=
41 kg
= 0.6098 m/s 2 [W] (two extra digits carried)
!
a = 0.61 m/s 2 [W]
Statement: The child’s acceleration across the ice is 0.61 m/s2 [W].
!
(b) Given: vi = 4.2 m/s [E]; a = 0.61 m/s 2 [W]
Required: Δt
!
! !v
Analysis: a =
. Choose east as positive.
!t
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.2-7
!v
!t
!v
!t =
a
m/s
0 m/s &quot;
=
&quot;0.6098 m/s 2
!t = 6.9 s
Statement: It will take the child 6.9 s to stop.
!
!
4. Given: a = a = 12 m/s 2 ; F = F = 2.2 ! 102 N
Solution: a =
Required:!m
!
Analysis: F!&quot; = ma&quot;
Solution: F = ma
!&quot;
F
m= &quot;
a
2.2 ! 102 N
=
12 m/s 2
m = 18 kg
Statement: The mass of the object is 18 kg.
5. (a) Given: vi = 0 m/s; vf = 2.5 m/s [forward]; Δt1 = 1.0 min = 60 s; m = 1.2 &times; 103 kg
!
Required: FN
!
!
Analysis: F = a . Choose forward as positive.
! !v
Solution: a =
!t
2.5 m/s &quot; 0 m/s
=
60 s
!
a = 0.0417 m/s 2 (two extra digits carried)
!&quot;
&quot;
F N = ma
= (1.2 ! 103 kg)(0.0417 m/s)
!&quot;
F N = 50 N
Statement: The normal force between the two bumpers is 50 N.
(b) Given: v2 = vf = 2.5 m/s; Δd = 2.0 km = 2000 m
Required: Δt = Δt1 + Δt2
Analysis: Δd = Δd1 + Δd 2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.2-8
⎛v +v ⎞
Solution: Δd1 = ⎜ i f ⎟ Δt
⎝ 2 ⎠
⎛ 0 m/s + 2.5 m/s ⎞
=⎜
⎟ (60 s)
2
⎝
⎠
(2.5 m/ s )
(60 s )
2
Δd1 = 75 m
=
Δd = Δd1 + Δd 2
Δd 2 = Δd − Δd1
= 2000 m − 75 m
Δd 2 = 1925 m
The second time interval is:
Δd
v2 = 2
Δt2
Δt2 =
Δd 2
v2
1925 m
2.5 m /s
= 770 s
=
Δt2 = 7.7 &times;102 s
Statement: It takes 7.7 !! 102 s to reach!the repair shop.
6. Given: m = 250 kg ; F1 = 150 N [E] ; F2 = 350 N [S 45&deg; W]
!
Required: a!
!
Analysis: !F = a . Choose east and north as positive.
Solution: ! For !the x-components
of the force:
!
!Fx = F1x + F2 x
= (150 N) + (&quot;350 N)sin 45&deg;
= 150 N &quot; 247.49 N
!
!Fx = &quot;97.49 N (one extra digit carried)
For the y-components
!&quot;
!&quot;
!&quot;of the force:
! F y = F 1y + F 2 y
= (0 N) + (&quot;350 N)cos 45&deg;
!&quot;
! F y = &quot;247.49 N (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.2-9
!
Construct !F :
!
!F = (!Fx )2 + (!Fy )2
= (97.49 N)2 + (247.49 N)2
!
!F = 266 N
\$ #F '
! = tan &quot;1 &amp; y )
% #Fx (
\$ 247.49 N '
= tan &quot;1 &amp;
% 97.49 N )(
! = 68&deg;
!
Calculate !a :
!
!F = a
!
! !
a=
=
266 N [W 68&deg; S]
250 kg
!
a = 1.1 m/s 2 [W 68&deg; S]
Statement: The acceleration of the mass is 1.1 m/s2 [W 68&deg; S] or 1.1 m/s2 [S 22&deg; W].
(a) When a tennis racquet hits a tennis ball, exerting a force on the ball, the tennis racquet pushes
forward on the tennis ball and the tennis ball pushes back on the racquet.
(b) When a car is moving at high speed and runs into a tree, exerting a force on the tree, the car
pushes forward on the tree and the tree pushes back on the car.
(c) When two cars are moving in opposite directions and collide head-on, the first car pushes the
second car in the direction of the first car’s initial velocity. The second car pushes the first car in
the opposite direction.
(d) When a person leans on a wall, exerting a force on the wall, the person pushes forward on the
wall and the wall pushes back on the person.
(e) When a mass hangs by a string attached to the ceiling, and the string exerts a force on the
mass, the mass pulls down on the string and the string pulls up on the mass.
(f) When a bird sits on a telephone pole, exerting a force on the pole, the bird pushes down on
the telephone pole and the telephone pole pushes up on the bird.
8. Given: m1 = m2 = 5.2 kg
! !
Required: F1 ; F2
!
Analysis: !F = 0 N . Choose right and up as positive.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.2-10
Solution:
!
!F = 0 N
!
F1 &quot; m1g = 0 N
!
F1 = (5.2 kg)(9.8 m/s 2 )
!
F1 = 51 N
Equation for second
mass:
!
!F = 0 N
!
F2 &quot; m2 = 0 N
!
F2 = (5.2 kg)(9.8 m/s 2 )
!
F2 = 51 N
Statement: The tension is each string is 51 N.
(b) The spring scale reads 51 N, the string tension on the left side that is pulling the hook.
(c) The answers would remain the same if you removed one mass and held everything in place.
A weight of 51 N would be exerted on the remaining mass, balanced by a string tension of 51 N.
The spring scale is at rest so the other string also would also have a tension of 51 N. You have to
pull down with a force of 51 N, effectively replacing the weight of the mass you removed.
!
9. Given: m = 62 kg ; Fground = 1.1! 103 N [backward 55&deg; up]
!
Required: a!
!
Analysis: !F = a . Choose forward and up as positive.
Solution:
For the x-components
of
!
!
! the force:
!Fx = Fearth x + Fgx
= (&quot;1100 N)cos55&deg; + (0 N)
= 150 N &quot; 247.49 N
!
!Fx = &quot;630.9 N (two extra digits carried)
For the y-components
of! the force:
!
!
!Fy = Fground y + Fgy
= (1100)sin55&deg; &quot; (62 kg)(9.8 m/s 2 )
= 901.1 N &quot; 607.6 N
!
!Fy = 293.5 N (two extra digits carried)
!
!F = (!Fx )2 + (!Fy )2
= (630.9 N)2 + (293.5 N)2
!
!F = 695.8 N (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.2-11
\$ #F '
! = tan &quot;1 &amp; y )
% #Fx (
\$ 293.5 N '
= tan &quot;1 &amp;
% 630.9 N )(
! = 25&deg;
!
!
!F = a
!
! !
a=
=
695.8 N [backward 25&deg; up]
62 kg
!
a = 11 m/s 2 [backward 25&deg; up]
Statement: The acceleration of the athlete is 11 m/s2 [backward 25&deg; up].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.2-12
Section 2.3: Applying Newton’s Laws of Motion
Tutorial 1!&quot;Practice, page
!&quot; 79
!&quot;
1. Given: F gB = 2.8 N ; F gA = 6.5 N ; F f = 1.4 N
(a) Required: F1
!
Analysis: !F = 0 N
Solution: The FBD for block B is shown below.
Equation for block
B:
!
!F = 0 N
! !
F1 &quot; FgB = 0 N
! !
F1 = FgB
!
F1 = 2.8 N
Statement: !The tension in
! the vertical rope is 2.8 N.
(b) Given: FgA = 6.5 N ; Ff = 1.4 N
! !
Required: F2 ; FN
!
Analysis: !F = 0 N
Solution: The FBD for block A is shown below.
Equations for!block A:
!F = 0 N
! !x
F2 &quot; Ff = 0 N
!
!
F2 = Ff
!
F2 = 1.4 N
!
!F = 0 N
!
!
FN &quot; FgA = 0 N
!
!
FN = FgA
!
FN = 6.5 N
Statement: The tension in the horizontal rope is 1.4 N. The normal force acting on block A
is 6.5 N.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.3-1
!
(c) Given: F2 = 1.4 N
!
Required: F3
!
Analysis: !F = 0 N
Solution: The FBD for point P is shown below.
Equations for point P.
For the x-components
of the force:
!
!Fx = 0 N
!
!
F3x &quot; F2 = 0 N
!
!
F3x = F2
!
F3x = 1.4 N
For the y-components
of the force:
!
!F = 0 N
!
!
F3y &quot; F1 = 0 N
!
!
F3y = F1
!
F3y = 2.8 N
!
Construct the vector F3 from its components:
!
!
!
F3 = ( F3x )2 + ( F3 y )2
= (1.4 N)2 + (2.8 N)2
!
F3 = 3.1 N
!
# F3 y &amp;
! = tan % ! (
\$ F3x '
&quot;1
# 2.8 N &amp;
= tan &quot;1 %
\$ 1.4 N ('
! = 63&deg;
Statement: The tension in the third rope is 3.1 N [right 63&deg; up].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.3-2
!
2. Given: m = 62 kg ; FT = 7.1! 102 N [right 32&deg; up]
!
Required: FW
!
Analysis: !F = 0 N
Balance the x-components
of the forces:
!
!Fx = 0 N
!
!
FWx + FT cos&quot; = 0 N
!
!
FWx = # FT cos&quot;
= #(710 N)cos32&deg;
!
FWx = #602.1 N (two extra digits carried)
Balance the y-components
of the forces:
!
!Fy 0 N
!
!
!
FWy + FTy Fg 0 N
!
!
FWy
FT sin # + mg
!
FWy
(710 N)sin32&deg; + (62 kg)(9.8m/s 2 )
= 376.2 N + 607.6 N
!
FWy = 231.4 N (two extra digits carried)
!
Construct the vector FW from its components:
!
!
!
FW = ( FWx )2 + ( FWy )2
= (602.1 N)2 + (231.4 N)2 (two extra digits carried)
!
FW = 6.5 ! 102 N
#F &amp;
! = tan &quot;1 % Wy (
\$ FWx '
# 231.4 N &amp;
= tan &quot;1 %
\$ 602.1 N ('
! = 21&deg;
Statement: The force exerted by the wall on the climber’s feet is 6.5 ! 102 N [left 21&deg; up].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.3-3
!
!
!
3. Given: F1 = 60.0 N [E 30.0&deg; S] ; F2 = 50.0 N [E 60.0&deg; N] ; !F = 0 N
!
Required: F3
! ! !
!
Analysis: !F = 0 N ; F1 + F2 + F3 = 0 N . Choose east and north as positive.
Solution:! For!the x-components
of the force:
!
F1x + F2 x + F3x = 0N
!
F3x = !(60.0 N)cos30.0&deg; ! (50.0 N)cos60.0&deg;
= !51.96 N ! 25.0 N
!
F3x = !76.96 N (two extra digits carried)
For the y-components
!
!
! of the force:
F1y + F2 y + F3 y = 0 N
!
F3 y = !(!60.0 N)sin30.0&deg; ! (50.0 N)sin60.0&deg;
= 30.0 N ! 43.30 N
!
F3 y = !13.30 N (two extra digits carried)
!
Construct the vector F3 from its components:
!
!
!
F3 = ( F3x )2 + ( F3 y )2
= (76.96 N)2 + (13.30 N)2
!
F3 = 78 N
!
&amp;
#
F
! 3 = tan &quot;1 % !3 y (
\$ F3x '
# 13.30 N &amp;
= tan &quot;1 %
\$ 76.96 N ('
! 3 = 9.8&deg;
Statement: The magnitude of the force is 78 N, at an angle [W 9.8&deg; S].
Tutorial 2 Practice, pages 81–82
1. (a) Solutions may vary. Sample solution:
!
Given: m1 = 1.2 kg ; m2 = 1.8 kg ; a = 1.2 m/s 2 [up]
! !
Required: F1 ; F2
!
!
Analysis: !F = ma
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.3-4
Solution: Equation !for top block (mass m1):
!
F = m1a
! !
!
F1 = F2 = m1g = m1a
! !
!
F1 = F2 + m1g + m1a
! !
!
F1 = F2 + m1 (g + a) (Equation 1)
Equation for bottom
block (mass m2):
!
!
!F = m2 a
!
!
F2 &quot; m2 g = m2 a
!
!
F2 = m2 g + m2 a
!
= m2 (g + a) (Equation 2)
= (1.8 kg)(9.8 m/s 2 + 1.2 m/s 2 )
!
F2 = 20 N
!
To calculate F1 , substitute Equation 2 into Equation 1:
! !
!
F1 = F2 + m1 (g + a)
!
!
!
F1 = m2 (g + a ) + m1 (g + a)
!
= (m2 + m1 )(g + a) (Equation 3)
= (3.0 kg)(11.0 m/s 2 )
!
F1 = 33 N
Statement: The tension in the top string is 33 N, and the tension in the bottom string is 20 N.
(b) Given: m1 = 1.2 kg ; m2 = 1.8 kg ; maximum string tension is 38 N
!
Required: maximum a that will not break the string
Analysis: (m2 + m1 )( g + a) ≤ 38 N
!
Solution: (m2 + m1 )(g + a) ! 38 N
38 N
!
g+a!
m2 + m1
! 38 N
a!
&quot;g
3.0 kg
!
a ! 12.67 m/s 2 &quot; 9.8 m/s 2
!
a ! 2.9 m/s 2
Statement: The maximum acceleration of the elevator that will not break the strings is
2.9 m/s2 [up].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.3-5
!
2. (a) Given: m = 63 kg ; Ff = 0 N ; ! = 14&deg; [above the horizontal]
!
Required: F N
!
Analysis: !Fy = 0 N
!
Solution:
!Fy = 0 N
!
!
FN &quot; Fgy = 0 N
!
FN = mg cos#
= (63 kg)(9.8 m/s 2 )cos14&deg;
!
FN = 6.0 \$ 102 N
Statement: The magnitude of the normal force on the skier is 6.0 &times;102 N .
(b) Given: θ =14&deg; [above the horizontal] ; g = 9.8 m/s2
! !
Required: a = a
!
!
Analysis: !F = ma . Choose +x-direction as the direction of acceleration, parallel to the hillside.
!
!
Solution: !Fx = ma
!
!
Fgx = ma
!
! Fgx
a=
m
m g sin &quot;
=
m
= (9.8 m/s 2 )sin14
!
a = 2.4 m/s 2
Statement: The magnitude of the skier’s acceleration is 2.4 m/s2.
!
!
3. Given: a = 1.9 m/s 2 [down hill] ; Ff = 0 N
Required: θ
!
Analysis: = sin !
!
Solution: = sin !
!
# &amp;
! = sin &quot;1 % (
\$ '
# 1.9 m/s 2 &amp;
= sin &quot;1 %
(
2
\$ 9.8 m/s '
! = 11&deg;
Statement: The angle between the hill and the horizontal is 11&deg;.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.3-6
!
!
!
4. (a) Given: Fa = 82 N [right 17&deg; up] ; FN = 213 N ; a = 0.15 m/s 2 [right]
Required: m!
Analysis: !Fy = 0 N . Choose right and up as positive.
Solution: For the y-components
of the forces:
!
!Fy = 0 N
!
!
FN + Fa sin &quot; # mg = 0 N
!
!
FN + Fa sin &quot;
m=
g
213 N + (82 N)sin17&deg;
=
9.8 m/s 2
m = 24.18 kg (two extra digits carried)
Statement: The mass of the desk is 24 kg.
!
!
!
(b) Given: Fa = 82 N [right 17&deg; up] ; FN = 213 N ; a = 0.15 m/s 2 [right]
!
Required: F
!
!
Analysis: !F = m
!
!
Solution:
!F = m
!
!
!
Fa &quot; Ff = m
!
!
!
Ff = Fa &quot; m
= (82 N)cos17&deg; &quot; (24.18 kg)(0.15 m/s 2 )
!
Ff = 75 N
Statement: The magnitude of the friction force on the
! desk is 75 N.
5. (a) Given: m1 = 9.1 kg ; m2 = 12 kg ; m3 = 8.7 kg ; F3 = 29 N [right 23&deg; up]
!
Required: a
Analysis: !F = ma . Choose right and up as positive.
Solution: For the x-components of the force:
!
!
!F = mT
!
!
F3 = mT
!
! F3 cos&quot;
=
mT
=
!
(29 N)cos 23&deg;
29.8 kg
= 0.8958 m/s 2 (two extra digits carried)
Statement: The carts accelerate at 0.90 m/s2 to the right.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.3-7
!
(b) Given: m3 = 8.7 kg ; a = 0.8958 m/s 2
!
Required: F1
!
!
Analysis: !F = ma
Solution: !For the x-components of the force:
!
!F = m3a
!
!
F1 = m3a
= (8.7 kg)(0.8958 m/s 2 )
= 7.793 N (two extra digits carried)
!
F1 = 7.8 N
Statement: The tension in the cord between m3 and m2 is 7.8 N.
!
!
(c) Given: a = 0.90 m/s 2 ; F1 = 7.793 N
!
Required: F2
!
!
Analysis: !F = m
Solution: Using
of the force:
! the x-components
!
!F = m2 a
! !
!
&quot; F1 + F2 = m2 a
!
!
F2 = F1 + (12 kg)(0.8958 m/s 2 )
= 7.793 N + 10.75 N
!
F2 = 19 N
Statement: The tension in the cord between m2 and m1 is 19 N.
6. (a) Given: mA = 4.2 kg ; mB = 1.8 kg ; θ = 32&deg;
!
Required: a!
!
Analysis: !F = ma
Solution: Equation
for block A:
!
!
!Fy = mA a y
!
!
!
FgA &quot; FT = mA a
!
!
mA g &quot; FT = mA a (Equation 1)
Equation for block B:
!
!
!Fx = mB ax
!
!
!
FT + FgBx = mB a
!
!
FT &quot; mB g sin # = mA a (Equation 2)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.3-8
Solve for acceleration
by
(2):
!
!
(mA g ! FT ) + ( FT ! mB g sin &quot; ) = mA a + mB a
!
(mA ! mB sin &quot; )g = (mA + mB ) a
! (m ! mB sin &quot; )g
a= A
mA + mB
=
(4.2 kg ! (1.8 kg)sin32&deg;)(9.8 m/s 2 )
4.2 kg + 1.8 kg
= 5.302 m/s 2 (two extra digits carried)
!
a = 5.3 m/s 2
Statement: The blocks accelerate at 5.3 m/s2. !
(b) Given: mA = 4.2 kg ; mB = 1.8 kg ; θ = 32&deg; ; a = 5.302 m/s 2
Required: tension in the string, FT
Analysis: We can substitute the value of acceleration into either of the equations from part (a) to
solve for FT . We will use Equation (1) because it is a bit simpler.
!
!
Solution: mA g ! FT = mA a
!
!
FT = mA g + mA a
!
= mA (g + a)
= (4.2 kg)(9.8 m/s 2 + 5.302 m/s 2 )
!
FT = 19 N
Statement: The tension in the string is 19 N.
Section 2.3
! Questions, page! 83
1. Given: F1 = 30 N [E 30&deg; N] ; F2 = 40 N [E 50&deg; S]
!
Required: !F
!
\$
'
! 2
! 2
!
#
F
Analysis: !F = (!Fx ) + (!Fy ) ; ! = tan &quot;1 &amp; !y ) . Choose east and north as positive.
% #Fx (
Solution: !For the
of the force:
! x-components
!
!Fx = F1x + F2 x
= (30 N)cos30&deg; + (40 N)cos50&deg;
= 25.98 N + 25.71 N
!
!Fx = 51.69 N (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.3-9
For the y-components
!
!
! of the force:
!Fy = F1y + F2 y
= (30 N)sin30&deg; + (&quot;40 N)sin50&deg;
= 15 N &quot; 30.64 N
!
!Fy = &quot;15.64 N (two extra digits carried)
!
Construct !F :
!
!
!
!F = (!Fx )2 + (!Fy )2
= (51.69 N)2 + (15.64 N)2
= 54.00 N
!
!F = 54 N
!
\$
'
#
F
! = tan &quot;1 &amp; !y )
% #Fx (
\$ 15.64 N '
= tan &quot;1 &amp;
% 51.69 N )(
! = 17&deg;
Statement: The total force exerted by the ropes on the skater is 54 N [E 17&deg; S].
2. Given: m = 45 kg
!
!
!
Required: FT1 ; FT2 ; FT
!
Analysis: !F = 0 N
Solution: For the forces on the mass,
!
FT2 ! m = 0 N
!
FT2 = m
For the y-components
!
! of the force:
FT2y + FT3y = 0 N
!
!
! FT2 + FT3 sin &quot; = 0 N
!
!
FT2
FT3 =
sin &quot;
!
mg
FT3 =
sin &quot;
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Chapter 2: Dynamics
2.3-10
For the x-components
!
! of the force:
F1 +F =( 2
!
!
! F 1 + F cos&quot; = 0 N
!
!
FT1 = FT3 cos&quot;
# mg &amp;
=%
cos&quot;
\$ sin &quot; ('
!
mg
FT1 =
tan &quot;
Calculate the tensions in the three cables.
!
mg
FT1 =
tan !
45 kg (9.8 m/s 2 )
=
tan60.0&deg;
!
FT1 = 250 N
!
FT2 = mg
= 45 kg (9.8 m/s 2 )
!
FT2 = 440 N
!
mg
FT3 =
sin !
45 kg (9.8 m/s 2 )
=
sin60.0&deg;
!
FT3 = 510 N
!
!
!
Statement: FT1 is 250 N, FT2 is 440 N, and FT is 510 N.
!
!
3. Given: m = 2.5 kg ; Fair = 12 N [right] ; !F = 0 N
Required: θ
!
&quot;!
&quot;!
#
&amp;
!
!
F
Analysis: !F = 0 N ; FT = (F Tx )2 + (F Ty )2 ; ! = tan &quot;1 % !Ty ( . Choose right and up as positive.
\$ FTx '
Solution: For the
! x-components of the force:
!Fx = 0 N
!
!
FTx + Fair = 0 N
!
!
FTx = &quot; Fair
!
FTx = &quot;12 N
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.3-11
For the y-components
of the force:
!
!Fy = 0 N
!
FTy &quot; mg = 0 N
!
FTy = mg
= (2.5 kg)(9.8 m/s 2 )
!
FTx = 24.5 N (one extra digit carried)
!
Construct F from its components:
!
!
!
FT = ( FTx )2 + ( FTy )2
= (12 N)2 + (24.5 N)2
!
FT = 27 N
!
#
&amp;
F
! = tan &quot;1 % ! y (
\$ Fx'
=t
&quot;1
# 24.5 N &amp;
%\$ 12 N ('
! = 64&deg;
Statement: The tension is the rope is 27 N. The rope makes an angle of 64&deg; with the horizontal.
!
4. (a) Given: θ = 15&deg; ; m = 1.41&times;103 kg ; !F = T
Required: FBD showing the forces on the car
Analysis: Choose [down the hill] as the positive x-direction and [up perpendicular to hill] as the
positive y-direction.
Solution: The FBD for the car is shown below.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.3-12
!
(b) Given: !F = 0 N
Required: equations for the conditions for static equilibrium along horizontal and vertical
directions !
!
Analysis: !Fx = 0 N ; !Fy = 0 N
Solution: For the x-components
of the force (horizontal):
!
!F = 0 N
!
!
F &quot;F =0 N
!
mg # &quot; F = 0 N
For the y-components
! of the force (vertical):
!Fy = 0 N
!
!
FN + Fgy = 0 N
!
FN &quot; mg cos# = 0 N
!
(c) Given: θ = 15&deg; ; m = 1.41&times;103 kg ; !F = 0 N
Required: FT
!
Analysis: mg 6 5 ! &quot; F = 0
!
Solution: m sin ! &quot; FT = 0 N
!
FT = m sin !
= (1410 kg)(9.8 m/s 2 )sin15&deg;
!
FT = 3.6 # 103 N
Statement: The tension in the cable is 3.6 &times;103 N .
!
5. (a) Given: Fa = 42 N [right 35&deg; down] ; m = 18 kg ; Δd = 5.0 m
Required: FBD for the mower
Analysis: Choose forward and up as positive
Solution: The FBD for the lawn mower is shown below.
!
(b) Given: Fa = 42 N [right 35&deg; down] ; m = 18 kg ; Δd = 5.0 m
Required: a !
!
Analysis: !F = m
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.3-13
!
!
Solution: !Fx = ma
!
!
Fax = ma
!
! Fax
a=
m
(42 N)cos35&deg;
=
18 kg
= 1.911 m/s 2 (two extra digits carried)
!
a = 1.9 m/s 2
Statement:!The acceleration of the mower is 1.9 m/s2 [forward].
(c) Given: Fa = 42 N [right 35&deg; down] ; m = 18 kg
!
Required: F
!
Analysis: Use the FBD to identify the forces with vertical components. Use !Fy = 0 N to solve
for the normal force. !
Solution:
!Fy = 0 N
!
!
FN + Fay &quot; mg = 0 N
!
!
FN = mg &quot; Fay
= (18 kg)(9.8 m/s 2 ) &quot; (&quot;42 N)sin35&deg;
= 176.4 N + 24.09 N
!
FN = 200 N
Statement: The normal force is 2.0 &times;102 N [up].
!
(d) Given: Δd = 5.0 m ; a = 1.911 m/s 2
!
Required: v
!
Analysis: vf2 = vi2 + 2a!d
!
Solution: f2 = i2 + 2a!
2
f
= 2(1.911 m/s 2 )(5.0 m)
= 4.4 m/s
Statement: The velocity of the mower
! when it reaches the lawn is 4.4 m/s [forward].
6. (a) Given: m = 1.3 kg ; θ = 25&deg; ; !F = 0 N
!
Required: F
!
Analysis: !Fx = 0 N
f
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.3-14
Solution:
!
!F = 0 N
!
Fa &quot; mg sin # = 0 N
!
Fa = mg sin #
= (1.3 kg)(9.8 m/s 2 )sin 25&deg;
!
Fa = 5.4 N
Statement: A force of 5.4 N is required to pull the cart up the ramp at a constant velocity.
!
(b) Given: = 1.3 kg ; θ = 25&deg; ; a = 2.2 m/s 2 [up the ramp]
!
Required: F
!
!
Analysis: !F = m
!
!
Solution:
!Fx = ma
!
!
Fa &quot; mg sin # = ma
!
!
Fa = ma + mg sin #
= (1.3 kg)(2.2 m/s 2 ) + (1.3 kg)(9.8 m/s 2 )sin 25&deg;
= 2.86 N + 5.384 N
!
Fa = 8.2 N
Statement: A force of 8.2 N is required to pull the cart up the ramp at an acceleration of 2.2 m/s.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.3-15
Section 2.4: Forces of Friction
Mini Investigation: Light from Friction, page 86
A. When we crushed the mints, the candy briefly gave off light.
B. The crystals of sugar in the mints rubbed together to create the friction that produces the light.
Tutorial 1 Practice, page 89
1. (a) An FBD of the top book during its acceleration is shown below.
(b) The force of static friction causes the top book to accelerate horizontally.
2. Given: a = 2.7 m/s 2
Required: &micro;S
!
!
!
!
Analysis: FS = &micro;S FN ; !F = ma
Solution: Equation
for y-components of the force:
!
!Fy = 0 N
!
FN &quot; mg = 0 N
!
FN = mg
Equation for! x-components of the force:
!
!Fx = ma
!
!
FS = ma
!
!
&micro;S FN = ma
!
&micro;S mg = ma
!
a
&micro;S =
g
=
2.7 m/s 2
9.8 m/s 2
&micro;S = 0.28
Statement: The smallest coefficient of static friction between dinner plates that will prevent
slippage is 0.28.
!
!
3. Given: FT = 28 N [forward 29&deg; up] ; &micro;S = 0.45 ; &micro;K = 0.45 ; !F = 0 N
Required: m!
!
Analysis: !F = ma
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.4-1
Solution: Equation for y-components
of the force:
!
!Fy = 0 N
!
!
FN + FT sin &quot; # mg = 0 N
!
!
FN = mg # FT sin &quot; (Equation 1)
Equation for x-components
of the force:
!
!
!Fx = ma
!
!
!
FT cos&quot; # FS = ma
!
!
!
FT cos&quot; # &micro;S FN = ma (Equation 2)
Substitute Equation (1)
(2) and solve for m:
! into Equation
!
FT cos! &quot; &micro;S FN = 0 N
!
!
FT cos! &quot; &micro;S ( mg &quot; FT sin ! ) = 0 N
!
FT (cos! + &micro;S sin ! ) = &micro;S mg
!
FT (cos! + &micro;S sin ! )
m=
&micro;S g
=
(28 N)(cos 29&deg; + 0.45sin 29&deg;)
0.45(9.8 m/s 2 )
m = 6.9 kg
Statement: The smallest possible mass for the box is 6.9 kg.
!
4. Given: θ = 6.0&deg; ; vi = 12 m/s [down slope] ; vf = 0 m/s ; &micro;K = 0.14
Required: Δ!d
!
Analysis: !F = ma ; vf 2 = vi 2 + 2aΔd
Solution: Equation!for y-components of the force:
!Fy = 0 N
!
FN &quot; mg cos# = 0 N
!
FN = mg cos#
Equation for x-components of
! the force:
!
!Fx = max
!
!
mg sin &deg; FK = max
!
mg sin &deg; &micro; K mg cos&deg; = max
!
ax = g(sin &deg; &micro; K cos&deg; )
= (9.8 m/s 2 )(sin6.0&deg; 0.14cos6.0&deg;)
!
ax = 0.3401 m/s 2 (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.4-2
Solve for the distance travelled:
!
vf2 = vi2 + 2 ax !d
vf2 &quot; vi2
!d =
!
2 ax
=
(0 m/s)2 &quot; (12 m/s)2
2(&quot;0.3401 m/s 2 )
!d = 2.1# 102 m
Statement: The sled will slide for 2.1 ! 102 m before coming to rest. !
5. Given: m = 39 kg ; direction of rope [forward 21&deg; up] ; &micro;K = 0.23 ; !F = 0 N
!
Required: FT
!
Analysis: !F = 0 N
Solution: Equation !for y-components of the force:
&micro;Fy = 0 N
!
!
FN + FTy &quot; mg = 0 N
!
!
FN = mg &quot; FT sin #
Equation for x-components of the!force:
&micro;Fx = 0 N
!
!
FTx &quot; FK = 0 N
!
!
FT cos# &quot; &micro; K FN = 0 N
!
!
FT cos# &quot; &micro; K (mg &quot; FT sin # ) = 0 N
!
FT (cos# + &micro; K sin # ) = &micro; K mg
!
&micro; K mg
FT =
cos# + &micro; K sin #
(0.23)(39 kg)(9.8 m/s 2 )
cos 21&deg; + 0.23sin 21&deg;
FT = 87 N
Statement: A tension of 87 N in the rope is needed to keep the box moving at a constant velocity.
!
6. (a) Given: m1 = 24 kg ; m2 = 14 kg ; &micro;K = 0.32 ; Fa = 1.8 ! 102 N [forward 25&deg; up]
!
Required: a!
!
!
Analysis: !Fy = 0 N ; !Fx = 1a
=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.4-3
Solution: Equation for
! y-components of the force:
!Fx = 0 N
!
!
FN1 + Fax &quot; m1g = 0 N
!
!
FN1 = m1g &quot; Fa sin #
= (24 kg)(9.8 m/s 2 ) &quot; (180 N)sin 25&deg;
!
FN1 = 159.1 N (two extra digits carried)
!
!
FK1 = &micro; K FN1
= 0.25(159.1 N)
!
FK1 = 39.78 N (two extra digits carried)
Equation for x-components
of the force:
!
!
Fx = 1a
!
!
!
!
=
=
= 1a (Equation 1)
ax
K1
T
Equation for y-components
of the force:
!
!Fy = 0 N
!
FN2 &quot; m2 g = 0 N
!
FN2 = m2 g
= (14 kg)(9.8 m/s 2 )
!
FN2 = 137.2 N (two extra digits carried)
!
!
FK2 = &micro; K FN2
= 0.25(137.2 N)
!
FK2 = 34.3 N
Equation for x-components
of the force:
!
!
!Fx = 2 a
!
!
!
&quot;
= 2 a (Equation 2)
T
K2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.4-4
Add equations (1) and (2) to eliminate the tension of the rope. Solve for a.
( FT ! FK2 ) + ( Fax ! K1 ! T ) = 1a + 2 a
ax
!
K1
!
K2
=(
a=
=
1
+
2
ax
!
K1
+
1
)a
!
K2
2
(180 N)cos 25&deg; ! 39.78 N ! 34.3 N
38 kg
= 1.797 m/s 2 (two extra digits carried)
a = 1.8 m/s 2
Statement: The acceleration of the boxes is 1.8 m/s2.
!
(b) Given: m2 = 14 kg ; ; &micro; K = 0.32; Fa = 1.8 ! 102 N [forward 25&deg; up]
!
Required: FT
!
!
!
Analysis: FT ! FK2 = m2 a
!
!
!
Solution: FT ! FK2 = m2 a
!
!
!
FT = &micro; K FN2 + m2 a
= 0.25(137.2) + (14 kg)(1.797 m/s 2 )
!
FT = 59 N
Statement: The tension in the rope is 59 N.
7. Given: &micro;K = 0.20 ; &micro;S = 0.25 ; m = 100.0 kg
Required: a !
!
!
Analysis: !F = 0 N ; !Fx = max
Solution: For !the y-components of the force:
!Fy = 0 N
!
FN &quot; mg = 0 N
!
FN = mg
= (100.0 kg)(9.8 m/s 2 )
!
FN = 980 N
For the x-components
of the force:
!
!F = 0 N
! !x
Fa &quot; FS = 0 N
!
!
Fa = FS
!
= &micro;s FN
= 0.25(980 N)
!
Fa = 245 N
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.4-5
Once the refrigerator
is moving, an applied for 245 N produces acceleration, a:
!
!
!F = a
! !x
!
&quot; K= a
a
!
!
&quot; &micro;K N
!
a
a=
=
245 N &quot; 0.20(980 N)
100.0 kg
!
a = 0.49 m/s 2
Statement: The acceleration when you apply minimum force needed to move the refrigerator
is 0.49 m/s2.
8. Given: &micro;S = 0.25 ; m = 110 kg
!
Required: Fa
!
!
!
Analysis: FS = &micro;S FN ; !F = 0 N
Solution: For !the y-components of the force:
!Fy = 0 N
!
FN &quot; mg = 0 N
!
FN = mg
= (110 kg)(9.8 m/s 2 )
!
FN = 1078 N (two extra digits carried)
For the x-components
of the force:
!
&micro;F = 0 N
! !x
Fa &quot; FS = 0 N
!
!
Fa = FS
!
= &micro;s FN
= 0.25(1078 N)
!
Fa = 2.7 # 102 N
Statement: The minimum force required to just set the stage prop into motion is 2.7 ! 102 N.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.4-6
Section 2.4 Questions, page 90
1. Given: vi = 20 m/s; vf = 0 m/s; Δd = 40 m
Required: coefficient of friction between the tires and the road, &micro; K
!
!
Analysis: vf 2 = vi 2 + 2aΔd ; !Fx = max . Choose forward as positive.
!
Solution: Find the acceleration a :
!
2
= i2 + 2 =
f
!
!
==
2
i
2=
(20 m/s)2
==
2(40 m)
= =5.0 m/s 2
Calculate &micro; K :
!
!
!Fx = ma
!
!
&quot; FK = ma
!
&quot; &micro;K m g = m a
!
a
&micro;K = &quot;
g
=
5.0 m/s 2
9.8 m/s 2
&micro; K = 0.5
Statement: The coefficient of friction between the tires and the road is 0.5.
2. Given: vi = 50.0 m/s ; Δt = 10.0 s ; &micro;K = 0.030
Required: vf
!
!
Analysis: !Fx = a . Choose forward as positive.
Solution: Find !the acceleration a:
!
!Fx = ma
!
!
&quot; FK = ma
!
&quot; &micro;K m g = m a
!
a = &quot; &micro;K g
= &quot;0.030(9.8 m/s 2 )
!
a = &quot;0.294 m/s 2 (one extra digit carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.4-7
Calculate vf :
vf = vi + a=t
= 50.0 m/s + (=0.294 m/s 2 )(10.0 s)
vf = 47 m/s
Statement: The speed of the puck after 10.0 s is 47 m/s.
!
3. (a) Given: m = 2.0 &times;102 kg ; Fa = 3.5 ! 102 N
Required: &micro;S
!
!
!
Analysis: !F = 0 N ; FS = &micro;S FN
Solution: For !the y-components of the force:
!Fy = 0 N
!
FN &quot; mg = 0 N
!
FN = mg
For the x-components
of the force:
!
!Fx = 0 N
! !
&quot; S =0 N
a
!
!
= a
S
!
!
&micro;S N = a
!
&micro;S =
=
a
350 N
(200 kg )(9.8 m/s 2 )
&micro;S = 0.18
Statement: The coefficient of static friction between the floor and the sofa is 0.18.
!
(b) Given: m = 2.0 &times;102 kg ; Fa = 3.5 ! 102 N ; vi = 0 m/s ; vf = 2.0 m/s ; Δt = 5.0 s
Required: &micro; K
!
!
Analysis: !F = ma
!
Solution: Find the acceleration, :
! !v
a=
!t
2.0 m/s
=
5.0 s
!
a = 0.40 m/s 2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.4-8
Using the x-components
of the force:
!
!
!F = ma
! !x
!
Fa &quot; FK = ma
!
!
!
FK = Fa &quot; ma
!
!
!
&micro; K FN = Fa &quot; ma
!
!
Fa &quot; ma
&micro;K =
mg
=
350 N &quot; (200 kg )(0.40 m/s 2 )
(200 kg )(9.8 m/s 2 )
&micro; K = 0.14
Statement: The coefficient of kinetic friction between the sofa and the floor is 0.14.
4. (a) Given: &micro;S = 0.29
Required: !
!
Analysis: !F = 0 N
Solution: For the y-components
of the force:
!
!Fy = 0 N
!
FN &quot; mg cos# = 0 N
!
FN = mg cos#
For the x-components
! of the force:
!Fx = 0 N
!
sin # S = 0 N
!
sin &quot; = &micro;S N
sin &quot; = &micro;S
cos&quot;
tan &quot; = &micro;S
&quot; = tan #1 (0.29)
&quot; = 16&deg;
Statement: The crate just begins to slip when the angle of inclination, θ , is 16&deg;.
(b) Given: &micro;K = 0.26 ; θ = 16.17&deg;
!
Required: !
!
Analysis: !Fx = ma
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.4-9
Solution:
!
!
!Fx = ma
!
!
mg sin # FK = ma
!
mg sin &quot; # &micro; K mg cos&quot; = ma
!
a = g(sin &quot; # &micro; K cos&quot; )
= (9.8 m/s 2 )(sin16.17&deg; # 0.26cos16.17&deg;)
!
a = 0.28 m/s 2
Statement: The crate accelerates at 0.28 m/s2 [down the incline] when the coefficient of kinetic
friction is 0.26.
5. (a) Two situations in which friction is helpful for an object moving on a horizontal surface are
when running, so you can push yourself forward, and when walking on a snowy field so you can
use traction from the snow to move yourself forward.
(b) Two situations in which it would be ideal if there were no friction when an object moves
across a horizontal surface are: shooting the puck across the ice when playing hockey; and when
trying to move a sled over a snowy sidewalk.
6. (a) Given: m1 = 45 kg ; m2 = 12 kg ; &micro;S = 0.45 ; &micro;K = 0.35
!
!
Required: Fg2 ! &micro; FN
Analysis: To determine if this system is in static equilibrium, you need to determine the !
magnitude of the static friction, FS , for mass m1 and compare it to the tension in the string, Fg2 , for
mass m2 .
Find FS for mass m1 .
!
!
FS = &micro;S FN
= &micro;S m1g
= 0.45(45 kg)(9.8 m/s 2 )
!
FS = 200 N
!
Find Fg2 for mass m2 .
!
Fg2 = m2 g
= (12 kg)(9.8 m/s 2 )
!
Fg2 = 120 N
Statement: Yes, the system is in static equilibrium. The tension in the string for mass m2 is not
sufficient to break the force of static friction for mass m1 , so there is no acceleration.
!
(b) Given: Fg2 = 120 N
!
Required: FT
Analysis: As long as the! two !masses remain at rest, the tension in the string is equal to the force
of gravity on mass m2 , FT = Fg2 = m2 g .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.4-10
!
!
Solution: FT = Fg2
!
FT = 120 N
Statement: The tension in the string is 120 N.
(c) Given: m1 = 45 kg; m2 = 32 kg; &micro;K = 0.35
!
Required: !
!
Analysis: !F = !ma
Solution: Find FT for mass m1 .
!
!
!Fx = m1a
!
!
!
FT &quot; FK = m1a
!
!
FT &quot; &micro; K m1g = m1a
!
Find FT for mass m2 .
!
!
!Fy = m2 a
!
!
m2 g &quot; F = m2 a
!
Solve for a :
!
!
( FT ! &micro;K m1g ) + ( m2 g ! FT ) = m1a! + 2 a!
!
( 2 ! &micro;K 1 ) = ( 1 + 2 )a
! (
a=
=
2
! &micro;K
+
1
1
)
2
(32 kg ! 0.35(45 kg))(9.8 m/s 2 )
77 kg
!
a = 2.1 m/s 2
Statement: The acceleration of the system is 2.1 m/s2, eliminating the string tension.
7. (a) Given: θ = 42&deg;
Required: &micro;S
!
Analysis: !F = 0 N
Solution: For the y-components
of the force:
!
!Fy = 0 N
!
FN &quot; mg cos# = 0 N
!
FN = mg cos#
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.4-11
For the x-components
! of the force:
!Fx = 0 N
!
sin &quot; # S = 0 N
!
sin &quot; = &micro;S N
sin &quot; = &micro;S
cos&quot;
tan &quot; = &micro;S
&micro;S = tan 42&deg;
&micro;S = 0.90
Statement: The coefficient of static friction is 0.90.
(b) Given: θ = 35&deg;
Required: &micro; K
!
Analysis: !Fx = 0 N
!
Solution:
!Fx = 0 N
!
mg sin &deg; # FK = 0 N
mg sin &quot; = &micro; K mg cos&quot;
&micro; K = tan &quot;
= tan35&deg;
&micro; K = 0.70
Statement: The coefficient of kinetic friction is 0.70.
8. Given: m1 = 8.0 kg ; m2 = 12 kg ; θ1 = 26&deg; ; θ 2 = 39&deg; ; &micro;K = 0.21
!
Required: !
!
Analysis: !F = ma
Solution:
For the y-components! of the force (mass m1):
!Fy = 0 N
!
FN1 &quot; m1g cos#1 = 0 N
!
FN1 = m1g cos#1
For the y-components!of the force (mass m2):
!Fy = 0 N
!
FN2 &quot; m2 g cos# 2 = 0 N
!
FN2 = m2 g cos# 2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.4-12
For the x-components of the force !(mass m1):
!
!Fx = m1a
!
!
!
FT &quot; m1g sin #1 &quot; FK1 = m1a
!
!
FT &quot; m1g sin #1 &quot; &micro; K m1g cos#1 = m1a
!
!
FT &quot; 49.17 N = m1a (two extra digits carried)
For the x-components of the force (mass
m2):
!
!
!Fx = m2 a
!
!
!
m2 g sin &quot; 2 # FK2 # FT = m2 a
!
!
m2 g sin &quot; 2 # &micro; K m2 g cos&quot; 2 # FT = m2 a
!
!
54.82 N # FT = m2 a (two extra digits carried)
! equations for mass m1 and
! mass m2 to!eliminate the string tension.
( FT ! 49.17 N) + (54.82 N ! FT ) = (m1 + m2 ) a
! 54.82 N ! 49.17 N
a=
20 kg
!
a = 0.28 m/s 2
Statement: The acceleration of the two masses (as a system) is 0.28 m/s2 [clockwise].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 2: Dynamics
2.4-13
Section 3.1: Inertial and Non-inertial Frames of Reference
Tutorial 1 Practice, page 110
1. (a) When the car is moving with constant velocity, I see the ball lie still on the floor. I would
see the same situation when the car is at rest.
(b) To an observer on the sidewalk, the ball appears to be moving with a constant velocity of
14 m/s [E].
(c) If the car accelerates forward, I see the ball roll backward on the floor.
(d) As observed from the frame of reference of the car:
As observed from the frame of reference of the sidewalk:
The car’s frame of reference is non-inertial. I observe a fictitious force (the force pushing the ball
backward, west) in the car’s frame of reference.
2. (a) Given: m = 22.0 g = 0.0220 kg; ! = 32.5&deg;
Required: a
Analysis: Look at the situation from an Earth (inertial) frame of reference. The horizontal
component of the tension FT balances the acceleration and the vertical component of the tension
FT balances the gravitational force. Express the components of the tension in terms of the
horizontal and vertical applied forces. Then calculate the magnitude of the acceleration.
Solution: Vertical component of force:
!Fy = 0
FT cos&quot; # mg = 0
mg
FT =
cos&quot;
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.1-1
Horizontal component of force:
ΣFx = ma
FT sin &quot; = ma
# mg &amp;
%\$ cos&quot; (' sin &quot; = ma
# sin &quot; &amp;
g%
=a
\$ cos&quot; ('
g tan &quot; = a
m/s 2 ) tan32.5&deg;
a=(
m/s 2
a=
Statement: The magnitude of the boat’s acceleration is 6.2 m/s2. I did not need to know the mass
of the ball to make the calculation because that value was cancelled out to obtain the forces.
(b) Given: m = 22.0 g = 0.0220 kg; ! = 32.5&deg;
Required: FT
mg
.
Analysis: From part (a), FT =
cos!
mg
Solution: FT =
cos!
m/s 2
0.0220 kg
=
cos32.5&deg;
FT = 0.26 N
Statement: The magnitude of the tension in the string is 0.26 N. I need to know the mass to
make this calculation because the force is the mass multiplied by the acceleration.
3. (a) When the subway is moving at a constant velocity, there is no tension in the strap.
(b) Given: m = 14 kg; ! = 35&deg;; a = 1.4 m/s2
Required: FT
Analysis: Look at the situation from an Earth (inertial) frame of reference. The horizontal
component of the tension FT balances the acceleration, so express the x-components of the
tension in terms of the horizontal applied forces. The vertical forces of gravity and the normal
force balance each other.
ΣFx = ma
FT sin &quot; = ma
ma
FT =
sin &quot;
ma
Solution: FT =
sin !
14 kg ) 1.4 m/s 2
(
=
tan35&deg;
FT = 34 N
Statement: The tension on the strap during acceleration is 34 N.
)(
(
(
)
)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.1-2
4. Given: &micro;s = 0.42
Required: maximum a
Analysis: The force due to the acceleration of the train, F = ma, must be less than the force of
static friction, Fs = &micro;s FN , where FN = mg FN = mg. So the maximum acceleration occurs when
F = Fs.
F = Fs
ma = !s FN
ma = !s ( mg)
a = !s g
Solution: a = !s g
m/s 2 )
= ( 0.42 )(
a = 4.1 m/s 2
Statement: The maximum acceleration of the train before the passenger begins to slip along the
floor is 4.1 m/s2.
Tutorial 2 Practice, page 112
!
1. (a) Given: m = 55 kg; a = 2.9 m/s2 [up]
Required: FN
Analysis: Use up as positive and solve for the normal force when +FN + (–mg) = ma.
= FN = (#mg) = ma
FN = ma = mg
Solution: FN = ma = mg
(
)
= (55 kg ) 2.9 m/s 2 = (55 kg )
FN =
! 10 N
2
(
m/s 2
)
Statement: The student’s apparent weight is 7.0 &times; 102 N.
!
(b) Given: m = 55 kg; a = 2.9 m/s2 [down]
Required: FN
Analysis: Use down as positive and solve for the normal force when –FN + (mg) = ma.
– FN = (mg) = ma
FN = ma – mg
Solution: FN = mg # ma
= (55 kg )
FN =
(
&times; 10 N
2
)
(
m/s 2 # (55 kg ) 2.9 m/s 2
)
Statement: The student’s apparent weight is 3.8 &times; 102 N.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.1-3
2. (a) Given: m1 = 9.5 kg; m2 = 2.5 kg; FN = 70.0 N
!
Required: a
Analysis: Use up as positive and solve for the acceleration when +FN + (–mg) = ma.
= FN = (#mg) = ma
F # mg
a= N
m
FN # ( m1 = m2 ) g
=
m1 = m2
Solution: a =
FN # ( m1 = m2 ) g
m1 = m2
m
70.0 kg &quot; 2 ! 9.5 kg = 2.5 kg 9.8 m/s 2
s
=
9.5 kg = 2.5 kg
= !3.967 m/s 2 (two extra digits carried)
a = !4.0 m/s 2
Statement: The acceleration of the elevator is 4.0 m/s2 [down].
!
(b) Given: m = 2.5 kg; a = 3.967 m/s2 [down]
!
Required: FN
Analysis: Use down as positive and solve for the normal force when –FN + (mg) = ma.
Solution: # FN = (mg) = ma
FN = mg # ma
(
)(
(
)
)
(
= ( 2.5 kg ) 9.8 m/s 2 # ( 2.5 kg ) 3.967 m/s 2
)
FN = 15 N
Statement: The force on the smaller box is 15 N [up].
3. (a) Given: m1 = 4.2 kg; m2 = 2.6 kg; a = 0
Required: FTA; FTB
Analysis: Since this is an inertial frame of reference, the tension on rope B balances the force of
gravity on block 2. The tension on rope A balances the force of gravity on block 1 and the
tension on rope B. Fg = mg
Solution: Determine the force on rope B:
FTB = m2 g
(
)(
)
(
)(
)
= 2.6 kg 9.8 m/s 2
= 25.48 N (two extra digits carried)
FTB = 25 N
Determine the force on rope A:
FTA = m1 g = FTB
= 4.2 kg 9.8 m/s 2 = 25.48 N
FTA = 67 N
Statement: The tension on the rope A is 67 N and the tension on the rope B is 25 N.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.1-4
!
(b) Given: m1 = 4.2 kg; m2 = 2.6 kg a = 1.2 m/s2 [up]
Required: FTA; FTB
Analysis: This is now a non-inertial frame of reference, so instead of just gravity, the
acceleration is g – (–a), or g + a, where down is positive.
Solution: Determine the force on rope B:
FTB = m2 g = a
(
(
)(
)
= 2.6 kg 9.8 m/s 2 = 1.2 m/s 2
)
= 28.6 N (one extra digit carried)
FTB = 29 N
Determine the force on rope A:
FTA = m1 g = a = FTB
(
(
)
)(
)
= 4.2 kg 9.8 m/s 2 = 1.2 m/s 2 = 28.6 N
FTA = 75 N
Statement: The tension on the rope A is 75 N and the tension on the rope B is 29 N.
!
4. Given: a = 0.98 m/s2 [down]; m = 61 kg
Required: FN
Analysis: Use down as positive and solve for the normal force when –FN + (mg) = ma.
Solution: # FN = (mg) = ma
FN = mg # ma
(
)
(
= ( 61 kg ) 9.8 m/s 2 # ( 61 kg ) 0.98 m/s 2
FN = 5.4 &times; 10 N
2
)
Statement: The passenger’s apparent weight is 5.4 &times; 102 m/s2.
Section 3.1 Questions, page 113
1. (a) The ball would appear to move straight up and down because I am moving with the same
velocity as the other train. From my viewpoint, the other train and passenger are standing still,
and the ball is not affected by the train’s motion.
(b) If the trains moved in opposite directions, the ball would appear to have horizontal motion, so
I would see the path as a parabola.
2. Given: a = 1.5 m/s2
Required: !
Analysis: The horizontal component of the tension FT balances the acceleration, and the vertical
component of the tension FT balances the gravitational force. Express the tangent ratio of the
angle in terms of the applied force and the gravitational force, then solve for the angle.
F
tan &quot; = a
Fg
# ma &amp;
&quot; = tan #1 %
\$ mg ('
# a&amp;
&quot; = tan #1 % (
\$ g'
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.1-5
# a&amp;
Solution: &quot; = tan #1 % (
\$ g'
#
m &amp;
% 1.5 s 2 (
(
= tan #1 %
% 9.8 m (
%\$
s 2 ('
&quot; = 8.7&deg;
Statement: The string makes an 8.7&deg; angle with the vertical.
3. Given: vf = 255 km/h; !t = 10.0 s
Required: !
v
Analysis: Determine the acceleration using vf = a!t or a = f . The horizontal component of the
!t
tension FT balances the acceleration and the vertical component of the tension FT balances the
gravitational force. Express the tangent ratio of the angle in terms of the applied force and the
gravitational force, then solve for the angle.
Solution: Determine the acceleration of the plane:
v
a= f
!t
km
255
h ! 1000 m ! 1 h ! 1 min
=
10.0 s
60 s
60 min
1 km
2
= 7.0833 m/s (two extra digits carried)
a = 7.08 m/s 2
Determine the angle the string makes with the vertical:
F
tan &quot; = a
Fg
# ma &amp;
&quot; = tan #1 %
\$ mg ('
# a&amp;
= tan #1 % (
\$ g'
#
m&amp;
% 7.0833 s 2 (
(
= tan #1 %
% 9.8 m (
%\$
s 2 ('
&quot; = 35.9&deg;
Statement: The string makes an 35.9&deg; angle with the vertical.
4. Given: ! = 16&deg;
Required: a
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.1-6
Analysis: Look at the situation from an Earth (inertial) frame of reference. The horizontal
component of the tension FT balances the acceleration and the vertical component of the tension
FT must balance the gravitational force since the cork ball does not move. Express the
components of the tension in terms of the horizontal and vertical applied forces. Then calculate
the magnitude of the acceleration.
Vertical component of force:
!Fy = 0
FT cos&quot; # mg = 0
mg
FT =
cos&quot;
Horizontal component of force:
ΣFx = ma
FT sin &quot; = ma
# mg &amp;
%\$ cos&quot; (' sin &quot; = ma
# sin &quot; &amp;
g%
=a
\$ cos&quot; ('
g tan &quot; = a
Solution: a = g tan !
= 9.8 m/s 2 tan16&deg;
(
)
a = 2.8 m/s
Statement: The magnitude of the car’s acceleration is 2.8 m/s2.
5. Given: vf = 6.0 m/s; !t = 10.0 s; m = 64 kg
Required: FN
2
Analysis: Determine the upward acceleration using vf = a!t or a =
vf
. Use up as positive and
!t
solve for the normal force when +FN + (–mg) = ma.
= FN = ( #mg ) = ma
FN = ma = mg
Solution: Determine the upward acceleration:
v
a= f
!t
6.0 m/s
=
10.0 s
a = 0.60 m/s 2
Determine the apparent weight:
FN = ma = mg
(
)
(
= ( 64 kg ) 0.6 m/s 2 = ( 64 kg ) 9.8 m/s 2
FN = 6.7 ! 10 N
2
)
Statement: The passenger’s apparent weight is 6.7 &times; 102 N.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.1-7
6. Given: FN = 255 N; m = 52 kg
Required: a
Analysis: Use up as positive and solve for the acceleration when +FN + (–mg) = ma.
= FN = ( #mg ) = ma
F # mg
a= N
m
F # mg
Solution: a = N
m
m
255 kg &quot; 2 ! 52 kg 9.8 m/s 2
s
=
52 kg
a = !4.9 m/s 2
Statement: The acceleration of the ride is 4.9 m/s2 [down].
7. (a) At rest:
Accelerating downhill:
(
)(
)
When the car is at rest, the dice are aligned with the vertical (with respect to level ground), and
make an angle of 17&deg; with respect to the normal (perpendicular to the roof). The only forces
acting on the dice are gravity and tension. When the car is accelerating, a horizontal fictitious
force pointing to the rear of the car deflects the dice so that they are aligned with the normal, and
make an angle of 17&deg; with respect to the vertical, as viewed from the level ground.
(b) Given: ! = 17&deg;
Required: a
Analysis: Look at the situation from an inertial frame of reference on the same angle as the hill.
The “vertical” component of the gravitational force Fg balances the tension and the “horizontal”
component of the gravitational force Fg balances the force applied by the acceleration. Express
the horizontal component of Fg in terms of the applied forces. Then calculate the magnitude of
the acceleration.
!Fx = 0
Fg sin &quot; # ma = 0
mg sin &quot; # ma = 0
g sin &quot; # a = 0
a = g sin &quot;
Solution: a = g sin !
= ( 9.8 m/s 2 ) sin17&deg;
a = 2.9 m/s 2
Statement: The magnitude of the car’s acceleration is 2.9 m/s2.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.1-8
8. (a) Given: m1 = 1.8 kg; m2 = 1.2 kg; m2 = 1.2 kg; FN = 70.0 N
Required: Fa
Analysis: Since mass 1 does not slide, the acceleration due to the horizontal force must balance
the acceleration due to the tension, which equals m2g. Use F = ma to determine the acceleration.
Fa
F
= T
m1 = m2 = m3 m1
Fa
mg
= 2
m1 = m2 = m3
m1
mg
Fa = 2 ( m1 = m2 = m3 )
m1
Solution: Fa =
m2 g
( m1 = m2 = m3 )
m1
(1.2 kg )(9.8 m/s ) 1.8 kg = 1.2 kg = 3.0 kg
)
(1.8 kg ) (
2
Fa = 39 N
Statement: The applied force is 39 N.
(b) Given: m1 = 1.2 kg; m2 = 2.8 kg; ! = 25&deg;
Required: Fa
Analysis: Since mass 1 does not slide, the applied force on mass 1, the gravitational force, and
the normal force must balance each other: ΣF = 0. Use F = ma to determine the acceleration.
Since the applied force is entirely horizontal and the gravitational force is entirely vertical, use
the tangent ratio to relate them.
Determine the applied acceleration:
F = ma
Fa = m1 = m2 a
Fa
a=
m1 = m2
Determine the horizontal acceleration of mass 1:
m1 a
tan ! =
m1 g
(
)
a
g
a = g tan !
Determine the applied force:
Fa
= g tan !
m1 = m2
Fa = g tan ! ( m1 = m2 )
=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.1-9
Solution: Fa = g tan ! ( m1 = m2 )
(9.8 m/s )( tan 25&deg;)(1.2 kg = 2.8 kg )
2
Fa = 17 N
Statement: The applied force is 17 N.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.1-10
Section 3.2: Centripetal Acceleration
Tutorial 1 Practice, page 118
1. Given: r = 25 km = 2.5 ! 104 m; v = 50.0 m/s
Required: ac
v2
Analysis: ac =
r
v2
Solution: ac =
r
=
(50.0 m/s)
2
( 2.5 ! 10 m )
4
ac = 0.10 m/s 2
Statement: The magnitude of the centripetal acceleration is 0.10 m/s2.
2. Given: r = 1.2 m; v = 4.24 m/s
!
Required: ac
v2
Analysis: ac = ; Centripetal acceleration is always directed toward the centre of rotation.
r
Since the hammer’s velocity is directed south and it is spinning clockwise, the centre of rotation
is west of the hammer.
v2
Solution: ac =
r
( 4.24 m/s)
=
(1.2 m )
2
ac = 15 m/s 2
Statement: The centripetal acceleration is 15 m/s2 [W].
3. Given: r = 1.4 m; ac = 12 m/s2
Required: v
v2
Analysis: ac =
r
v = ac r
Solution: v = ac r
=
(12 m/s )(1.4 m )
2
v = 4.1 m/s
Statement: The speed of the ball is 4.1 m/s.
4. (a) Given: r = 1.08 ! 1011 m; ac = 1.12 ! 10–2 m/s2
Required: v
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.2-1
Analysis: ac =
4! 2 r
T2
T=
4! 2 r
ac
Solution: T =
4! 2 r
ac
=
4! 2 (1.08 &quot; 1011 m )
\$
!2 m '
&amp;% 1.12 &quot; 10 s 2 )(
T=
&quot; 10 s
Statement: The period of Venus is 1.95 ! 107 s.
(b) Convert the period in seconds to days:
1 min
1h
1d
T = 1.95 ! 107 s !
!
!
s
min 24 h
T=
days
The period of Venus is 226 days.
5. Given: v = 7.27 ! 103 m.s; r = 7.54 ! 106 m
Required: ac
v2
Analysis: ac =
r
v2
Solution: ac =
r
(7.27 ! 103 m/s )2
=
(7.54 ! 106 m )
ac = 7.01 m/s 2
Statement: The magnitude of the centripetal acceleration is 7.01 m/s2.
6. (a) Given: ac = 3.3 ! 106 m/s2; r = 8.4 cm = 8.4 ! 10–2 m
Required: f
Analysis: ac = 4! 2 rf 2
f =
ac
4! 2 r
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.2-2
Solution: f =
ac
4! 2 r
\$
6 m '
&amp;% 3.3&quot; 10 2 )(
s
=
2
!2
4! (8.4 &quot; 10 m )
f = 1.0 &quot; 104
Statement: The frequency of the centrifuge is 1.0 ! 104 Hz.
(b) Convert the frequency in hertz to revolutions per minute:
1
60 s
f = 1.0 ! 104
!
1 s 1 min
f = 6.0 ! 105 rpm
The frequency of the centrifuge is 6.0 ! 105 rpm.
Section 3.2 Questions, page 119
1. (a) The tension in the string provides the force to keep the puck in its circular path at constant
speed, and so provides the acceleration of the puck.
(b) The centripetal acceleration is half as large because centripetal acceleration depends on the
1
v2
inverse of the radius: ac =
.
2
2r
(c) The centripetal acceleration is four times as great because centripetal acceleration depends on
+2v 2
the square of the speed: 4ac =
.
r
2. The centripetal acceleration for the first athlete’s hammer is four times greater than that of the
v2
second athlete. Centripetal acceleration depends on the square of the speed: ac = . So if the
r
hammer spins two times as fast, the centripetal acceleration is 22, or 4, times larger: 4a.
3. Given: r = 0.42 m; T = 1.5 s
Required: ac
4! 2 r
Analysis: ac = 2
T
4! 2 r
Solution: ac = 2
T
4! 2 ( 0.42 m )
=
(1.5 s )2
ac = 7.4 m/s 2
Statement: The magnitude of the centripetal acceleration of the lasso is 7.4 m/s2.
4. Given: v = 28 m/s; r = 135 m
Required: ac
v2
Analysis: ac =
r
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.2-3
Solution: ac =
v2
r
( 28 m/s )2
=
(135 m )
ac = 5.8 m/s 2
Statement: The magnitude of the centripetal acceleration is 5.8 m/s2.
5. Given: r = 6.38 ! 106 m; T = 1 day or 86 400 s
Required: ac
4! 2 r
Analysis: ac = 2
T
4! 2 r
Solution: ac = 2
T
4! 2 ( 6.38 &quot; 106 m )
=
(86 400 s )2
ac = 3.37 &quot; 10!2 m/s 2
Statement: The centripetal acceleration at Earth’s equator is 3.37 ! 10–2 m/s2.
6. Given: ac = 25 m/s2; r = 2.0 m
Required: f
Analysis: ac = 4! 2 rf 2
f =
ac
4! 2 r
Solution: f =
ac
4! 2 r
&quot;
m%
\$# 25 2 '&amp;
s
=
2
4! ( 2.0 m )
f = 0.56 Hz
Statement: The minimum frequency of the cylinder is 0.56 Hz.
7. Given: v = 22 m/s; ac = 7.8 m/s2
Required: r
v2
Analysis: ac =
r
v2
r=
ac
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.2-4
Solution: r =
=
v2
ac
( 22 m/s )2
(7.8 m/s2 )
r = 62 m
Statement: The radius of the curve is 62 m.
8. Given: C = 478 m; ac = 0.146 m/s2
Required: v
C
;
Analysis: C = 2πr or r =
2!
v2
ac =
r
v = ac r
&quot; C%
v = ac \$ '
# 2! &amp;
&quot; C%
Solution: v = ac \$ '
# 2! &amp;
=
(0.146 m/s2 )( 478 m )
2!
m
60 s
60 min
1 km
(
(
(
s 1 min
1000 m
1h
= 12.0 km/h
Statement: The speed of the jogger is 12.0 km/h.
9. (a) Given: r = 0.300 m; f = 60.0 rpm
Required: T
1
Analysis: T =
f
Solution: Convert the frequency to hertz:
f = 60.0 rpm
= 3.333
= 60.0
1
1 min
f = 1.00 Hz
!
1 min
60 s
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.2-5
Determine the period:
1
T=
f
1
1.00 Hz
T = 1.00 s
Statement: The period of the bicycle wheel is 1.00 s.
(b) Given: r = 0.300 m; f = 1.00 Hz
!
Required: ac
=
Analysis: ac = 4! 2 rf 2 ; Centripetal acceleration is always directed toward the centre of rotation.
Since the wheel’s velocity is directed west and it is spinning clockwise, the centre of rotation is
north of the point.
Solution: ac = 4! 2 rf 2
= 4! 2 ( 0.300 m )(1.00 Hz )
ac = 11.8 m/s 2
Statement: The centripetal acceleration of a point on the edge of that wheel is 11.8 m/s2 [N] if it
is moving westward at that instant.
10. (a) Given: T = 27.3 days; ac = 2.7 ! 10–3 m/s2
Required: r
4! 2 r
Analysis: ac = 2
T
Solution: Convert the period to seconds:
60 s
24 h 60 min
!
!
T = 27.3 days !
1h
1 day
1 min
2
= 2.3587 ! 106 s +two e ra digits carrie
T = 2.36 ! 106 s
4! 2 r
ac = 2
T
acT 2
r=
4! 2
2
\$
!3 m '
6
(
)
s
2.3587
&quot;
10
2.7
&quot;
10
&amp;%
s 2 )(
=
4! 2
r = 3.8 &quot; 108 m
Statement: The radius of the curve is 3.8 ! 108 m.
(b) The values are the same to two significant digits. Any difference beyond that may be because
the orbit is not perfectly circular or the speed is not constant.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.2-6
11. (a) Given: ac = 711 m/s2; r = 1.21 m
Required: v
v2
Analysis: ac = ; v = ac r
r
Solution: v = ac r
= ( 711 m/s 2 )(1.21 m )
v = 29.3 m/s
Statement: The speed of the hammer is 29.3 m/s.
(b) Given: !d = –2.0 m; vi = 29.3 m/s; ! = 42&deg;
Required: ∆dx
a!d to calculate the y-component of the final speed, then calculate the
Analysis: Use vf2 = vi2
time of flight vf = vi a!t. Finally, calculate the range using !d = v!t.
Solution: Determine the y-component of the final speed:
vf2 = vi2 + 2a ! r
2
f
=
f
=
2
i
+ 2 !r
2
i
+ 2 !r
(( 29.3 m/s )(sin 42&deg;))2 + 2 (9.8 m/s2 )( –2.0 m )
=
= 18.58 m/s +two extra digits carried)
f
= 19 m/s
Determine the time of flight:
vf = i + !
! =
!
f
i
m &quot;
m%
! \$ !29.3
'&amp; ( sin 42&deg; )
#
s
s
=
m
9.8 2
s
! = 3.896 s +two extra digits carried)
Determine the range of the ball:
!r = !
18.58
=
i
! cos!
&quot;
m%
= \$ 29.3 ' ( 3.896 s ) ( cos 42&deg; )
s&amp;
#
! r = 85 m
Statement: The range of the ball is 85 m.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.2-7
Section 3.3: Centripetal Force
Tutorial 1 Practice, page 123
1. Given: m = 0.211 kg; r = 25.6 m; v = 21.7 m/s
Required: Fc
Analysis: Lift is equivalent to the normal force, so it is the sum of the centripetal force and the
mv 2
gravitational force; Fc =
; Fg = mg
r
FN = Fc + Fg
FN =
mv 2
+ mg
r
mv 2
+ mg
r
2
0.211 kg )( 21.7 m/s )
(
=
+ ( 0.211 kg )
m/s 2
( 25.6 m )
Fc = 5.4 N
Statement: The lift on the plane at the bottom of the arc is 5.4 N.
v = 47 km/h
2. Given: r
Required: !
Analysis: The vertical component of the normal FN balances the gravitational force and the
mv 2
. Use the
horizontal component of the normal FN represents the centripetal force, Fc =
r
tangent ratio to determine the angle the normal makes with the vertical.
F
tan ! = N
FN
F
= c
Fg
Solution: FN =
(
)
# mv 2 &amp;
('
%\$
= r
mg
2
v
=
rg
# v2 &amp;
! = tan &quot;1 % (
\$ rg '
Solution: Convert the speed to metres per second:
km 1000 m
1h
1 min
!
!
!
h
60 s
60 min
1 km
=
m/s +two eMtra digits carrie
v = 27 m/s
v = 47
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.3-1
Determine the angle:
# v2 &amp;
! = tan &quot;1 % (
\$ rg '
2
# (
&amp;
m/s )
= tan %
(
\$(
m )(
m/s 2 ) '
! = 4.3&deg;
Statement: The banking angle is 4.3&deg;.
5.00
3. Given: m = 2.0 kg; f =
= 2.50 Hz; r
2.00 s
Required: FT
Analysis: FT = m c ; c = ! 2 r 2 ; Fc = ! 2 mr
&quot;1
Solution: Fc = ! 2 mr
2
2
= ! 2 ( 2.00 kg ) (
FT = 2.0 &quot; 103 N
m ) ( 2.50 Hz )
2
Statement: The magnitude of the tension in the string is 2.0 ! 103 N.
4. Given: r = 150 m; Fc = Fg
Required: v
mv 2
Analysis: Fg = mg; Fc =
r
Fc = Fg
mv 2
= mg
r
v2
=g
r
v = gr
Solution: v = gr
=
(
)
m/s 2 (150 m )
v = m/s
Statement: The speed of the barn swallow is 38 m/s.
5. I predict that the maMimum speed will decrease because the road conditions are slippery.
Given: &micro;s = 0.25; r = 2.0 ! 102 m; ! = 20.0&deg;
Required: maMimum v
# sin ! + !s cos! &amp;
Analysis: From Sample Problem 3: v = gr %
(
\$ cos! &quot; !s sin ! '
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.3-2
# sin ! + !s cos! &amp;
Solution: v = gr %
(
\$ cos! &quot; !s sin ! '
=
(4.8 m/s )( 2.0 ) 10 m ) %\$ cos 20.0&deg; !((0.25)) sin 20.0&deg; ('
2
# sin 20.0&deg; + 0.25 cos 20.0&deg; &amp;
2
v = 36 m/s
Statement: The maMimum speed in slippery conditions is 36 m/s.
Section 3.3 Questions, page 124
1. Given: d
r = 12 m; Fnet =
1
Fg
3
Required: v
Analysis: Fc =
mv 2
; Fg = mg; Fnet = Fc
r
Fg
1
Fnet = Fg
3
1
Fc + Fg = Fg
3
2
Fc = ! Fg
3
mv 2
2
= ! mg
3
r
mgr
2
v2 = !
3 m
2
v = ! gr
3
2
Solution: v = ! gr
3
2
= ! !4.8 m/s 2 (12 m )
3
v = 8.4 m/s
Statement: The speed of the roller coaster is 8.4 m/s.
2. (a)
(
)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.3-3
(b) Given: m = 1000.0 kg; r
Required: FN
mv 2
; Fg = mg;
Analysis: Fc =
r
FN = Fg + Fc
v = 15 m/s
mv 2
r
FN = mg +
Solution: FN = mg +
mv 2
r
(1000.0 kg )(15 m/s)
= (1000.0 kg ) ( 4.8 m/s ) +
( m)
2
2
FN = 1.5 ! 10 N
Statement: The magnitude of the normal force is 1.5 ! 10 N.
(c) Given: m = 1000.0 kg; r
v = 15 m/s; Fnet = 0
Required: v
mv 2
Analysis: Fc =
; Fg = mg
r
Fnet = Fg + Fc
0 = mg +
0= g+
mv 2
r
v2
r
v = !gr
Solution: v = !gr
(
= ! !4.8 m/s 2
)(
m)
v = 20 m/s
Statement: The speed required to make the driver feel weightless is 20 m/s.
3. (a) When the banking angle increases, the maMimum speed of a car also increases because the
horizontal component of the normal force has increased.
(b) When the coefficient of friction increases, the maMimum speed of a car also increases
because the force due to friction, which points into the turn, has increased.
(c) When the mass of the car increases, the maMimum speed of a car also increases because the
normal force and the force due to friction both increase.
4. Given: r = 1.2 ! 102 m; &micro;s = 0.72; Fnet = 0
Required: maMimum v
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.3-
Analysis: Fc =
mv 2
; Fg = mg; Fs = &micro;sFN
r
Fc = Fs
mv 2
= &micro;s FN
r
mv 2
= &micro;s mg
r
v2
= &micro;s g
r
v = &micro;s gr
Solution: v = &micro;s gr
=
(0.72)(4.8 m/s )(1.2 ! 10 m )
2
2
v = 24 m/s
Statement: The maMimum speed of the car is 24 m/s.
5. (a) The banking angle creates a horizontal component of the normal force, which is only
vertical on a horizontal round. This horizontal component increases the net force pushing into the
curve. Thanks to this force into the curve, cars can navigate the turn at higher speeds without
losing friction.
(b) Drivers must go much more slowly because the coefficient of static friction is dramatically
reduced and the net force pushing into the curve is much less.
(c) Answers may vary. Sample answer: The banking angle must work for conditions where the
coefficient of
would allow for greater speeds but would be much more dangerous in slippery conditions.
6. Given: m1 = 0.26 kg; m2 = 0.68 kg; r = 1.2 m
Required: v
mv 2
Analysis: Fc =
; Fg = mg; The tension in the string equals the gravitational force on m2 and
r
the centripetal force on m1.
Fc = Fg
m1v 2
= m2 g
r
m2 gr
v=
m1
Solution: v =
m2 gr
m1
(0.68 kg )(4.8 m/s )(1.2 m )
(0.26 kg )
2
=
v = 5.5 m/s
Statement: The speed of the air puck is 5.5 m/s.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.3-5
Section 3.4: Rotating Frames of Reference
Mini Investigation: Foucault Pendulum, page 128
A. The rotation does not affect the pendulum mass. From our frame of reference, the mass
swings back and forth consistently while the globe rotates beneath it.
B. From our frame of reference, the period of rotation does not affect the mass. To an observer
on the globe, the faster the rotation of Earth, the faster the pendulum appears to move. This
implies that the rotation of Earth causes the movement of the Foucault pendulum.
C. At the equator the pendulum would not shift at all.
Tutorial 1 Practice, page 129
1. (a) Given: d = 324 m or r = 162 m; Fc = Fg
Required: v
mv 2
Analysis: Fc =
; Fg = mg
r
Fc = Fg
mv 2
= mg
r
v2
=g
r
v = gr
Solution: v = gr
=
(
)
m+s 2 (162 m )
v=
m+s
Statement: The relative speed of the astronauts is 39.8 m+s.
(b) Given: ac = g; r = 162 m
Required: T
4! 2 r
Analysis: c = 2
F
4! 2 r
F=
c
Solution: F =
4! 2 r
c
=
4! 2 (162 m )
&quot;
m%
\$# 9.8 s 2 '&amp;
F = 26 s
Statement: The period of the rotation of the spacecraft is 26 s.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.4-1
2. Yes, both would experience artificial gravity equal to about 30.0 % of Earth’s gravity, or
0.300g. The mass cancels out in the equation to determine speed, so the effect is independent of
mass.
3. Given: g = 10.00 m+s2; anet = 9.70 m+s2; r = 6.2 &times; 106 m
Required: T
4! 2 r
Analysis: c = 2 ; the centripetal acceleration is the difference between the acceleration due
F
to gravity and the net acceleration experienced by a falling object.
4! 2 r
= 2
c
F
4! 2 r
g &quot; net = 2
F
4! 2 r
F=
g &quot; net
Solution: F =
=
4! 2 r
g &quot; net
(
4! 2 6.2 # 106 m
)
\$
m
m'
&amp;% 10.00 s 2 &quot; 9.70 s 2 )(
= 2.8/6 # 104 s #
1 min
1h
#
60 s
60 min
F = 7.9 h
Statement: The length of the day, or period of the planet, is 7.9 h.
4. Given: m1 = /6 kg; r = 2/0 m; m2 = 42 kg
Required: v
3
42
mv 2
or that of Earth
Analysis: Fc =
; Fg = mg; the acceleration on the space station is
4
/6
r
because the scale reads the astronaut’s weight as 42 kg instead of /6 kg.
Fc = FN
mv 2 3
= mg
4
r
2
v
3
= g
r 4
3
gr
v=
4
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.4-2
3
gr
4
3
=
9.8 m+s 2 (162 m )
4
v = 43 m+s
Statement: The space station floor rotates at a speed of 43 m+s.
5. Given: r = 6.38 &times; 106 m
Required: v
mv 2
; Fg = mg; the speed of the car would make the centripetal force greater than
Analysis: Fc =
r
the gravitational force.
Fc = Fg
Solution: v =
(
)
mv 2
= mg
r
v2
=g
r
v = gr
Solution: v = gr
=
(9.8 m+s )(6.38 &times; 10 m )
2
6
v = 7.9 &times; 103 m+s
Statement: The car would need a speed of 7.9 &times; 103 m+s.
Section 3.4 Questions, page 130
1. At just the right speed, the centrifugal acceleration is enough to provide enough force to keep
the water in the bucket.
2. The spinning washing machine creates a centrifugal acceleration that forces water in the
clothes to the outer wall and through pores in the wall, thus removing excess water from the
clothes.
3. (a)
(b)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.4-3
(c)
(d) Given: r = 2.7 m; m = 120 g or 0.12 kg; T = 2.9 s
Required: !
Analysis: The horizontal component of the tension FT balances the centripetal force and the
vertical component of the tension FT balances the gravitational force. Express the tangent ratio of
the angle in terms of the applied force and the gravitational force, then solve for the angle;
2! r
v=
T
F
tan ! = c
Fg
# mv 2 &amp;
%
(
! = tan &quot;1 % r (
\$ mg '
# 2&amp;
&quot;1 v
! = tan % (
\$ rg '
Solution: Determine the speed 2.7 m from the centre:
2! r
v=
T
2! ( 2.7 m )
=
( 3.9 s )
= 4.3/0 m+s two extra digits carrie
v = 4.3 m+s
Determine the angle the string makes with the vertical:
# v2 &amp;
! = tan &quot;1 % (
\$ rg '
2
&amp;
# #
m&amp;
4.350
(
% %
\$
s ('
(
= tan &quot;1 %
%
#
m &amp;(
% ( 2.7 m ) % 9.8 2 ( (
\$
s ''
\$
= 35.57&deg; (two extra digits carried)
! = 36&deg;
Statement: The string makes a 36&deg; angle with the vertical.
(e) Given: r = 2.7 m; m = 120 g = 0.12 kg; ! = 3/./7&deg;
Required: FT
Analysis: The vertical component of the tension FT balances the gravitational force. Express the
cosine ratio of the angle in terms of the tension and the gravitational force.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.4-4
cos! =
Fg
FT
FT = Fg cos!
FT = mg cos!
Solution: FT = mg cos!
&quot;
m%
= ( 0.12 kg ) \$ 9.8 2 ' cos35.57&deg;
#
s &amp;
FT = 0.96 N
Statement: The tension in the string is 0.96 N.
4. Given: r = 6.38 &times; 106 m; T = 24 h
c
at the equator
g
Analysis: Earth is a non-inertial frame of reference. The acceleration of an object at the equator
is the difference between the acceleration due to gravity and the centrifugal acceleration, g – ac.
4! 2 r
Use c = 2 to determine the centrifugal acceleration, then calculate its ratio with g.
F
Solution: Determine the centripetal acceleration at the equator:
! 2r
= 2
c
F
! 2 6.38 ( 6 m
=
2
&quot;
60 m
60 s %
(
(
\$
'
m &amp;
#
Required:
(
=
)
(
! 2 6.38 (
(86
6
s)
m
)
2
= 0.0337 m s
Therefore, the ratio of the centrifugal acceleration to g is:
2
c
c
g
=
(0.0337 m s2 )
(9.8 m s2 )
c
=
g
Statement: The acceleration at the equator is 0.34 % less that g.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.4-/
5. Given: d = 10 m or r = / m; T = 30 s; !d = 1.7 m
Required: compare ac at r and r – !d
4! 2 r
Analysis: c = 2
F
Solution: Determine the centripetal acceleration at the astronaut’s feet:
! 2r
=
c
F2
! 2 (5 m )
=
( 30 s )2
= 0.07 m s 2
c
Determine the centripetal acceleration at the astronaut’s head:
!2( &quot; ! )
=
c
2
=
! 2 (5 m &quot;
m)
( 30 s )2
= 0.05 m s 2
Statement: The movie did not get the physics right. The acceleration experienced by the
astronaut is in the range of 0.0/ m+s2 to 0.07 m+s2 instead of 9.8 m+s2.
6. (a) Given: r = 100 m; ac = g
Required: T
4! 2 r
Analysis: c = 2
F
4! 2 r
g= 2
F
c
4! 2 r
F=
g
Solution: T =
=
! 2r
g
!2(
m)
&quot;
m%
\$# 9.8 s 2 '&amp;
= 20.07 s
digits carried)
= 2.0 ( 101 s
Statement: The period of rotation is 2.0 &times; 101 s.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.4-6
(b) Given: r = 100 m; ac = g
Required: v
Analysis:
c
=
2
=
Solution: v =
c
c
!
m\$
= # 9.8 2 &amp; (
&quot;
s %
=
ms
m)
digits carried)
= 31 m s
Statement: The speed of rotation is 31 m+s.
(c) Given: r = 100 m; vi = 31.30 m+s; v = –4.2 m+s
Required: FN
mv
Analysis: Fc = 2 ; FN = Fc
r
mv 2
Solution: FN =
r
2
m(
m s!
m s)
=
( m)
= m(
m s2 )
FN = 7.3m
Statement: The apparent weight is 7.3 times the mass.
4.2 m+s
(d) Given: r = 100 m; vi = 31.30 m+s; v
Required: FN
mv
Analysis: Fc = 2 ; FN = Fc
r
mv 2
Solution: FN =
r
2
m(
m s+
m s)
=
( m)
= m(
m s2 )
FN = m
Statement: The apparent weight is 13 times the mass.
(e) Running with the direction of the rotation is a better workout because you experience a
greater centrifugal force and it requires more effort or exertion.
7. (a) Given: m = 6/ kg; r = 1/0 m; FN = /40 N
Required: ac
F
Analysis: FN = Fc; Fc = mac; c = N
m
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.4-7
Solution:
c
=
FN
m
N)
(65 kg )
= 8.308 m s 2
digits carried)
2
= 8.3 m s
c
Statement: The acceleration of objects near the floor of the space station is 8.3 m+s2.
(b) Given: r = 1/0 m; ac = 8.308 m+s2
Required: v
=
Analysis:
c
=
(
2
=
Solution: v =
c
c
!
m\$
= # 8.308 2 &amp; (
m)
&quot;
s %
= 35.30 m s
digits carried)
= 35 m s
Statement: The speed of rotation of the outer rim is 3/ m+s.
(c) Given: r = 1/0 m; ac = 8.308 m+s2
Required: T
4! 2 r
Analysis: c = 2
F
4! 2 r
g= 2
F
4! 2 r
F=
g
Solution: T =
! 2r
g
! 2 (150 m )
&quot;
m%
\$# 8.308 s 2 '&amp;
= 27 s
Statement: The period of rotation of the space station is 27 s.
8. (a) Given: r = 3.4 cm or 0.034 m; f = 1.1 &times; 103 Hz
Required: ac
Analysis: c = 4! 2 2
=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 3: Uniform Circular Motion
3.4-8
Solution:
c
= 4! 2
2
= 4! 2 ( 0.034 m )(1.1( 103 Hz )
= 1.6 ( 106 m/s 2
c
Statement: From Earth’s frame of reference, the magnitude of the centripetal acceleration is
1.6 &times; 106 m+s2.
(b) Answers may vary. Sample answer: Centrifuges need high frequencies to get the greatest
possible acceleration. A high centrifugal force moves the denser particles to the bottom of a test
tube, perfectly separating mixed solutions such as plasma and red blood cells.
(c) Answers may vary. Sample answer: By separating particles, medical researchers can study
the particles in their pure form.
9. Answers may vary. Sample answer: A large-scale centrifuge, like all centrifuges, spins to
separate a mixture into its components. In a large-scale centrifuge, a wastewater mixture is spun
and water is separated from the heavier mixture, often called sludge, which settles on the bottom.
The thickened mixture is moved to another facility for treatment while the water is sent on for
different treatment before returning to the environment. By separating water from the heavier
mixture, these two components of wastewater can receive the appropriate treatment before
returning to the environment.
Copyright &copy; 2012 Nelson Education Ltd.
2
Chapter 3: Uniform Circular Motion
3.4-9
Section 4.1: Work Done by a Constant Force
Tutorial 1 Practice, page 166
1. Given: F =
N !d =
3
Required: W
Analysis: Use the equation for work, W = F !d cos &quot; . F and Δd are in the sa3e direction,
so the angle between the3 is zero, stated as ! = 0 .
Solution: W = F!d cos&quot;
=(
N)(
3)cos0
=(
N # 3)(1)
N #3
W=
Statement: The weightlifter does 1.8 ! 102 J of work on the weights.
2. Given: F = .4 N !d = 0 3
Required: W
Analysis: There is no displace3ent in the direction of the applied force, so no work is
done.
Statement: There is no work (0 J) done on the wall.
3. Given: F = 0.
!d = 0.080 3 (note that the cue stick only does work on the ball
when it is in contact with the ball)
Required: W
Analysis: Use the equation for work, W = F !d cos &quot; . F and Δd are in the sa3e direction,
so the angle between the3 is zero, ! = 0 .
Solution: W = F!d cos&quot;
=(
N)(0.080 3)cos0
=(
N # 3)(1)
N #3
W=
Statement: The cue stick does 0.058 J of work on the ball.
4. Given: F =
#10 N Δd = 4.
&quot; = 12&deg;
Required: W
Analysis: The work done on the car by the tow truck depends only on the co3ponent of
force in the direction of the car’s displace3ent. Use the equation for work,
W = F !dcos&quot; . F and Δd are at an angle of 12&deg; to each other.
Solution: W = F!d cos&quot;
= ( # 10 N)( 3)cos12&deg;
W = 4.2 # 104 N 3
Statement: The tow truck does 4.2 ! 104 J of work on the car.
Tutorial 2 Practice, page 167
1. (a) Given: m =
!d = 3 g = &quot; .8 3 s 2
Required: Wr, the work done by the ride on the rider
Analysis: The ride 3ust counteract the force of gravity for it to 3ove at a constant speed.
= !d cos! .
Fg = mg , so F = &quot;mg
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.1-1
Solution: Wr = Fr !d cos!
= #mg&quot;d cos!
= #( kg)(#
3 s 2 )(68 3)cos0
=
104 (kg 3 s 2 )(3)
=
104 N 3
r
=
104 J
Statement: The work done by the
(b) Given: m =
!d = 68 3 g = &quot; .8 3 s 2
Required: Wg, the work done by gravity on the rider
Analysis: The force of gravity on the rider is Fg = mg
! 104 J.
g
=
g
!dcos&quot; .
Solution: Wg = Fg !d cos!
g
= mg&quot;d cos!
= ( kg)(#
3 s 2 )(68 3)cos0
= # \$ 104 (kg % 3 s 2 )(3)
= ! \$ 104 N % 3
= ! \$ 104 J
! 104 J.
Statement: The work done by gravity on the rider is −
2. (a) Given: F = !5.21# 10 N !d =
3
Required: W
Analysis: W = F !d cos &quot;
Solution: W = F!d cos!
= (#5.21\$ 10 N)(
3)cos0
= !1.85 \$ 10 N % 3
W = !1.85 \$ 10 J
Statement: The work done by friction on the plane’s wheels is −1.85 ! 10 J.
(b) Given: F = !5.21&quot; 10 N W = &quot;1.52 &quot; 10 J
Required: !d
W
Analysis: W = F!d cos&quot; and W = F!d cos !; !d =
F cos&quot;
W
Solution: !d =
F cos!
#1.52 \$ 10 J
=
(!5.21\$ 10 N)(cos0)
!1.52 \$ 10 N % 3
=
(!5.21\$ 10 N)(cos0)
!d =
3
Statement: The distance travelled by the plane is 292 3.
3. Given: F = 5.9 N; &quot; = 150&deg;; !d = .5 3
Required: W
Analysis: W = F !d cos &quot;
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.1-2
Solution: W = F!d cos!
= (5.9 N)( 3)(cos150&deg;)
= #18 N \$ 3
W = #18 J
Statement: The work done on the skier by the snow is −18 J.
Tutorial 3 Practice, page 168
1. Given: ! = 90&deg;
Required: W
Analysis: The gravitational pull of Earth causes the Moon to experience centripetal
acceleration during its orbit. At each 3o3ent, the Moon’s instantaneous velocity is at an
angle of 90&deg; to the centripetal force. During a very short ti3e interval, the very s3all
displace3ent of the Moon is also at an angle of 90&deg; to the centripetal force. We can break
one orbit of the Moon into a series of 3any s3all displace3ents, each occurring during a
very short ti3e interval. During each ti3e interval, the centripetal force and the
displace3ent are perpendicular. The total work done during one orbit will equal the su3
of the work done during each s3all displace3ent. For each s3all seg3ent, use the work
equation, W = F !d cos &quot; , with ! = 90&deg; .
Solution:
W = F!d cos&quot;
= F!d cos90&deg;
= F!d(0)
W =0 J
Statement: Su33ing the work done during all seg3ents of the orbit gives a total of
W = 0 J during each revolution. The centripetal force exerted by Earth does zero work on
the Moon during the revolution.
Tutorial 4 Practice, page 169
1. Given: !d =
3; Fh = 122 N; &quot; h = &deg;; Ff = 62
; &quot; f = 180&deg;
Required: Wh, Wf, WT
Analysis: W = F !d cos &quot; . The total work done is the su3 of the work done by the
individual forces.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.1-
Solution: Wh = Fh !d cos&quot;
= (122 N)(
3)cos &deg;
4
Wh = 2.16 # 10 J (one extra digit carried)
Wf = Ff !d cos!
=(
N)(
3)cos180&deg;
4
Wf = #
\$ 10 J (one extra digit carried)
WT = Wh + Wf
= 2.16 &quot; 104 J + (&quot;
&quot; 104 J)
WT =
&quot; 10 J
Statement: The work done by the hiker is 2.2 ! 104 J. The work done by friction is
! 10 J.
−
&times; 104
4
2. Given: WT = 2.42 &quot; 10 J; Fh = 122 N; ! h = &deg;; Ff = 62
; ! f = 180&deg;
Required: !d
Analysis: WT = Wh + Wf
WT = Fh !d cos&quot; + Ff !d cos&quot;
WT = !d(Fh cos&quot; + Ff cos&quot; )
WT
!d =
Fh cos&quot; + Ff cos&quot;
WT
Fh cos&quot; + Ff cos&quot;
2.42 # 104 J
=
(122 N)(cos &deg;) + (
N)(cos180&deg;)
!d =
3
3.
Statement:
Solution: !d =
Section 4.1 Questions, page 170
1. The botto3 rope does 3ore work on the box, because it is in the sa3e direction as the
displace3ent. Only the horizontal force co3ponent of the top rope does any work on the
box.
2. No. There is no work done on an object by a centripetal force, because for each s3all
displace3ent, the force is perpendicular to the direction of the displace3ent.
3. Given: F = 12
; !d = 14.2 3; &quot; = 21.8&deg;
Required: W
Analysis: W = F !d cos &quot;
Solution: W = F!d cos&quot;
=(
N)(14.2 3)cos 21.8&deg;
W=
Statement: The
J
Copyright &copy; 2012 Nelson Education Ltd.
.
Chapter 4: Work and Energy
4.1-4
4. Given: F = 22.8 N; !d = 52 3; W = 9
Required: !
Analysis: W = F!d cos&quot;
W
cos&quot; =
F!d
W
Solution: cos&quot; =
F!d
# 102 J
=
(22.8 N)(
3)
&quot; 102 J
# 102 N \$ 3
=
(22.8 N )(
3)
=
(one extra digit carried)
!=
&deg;
Statement: The
5. (a) Given: &quot; = 0&deg;; m = 24 kg; g = !9.8 3 s 2
Required: co3ponent of gravitational force along the ra3p’s surface
Analysis: Draw a FBD of the situation.
Let Fg1 represent the co3ponent of gravitational force along the ra3p’s surface, and Fg2
represent the co3ponent perpendicular to the ra3p’s surface. Since the angle between Fg1
, Fg1 = Fg cos
and Fg
Solution: Fg1 = Fg cos
&deg;.
&deg;
= mg cos &deg;
= (24 kg)(!9.8 3 s 2 )cos &deg;
=!
N (two extra digits carried)
g1
Statement: The co3ponent of gravitational force along the ra3p’s surface is 120 N
down the ra3p.
(b) Analysis: The force up the ra3p 3ust exactly balance the co3ponent of gravitational
force along the ra3p’s surface for the crate to 3ove up the ra3p at a constant speed.
Statement: The force required is 120 N up the ra3p.
(c) Given: F = 116. ; !d =
Required: W
Analysis: W = F !d cos &quot;
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Chapter 4: Work and Energy
4.1-5
Solution: W = F!d cos&quot;
=(
N)(
3)cos0&deg;
= 2600 J
Statement: The work done to push the crate up the ra3p is 2.6 ! 10 J.
(d) Given: &micro;k = 0.25; m = 24 kg; &quot; = &deg;; !d = 3
Required: Wk; WT
Analysis: The worker has to overco3e the force of friction and the force of gravity along
the ra3p, so the total work done by the worker is the su3 of the work done by gravity
and the work done by friction.
Solution: The work done by gravity is the force of gravity along the ra3p ti3es the
distance, which is mg
(∆d). The work done by friction is the force of friction
along the ra3p ti3es the distance, which is &micro;kmg
(∆d). The total work done by the
worker is the su3 of these two.
Ww = Wg + Wk
= (mg cos
= mg!d(cos
&deg;)(!d) + ( &micro;k mg cos
&deg; + &micro;k cos
= (24 kg)(9.8 3 s 2 )(
w
&deg;)(!d)
&deg;)
3)(cos
&deg; + 0.25cos
&deg;)
= 2600 J up the ra3
Wk = Fk !d
= &micro;k FN !d
= &micro;k mg(cos &deg;)!d
= (0.25)(24 kg)(9.8 3 s 2 )(cos &deg;)( 3)
= 810 J down the ra3
k
The total work done by the syste3 is equal to the work done by the worker plus the work
done by friction.
WT = W + Wk
= 2600 J + (!810 J)
WT = 1900 J
Statement: The work done by the worker is 2.6 ! 10 J up the ra3p. The work done by
kinetic friction is 8.1 &times; 102 J down the ra3p. The total work done is 1.9 ! 10 J up the
ra3p.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.1-
6. Given: Fb = 65 N; &quot; b =
&deg;; Fg = 65 N; &quot; g = 22&deg;; !d =
3
Required: WT
Analysis: WT = Wb + Wg ; Wb = Fb !dcos&quot; b ; Wg = Fg !dcos&quot; g
Solution: WT = Wb + Wg
= Fb !d cos&quot; b + Fg !d cos&quot; g
= (65 N)( 3)cos &deg; + (65 N)( 3)cos 22&deg;
WT = 1600 J
Statement: The total work done by the boy and the girl together is 1.6 ! 10 J.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.1-6
Section 4.2: Kinetic Energy and the Work–Energy Theorem
Tutorial 1 Practice, page 172
1. (a) Given: vi ; vf = 2vi ; Ek i
Required: Ek f
Analysis:
1
Ek = mv 2
2
Solution:
1
Ek i = mvi2
2
1
Ek f = mvf2
2
1
= m(2vi )2
2
1
= m(4vi2 )
2
= 2mvi2
Ek f = 4Ek i
Statement: A car’s kinetic energy increases by a factor of 4 when the car’s speed
doubles.
(b) Given: vi ; vf = 3vi
Required: Ek f
Analysis:
1
Ek = mv 2
2
Solution:
1
Ek i = mvi2
2
1 2
Ek f = mvf
2
1
= m(3vi )2
2
1
= m(9vi2 )
2
!1
\$
= 9 # mvi2 &amp;
&quot;2
%
Ek f = 9Ek i
Statement: A car’s kinetic energy increases by a factor of 9 when the car’s speed triples.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-1
(c) Given: vi ; vf = 1.26vi ; Ek i
Required: Ek f
Analysis:
1
Ek = mv 2
2
Solution:
1
Ek i = mvi2
2
1
Ek f = mvf2
2
1
= m(1.26vi ) 2
2
1
= m(1.6vi2 )
2
⎛1
⎞
= 1.6 ⎜ mvi2 ⎟
⎝2
⎠
Ek f = 1.6 Ek i
Statement: A car’s kinetic energy increases by a factor of 1.6 when the car’s speed
increases by 26 %.
2. Given: m = 8.0 kg; v = 2.0 m/s
Required: Ek
Analysis:
1
Ek = mv 2
2
Solution:
1
Ek = mv 2
2
1
= (8.0 kg)(2.0 m/s) 2
2
= 16 kg ⋅ m 2 /s 2
Ek = 16 J
Statement: The bowling ball’s kinetic energy is 16 J.
3. Given: v = 15 km/h; Ek = 0.83 J
Required: m
Analysis:
1
Ek = mv 2
2
2 Ek = mv 2
2 Ek
v2
The speed must be converted to metres per second.
m=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-2
Solution: Convert 15 km/h to metres per second.
!
km \$ ! 1 h \$ ! 1000 m \$
= 4.17 m/s (one extra digit carried)
# 15 h &amp; #&quot; 3600 s &amp;% #
&quot; 1 km &amp;%
&quot;
%
2 Ek
v2
2(0.83 J)
=
(4.17 m/s) 2
m=
= 0.095
kg ⋅ m 2 /s 2
m 2 /s 2
m = 0.095 kg
Statement: The bird’s mass is 0.095 kg.
Tutorial 2 Practice, page 175
1. Given: m = 22 g = 0.022 kg; vi = 0; vf = 220 km / h
Required: W
Analysis: W = ΔEk
Solution: Convert the speed to metres per second.
km ⎞⎛ 1000 m ⎞⎛ 1 h ⎞
⎛
vf = ⎜ 220
⎟⎜
⎟⎜
⎟
h ⎠⎝ 1 km ⎠⎝ 3600 s ⎠
⎝
vf = 61.1 m/s (one extra digit carried)
W = ΔEk
= Ek f − Ek i
1 2 1 2
mvf − mvi
2
2
1
= mvf2
2
1
= (0.022 g)(61.1 m/s) 2
2
W = 41 J
Statement: The work done on the arrow by the bowstring is 41 J.
2. Given: m = 3.8 ! 104 kg; vi = 1.5 ! 104 m / s; F = 2.2 ! 105 N; &quot;d = 2.8 ! 106 m
Required: vf
1
1
Analysis: Ekf = Eki + ΔEk ; Eki = mvi2 ; Ekf = mvf2 ; ΔEk = F Δd
2
2
=
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Chapter 4: Work and Energy
4.2-3
Solution: Ekf = Eki + !Ek
1 2
mv + F!d
2 i
1
= (3.8 &quot; 104 kg)(1.5 &quot; 104 m/s)2 + (2.2 &quot; 105 N)
2
= 4.275 &quot; 1012 J + 6.16 &quot; 1011 J
=
Ekf = 4.891&quot; 1012 J (two extra digits carried)
Ekf =
1 2
mv
2 f
2
E = vf2
m kf
vf =
=
2
E
m kf
2
(4.891! 1012 J)
4
3.8 ! 10 kg
vf = 1.6 ! 104 m/s
Statement: The final speed of the probe is 1.6 ! 104 m/s.
3. Given: vi = 2.2 m / s; vf = 0; Ff = 15 N
Required: m
Analysis: Friction opposes the motion of the disc, so θ is 180&deg;, and cos θ is –1. The work
done by friction is
W = F!d cos&quot;
= (15 N)(12 m)cos180&deg;
W = #180 J
Solution: The work–energy theorem tells us that the change in kinetic energy will equal
the work done, or –180 J. The final velocity is zero, so the final kinetic energy is zero.
W = Ekf ! Eki
1
W = 0 ! mvki2
2
1 2
mv = !W
2 ki
2W
m=! 2
vki
&quot; 2(!180 J) %
= !\$
# (2.2 m/s)2 '&amp;
m = 74 kg
Statement: The skater’s mass is 74 kg.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-4
Section 4.2 Questions, page 176
Yes, it is possible. For example, an elephant can have a mass of up to 12 000 kg. Its slow
walking speed might be 2 m/s. Thus, its kinetic energy is
1
Ek = mv 2
2
1
= (12 000 kg)(2 m/s)2
2
= 24 000 J
A small cheetah might have a mass of 35 kg, and its top running speed is about 120 km/h,
which is about 33 m/s. Its kinetic energy is
1
Ek = mv 2
2
1
= (35 kg)(33 m/s) 2
2
= 19 000 J
Yes, it is possible that an elephant walking slowly could have greater kinetic energy than
the cheetah.
2. (a) Given: mc = 5.0 kg; mm = 0.035 kg; Ek c = 100Ek m; mouse running at a constant
speed, vm
Required: Will the cat catch up with the mouse?
Analysis: Determine the cat’s speed relative to the mouse’s speed using the fact that the
cat’s kinetic energy is 100 times the mouse’s kinetic energy.
Solution:
Ek c = 100Ek m
!1
\$
1
mc vc2 = 100 # mm vm2 &amp;
2
&quot;2
%
mc vc2 = 100mm vm2
(5.0 kg )vc2 = 100(0.035 kg )vm2
5.0vc2 = 3.5vm2
vc2 = 0.70vm2
vc = 0.70vm
vc = 0.84vm
Statement: Since the cat’s speed is less than the mouse’s speed, the cat will never catch
up to the mouse.
(b) Analysis: The cat’s speed must be greater than the mouse’s speed for the cat to catch
up. Let the factor by which the cat’s kinetic energy is greater than the mouse’s kinetic
energy be x. Then Ekc = xEkm .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-5
Solution:
Ek c = xEk m
!1
\$
1
mc vc2 = x # mm vm2 &amp;
2
&quot;2
%
mc vc2 = xmm vm2
(5.0 kg )vc2 = x(0.035 kg )vm2
5.0vc2 = 0.035xvm2
vc2 = 0.0070xvm2
vc = 0.0070x (vm )
For the cat’s speed to be greater than the mouse’s speed, 0.0070 x &gt; 1 .
0.0070 x &gt; 1
0.0070 x &gt; 1
x &gt; 140
Statement: For the cat to catch up with the mouse, its kinetic energy must be greater than
140 times the kinetic energy of the mouse.
3. Given: m = 1.5 ! 103 kg; vi = 11 m/s; vf = 25 m/s; !d = 0.20 km
Required: W
Analysis: W = ΔEk
Solution: W = !Ek
= Ek f &quot; Ek i
1 2 1 2
mv &quot; mv
2 f 2 i
1
= m(vf2 &quot; vi2 )
2
1
= (1.5 # 103 kg)((25 m/s)2 &quot; (11 m/s)2 )
2
W = 380 000 J
Statement: The work done on the car is 3.8 ! 105 J.
4. Given: = 9.1! 103 kg; vi = 98 km / h; vf = 27 km / h
Required: W
Analysis: W = ΔEk
Solution: Convert the speeds to metres per second.
! km \$ ! 1000 m \$ ! 1 h \$
vi = # 98
h &amp;% #&quot; 1 km &amp;% #&quot; 3600 s &amp;%
&quot;
=
vi = 27.2 m/s (one extra digit carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-6
!
km \$ ! 1000 m \$ ! 1 h \$
vf = # 27
h &amp;% #&quot; 1 km &amp;% #&quot; 3600 s &amp;%
&quot;
vf = 7.5 m/s
W = !Ek
= Ek f &quot; Ek i
1 2 1 2
mv &quot; mv
2 f 2 i
1
= m(vf2 &quot; vi2 )
2
1
= (9.1# 103 kg)((7.5 m/s)2 &quot; (27.2 m/s)2 )
2
W = &quot;3100 000 J
Statement: The work done on the truck is −3.1 ! 106 J.
5. Given: Ek1 = Ek2 ; v2 = 2.5v1
Required: m1 : m2
=
1 2
mv ;
2
Solution: Substitute the given values into the equation Ek1 = Ek2 .
Analysis: Ek =
Ek 1 = Ek 2
1
1
m1v12 = m2 v22
2
2
m1v12 = m2 (2.5v1 )2
m1 v12 = 6.25m2 v12
m1 = 6.2m2
m1
= 6.2
m2
m1 : m2 = 6.2 :1
Statement: The ratio of the slower mass to the faster mass is 6.2 : 1.
6. Given: mc = 1.2 &times; 103 kg; ms = 4.1 ! 103 kg; vc = 99 km/h; Ek c = Ek s
Required: vs
1
Analysis: Convert the speed to kilometres per hour. Substitute Ek = mv 2 into the
2
equation Ekc = Eks and solve for vs.
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Chapter 4: Work and Energy
4.2-7
! km \$ ! 1000 m \$ ! 1 h \$
Solution: vc = # 99
h &amp;% #&quot; 1 km &amp;% #&quot; 3600 s &amp;%
&quot;
vc = 27.5 m/s (one extra digit carried)
Ek c = Ek s
1
1
mc vc2 = ms vs2
2
2
mc vc2
v =
ms
2
s
vs2 =
(1.2 ! 103 kg)(27.5 m/s)2
(4.1! 103 kg)
vs =
(1.2 ! 103 kg )(27.5 m/s)2
(4.1! 103 kg )
= 14.9 m/s (one extra digit carried)
Convert the speed back to kilometres per hour.
!
m \$ ! 1 km \$ ! 3600 s \$
vs = # 14.9 &amp; #
s % &quot; 1000 m &amp;% #&quot; 1 h &amp;%
&quot;
vs = 54 km/h
Statement: The speed of the SUV is 54 km/h.
7. Given: ma = 0.020 kg; va = 250 km / h; mb = 0.14 kg; Eka = Ekb
Required: vb
1
Analysis: Ek = mv 2 ; convert speed to metres per second; substitute into E = E
2
km 1000 m 1 h
!
!
h 1 km 3600 s
va = 69.4 m/s (one extra digit carried)
Solution: va = 250
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-8
Ek a = Ek b
1
1
ma va2 = mb vb2
2
2
ma va2
= vb2
mb
vb =
=
ma va2
mb
(0.020 kg )(69.4 m/s)2
0.14 kg
vb = 26.2 m/s
Convert the speed back to kilometres per hour.
!
m \$ ! 1 km \$ ! 3600 s \$
vb = # 26.2 &amp; #
s % &quot; 1000 m &amp;% #&quot; 1 h &amp;%
&quot;
vb = 94 km/h
Statement: The speed of the baseball is 94 km/h.
8. Given: v = 150 km / h; m = 0.16 kg; !d = 0.25 m
Required: F
1 2
mv ; W = F!d ; = ! k ;
2
!Ek = E &quot; E . The puck has no initial velocity, so it has no initial kinetic energy, that
Analysis: Convert the speed to metres per second;
k
=
is, E = .
km 1000 m 1 h
!
!
h 1 km 3600 s
v = 41.7 m/s (one extra digit carried)
Solution: v = 150
W = F!d
W
F=
!d
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-9
W = !Ek
= Ek f &quot; Ek i
= Ek f
1 2
mv
2
1
= (0.16 kg)(41.7 m/s)2
2
W = 139 J (one extra digit carried)
=
W
!d
139 J
=
0.25 m
F = 560 N
Statement: The average force exerted on the puck by the player is 560 N.
9. Given: m = 5.31! 10&quot;26 kg; Ek = 6.25 ! 10&quot;21 J
Required: v
1
Analysis: Ek = mv 2 ; solve for v and substitute.
2
1
Solution: Ek = mv 2
2
2
Ek = v 2
m
F=
v=
=
2
E
m k
2
(6.25 ! 10&quot;21 J)
&quot;26
5.31! 10 kg
v = 485 m/s
Statement: The speed of the molecule is 485 m/s.
10. Given: Fa = 15 N;m = 3.9 kg; &micro;k = 0.25; vi = 0.0 m / s; !d = 12 m
Required: vf
Analysis: Ff = &micro;k FN ; FN = mg; F = Fa ! Ff ; W = &quot;Ek ; since vi = 0.0 m / s; &quot;Ek = Ekf ;
Ekf =
1 2
mv
2 f
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-10
Solution:
Ff = &micro;k FN
= 0.25mg
= 0.25(3.9 kg)(!9.8 m/s 2 )
Ff = !9.56 N (one extra digit carried)
F = Fa + Ff
= 15 N + (!9.56 N)
F = 5.44 N (one extra digit carried)
W = F&quot;d
= (5.44 N)(12 m)
W = 65.3 J
W = Ek
W=
1 2
mv
2 f
2
W = vf2
m
vf =
=
2
W
m
2
(65.3 J)
3.9 kg
vf = 5.8 m/s
Statement: The final speed of the block is 5.8 m/s.
11. Given: m = 5.55 ! 103 kg; v1 = 2.81 km / s or 2810 m / s; v2 = 3.24 km / s or 3240 m / s
Required: Wg
1
Analysis: Wg = !Ek ; !Ek = Ek 2 &quot; Ek1; Ek = mv 2
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-11
Solution: Wg = !Ek
= Ek 2 &quot; Ek 1
1 2 1 2
mv &quot; mv
2 2 2 1
1
= m(v22 &quot; v12 )
2
1
= (5.55 # 103 kg)((3240 m/s)2 &quot; (2810 m/s)2 )
2
Wg = 7.22 # 109 J
=
Statement: The work done by gravity on the satellite is 7.22 ! 109 J.
12. (a) It is a quadratic function.
(b) The graph passes through the origin because when the speed is zero, the kinetic
energy is zero.
(c) Given: From the graph, when the speed, v, is 2 m/s, the kinetic energy, Ek, is 4 J.
Required: m
1
Analysis: Ek = mv 2 ; solve for m.
2
1
Solution:
Ek = mv 2
2
2
E =m
v2 k
2
m = 2 Ek
v
2
=
(4 J)
(2 m/s)2
m = 2 kg
Statement: The mass of the robot is 2 kg.
(d) Substitute the mass of the robot into the equation for kinetic energy, omitting units.
1
Ek = mv 2
2
1
= (2)v 2
2
Ek = v 2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-12
Section 4.3: Gravitational Potential Energy
Tutorial 1 Practice, page 180
1. Given: m = 0.02 kg; ∆d = 8.0 m; g = 9.8 m/s2
Required: ∆Eg
Analysis: Use the gravitational potential energy equation, ΔEg = mg Δy . Let the y = 0
reference point be the ground.
Solution: !Eg = mg!y
= (0.02 kg)(9.8 m/s 2 )(8.0 m)
!Eg = 1.6 J
Statement: The change in potential energy between the branch and the ground is 1.6 J.
2. Given: ΔEg = 660 J; Δy = 2.2 m; g = 9.8 m / s2
Required: m
Analysis: Rearrange the gravitational potential energy equation, ΔEg = mg Δy , to solve
for m.
Solution: !Eg = mg!y
m=
=
!Eg
g!y
660 J
(9.8 m/s 2 )(2.2 m)
m = 31 kg
Statement: The mass of the loaded barbell is 31 kg.
3. Given: height of each book, h = 3.6 cm = 0.036 m ; number of extra books = 2
Required: W
Analysis: ΔEg = mg Δy
The 11th book is moved 10 ! 3.6 cm and the 12th book is moved 11 ! 3.6 cm.
Solution: !Eg = mg!y
= (1.6 kg)(9.8 m/s 2 )[10(0.036 m) + 11(0.036 m)]
!Eg = 12 J
Statement: The work done by the student to stack the two extra books is 12 J.
Section 4.3 Questions, page 181
1. (a) Given: m = 2.5 kg; g = 9.8 m / s 2 ; !y = 2.0 m
Required: Ek
Analysis: The kinetic energy of the wood when it hits the table is equal to the potential
energy of the wood before it falls. Ek = ΔEg = mg Δy
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.3-1
Solution: Ek = Eg
= mg!y
= (2.5 kg)(9.8 m/s 2 )(2.0 m)
Ek = 49 J
Statement: The kinetic energy of the piece of wood as it hits the table is 49 J.
(b) Given: m = 2.5 kg; Ek = 49 J
Required: v
1
2
Analysis: Ek = mv2 ; solve for v
Solution:
Ek =
1 2
mv
2
2Ek
= v2
m
v=
=
2Ek
m
2(49 J)
2.5 kg
v = 6.3 m/s
Statement: The speed of the wood as it hits the table is 6.3 m/s.
2. Given: g = 9.8 m / s2 ; m = 5.0 kg; Δy = 553 m
Required: Eg
Analysis: Eg = mg Δy
Solution: Eg = mg!y
= (5.0 kg)(9.8 m/s 2 )(553 m)
Eg = 2.7 &quot; 104 J
Statement: The gravitational potential energy of the Canada goose is 2.7 &times; 104 J.
3. (a) Given: m = 175 g = 0.175 kg; !y = 1.05 m ; g = – 9.8 m/s2
Required: gravitational potential energy of the puck, Eg
Analysis: Eg = mg!y
Solution: Eg = mg!y
= (0.175 kg)(9.8 m/s 2 )(1.05 m)
Eg = 1.8 J
Statement: The gravitational potential energy of the puck is 1.8 J.
(b) Given: Eg = 1.8 J
Required: change in gravitational potential energy of puck, !Eg
Analysis: Since the gravitational potential energy of the puck when it hits the ice is equal
to 0, it is expressed as ΔEg = − Eg .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.3-2
Solution: !Eg = &quot; Eg
!Eg = –1.8 J
Statement: The change in gravitational potential energy of the puck is −1.8 J.
(c) Given: !Eg = –1.8 J
Required: work done by the puck, W
Analysis: Since work and energy use the same units, W is equal to the change in
gravitational potential energy of the puck.
Solution: W = !Eg
W = &quot;1.8 J
Statement: The work done on the puck by gravity is 1.8 J.
4. The total work done is 0 J. The work done by gravity while you lift the cat is exactly
balanced by the work done by gravity while you lower the cat.
5. Given: !y = &quot;5.4 m; !Eg = &quot;3.1# 103 J; = 9.8 m / s 2
Required: m
Analysis: Eg = mg Δy
Solution: At the mat, the pole vaulter’s gravitational potential energy is 0 J.
Thus, ∆Eg = −Eg.
!Eg = &quot; Eg
!Eg = &quot;mg!y
m=&quot;
=
!Eg
g!y
&quot;3.1# 103 J
(9.8 m/s 2 )(&quot;5.4 m)
m = 59 kg
Statement: The pole vaulter’s mass is 59 kg.
6. Given: m = 0.46 kg; ΔEg = 155 J; g = 9.8 m / s 2
Required: ∆y
Analysis: ΔEg = mg Δy
Solution: !Eg = mg!y
!y =
=
!Eg
mg
155 J
(0.46 kg)(9.8 m/s 2 )
!y = 34 m
Statement: The maximum height of the ball above the tee is 34 m.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.3-3
7. Given: m =
Required: Eg
Analysis: sin! =
; !y = 1.3
&quot;y
= 1300 m; &quot; = 14 ; g = 9.8 m / s 2
; Eg = mg Δy
&quot; = sin!
Solution: !y = sin&quot;
= (1300 m) sin 14&deg;
! = 314 498 m (four extra digits carried)
Eg = mg!y
= (59 kg)(9.8 m/s 2 )(314.498 m)
Eg = 1.8 &quot; 105 J
Statement: The snowboarder’s gravitational potential energy is 1.8 ! 105 J.
8. (a) The work done on the first box is zero, because it doesn’t move. The second box is
lifted a height of ∆y, the third is lifted a height of 2!y , the fourth is lifted a height of
3!y , and so on until the Nth box, which is lifted a height of
! 1) &quot; . Therefore, the
work done to raise the last box to the top of the pile is expressed as mg ( N ! 1) &quot;y .
(b) As in Sample Problem 3 of Tutorial 1 on page 180, the gravitational potential energy
of the stack of boxes is the sum of the gravitational potential energies of the individual
boxes.
!Eg = mg[0 &quot; !y + 1&quot; !y + 2 &quot; !y + 3&quot; !y +!+ (N # 1)!y]
!Eg = mg!y[0 + 1+ 2 + 3+!+ (N # 1)]
= [2g1 + ( ! 1)m] . To
2
find the sum of the sequence 0 + 1+ 2 + 3+ ... + ( N ! 1) , substitute n = N , a1 = 0 , and
d = 1.
n
S n = [2a1 + (n ! 1)d]
2
N
S N = [2(0) + (N ! 1)(1)]
2
N (N ! 1)
SN =
2
Therefore, the gravitational potential energy that is stored in the entire pile is expressed
as:
mg!yN (N &quot; 1)
!Eg =
2
!y
!Eg = mgN (N &quot; 1)
2
The sum of an arithmetic sequence is given by the formula
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.3-4
Given: EJ = .3! 08 5 3
=
5 = 9.8 m / s 2 ; 30 students in class; average mass
of each student = 70 kg
Required: ∆y
Analysis: Find the chemical potential energy, Ec 1, in 1 L of gas by dividing Ec by 3.79.
Find the total mass of the class of students. Solve the equation Eg = mg Δy for ∆y.
Ec
3.79
1.3! 108 J
=
3.79
Ec 1 = 3.43! 107 J (one extra digit carried)
Solution: Ec 1 =
There are 3.43! 107 J of chemical potential energy in 1 L of gas. Assuming that there are
30 students in the class, each with an average mass of 70 kg, this equals a total mass of
30 ! 70 kg = 2100 kg .
Therefore,
Eg = mg!y
!y =
=
Eg
mg
3.43&quot; 107 J
(2100 kg)(9.8 m/s 2 )
!y = 1700 m
Statement: The chemical potential energy of the gas could lift the students 1700 m if it
could all be converted to gravitational potential energy.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.3-5
Section 4.5: The Law of Conservation of Energy
Mini Investigation: Various Energies of a Roller Coaster, page 185
A. The total energy graph is a straight line because total energy is conserved and it is
constant.
B. At any height, h, the sum of the energy values on the potential energy and kinetic
energy curves is equal to the value of the total energy at that height.
C. It is necessary to know the height at point A because it represents the potential energy
before the roller coaster starts. This is equal to the total mechanical energy. If the height
of point A were greater, the change in slopes of the potential and kinetic energy graphs
would be greater, and the total mechanical energy graph would be higher, but still
horizontal.
D. If the mass of the roller coaster car were greater, the total mechanical energy would be
greater, and the change in slopes for the graphs for potential and kinetic energy would be
greater.
Tutorial 1 Practice, page 187
1. (a) Given: m = 0.43 kg; !y = 18 m; g = 9.8 m / s 2 ; vi = 7.4 m / s
Required: vf
Analysis: The total energy at the top of the hill is equal to the total energy at the bottom
of the hill. At the top of the hill, the total energy is the gravitational potential energy,
mg Δy , plus the kinetic energy,
to the kinetic energy,
1 2
mvi . At the bottom of the hill, the total energy is equal
2
1 2
mvf , since there is no gravitational potential energy at ∆y = 0.
2
1
1
mvi2 = mvf2
2
2
2
2g!y vi = vf2
Solution: mg!y
vf = 2g!y vi2
= 2(9.8 m/s 2 )(18 m) (7.4 m/s)2
vf = 2.0 &quot; 101 m/s
Statement: The ball’s speed at the bottom of the hill is 2.0 ! 101 m/s.
(b) Given: m = 0.43 kg; !y = 18 m; vi = 4.2 m / s; g = 9.8 m / s 2
Required: vf
Analysis: The total energy when the ball is kicked up the hill is equal to the total energy
when the ball reaches the bottom of the hill. The energy at any time when the ball is on
1
its way up or down the hill does not matter. When the ball is kicked, ET = mg!y + mvi2 .
2
1
At the bottom of the hill, ET = mvf2 .
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-1
1
1
mvi2 = mvf2
2
2
2
2g!y + vi = vf2
Solution: mg!y +
vf = 2g!y + vi2
= 2(9.8 m/s 2 )(18 m) + (4.2 m/s)2
vf = 19 m/s
Statement: The ball’s speed as it reaches the bottom of the hill is 19 m/s.
2. (a) Given: m = 0.057 kg; !y = 1.8 m; g = 9.8 m / s 2 ; vf = 0 m / s
Required: vi
Analysis: Let the player’s hand be the y = 0 reference point. The total energy when the
1 2
mv . The total energy at the highest point of the
2 i
1
ball is all gravitational potential energy, mg Δy . Thus, mvi2 = mg!y .
2
1
Solution: mvi2 = mg!y
2
vi2 = 2g!y
ball is released is all kinetic energy,
vi = 2g!y
= 2(9.8 m/s 2 )(1.8 m)
vi = 5.9 m/s
Statement: The speed of the ball as it leaves the player’s hand is 5.9 m/s.
(b) Given: m = 0.057 kg; vi 2 = vi ; g = 9.8 m / s 2
Required: Δy2 : Δy1
Analysis:
1 2
mv = mg!y
2 i
Solution:
1
mvi21 = mg!y1
2
1 2
v = g!y1
2 i1
v2
!y1 = i 1
2g
1
mvi22 = mg!y2
2
1 2
v = g!y2
2 i2
v2
!y2 = i 2
2g
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-2
vi22
!y2 2g
=
!y1 vi21
2g
=
=
vi22
2g
&quot;
2g
vi21
vi22
vi21
#1 &amp;
%\$ 4 vi 1 ('
=
vi21
2
1 2
vi 1
16
=
vi21
!y2 1
=
!y1 16
Statement: The ratio of the maximum rise of the ball after leaving the player’s hand to
the maximum rise in (a) is 1:16.
Tutorial 2 Practice, page 190
1. (a) Given: v = 1.4 m / s; Δy = 5.0 m; m = 65 kg; g = 9.8 m / s 2
Required: P
Analysis: The work done to get to the top of the ladder is equal to the gravitational
potential energy at the top of the ladder, W = mg Δy . The time, t, taken to get to the top of
the ladder is the distance, ∆y, divided by the speed, v. P =
W
t
!y
v
5.0 m
=
1.4 m/s
t = 3.57 s (one extra digit carried)
Solution: t =
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-3
W
t
mg!y
=
t
(65 kg)(9.8 m/s 2 )(5.0 m)
=
3.57 s
P = 890 W
Statement: The firefighter’s power output while climbing the ladder is 890 W.
(b) From (a), it takes the firefighter 3.6 s to climb the ladder.
2. Given: vf 2 = 2vf 1; t2 = t1; vi = 0 m / s; m2 = m1
Required: ratio of power needed, P2 : P1
Analysis: We are told that the Grand Prix car accelerates to twice the speed of the car in
Sample 1, which can be expressed as vf 2 = 2vf 1 . We are also told that the Grand Prix car
accelerates to this speed in the same amount of time as the car in Sample 1, which is
stated as 7.7 s. We will assume that the two cars are equal in mass, at 1.1! 103 kg .
P=
Solution: P1 =
mvf21
2t
P2 =
mvf22
2t
mvf22
P2
2t
=
P1 mvf21
2t
=
2
f2
2
f1
v
v
(2vf 1 )2
=
vf21
=
4 vf21
vf21
P2 4
=
P1 1
Statement: The ratio of the power needed by the Grand Prix car to the power needed by
the car in Sample Problem 1 is 4:1.
3. Given: Δd = 190 m; t = 4 min 50 s = 290 s; m = 62 kg; g = 9.8 m / s 2
Required: P
W
Analysis: P = ; W = mg!y
t
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-4
W
t
mg!y
=
t
(62 kg)(9.8 m/s 2 )(190 m)
=
290 s
P = 0.40 kW
Statement: The racer’s average power output during the race is 0.40 kW.
Solution: P =
Section 4.5 Questions, page 191
1. (a) Given: vi = 11 m / s; g = 9.8 m / s 2
Required: maximum height that the ball will reach, ∆y
Analysis: The kinetic energy when the child tosses the ball is equal to the gravitational
1
potential energy at the ball’s maximum height, expressed as mvi2 = mg!y .
2
1
Solution: mvi2 = mg!y
2
1 2
v = g!y
2 i
v2
!y = i
2g
=
(11 m/s)2
2(9.8 m/s 2 )
!y = 6.2 m
Statement: The maximum height that the ball will reach is 6.2 m.
(b) As the ball leaves the child’s hand, the gravitational potential energy is zero.
It increases quadratically to its maximum when the ball reaches its maximum height.
It decreases quadratically to zero as the ball returns to the level of the child’s hand.
(c) The graph has this shape because, as the ball leaves the child’s hand, the kinetic
energy is at its maximum. It decreases quadratically to zero when the ball reaches its
maximum height. It increases quadratically to its maximum as the ball returns to the level
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-5
of the child’s hand. The total energy is conserved, so it is a constant horizontal line equal
to the maximum kinetic energy or potential energy.
(a) The kinetic energy is the greatest just before the apple hits the ground.
(b) The gravitational potential energy is the greatest as the apple leaves the branch.
3. (a) The law of conservation of energy states that energy can neither be created nor
destroyed in an isolated system; it can only change form. Assuming the puck and surface
form an isolated system, the energy of the hockey puck is conserved. The kinetic energy
of the puck is transformed to thermal energy by friction.
(b) The initial kinetic energy is transformed to thermal energy by friction as the puck
slows down to a stop.
4. (a) Given: = 110 kg; !y = 210 m; = 9.8 m / s 2
Required: W
Analysis: The work done by gravity is equal to the gravitational potential energy at the
top of the hill, expressed as Eg = mg Δy .
Solution: W = Eg
= mg!y
= (110 kg)(9.8 m/s 2 )(210 m)
W = 2.3&quot; 105 J
Statement: The work done by gravity on the skier is 2.3 ! 105 J.
(b) Given: m = 110 kg; !y = 210 m; g = 9.8 m / s 2 ; vi = 0 m / s
Required: vf
Analysis: Because the skier has no initial velocity, the total energy at the top of the hill is
all potential energy. The total energy at the bottom of the hill is all kinetic energy. The
total energy at the top of the hill is equal to the total energy at the bottom of the hill, or
1
mg!y = mvf2 . Solve for vf and substitute.
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-6
1
mv 2
2 f
2g!y = vf2
Solution: mg!y =
vf = 2g!y
= 2(9.8 m/s 2 )(210 m)
vf = 64 m/s
Statement: The skier’s speed when he reaches the bottom of the hill is 64 m/s.
5. Given: m = 62 kg; vi = 8.1 m / s; g = 9.8 m / s 2 ; !y = 3.7 m
Required: vf
Analysis: The total energy as the snowboarder leaves the ledge is the sum of her
gravitational potential energy, mg Δy , and her kinetic energy,
when she lands is all kinetic energy,
mg!y +
1 2
mvi . The total energy
2
1 2
mvf , if we take Δy = 0 at the landing point. Thus,
2
1 2 1 2
mv = mv . Solve for vf and substitute the given values.
2 i 2 f
1
1
mvi2 = mvf2
2
2
2
2g!y + vi = vf2
Solution: mg!y +
vf = 2g!y + vi2
= 2(9.8 m/s 2 )(3.7 m) + (8.1 m/s)2
vf = 12 m/s
Statement: The snowboarder’s speed at the moment she hits the ground is 12 m/s.
6. Given: !y = 3.5 m; &quot; = 40&deg;; g = 9.8 m / s 2
Required: the speed in the y-direction
1
Analysis: mg!y = mv 2
2
The energy equations give the speed in the direction along the jump, so we need to use
v
components to solve for the vertical velocity: v y =
.
1n !
1
Solution: mg!y = mv 2
2
2g!y = v 2
v = 2g!y
= 2(9.8 m/s 2 )(3.5 m)
v = 8.28 m/s (one extra digit carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-7
v
sin !
8.28 m/s
=
sin 40&deg;
v y = 13 m/s
vy =
Statement: The dolphin’s minimum speed is 13 m/s.
7. (a) Yes, the mechanical energy of the roller coaster is conserved because there is no
friction.
(b) Given: m = 640 kg; vi = 0;!yA = 30.0 m; g = 9.8 m / s 2
Required: ET, the total mechanical energy
Analysis: Because the car starts from rest, its total mechanical energy is equal to its
potential energy at point A: ET = mg ΔyA .
Solution:
T
=m !
A
= (640 kg)(9.8 m/s 2 )(30.0 m)
= 1.9 &quot; 105 J
Statement: The total mechanical energy at point A is 1.9 ! 105 J.
(c) The total mechanical energy is conserved, so it is the same at point B as it is at point
A: 1.9 ! 105 J.
(d) Given: = 640 kg;!yB = 15.0 m; = 9.8 m / s 2 ; !yA = 30.0 m
T
Required: v i vs
Analysis for vB: The total mechanical energy at point B is the sum of the kinetic
1
2
energy, mvB2 , and the potential energy,
!yB . The total mechanical energy is equal to
the potential energy at point A: mg!yA . Therefore,
Solution for vB:
1 2
mv + mg!yB = mg!yA .
2 B
1
mvB2 + mg!yB = mg!yA
2
vB2 + 2g!yB = 2g!yA
vB2 = 2g!yA &quot; 2g!yB
vB = 2g(!yA &quot; !yB )
= 2(9.8 m/s 2 )(30.0 m &quot; 15 m)
vB = 17 m/s
Statement for vB: The speed of the car when it reaches point B is 17 m/s.
1 2
mv .
2 s
The total mechanical energy is equal to the potential energy at point A: mg∆yA.
1
Thus, mv 2 = mg!yT .
2
Analysis for vC: The total mechanical energy at point C is all kinetic energy,
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-8
Solution for vC:
1
mv 2 = mg!yA
2 C
vC2 = 2g!yA
vC = 2g(!yA )
= 2(9.8 m/s 2 )(30.0 m)
vC = 24 m/s
Statement for vC: The speed of the car when it reaches point C is 24 m/s.
(e) Given: !yA = 30.0 m; !yB = 15.0 m; g = 9.8 m / s 2
Required: vB ; vC
Analysis: The total energy at A is equal to the total energy at B and at C. At A, the total
energy consists of kinetic energy and gravitational potential energy. At B, the total
energy also consists of kinetic energy and gravitational potential energy. At C, the total
1
energy consists of kinetic energy only: Ek = mv 2 ; Eg = mg!y
2
Solution:
ET B = ET A
mg!yB +
1
1
mvB2 = mg!yA + mvA2
2
2
1 2
1
vB = g!yA + vA2 &quot; g!yB
2
2
2
vB = 2g!yA + vA2 &quot; 2g!yB
vB = 2g!yA + vA2 &quot; 2g!yB
= 2(9.8 m/s 2 )(30.0 m) + (12 m/s)2 &quot; 2(9.8 m/s 2 )(15.0 m)
vB = 21 m/s
ET C = ET A
1
1
mvC2 = mg!yA + mvA2
2
2
1
1 2
vC = g!yA + vA2
2
2
2
vC = 2g!yA + vA2
vC = 2g!yA + vA2
= 2(9.8 m/s 2 )(30.0 m) + (12 m/s)2
vC = 27 m/s
Statement: The speed at B is 21 m/s, and the speed at C is 27 m/s.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-9
8. Given: m = 52 kg; t = 24 s; ! = 18 m; = 9.8 m / s 2
Required: P
W
Analysis: P = ; W = mg!y
t
W
Solution: P =
t
mg!y
=
t
(52 kg)(9.8 m/s 2 )(18 m)
=
24 s
P = 380 W
Statement: The power the woman exerts is 3.8 ! 102 W.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-10
Section 4.6: Elastic Potential Energy and Simple
Harmonic Motion
Mini Investigation: Spring Force, page 193
A. The relationship between Fg and ∆x is linear.
B. The slope of the best fit line of my graph is 50. This line represents the relationship
between Fg and ∆x, where the slope is the spring constant.
C. For the equipment used in this investigation, where k is the slope of the line of best fit,
the equation is Fg = 50!x .
Tutorial 1 Practice, page 195
1. (a) Given: m =
3 !x = 0.44 m3 g =
m ) s2
Required: k
Analysis: The force of gravity on the mass points down. The restorative spring force on
the mass points up because the spring is stretched down. To calculate the total force,
subtract !the magnitudes:
!
!
Fg = mg (dow = !mg ( 3 Fx = !k&quot;x = k&quot;x (
!
!
F
= 0 according to Newton’s second law.
Since the mass is not
accelerating,
!
Solution:
!F = 0
k&quot;x #
=0
k=
=
&quot;x
0.65
m)s 2
0.44 m
k = 14.5 N)m one e ra digit carrie
Statement: The spring constant is 14 N)m.
(b) Given: k = 14.5 N ) m3 !x =
m3 F =
m ) s2
Required: m
Analysis: Use the equation k&quot;x ! F = 0
Solution: k&quot;x ! F = 0
k&quot;x = F
k&quot;x
=
F
14.5 N)m
m
2
m)s
= 1.1 8g
Statement: The new mass is 1.1 8g.
2. Given: m = 5
3 k = 0 N ) m3 !x =
m
! !
Required: Fnet 3
Analysis: The free-body diagram for the mass is shown.
=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-1
The force of gravity on the mass points down. The spring force on the mass points down
because !the spring is compressed
upward:
!
!
Fg = mg (dow , Fx = !k&quot;x = k&quot;x (dow
!
! !
Fnet = Fg + Fx
!
!
Use the equation Fnet m to find the acceleration.
!
! !
Solution: Fnet = Fg + Fx
= mg (down9 + k!x (down9
!
net
=
8g
=
N one e ra digit carrie
m)s 2 (down9 +
N)m
m (down9
!
!
Fnet = m
!
!
= net
!
=
N (down9
8g
=
m)s 2 (down9
Statement:
2
(down9.
Tutorial 2 Practice, page 196
1. Given: mb = 2m3 k = 2 ! 10/ N ) m3 F =
Required: ratio of eb : e
Analysis:
1
= k !x
2
m ) s2
2
To find eb , we need !xb . Use the same equation as in Sample Problem 1,
k&quot;x (up9 ! mg (down9 = 0 .
Solution: k&quot;xb (up9 ! mb g (down9 = 0
k&quot;xb = mb g
mb g
k
2mg
&quot;xb =
k
&quot;xb =
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-2
eb
=
1
k !xb
2
2
1 &quot; 2mg %
= k\$
2 # k '&amp;
e
2
eb
e
eb
e
1
k !x
2
2
1 &quot; mg %
= k\$
2 # k '&amp;
&quot; 2
%
1 \$ 4 m2 g 2 '
= k
2 \$ k2 '
#
&amp;
2m2 g 2
=
eb
k
=
e
2
=
1 &quot; m2 g 2 %
k
2 \$# k 2 '&amp;
=
m2 g 2
2k
2 m2 g 2
k
=
2
m g2
2k
4
=
1
Statement: The ratio of the elastic potential energy,
!
2. Given: Fx = 220 N3 !x = 0.14 m
Required: Ee
!
1
!
Analysis: Fx = !k&quot;x = k&quot;x 3 e = k !x 2
2
!
Solution: Fx = k!x
!
Fx
k=
!x
220 N
=
0.14 m
k=
N)m one e ra digit carrie
eb
:
e
is 4:1.
1
k !x 2
2
1
=
N)m 0.14 m 2
2
= 15
e
Statement: The elastic potential energy of the toy is 15 J.
e
=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-/
Tutorial 3 Practice, page 199
1. Given: m = 105 8g3 k = .1! 10/ N ) m
Required: f, T
Analysis: Use the equations for simple harmonic motion period and frequency:
T = 2!
m
and
k
=
Solution: T = 2!
1
T
k
105 8g
&quot; 10/ N)m
= 2!
T=
m =
=
s
1
T
1
s
m = 1.4
Statement: The period of the vibrations is 0.74 s, and the frequency is 1.4 Hz.
2. The frequency of oscillations will change if passengers are added to the car because
when the mass increases, the period increases. This happens because mass is in the
numerator of the equation for the period. If the period increases, the frequency decreases,
because frequency is the reciprocal of period.
Section 4.6 Questions, page 200
1. Spring A!is more difficult to stretch because it has a greater spring constant.
2. Given: F = 5 N3 !x = 10 mm = 0.01 m
Required: k
!
Analysis: The spring force opposes the applied force, so Fx = !5 N . Rearrange the
!
formula Fx = !k&quot;x to solve for k.
!
Solution: Fx = !k&quot;x
!
Fx
k=!
&quot;x
!5 N
=!
0.01 m
k = 500 N)m
Statement: The spring constant is 500 N)m.
3. The elastic potential energy stored in a spring is the same whether it is stretched by
1.5 cm or compressed by 1.5 cm. The spring constant is exactly the same whether the
spring is stretched or compressed, so the elastic potential energy must also be the same.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-4
4. (a) Given: k = 5.5 &quot; 10/ N ) m3 !x = 2.0 cm = 0.020 m
Required: Ee
1
Analysis: e = k !x 2
2
1
Solution: e = k !x 2
2
1
= 5.5 &quot; 10/ N)m 0.020 m 2
2
= 1.1 J
e
Statement: The elastic potential energy of the spring when it stretches 2.0 cm is 1.1 J.
(b) Given: k = 5.5 &quot; 10/ N ) m3 !x = #/.0 cm = #0.0/0 m
Required: Ee
1
Analysis: e = k !x 2
2
1
Solution: e = k !x 2
2
1
= 5.5 &quot; 10/ N)m #0.0/0 m 2
2
= 2.5 J
e
Statement: The elastic potential energy of the spring when it compresses /.0 cm is 2.5 J.
5. (a) Given: m = 0.6/ 8g3 k = 65 N ) m3 g = .
) s2
Required: ∆x
Analysis: The force of gravity on the mass points down. The restorative force on the
mass
! points up since the spring!is compressed.
!
Fg = mg (down9 = !mg (up9 3 Fx = !k&quot;x = k&quot;x (up9
!
Solution: Since the mass is at rest, !F = 0 .
!
!F = 0
k&quot;x # mg = 0
mg
&quot;x =
k
0.6/ 8g
m)s 2
=
65 N)m
&quot;x =
m
Statement:
m from its equilibrium position.
(b) Given: m = 0.6/ 8g3 k = 65 N ) m3 g =
m ) s2
!
Required: a
Analysis: The force of gravity on the mass points down. The spring force on the mass
points
up because the
being compressed downward.
!
! mass is
!
=
(down9 3
= ! &quot; = &quot; (up9
g
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-5
!
net
=
!
!
+
g
Solution:!
net
3
=
!
net
!
= a
!
+
g
=
!
(down9 + ! (up9
= 0.6/ 8g
!
net
!
net
m)s 2 (down9 + 65 N)m 0.041 cm (up9
= /.5 N (down9
!
= a
!
!
a=
=
net
/.5 N (down9
0.6/ 8g
!
a = 5.6 m)s 2 (down9
Statement: The acceleration of the mass after it falls 4.1 cm is 5.6 m)s2 (down9.
6. Given: m = 5.2 8g3 = 1.2 s
Required: k
Analysis:
= 2!
= 2!
Solution:
2!
=
2
&quot; %
\$# 2! '&amp; =
=
=
&quot; %
\$# 2! '&amp;
2
5.2 8g
&quot; 1.2 s %
\$# 2! '&amp;
2
= 140 N)m
Statement: The spring constant is 140 N)m.
7. Given: = 1.5 ! 10/ N ) m3 e = 0.0 J
Required: ∆x
1
! 2
Analysis: e =
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-6
Solution:
e
2
e
=
1
2
!
= !
! =
2
2
2
e
J
2
/
1.5 &quot; 10 N)m
! = 0.// m
Statement: The spring should be stretched 0.//
8. Given: !x = 15 mm = 0.015 m3 = 400.0 N ) m
Required: W
Analysis: The wor8 done is equal to the elastic potential energy:
1
! 2
= e3 e =
2
1
Solution: e =
! 2
2
1
= 400.0 N)m 0.015 m 2
2
= 0.045 J
e
Statement: The wor8 done by the spring force acting on the spring is 4.5 ! 10−2 J.
9. Given: e = 7.50 J3 = 0.20 8g3 = 240 N ) m
Required: f3 ∆x
1
1
3 = 3 e = k !x 2
Analysis: = 2!
=
2
Solution: T = 2!
= 2!
T=
m =
=
m
k
0.20 8g
240 N)m
s one extra digit carrie
1
T
1
s
m = 5.5 Hz
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-7
e
2
e
k
=
1
k !x
2
= !x
!x =
2
2
2
e
k
2 7.50 J
240 N)m
!x = 0.25 m
Statement: The frequency of oscillation is 5.5 Hz and the amplitude of oscillation
is 0.25 m.
10. Given: m = 5.5 ! 102 8g 3 six cycles in 4.4 s
Required: k
m
Analysis: Divide 4.4 s by 6 to get T. Use the formula for the period, T = 2!
, to
k
calculate k.
4.4 s
Solution: T =
6
=
T = 2!
m
k
T
m
=
2!
k
2
T
m
=
2
k
4!
4! 2 m
k=
T2
4! 2 5.5 &quot; 102 8g
=
2
# 4.4 s &amp;
%\$ 6 ('
k = 4.0 &quot; 104 N)m
Statement: The spring constant of either spring is 4.0 &times; 104 N)m.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-
11. Answers may vary. Sample answer: Pyon pyon shoes strap onto the outside of your
regular shoes. They are made of two curved springy pieces of material joined together in
a shape similar to that of a football. When you jump, the springy material increases the
height you can attain. When your mass presses down on the springs, the springs press
bac8 up, causing you to jump up in the air much higher than normal.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-
Section 4.7: Springs and Conservation of Energy
Tutorial 1 Practice, page 205
1. If the ramp is not frictionless, some of the kinetic energy of the block will be
transformed to thermal energy by friction as the block slides down the ramp. Thus, the
block will have less kinetic energy to compress the spring, and the amount of
compression will be less.
2. Given: m = 3.5 kg; !y = 2.7 m; !x = 26 cm = 0.26 m
Required: k
1
Analysis: Eg = mg∆y; Ee = k(!x)2
2
Since energy is conserved, the change in potential energy of the mass must equal the
change in elastic potential energy when the spring is compressed.
Solution: If we choose the bottom of the ramp to be the y = 0 reference point, the mass
will have no gravitational potential energy at the bottom of the ramp. The initial
gravitational potential energy has transformed into kinetic energy. When the spring is
fully compressed, the kinetic energy has transformed into elastic potential energy.
Therefore, the spring’s initial gravitational potential energy must equal its final elastic
potential energy.
Ee = Eg
1
k(!x)2 = mg!y
2
2mg!y
k=
(!x)2
=
2(3.5 kg)(9.8 m/s 2 )(2.7 m)
(0.26 m)2
k = 2700 N/m
Statement: The spring constant is 2.7 ! 103 N/m.
3. Given: m = 43 kg; k = 3.7 kN / m = 3700 N / m; !x = 37 cm = 0.37 m
Required: Δy
1
Analysis: !Eg = mg!y ; Ee = k(!x)2
2
Solution: Choose the lowest point of the bounce as the y = 0 reference point. At the
maximum height, ∆y, all of the elastic potential energy has converted to gravitational
potential energy.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-1
!Eg = Ee
1
k(!x)2
2
k(!x)2
!y =
2mg
mg!y =
=
=
(3700 N/m)(0.37 m)2
2(43 kg)(9.8 m/s 2 )
#
m&amp;
3700
kg
&quot;
(0.37 m)
%
s 2 ('
\$
2(43 kg )(9.8 m/s 2 )
!y = 0.60 m
Statement: The maximum height he reaches on the following jump is 0.60 m above the
compressed point.
4. Given: m = 0.35 kg; h = 2.6 m; !x = 0.14
Required: k
1
Analysis: Eg = mg!y ; Ee = k(!x)2
2
The initial gravitational potential energy of the branch is equal to the final elastic
potential energy of the trampoline at its lowest point.
Solution: Choose the lowest point of the trampoline as the y = 0 reference point. At this
point, all of the gravitational potential energy has transformed to elastic potential energy.
Since the lowest point represents y = 0 , the change in y is the height above the
trampoline surface, 2.6 m, plus the maximum compression of the trampoline, 0.14 m.
Therefore, Δy = 2.6 m + 0.14 m = 2.74 m .
Ee = Eg
1
k(!x)2 = mg!y
2
2mg!y
k=
(!x)2
=
2(0.35 kg)(9.8 m/s 2 )(2.74 m)
(0.14 m)2
k = 960 N/m
Statement: The spring constant is 960 N/m.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-2
5. (a) Given: m = 4.0 kg; ∆y = 0.308 m
Required: v
1
Analysis: Eg = mg!y ; Ek = m 2
2
When the mass is doubled at the top of the ramp, it is at rest, so it has only gravitational
potential energy. At the bottom of the ramp, on the horizontal segment, all of the
gravitational potential energy has transformed to kinetic energy.
Solution: Let the bottom of the ramp be the y = 0 reference point. Therefore, the
gravitational potential energy at the top of the ramp is equal to the kinetic energy along
the horizontal part of the ramp.
Ek = Eg
1
mv 2 = mg!y
2
v = 2g!y
= 2(9.8 m/s 2 )(0.308 m)
v = 2.5 m/s
Statement: The speed of the block as it returns along the horizontal surface is 2.5 m/s.
(b) No, the block does not have the same kinetic energy as before along the horizontal
surface. The kinetic energy is equal to the gravitational potential energy, which is greater
than before.
(c) Given: md = 2m
Required: Δxd
Analysis: The elastic potential energy of the spring at its greatest compression is equal to
the gravitational potential energy at the top of the ramp.
Solution: Block in Sample Problem 3: Block with Mass Doubled
1
mg!y = k(!x)2
2
!x =
2mg!y
k
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-3
1
k(!xd )2
2
1
2mg!y = k(!xd )2
2
md g!y =
4mg!y
k
!xd =
2(2mg!y)
k
=
2mg!y
k
= 2
!xd = 2(!x)
Since the gravitational potential energy is doubled, the compression increases, but it is
not doubled. It increases by a factor of 2. So the new value of ∆x is
0.31 m.
Statement: The new value of ∆x is 31 cm.
(d) Given: &micro;k = 0.15; ! = 0.308 m;! = 0.62 m; = 250 N / m
Required: ∆x
!
!
Analysis: f = &micro; N = &micro; m ; f = f ! ; Ek = Eg ! Wf ; Ee = Ek
2(0.22 m) or
Solution: Let the bottom of the ramp be the y = 0 reference point.
!
= &micro;m
f
= 0.15(4.0 kg)(9.8 m/s 2 )
!
f
= 5.88 N (one extra digit carried)
f
=
f
!
= (5.88 N)(0.62 m)
f
= 3.65 J (one extra digit carried)
Ek = Eg ! Wf
= mg&quot;y ! Wf
= (4.0 kg)(9.8 m/s 2 )(0.308 m) ! 3.65 J
Ek = 8.42 J (one extra digit carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-4
Ee = Ek
1
k(!x)2 = 8.42 J
2
!x =
=
2(8.42 J)
k
2(8.42 N &quot; m)
250 N/m
!x = 0.26 m
Statement: The new value of the compression is 0.26 m.
Research This: Perpetual Motion Machines, page 206
A. I chose a metronome. A spring is wound tight, and as it unwinds, the elastic potential
energy is converted to kinetic energy, forcing the metronome wand to swing. As the
wand swings to its highest point, kinetic energy transforms to potential energy. As the
wand swings down again, the gravitational potential energy converts back to kinetic
energy. The spring contributes kinetic energy at the bottom of each swing, until the
spring is fully unwound. Eventually the metronome stops due to air resistance.
B. The design of the machine has been improved over time by creating quartz
metronomes, electronic metronomes, computer metronomes, and even metronome apps
for smart phones.
C. The improvements have been the results of all three developments: new materials;
new technology; and new scientific discoveries.
Section 4.7 Questions, page 208
1. The total mechanical energy of the system increases. Energy has been added by the
person outside the system of the mass and spring.
2. Given: k = 520 ; m6 m = 4.5 kg6 !x = 0.35 m
Required: v
Analysis: When the spring is compressed, but the mass is at rest, the mass has only
elastic potential energy. As the spring is released, the elastic potential energy transforms
to kinetic energy. When the mass is no longer touching the spring, all of the energy is
kinetic.
1
1
Ee = k(!x)2 ; Ek = m 2
2
2
Solution: The kinetic energy when the mass leaves the spring is equal to the elastic
potential energy when the spring is at its maximum compression.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-5
Ek = Ee
1
m
2
2
=
1
(! )2
2
=
(! )2
m
=
(520 N/m)(0.35 m)2
4.5 kg
= 3.8 m/s
Statement: The speed of the mass when it leaves the spring is
3.8 m/s [away from the spring].
3. (a) Given: k = 5.2 &times;102 N / m; Δx = 5.2 cm = 0.052 m
Required: Ee
1
Analysis: Ee = k(!x)2
2
1
Solution: Ee = k(!x)2
2
1
= (5.2 &quot; 102 N/m)(0.052 m)2
2
Ee = 0.703 J (one extra digit carried)
Statement: The elastic potential energy of the compressed spring is 0.70 J.
(b) Given: Ee = 0.703 J; m = 8.4 g = 0.0084 kg; ∆x = 5.2 cm = 0.052 m
Required: v
1
Analysis: Eg = mg∆y; Ek = m 2
2
All of the elastic potential energy of the compressed spring has transformed to
gravitational potential energy and kinetic energy as the pilot ejects.
Solution: The kinetic energy of the pilot as it ejects and the additional gravitational
potential energy is equal to the elastic potential energy of the compressed spring.
Eg + Ek = Ee
1
mg ! + m
2
2
= Ee
=
2Ee ! 2mg !
m
2(0.703 J) ! 2(0.084 kg)(9.8 m/s 2 )(0.052 m)
=
0.0084 kg
=
2(0.660 194 kg &quot; m 2 /s 2 )
0.0084 kg
= 13 m/s
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-6
Statement: The speed of the pilot as it ejects upward from the airplane is 13 m/s above
the launch point of the compressed spring.
(c) Given: m = 8.4 g = 0.0084 kg; = 5.2 ! 102 N / m; &quot;x = 5.2 cm = 0.052 m
Required: Δy
Analysis: Let the point where the pilot ejects be the y = 0 reference point. All of the
elastic potential energy of the compressed spring has transformed to gravitational
potential energy at the pilot’s maximum height.
1
Ee = k (Δx)2 ; Eg = mg!y
2
Solution: The elastic potential energy of the compressed spring is equal to the
gravitational potential energy at the pilot’s maximum height.
Eg = Ee
1
k(!x)2
2
k(!x)2
!y =
2mg
mg!y =
=
\$
m'
2
(0.052 m) 2
5.2
&quot;
10
kg
#
&amp;
2 )
s (
%
2(0.0084 kg )(9.8 m/s 2 )
!y = 8.5 m
Statement: The maximum height that the pilot will reach is 8.5 m above the launch point
of the compressed spring.
4. Given: k = 1.2 ! 102 N / m; m = 82 g = 0.082 kg; &quot; = 3.4 cm = 0.034 m
Required: Δx
Analysis: The elastic potential energy of the spring transforms to kinetic energy and then
to gravitational potential energy as it comes to rest at the top of the ramp.
1
Ee = k(!x)2 ; Eg = mg!y
2
Solution: Let the bottom of the ramp represent the y = 0 reference point. The elastic
potential energy of the spring is equal to the gravitational potential energy at the top of
the ramp.
Ee = Eg
1
k(!x)2 = mg!y
2
!x =
2mg!y
k
2(0.082 kg)(9.8 m/s 2 )(0.034 m)
1.2 &quot; 102 N/m
!x = 0.021 m
Statement: The distance of the spring’s compression is 0.021 m.
=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-7
5. Given: m = 75 kg; k = 6.5 N / m
Required: v
Analysis: Let the y = 0 reference point be 19 m below the platform. Since the unstretched
bungee cord is 11 m long, and the cord is stretched 19 m below the
platform, Δx = 19 m − 11 m = 8 m . The gravitational potential energy is transformed to
elastic potential energy and kinetic energy at this point.
1
1
Eg = mg Δy ; Ek = m 2 ; Ee = k(!x)2
2
2
Solution: The gravitational potential energy at the platform is equal to the sum of the
kinetic energy and the elastic potential energy 19 m below the platform.
Eg = Ek + Ee
1 2 1
mv + k(!x)2
2
2
1
1 2
mv = mg!y &quot; k(!x)2
2
2
mg!y =
v=
=
&amp;
2#
1
mg!y &quot; k(!x)2 (
%
m\$
2
'
,
2 )
1
(75 kg)(9.8 m/s 2 )(19 m) &quot; (65.5 N/m)(8.0 m)2 .
+
(75 kg) *
2
-
v = 18 m/s
Statement: The speed of the bungee jumper at 19 m below the bridge is 18 m/s.
6. Given: k = 5.0 N / m; m = 0.25 kg; !x = 14 cm = 0.14 m
Required: hmax ; max ; max
Analysis: Let the y = 0 reference point be the rest position of the spring. In simple
harmonic motion, the maximum height is the opposite of the lowest point.
The maximum velocity occurs as the box passes through the rest position. At this point,
1
1
there is only kinetic energy. Thus, k(!x)2 = mv 2 .
2
2
The maximum acceleration occurs at the maximum height. At this point, the spring force
is equal to the applied force, so k!x = ma .
Solution: The maximum height is 14 cm [above rest position].
For the maximum velocity,
1
1
k(!x)2 = mv 2
2
2
v=
k(!x)2
m
(5.0 N/m)(0.14 m)2
=
0.25 kg
v = 0.63 m/s
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-8
For the maximum acceleration,
k!x = ma
k!x
m
(5.0 N/m)(0.14 m)
=
0.25 kg
a=
a = 2.8 m/s 2
Statement: The maximum height is 14 cm [above rest position]. The maximum speed is
0.63 m/s. The maximum acceleration is 2.8 m/s2 [toward rest position].
7. Given: m = 0.22 kg6 = 280 N / m; !x = 11 cm = 0.11 m
Required: h
Analysis: Let the y = 0 reference point be the fully compressed position of the spring.
When the block is dropped, all the energy is in the form of gravitational potential energy.
At the full compression of the spring, all the energy has transformed to elastic potential
energy.
1
= m ! ; Ee = k(!x)2
g
2
Solution: The gravitational potential energy as the block is dropped is equal to the elastic
potential energy at full compression of the spring.
Eg = EN
1
k(!x)2
2
k(!x)2
!y =
2mg
mg!y =
=
(240 /m)(0.11 m)2
2(0.22 kg)(5.4 m/s 2 )
!y = 0.79 m
This height represents the total distance from where the block was dropped to the lowest
point of the spring, so subtract the maximum compression of the spring, 0.11 m, to get
the height from which the block was dropped.
h = ! &quot; !x
= 0.79 m &quot; 0.11 m
h = 0.68 m
Statement: The block was dropped from a height of 0.68 m.
8. (a) Given: m = 1.0 kg; = 1.0 m / s; = 1000.0 N / m
Required: ∆x
Analysis: The kinetic energy of the block transforms fully to elastic potential energy at
the maximum compression of the spring, because the block is at rest at this point.
1
1
Ek = mv 2 ; Ee = k (Δx)2
2
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-9
Solution: The kinetic energy of the block before it hits the spring is equal to the elastic
potential energy of the spring at its maximum compression.
Ee = Ek
1
1
(! )2 = m
2
2
! =
m
2
2
(1.0 kg)(1.0 m/s)2
=
1000.0 N/m
! = 0.032 m
Statement: The maximum compression of the spring is 0.032 m.
(b) The block will travel to the maximum compression of the spring, 0.032 m before
coming to rest.
9. (a) Given: m = 8.0 ; = 3.0 m / ; =
0 N/m
Required: Δx
Analysis: The kinetic energy of the block transforms fully to elastic potential energy at
the maximum compression of the spring, because the block is at rest at this point.
1
1
Ek = m 2 ; Ee = k(!x)2
2
2
Solution:
Ee = Ek
1
1
(! )2 = m
2
2
! =
m
2
2
(6.0 kg)(3.0 m/s)2
1250 N/m
! = 0.208 m (one extra digit carried)
Statement: The maximum distance the spring is compressed is 0.21 m.
(b) Given: !x =
m = 0.14 m m = .0 ke vg = 3.0 m / s k = 1250 N / m
=
Required: vf ; a
Analysis: Some of the kinetic energy of the block transforms to elastic potential energy
as the spring compresses. So the initial kinetic energy transforms to final kinetic energy
and elastic potential energy.
1
1
Ek = m 2 ; Ee = k(!x)2
2
2
For the acceleration, the only force acting on the block is the spring force, which is equal
to! k∆x away from the spring.
=m
!
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-10
Solution: The initial kinetic energy is equal to the sum of the final kinetic energy and the
elastic potential energy.
Ek = Ek N + E
1
m
2
m
1
m
2
2
=
2
N
=m
N
=
=
2
m
2
N
+
1
(! )2
2
&quot; (! )2
2
&quot; (! )2
m
(6.0 kg)(3.0 m/s)2 &quot; (
6.0 kg
/m)(0.14 m)2
= 2.2 m/s
For the acceleration:
N
!
!
F = ma
!
k!x [away from the spring] = ma
! k!x [away from the spring]
a=
m
(1250 N/m)(0.14 m) [away from the spring]
=
6.0 kg
!
2
a = 29 m/s [away from the spring]
Statement: When the spring is compressed, the speed of the block is 2.2 m/s, and the
acceleration is 29 m/s2 [away from the spring].
10. Given: k = 440 N / m; !x = 45 cm = 0.45 m; m = 57 g = 0.057 kg; y = 1.2 m
Required: d x , the horizontal distance
Analysis: Find the speed of the ball as it leaves the machine, and then use projectile
motion equations to determine when it will hit the ground.
Let the y = 0 reference point be the height at which the ball leaves the machine. Thus,
there is no gravitational potential energy at this point. The elastic potential energy of the
spring when it is at its maximum compression transforms to kinetic energy when the ball
leaves the machine.
1
1
Ek = m 2 ; Ee = k(!x)2
2
2
Once v has been determined, use projectile motion equations to determine the distance,
dx, the ball travels. Since the ball is projected horizontally, the initial launch angle is 0&deg;,
1 2
and there is no vertical component of velocity: v y = . Use the equation y = y !
2
to determine t, and then use the equation x = x to determine d x .
Solution: The elastic potential energy of the spring when it is at its maximum
compression is equal to the kinetic energy when the ball leaves the machine.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-11
Ek = Ee
1
m
2
2
=
1
(! )2
2
=
(! )2
m
=
(440 N/m)(0.45 cm)2
0.057 g
= 39.5 m/s (one extra digit carried)
Now, use the projectile motion equations.
1 2
=
!
2
1 2
=!
2
!2
=
!2(!1.2 m)
9.8 m/s 2
= 0.495 s (one extra digit carried)
=
x
=
x
= (39.5 m/s)(0.495 s)
= 2.0 ! 101 m
Statement: The horizontal distance that the tennis ball can travel before hitting the
ground is 2.0 ! 101 m.
x
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-12
Section 5.1: Momentum and Impulse
Tutorial 1 Practice, page 223
!
1. Given: m = 160 g = 0.16 kg; v = 140 m/s [E]
!
Required: p ; Ek
1
!
!
Analysis: p = mv ; Ek = mv 2
2
!
!
Solution: p = mv
= (0.16 kg)(40.0 m/s [E])
!
p = 6.4 kg ! m/s [E]
1 2
mv
2
1
= (0.16 kg)(40.0 m/s)2
2
Ek = 130 J
Statement: The momentum of the puck is 6.4 kg&middot;m/s [E], and the kinetic energy is 130 J.
2. Given: m1 = 6.2 kg; v1 = 1.6 m/s [E]; m2 = 160 g = 0.16 kg; v2 = 40.0 m/s [E]
!
!
Required: p2 ! p1
!
!
Analysis: p = mv
! !
!
!
Solution: p1 ! p2 = m1v1 ! m2 v2
Ek =
= (6.2 kg)(1.6 m/s [E]) ! (0.16 kg)(40.0 m/s [E])
! !
p1 ! p2 = 3.5 kg &quot; m/s [E]
Statement: The difference in the momenta is 3.5 kg&middot;m/s [E].
Tutorial 2 Practice,
page 226
!
1. (a) Given:
= 250 N [forward]; t = 0.0030 s
!
Required: !p
! !
Analysis: !p = !
! !
Solution: !p = !
= (250 N [forward])(0.0030 s)
!
! = 0.75 N &quot; s [forward]
Statement: The
is 0.75 N&middot;s [forward].
! impulse imparted by the hockey stick
!
(b) Given: F!t = 0.75 N &quot; s ; m = 180 g = 0.18 kg; vi = 0
!
Required: vf
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.1-1
!
!
Analysis: F!t = !p
!
! !
!p = m( vf + vi )
!
! !
F!t = m( vf + vi )
!
!
F!t = mvf
!
F!t !
!
vf =
+ vi
m
!
F t !
!
+ vi
Solution: vf =
m
0.75 kg = m/s [forward]
=
+ 0 m/s
0.18 kg
!
vf = 4.2 m/s [forward]
Statement: The final velocity is 4.2 m/s [forward].
2. Given: average force = 468 !N; ∆t = 2.9 s
Required: force–time graph; F!t
Analysis: The average force is 468 N, so the graph is a straight line at F = 468. Calculate the
area under the curve.
Solution: Draw the graph.
!
F!t = bh
= (2.9 s)(468 N)
!
F!t = 1400 N &quot; s
Statement: The impulse of the collision is 1400 N&middot;s [away from the wall].
Section 5.1 Questions, page 227
!
1. (a) Given: m = 4.25 &times; 102 kg; v = 6.9 m/s [N]
!
Required: p
!
!
Analysis: p = mv
!
!
Solution: p = mv
= (4.25 ! 102 kg)(6.9 m/s [N])
!
p = 2900 kg &quot; m/s [N]
Statement: The momentum of the moose is 2.9 &times; 103 kg&middot;m/s [N].
!
(b) Given: m = 9.97 &times; 103 kg; v = 5 km/h [forward]
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.1-2
!
Required: p
!
!
Analysis: p = mv
Convert the velocity to metres per second.
km
1h
1000 m
!
v =5
[forward] !
!
3600 s 1 km
h
!
v = 1.4 m/s [forward]
!
!
Solution: p = mv
= (9.97 ! 103 kg)(1.4 m/s [forward])
!
p = 1.4 ! 104 kg &quot; m/s [forward]
Statement: The momentum of the bus is 1.4 &times; 104 kg&middot;m/s [forward].
!
(c) Given: m = 995 g = 0.995 kg; v = 16 m/s [S]
!
Required: p
!
!
Analysis: p = mv
!
!
Solution: p = mv
= (0.995 kg)(16 m/s [S])
!
p = 16 N !s [S]
Statement: The momentum of the squirrel is 16 N&middot;s [S].
2. Answers may vary. Sample answer: Impulse is the change in momentum of an object during a
certain time. Its units are newton seconds (N&middot;s).
!
3. Given: m = 79.3 kg; p = 2.16 ! 103 kg &quot; m/s [W]
!
Required: v
!
!
Analysis: p = mv
!
! p
v=
m
!
! p
Solution: v =
m
2.16 ! 103 kg &quot; m/s [W]
=
79.3 kg
!
v = 27.2 m/s [W]
Statement: The velocity of the bicycle is 27.2 m/s [W].
!
!
4. Given: v = 9.0 ! 102 m/s [W] ; p = 4.5 kg ! m/s [W]
Required: m
!
!
Analysis: p = mv
!
p
m= !
v
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.1-3
!
p
Solution: m = !
v
4.5 kg ! m/s [W]
=
9.0 &quot; 102 m/s [W]
m = 5.0 &quot; 10#3 kg
Statement: The mass of the projectile is 5.0 ! 10+3 kg or 5.0 g.
!
!
5. Given: v = 29.5 m/s [forward] ; p =
! 1 kg &quot; m/s [
5
Required: m
!
!
Analysis: p = mv
!
p
m= !
v
!
p
Solution: m = !
v
2.31 ! 103 kg &quot; m/s [forward]
=
29.5 m/s [forward]
m = 78.3 kg
Statement: The mass of the skier is 78.3 kg.
6. Answers may vary. Sample answer: In lacrosse, when a player follows through when striking
a ball, this increases the time interval over which the collision between the ball and lacrosse stick
occurs and thus contributes to an increase in the velocity change of the ball. Following through, a
hitter can hit the ball in such a way that it is moving faster when it leaves the stick. This can
improve performance in lacrosse by making the ball travel farther in a given amount of time and
by making it harder for an opposing player to catch up with or react to the ball in time.
7. Both balls will hit the floor with the same speed, and both will rebound with this same speed.
Since the basketball has more mass, the same speed means it has more momentum. To reverse its
larger momentum
! requires a larger impulse on the basketball. −3
8. (a) Given: = 1100.0 N [forward] ; ∆t = 5.0 ms = 5.0 &times; 10 s
!
Required: !p
! !
Analysis: !p = !
! !
Solution: !p = !
= (1100.0 N [forward])(5.0 &quot; 10#3 s)
!
! = 5.5 N \$ s [forward]
Statement: The impulse is 5.5 N&middot;s [forward].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.1-4
(b) Given: m = 0.12 kg; vi = 0 m/s;
!
! = 5.5 N &quot; s [forward]
Required: vf
!
!
Analysis: F!t = !p
!
! !
!p = m( vf + vi )
!
! !
F!t = m( vf + vi )
!
F!t !
!
vf =
+ vi
m
F!t
+ vi
Solution: vf =
m
5.5 N &quot; s
=
+ 0 m/s
0.12 kg
!
vf = 46 m/s
Statement: The speed of the puck just after it leaves the stick is 46 m/s.
9. (a) Given: m = 225 g = 0.225 kg; ∆y = 74 cm = 0.74 m
!
Required: p
Analysis: The potential energy as you drop the phone is equal to the kinetic energy when it hits
1
1
the ground. Use k = g , Ek = mv 2 and Eg = mg!y to obtain mv 2 = m ! . Isolate v in this
2
2
!
!
! . Calculate the speed, and then use p = mv to calculate the momentum.
equation: v =
Solution: Calculate the speed when the phone hits the ground.
v= 2 !
= 2(9.8 m/s 2 )(0.74 m)
v = 3.81 m/s (one extra digit carried)
Calculate the momentum.
!
!
p = mv
= (0.225 kg)(3.81 m/s [down])
!
p = 0.86 kg ! m/s [down]
Statement: The cellphone’s momentum at the moment of impact is 0.86 kg&middot;m/s [down].
(b) The surface of the impact makes no difference to the momentum. The speed and mass do not
change, so the momentum does not change.
!
10. (a) Given: m = 0.25 kg; ∆y = 1.5 m; vf = 4.0 m/s [up]
!
Required: impulse of the ball, !p
Analysis: The potential energy as you drop the ball is equal to the kinetic energy when it hits the
1
1
ground. Use k = g , Ek = mv 2 and Eg = mg!y to obtain mv 2 = m ! . Isolate v in this
2
2
!
! !
! . Calculate the velocity, and then use !p = m( vf + vi ) to calculate the
equation: v =
impulse imparted by the floor to the ball.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.1-5
Solution: Calculate the velocity of the ball just before it hits the ground.
v = 2g!y
= 2(9.8 m/s 2 )(1.5 m)
v = 5.42 m/s (one extra digit carried)
!
The velocity is downward, so vi = 5.42 m/s [down] .
!
! !
!p = m( vf &quot; vi )
= (0.25 kg)(4.0 m/s [up] &quot; 5.42 m/s [down])
= (0.25 kg)(4.0 m/s [up] + 5.42 m/s [up])
= 2.36 N #s [up] (one extra digit carried)
!
!p = 2.4 N #s [up]
Statement: The impulse imparted!by the floor to the ball is 2.4 N&middot;s [up].
!
(b) Given: !p = 2.36 N &quot; s [up] ; F = 18 N [up]
Required: ∆t
!
!
!
!
Analysis: ! = ! , so ! = ! .
!
!
Solution: ! = !
2.36 N &quot; s
18 N
! = 0.13 s
Statement: The ball is in contact with the floor for 0.13 s.
!
11. (a) Given: m = 0.030 kg; v = 88 . [forward]
!
Required: !p
!
!
Analysis: The initial velocity of the arrow is 0 m/s. Use !p = m!v .
!
!
Solution: !p = m!v
=
= (0.030 kg)(88 m/s [forward])
= 2.64 N &quot;s [forward] (one extra digit carried)
!
!p = 2.6 N &quot;s [forward]
Statement: The impulse is 2.6 N&middot;s [forward].
!
(b) Given: ∆t = 0.015 s; !p = 2.64 N &quot; s [forward]
!
Required:
!
! !!
!
Analysis: ! = ! , so =
.
!
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.1-6
!
!
Solution: =
!
2.64 kg &quot; m/s [forward]
=
0.015 s
!
2
= 1.8 # 10 N [forward]
!
Statement: The average force of the bowstring on the arrow is 1.8 ! 102 N [forward] .
!
!
12. (a) Given: vi = 63 m/s [W] ; m = 0.057 kg; vf = 41 m/s [E]
!
Required: !p
!
! !
Analysis: !p = m( vf + vi )
!
! !
Solution: !p = m( vf &quot; vi )
= (0.057 kg)(41 m/s [E] &quot; 63 m/s [W])
= (0.057 kg)(41 m/s [E] + 63 m/s [E])
!
!p = 5.93 N # s [E] (one extra digit carried)
Statement: The magnitude of the impulse is 5.9 N&middot;s.
!
(b) Given: !p = 5.93 N &quot; s [E] ; ∆t = 0.023 s
!
Required:
!
! !!
!
Analysis: ! = ! , so =
.
!
! !!
Solution: =
!
5.93 N &quot; s [E]
=
0.023 s
!
= 2.6 # 102 N [E]
Statement: The average force is 2.6 ! 102 N [E] .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.1-7
Section 5.2: Conservation of Momentum in One Dimension
Tutorial 1 Practice, page 231
!
!
1. Given: m1 = 1350 kg; v1 = 72 km/h [S] ; m2 = 1650 kg; vf = 24 km/h [S]
!
Required: v2
Analysis: The cars stick together after the collision, so they can be treated as a single object with
!
!
mass m1 + m2, velocity vf , and momentum pf .
pf = (m1 + m2 )vf
By conservation of momentum,
!
! !
pf = p1 + p2
!
!
and p = mv ; therefore,
!
! !
pf = p1 + p2
!
!
!
(m1 + m2 )vf = m1v1 + m2 v2
!
!
! (m1 + m2 )vf ! m1v1
v2 =
m2
!
!
! (m1 + m2 )vf ! m1v1
Solution: v2 =
m2
=
(1350 kg + 1650 kg )(24 km/h [S]) ! (1350 kg )(72 km/h [S])
1650 kg
!
v2 = !15 km/h [S]
!
v2 = 15 km/h [N]
Statement: The initial velocity of the second car is 15 km/h [N].
!
!
2. Given: m1 = 28 g = 0.028 kg; v1 =
m/ [ orward] ; v2 =
1 ma sward]
Required: m2
Analysis: The momentum of the arrow is the opposite of the momentum of the archer.
!
!
m v = !m v
!
mv
m2 = ! !
v
!
mv
Solution: m2 = ! !
v
1 orward]
1 ma sward]
=!
=!
1
!
1
orward]
orward]
m = ff
Statement: The mass of the archer and the bow is 6.6 ! 101 kg.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.2-1
Section 5.2 Questions, page 232
1. The total momentum of a system is conserved if there is no net force applied on the system.
2. Given: mass of student and surfboard, m = 59.6 kg; mass of student, m2 = 55 kg; velocity of
!
surfboard relative to water, v = 2.0 m/s [E]; velocity of student relative to surfboard,
!
v2 = 1.9 m/s [E]
!
Required: resultant velocity, v
!
Analysis: The momentum of the student and surfboard before the student starts walking is m v .
The momentum of the system after the student starts walking is the momentum of the student,
!
!
m2 v2 , plus the momentum of the student and surfboard, m v .
!
!
!
mv = m v + mv
!
!
! mv ! m v
v=
m
!
!
! mv ! m v
Solution: v =
m
=
59.6 kg )(2.0 m/s [E]) ! (55 kg )(1.9 m/s [E])
59.6 kg
!
v = 0.25 m/s [E]
Statement: The final velocity of the surfboard is 0.25 m/s [E].
3. Given: m1 = 35.6 kg; v1 = 2.42 m/s; v2 = 3.25 m/s
Required: m2
Analysis: Assume the first hockey player is moving left after the collision and the second player
is moving right. The momentum before the push is zero, so the total momentum after the push is
also zero.
!
!
m1v1 + m2 v2 = 0
!
m1v1
m2 = ! !
v2
!
m1v1
Solution: m2 = ! !
v2
=!
=!
(
(
m/s [
m/ [
]
kg)(!
m/
]
m/
[
[
]
])
m2 =
kg
Statement: The mass of the other player is 26.5 kg.
!
4. Given: m1 = 0.14 kg; v1 = 50 m/s [forward] ; m2 = 80 kg
!
Required: v2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.2-2
Analysis: The momentum before the throw is zero, so the total momentum after the throw is also
zero.
!
!
m1v1 + m2 v2 = 0
!
m1v1
!
v2 = !
m2
!
m1v1
!
Solution: v2 = !
m2
=!
1
(
&quot;
=! &quot;
!
v = &quot;
&quot;
1
orward]
orward]
1 ma sward]
Statement: The recoil velocity of the pitcher is 9 &quot; 10&quot;2 m/s [ma sward] .
!
!
!
5. Given: m1 = 4.5 kg; m2 = 6.2 kg; v =
1 9] ; v = 0 m/ ; vf = 10 m/s [E]
2
2
!
Required: vf
1
Analysis: Momentum is conserved.
!
!
!
!
m1v + m2 v = m1v + m2 v
1
2
1
2
!
!
!
m
v
v
+
m
! m2 v
!
1 1
2 2
2
v =
1
m1
!
!
!
m1vf + m2 vf ! m2 vf
!
1
2
2
Solution: vf =
1
m1
=
(4.5 kg )(16 m/s [E]) + (6.2 kg )(0) ! (6.2 kg )(10 m/s [E])
!
vf = 2.2 m/s [E]
4.5 kg
1
Statement: The final velocity of the smaller object is 2.2 m/s [E].
!
!
6. Given: m1 = m m2 = [m vf = 3v v = !v
1
2
! !
Required: vf ; vf
1
2
Analysis: Use conservation of momentum.
Solution: Assume the lighter mass is initially moving to the right. Consider the conservation of
momentum.
!
!
!
!
m1vf + m2 vf = m1vf + m2 vf
1
2
1
2
!
!
m ([v) + [m ( v) = mvf + mvf
1
2
!
!
0 = vf + vf
1
2
!
!
vf = ! vf
1
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.2-3
Statement: The final speed of the larger mass will be three times the final speed of the smaller
mass, and in the opposite direction.
7. Given: m1 = 2.5 kg; m2 = 7.5 kg; v1 f = +6.0 m/s ; v2 f = &quot;15 m/s
!
Required: vf
Analysis: The two objects stick together after the collision, with a mass of m1 + m2 . Momentum
is conserved.
!
!
!
m1vf + m2 vf = (m1 + m2 )vf
1
2
!
!
! m1vf1 + m2 vf2
vf =
m1 + m2
!
!
! m1vf1 + m2 vf2
Solution: vf =
m1 + m2
=
(2.5 kg )(6.0 m/s) + (
2.5 kg +
kg )(&quot;15 m/s)
kg
!
vf = &quot;9.8 m/s
Statement: The final velocity of the two objects is 9.8 m/s [left].
8. Answers may vary. Sample answer: The astronaut can throw the tool bag in a direction away
from the International Space Station. Conservation of momentum means she will move in the
opposite direction, toward the ISS.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.2-4
Section 5.3: Collisions
Mini Investigation: Newton’s Cradle, page 234
A. In Step 2, releasing one end ball caused the far ball on the other end to swing out at the same
speed as the original ball, while the middle balls appeared to remain still. Changing the setup did
not change the outcome as long as all balls were touching. When the middle ball was removed,
momentum was not transferred all the way to the end of the line.
B. Yes, the collisions appear to conserve momentum, although the balls slow down after a while.
C. Yes, the collisions appear to conserve kinetic energy. The end ball moves at the same speed as
the beginning ball, so kinetic energy is conserved (ignoring external forces).
D. The device as a whole does not appear to conserve mechanical energy. Energy is lost in the
sound of the collisions, and friction between the balls and between the string and the supports.
Tutorial 1 Practice, page 236
!
!
1. Given: m1 = 3.5 kg; vi = 5.4 m/s [right] ; m2 = 4.8 kg; vi = 0 m/s
1
2
!
Required: vf
2
Analysis: The collision is perfectly elastic, which means that kinetic energy is conserved. Apply
conservation of momentum and conservation of kinetic energy to construct and solve a linearquadratic system of two equations in two unknowns. First, use conservation of momentum to
!
!
solve for vf in terms of vf , remembering that vi = 0 m/s . This is a one-dimensional problem,
2
1
2
so omit the vector notation for velocities, recognizing that positive values indicate motion to the
right and negative values indicate motion to the left.
m1v i + m2 v i = m1 vf + m2 v f
1
2
1
2
m1v i = m1 vf + m2 v f
1
vf =
1
m1v i ! m2 v f
1
2
2
m1
Then, substitute the resulting equation into the conservation of kinetic energy equation to solve
for the final velocity of ball 2.
Solution:
m1vi ! m2 vf
1
2
vf =
1
m1
1
=
vf =
(3.5 kg )(5.4 m/s) ! (4.8 kg )vf
2
3.5 kg
18.9 m/s ! 4.8vf
2
3.5
The conservation of kinetic energy equation can be simplified by multiplying both sides of the
!
equation by 2 and noting that vi = 0 m/s .
1
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.3-1
1
1
1
1
m1vi2 + m2 vi2 = m1vf2 + m2 vf2
1
2
1
2
2
2
2
2
m1vi2 = m1vf2 + m2 vf2
2
1
(3.5 kg )(5.4 m/s) = ( 3.5 kg )
2
2
(18.9 m/s ! 4.8vf )2
2
(3.5) 2
+ (4.8 kg )vf2
(3.5)(3.5)(5.4 m/s)2 = (18.9 m/s ! 4.8vf )2 + (3.5)(4.8)vf2
2
2
2
357.21 m /s = 357.21 m /s ! (181.44 m/s)vf + 23.04vf2 + 16.8vf2
2
2
2
2
2
0 = 39.84v ! (181.44 m/s)vf
2
f2
2
2
2
0 = vf (39.84vf ! 181.44 m/s)
2
2
The factor of vf means the equation has a solution vf = 0 m/s. This solution describes the
2
2
system before the collision. The equation has a second solution describing the system after the
collision.
0 = 39.84 v f ! 181.44 m/s
2
181.44 m/s
39.84
vf = 4.6 m/s
vf =
2
2
Statement: The final velocity of ball 2 is 4.6 m/s [right].
!
! !
2. Given: vi = v1 ; vi = 0 m/s ; m1 = m; m2 = m
1
2
Required: vf ; vf
1
2
1
1
1
1
m1 vi2 + m2 v i2 = m1 vf2 + m2 v f2
2
2
2
2
Analysis: m1v i + m2 v i = m1 vf + m2 v f ;
1
Solution:
2
1
1
2
m1 vi + m2 v i = m1v f + m2 v f
1
2
1
mv1 + m (0 m/s) = mv f + mvf
1
v1 = vf + vf
1
vf = v1 ! v f
2
2
1
2
2
2
2
1
1
1
1
1
m1v i2 + m2 vi2 = m1v f2 + m2 vf2
2
2
2
2
1
2
1
2
mv12 + m (0 m/s) 2 = mvf2 + m ( v1 ! v f ) 2
1
1
v = v + ( v1 ! v f )
2
1
2
f1
2
1
v = v + v ! 2 v1 vf + v f2
2
1
2
f1
2
1
0 = 2 v ! 2v1 vf
2
f1
1
1
1
0 = 2 vf ( v f ! v1 )
1
1
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.3-2
vf = 0 or vf = v1
1
1
The final speed of the first stone cannot be the same as its initial speed, so vf = 0. Substitute
1
vf = 0 in the equation for vf .
1
2
vf = v1 ! v f
2
1
= v1 ! 0
vf = v1
2
Statement: The final speed of the first stone is 0 m/s. The final speed of the second stone is v1.
Tutorial 2 Practice, page 238
!
!
1. Given: m1 = 4.0 kg; m2 = 2.0 kg; vi = 6.0 m/s [forward] ; vi = 0 m/s
1
2
!
Required: vf
!
!
! m1vi 1 + m2 vi 2
Analysis: Use vf =
.
m1 + m2
!
!
! m1vi 1 + m2 vi 2
Solution: vf =
m1 + m2
=
(4.0 kg )(6.0 m/s [forward]) + (2.0 kg )(0 m/s)
4.0 kg + 2.0 kg
!
vf = 4.0 m/s [forward]
Statement: The velocity of the balls is 4.0 m/s [forward] after the collision.
!
!
2. (a) Given: m1 = 2200 kg; vi = 60.0 km/h [E] ; m2 = 1300 kg; vi = 30.0 km/h [E]
1
2
!
Required: vf
!
!
! m1vi 1 + m2 vi 2
Analysis: Convert the velocities to metres per second and use vf =
.
m1 + m2
km 1000 m
1h
!
vi = 60.0
!
!
[E]
1
h
3600 s
1 km
!
vi = 16.7 m/s [E] (one extra digit carried)
1
km 1000 m
1h
!
vi = 30.0
!
!
[E]
2
h
3600 s
1 km
!
vi = 8.33 m/s [E] (one extra digit carried)
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.3-3
!
!
! m1vi 1 + m2 vi 2
Solution: vf =
m1 + m2
=
(2200 kg )(16.7 m/s [E]) + (1300 kg )(8.33 m/s [E])
2200 kg + 1300 kg
!
vf = 13.6 m/s [E] (one extra digit carried)
Statement: The final velocity of the vehicles is 14 m/s [E].
!
(b) Given: m1 = 2200 kg; m2 = 1300 kg; vf = 13.6 m/s [E]
!
Required: p
Analysis: The momentum before the collision is equal to the momentum after the collision. Use
the answer from (a) to determine the momentum after the collision.
!
!
p = (m1 + m2 )vf
!
!
Solution: p = (m1 + m2 )vf
= (2200 kg + 1300 kg)(13.6 m/s [E])
!
p = 4.8 ! 104 kg &quot; m/s
Statement: The momentum before and after the collision is 4.8 &times; 104 kg&middot;m/s.
!
(c) Given: m1 = 2200 kg; vi = 60.0 km/h [E] = 16.7 m/s [E]; m2 = 1300 kg;
1
!
!
vi = 30.0 km/h [E] = 8.33 m/s [E]; vf = 13.6 m/s [E]
2
Required: ∆Ek
Analysis: !E k = Ek &quot; E k ; Ek =
f
Solution:
!Ek = Ek &quot; Ek
f
i
1 2
mv
2
i
#1
&amp;
1
1
= (m1 + m2 )vf2 &quot; % m1vi2 + m2 vi2 (
1
2
2
2
\$2
'
1
1
= (2200 kg + 1300 kg)(13.6 m/s)2 &quot; [(2200 kg)(16.7 m/s)2 + (1300 kg)(8.33 m/s)2 ]
2
2
4
!Ek = &quot;2.8 ) 10 J
Statement: The decrease in kinetic energy is 2.8 &times; 104 J.
3. Given: m1 = 66 kg; ∆y = 25 m; m2 = 72 kg; vi = 0 m/s
2
Required: vf
Analysis: Use conservation of energy to determine the speed of the snowboarder at the bottom
of the hill.
1
m1 g !y = m1 vk2
2
!
!
! m1vi 1 + m2 vi 2
Then, use vf =
to calculate the final velocity of both people after the collision.
m1 + m2
1
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.3-4
Solution: m1 g !y =
1
m1 v i2
2
1
v i = 2 g !y
1
=
2(9.8 m/s 2 )( 25 m)
v i = 22.1 m/s (one extra digit carried)
!
!
m1vi + m2 vi
1
!
vf =
=
1
2
m1 + m2
(66 kg )(22.1 m/s) + (72 kg )(0 m/s)
66 kg + 72 kg
vf = 11 m/s
Statement: The final speed of each person after the collision is 11 m/s.
Section 5.3 Questions, page 239
(a) Since the boxes stick together after the collision, we know this is an inelastic collision.
Momentum is conserved in an inelastic collision. Momentum is always conserved if there are no
external forces acting on the system.
(b) Kinetic energy is not conserved in an inelastic collision. In an inelastic collision, some
kinetic energy is absorbed by one or both objects, causing the kinetic energy after the collision to
be less than the kinetic energy before the collision.
!
!
2. Given: m1 = 85 kg; m2 = 8.0 kg; vi = 0 m/s ; v = 3.0 m/s [forward]
2
f
Required: vi 1
Analysis: Use vf =
vf =
m1vi + m2 vi
1
2
m1 + m2
m1vi + m2 vi
1
1
2
m1 + m2
(m1 + m2 )vf = m1vi + m2 vi
1
2
m1vi = (m1 + m2 )vf ! m2 vi
2
(m1 + m2 )vf ! m2 vi
2
1
vi =
, rearranged to isolate vi .
m1
1
Solution: vi =
(m1 + m2 )vf ! m2 vi
=
2
m1
1
(85 kg + 8.0 kg )(3.0 m/s) ! (8.0 kg )(0 m/s)
85 kg
vi = 3.3 m/s
1
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.3-5
Statement: The speed of the skateboarder just before he landed on the skateboard was 3.3 m/s.
!
!
3. Given: mT = 3.0 kg; m1 = 2.0 kg; v = m/s ; vf = 2.5 m/s [S]
1
!
Required: vf 2
Analysis: Since the total mass is 3.0 kg, and the mass of the first cart is 2.0 kg, the mass of the
second cart, m2, is 1.0 kg. The carts are at rest before they are released, so the initial momentum
of the system is zero. The final momentum must equal the initial momentum.
!
!
p = mv
!
!
!
Solution: pf = m1vf + m2 vf
1
2
!
0 = (2.0 kg )(2.5 m/s [S]) + (1.0 kg )vf
2
!
vf = !5.0 m/s [S]
2
!
vf = 5.0 m/s [N]
2
Statement: The final velocity of the other cart is 5.0 m/s [N].
4. (a) Yes. Both momentum and kinetic energy are conserved in a perfectly elastic collision.
!
!
!
(b) Given: m1 = 85 kg; vi = 6.5 m/s ; m2 = 120 kg; vi = 0 m/s ; vf = !1.1 m/s
1
2
1
!
Required: vf
2
!
Analysis: Use the conservation of momentum equation to solve for vf .
2
!
!
!
!
m1vi + m2 vi = m1vf + m2 vf
1
2
1
2
!
!
!
!
Solution:
m1vi + m2 vi = m1vf + m2 vf
1
2
1
2
!
(85 kg )(6.5 m/s) + (120 kg )(0 m/s) = (85 kg )(!1.1 m/s) + (120 kg )vf
2
!
552.5 m/s = !93.5 m/s + 120vf
2
!
vf = 5.4 m/s
2
Statement: The final velocity of the second person is 5.4 m/s in the direction that the first person
was originally travelling.
5. (a) It is an inelastic collision because the two skaters stick together after the collision.
(b) Given: m1 = 95 kg; vi = 5.0 m/s; m2 = 130 kg; vi = 0 m/s
1
2
Required: vf
Analysis: Use vf =
Solution: vf =
=
m1vi + m2 vi
1
m1 + m2
m1vi + m2 vi
1
2
.
2
m1 + m2
(95 kg )(5.0 m/s) + (130 kg )(0 m/s)
95 kg + 130 kg
vf = 2.1 m/s
Statement: The final speed of the skaters is 2.1 m/s [initial direction of the first skater].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.3-6
!
!
6. Given: m1 = m2 = 1250 kg; vi = 12 m/s [E] ; vi = 12 m/s [W]
1
2
!
Required: vf
!
!
! m1vi 1 + m2 vi 2
Analysis: Use vf =
.
m1 + m2
!
!
! m1vi 1 + m2 vi 2
Solution: vf =
m1 + m2
=
=
(1250 kg)(12 m/s [E]) + (1250 kg)(12 m/s [W])
1250 kg + 1250 kg
(1250 kg )(12 m/s [E]) ! (1250 kg )(12 m/s [E])
1250 kg + 1250 kg
!
vf = 0 m/s
Statement: The velocity of both cars is 0 m/s after the collision. They come to a complete stop.
7. Answers may vary. Sample answer: No, if a moving object collides with a stationary object in
a perfectly elastic collision, is it not possible for both objects to be at rest after the collision.
Because kinetic energy is conserved in a perfectly elastic collision, and there was kinetic energy
before the collision, there must also be kinetic energy after the collision. Therefore, one of the
objects must be moving after the collision.
!
8. (a) Given: m1 = 1.3 &times; 104 kg; vi =
!
km/h [ ] ; m2 = 1.1 &times; 103 kg;
1
!
vi = 3.0 ! 101 km/h [N]
2
!
Required: vf
!
!
! m1vi 1 + m2 vi 2
Analysis: Convert the velocities to metres per second and then use vf =
.
m1 + m2
km 1000 m
1h
!
vi = 9.0 ! 101
!
!
[N]
1
h
3600 s
1 km
!
vi = 25 m/s [N]
1
km 1000 m
1
!
vi = 3.0 ! 101
!
!
2
1 km
!
vi = 8.33 m/s [N]
2
!
!
! m1vi 1 + m2 vi 2
Solution: vf =
m1 + m2
=
s
(1.3! 104 kg )(2.5 ! 101 m/s [N]) + (1.1! 103 kg )(8.33 m/s [N])
!
vf = 23.7 m/s [N]
1.3! 104 kg + 1.1! 103 kg
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.3-7
Convert back to kilometres per hour.
m
1 km
3600 s
!
vf = 23.7
!
!
s 1000 m
1h
!
vf = 85 km/h
Statement: The velocity of the vehicles after the collision is 85 km/h [N].
!
!
(b) Given: m1 = 1.3 &times; 104 kg; vi = 25 m/s [N] ; m2 = 1.1 &times; 103 kg; vf = 8.33 m/s [N] ;
1
2
!
v =
m/s [N]
Required: Ek ; Ek
i
Analysis: Ek =
f
1 2
mv
2
Solution:
1
Ek = (m1vi2 + m2 vi2 )
2
1
i
2
1
= [(1.3! 104 kg)(25 m/s)2 + (1.1! 103 kg)(8.33 m/s)2 ]
2
Ek = 4.1! 106 J
i
1
( m + m2 )v f2
2 1
1
= (1.3 ! 10 4 kg + 1.1 ! 10 3 kg)(23.7 m/s) 2
2
Ek = 3.96 ! 10 6 J (one extra digit carried)
Ek =
f
f
Statement: The total kinetic energy before the collision was 4.1 &times; 106 J, and the total kinetic
energy after the collision was 4.0 &times; 106 J.
(c) Given: Ek = 4.1 &times; 106 J; Ek = 3.96 &times; 106 J
i
f
Required: ∆Ek
Analysis: !E k = E k &quot; E k
f
Solution: !E k = E k &quot; Ek
f
i
i
= (3.96 #10 6 J) &quot; (4.1 # 10 6 J)
!E k = &quot;1.4 # 10 J
6
Statement: The decrease in kinetic energy during the collision is 1.4 &times; 105 J.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.3-8
Section 5.4: Collisions
Tutorial 1 Practice, page 243
!
!
1. Given: m1 = 80.0 g = 0.0800 kg; vi = 7.0 m/s [W] ; m2 = 60.0 g = 0.0600 kg; vi 2 = 0 m/s
1
! !
Required: vf ; vf
1
2
Analysis: Since the second ball is initially at rest in the head-on elastic collision, use the
&quot; m ! m2 % !
! 2m1 \$ !
!
!
simplified equations, vf = \$ 1
vi and vf = #
'
&amp; vi .
1
2
# m1 + m2 &amp; 1
&quot; m1 + m2 % 1
&quot; m1 ! m2 % !
!
Solution: vf 1 = \$
' vi
# m1 + m2 &amp; 1
&quot; 0.0800 kg ! 0.0600 kg %
=\$
' (7.0 m/s [W])
# 0.0800 kg + 0.0600 kg &amp;
!
vf = 1.0 m/s [W]
1
! 2m1 \$ !
!
vf = #
&amp; vi
2
&quot; m1 + m2 % 1
\$
!
2(0.0800 kg )
=#
&amp; (7.0 m/s [W])
&quot; 0.0800 kg + 0.0600 kg %
!
vf = 8.0 m/s [W]
1
Statement: The final velocity of ball 1 is 1.0 m/s [W], and the final velocity of ball 2 is
8.0 m/s [W].
!
2. Given: m1 = 1.5 kg; vi 1 = 36.5 cm/s [E] = 0.365 m/s [E] ; m2 = 5 kg;
!
vi = 42.8 cm/s [W] = !0.428 m/s [E]
2
! !
Required: vf 1 ; vf 2
&quot; m ! m2 % !
&quot; 2m2 % !
&quot; m ! m1 % !
&quot; 2m1 % !
!
!
Analysis: vf = \$ 1
vi + \$
vi ; vf = \$ 2
vi + \$
'
'
'
' vi
1
2
# m1 + m2 &amp; 1 # m1 + m2 &amp; 2
# m1 + m2 &amp; 2 # m1 + m2 &amp; 1
&quot; m1 ! m2 % !
&quot; 2m2 % !
!
vi + \$
Solution: vf 1 = \$
'
' vi
# m1 + m2 &amp; 1 # m1 + m2 &amp; 2
&quot; 1.5 kg ! 5 kg %
&quot;
2(5 kg ) %
=\$
' (0.365 m/s [E]) + \$
' (!0.428 m/s [E])
# 1.5 kg + 5 kg &amp;
# 1.5 kg + 5 kg &amp;
!
vf = !0.9 m/s
1
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.4-1
&quot; m ! m1 % !
&quot; 2m1 % !
!
vf = \$ 2
vi + \$
'
' vi
2
# m1 + m2 &amp; 2 # m1 + m2 &amp; 1
&quot; 5 kg ! 1.5 kg %
&quot; 2(1.5 kg ) %
=\$
' (!0.428 m/s [E]) + \$
' (0.365 m/s [E])
# 1.5 kg + 5 kg &amp;
# 1.5 kg + 5 kg &amp;
!
vf 2 = !0.06 m/s [E]
Statement: The final velocity of cart 1 is 90 cm/s [W], and the final velocity of cart 2 is
6 cm/s [W].
Tutorial 2 Practice, page 247
!
!
1. (a) Given: m1 = 1.2 kg; vi 1 = 3.0 m/s [iight] ; m2 = 1.2 kg; vi 1 = 3.0 r/s [m t] ;
!
vf = 1.5 m/s [right]
2
Required: x
Analysis: Use the conservation of momentum equation to determine the velocity of glider 1
during the collision, when glider 2 is moving at 1.5 m/s [right]. Rearrange this equation to
express the final velocity of glider 1 in terms of the other given values.
m1vi + m2 vi = m1vf + m2 vf
1
2
1
vf =
2
m1vi + m2 vi ! m2 vf
1
2
2
m1
Then, apply conservation of mechanical energy to determine the compression of the spring at
this particular moment during the collision. Consider right to be positive and left to be negative,
and omit the vector notation. Clear fractions first, and then isolate x.
m1vi + m2 vi ! m2 vf
1
2
2
Solution: vf 1 =
m1
1
=
( 1.2 kg )(3.0 m/s) + ( 1.2 kg )(!3.0 m/s) ! ( 1.2 kg )(1.5 m/s)
1.2 kg
vf = !1.5 m/s
1
1
1
1
1
m1vi2 + m2 vi2 = (m1vf2 + m2 vf2 ) + kx 2
1
2
1
2
2
2
2
2
m1vi2 + m2 vi2 ! (m1vf2 + m2 vf2 ) = kx 2
1
x=
1
2
2
m v + m v ! (m v + m2 vf2 )
2
1 i1
2
2 i2
2
1 f1
2
k
(1.2 kg)(3.0 m/s)2 + (1.2 kg)(!3.0 m/s)2 ! (1.2 kg)(!1.5 m/s)2 + (1.2 kg)(1.5 m/s)2
6.0 &quot; 104 N/m
x = 0.016 m
Statement: The compression of the spring when the second glider is moving at 1.5 m/s [right] is
1.6 cm.
=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.4-2
!
!
(b) Given: m1 = 1.2 kg; vi 1 = 3.0 m/s [ ight] ; m2 = 1.2 kg; vi 1 = 3.0 r/s [meft] ;
!
vf = 1.5 m/s ght]
2
Required: x
Analysis: At the beginning of the collision, as the gliders come together and the spring is being
compressed, glider 1 and glider 2 are moving at the same speed, in opposite directions.
Immediately after the collision, the gliders will reverse direction, but still have the same speed.
At the point of maximum compression of the spring, the two gliders will have the same velocity,
!
vf . Use the conservation of momentum equation to determine this velocity.
m1vi + m2 vi = (m1 + m2 )vf
1
2
vf =
m1vi + m2 vi
1
2
m1 + m2
Then, apply the conservation of mechanical energy to calculate the maximum compression of the
spring.
m[vl + m vl
[
Solution: vf =
m[ + m
1
=
.
s2 / + .
mi /.
mi +
mi /.!
s2 /
mi
vf = 3 s2
Now use the law of conservation of mechanical energy to determine the maximum compression
of the spring, using the fact that vf = 0 m/s.
1
1
1
1
m1vi2 + m2 vi2 = (m1 + m2 )vf2 + kx 2
1
2
2
2
2
2
m1vi2 + m2 vi2 = kx 2
1
2
x=
m1vi2 + m2 vi2
1
2
k
(1.2 kg)(3.0 m/s)2 + (1.2 kg)(!3.0 m/s)2
6.0 &quot; 104 N/m
x = 0.019 m
Statement: The maximum compression of the spring is 1.9 cm.
=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.4-3
!
!
2. Given: m1 = 4.4 &times; 102 kg; vi 1 = 3.0 m/s [E] ; m2 = 4.0 &times; 102 kg; vi 2 = 3.3 m/s [W] ;
∆x = 44 cm = 0.44 m
Required: k
Analysis: At the point of maximum compression of the spring, the two carts will have the same
!
velocity, vf . Use the conservation of momentum equation to determine this velocity.
m1vi + m2 vi = (m1 + m2 )vf
1
2
vf =
m1vi + m2 vi
1
2
m1 + m2
Then apply the conservation of mechanical energy to calculate the spring constant.
m1vi + m2 vi
1
2
Solution: vf =
m1 + m2
1
=
(4.4 ! 102 kg )(3.0 m/s) + (4.0 ! 102 kg )(&quot;3.3 m/s)
4.4 ! 102 kg + 4.0 ! 102 kg
vf = 0 m/s
Now use the law of conservation of mechanical energy to determine the spring constant, using
the fact that vf = 0.
1
1
1
1
m1vi2 + m2 vi2 = (m1 + m2 )vf2 + k(!x)2
1
2
2
2
2
2
m1vi2 + m2 vi2 = k(!x)2
1
2
k=
m1vi2 + m2 vi2
1
(!x)
2
2
(4.4 &quot; 102 kg)(3.0 m/s)2 + (4.0 &quot; 102 kg)(#3.3 m/s)2
=
(0.44 m)2
k = 4.3 &quot; 104 N/m
Statement: The spring constant is 4.3 &times; 104 N/m.
Section 5.4 Questions, page 248
1. Answers may vary. Sample answer: Elastic collision: No, it is not possible for two moving
masses to undergo an elastic head-on collision and both be at rest immediately after the collision.
In an elastic collision, kinetic energy is conserved. Therefore, if the objects were moving before
the collision, at least one of the objects has to be moving after the collision.
Inelastic collision: Yes, it is possible for two moving masses to undergo an elastic head-on
collision and both be at rest immediately after the collision. Kinetic energy is not conserved in an
inelastic collision. If the masses have equal but opposite momentum before the collision, then the
total momentum is zero. After the collision, they could both be at rest and still conserve
momentum.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.4-4
2. Answers may vary. Sample answer: The two curling stones have the same mass. In an elastic
collision, the momentum of the first object can transfer completely to the other object if the
objects have the same mass. Thus, the speed of the first object is zero, and the speed of the
second object is equal to the initial speed of the first object.
!
3. Given: m1 = 1.5 g = 0.0015 kg; m2 = 3.5 g = 0.0035 kg; vf = 12 m/s [right];
1
!
vi = 7.5 m/s [left]
1
! !
Required: vf ; vf
1
2
&quot; m ! m2 % !
&quot; 2m2 % !
&quot; m2 ! m1 % !
&quot; 2m1 % !
!
!
Analysis: vf = \$ 1
;
v
+
v
v
=
v
+
' i \$
' i
\$ m + m ' i 2 \$ m + m ' vi 1
f2
1
# m1 + m2 &amp; 1 # m1 + m2 &amp; 2
# 1
# 1
2&amp;
2&amp;
&quot; m1 ! m2 % !
&quot; 2m2 % !
!
v
+
Solution: vf 1 = \$
' i \$
' vi
# m1 + m2 &amp; 1 # m1 + m2 &amp; 2
&quot; 0.0015 kg ! 0.0035 kg %
&quot;
%
2(0.0035 kg )
(12
m/s)
+
=\$
'
\$
' (!7.5 m/s)
# 0.0015 kg + 0.0035 kg &amp;
# 0.0015 kg + 0.0035 kg &amp;
!
vf = !15 m/s
1
&quot; m ! m1 % !
&quot; 2m1 % !
!
vf = \$ 2
vi + \$
'
' vi
2
# m1 + m2 &amp; 2 # m1 + m2 &amp; 1
%
&quot; 0.0035 kg ! 0.0015 kg %
&quot;
2(0.0015 kg )
=\$
' (!7.5 m/s) + \$
' (12 m/s)
# 0.0015 kg + 0.0035 kg &amp;
# 0.0015 kg + 0.0035 kg &amp;
!
vf = 4.2 m/s
2
Statement: The velocity of particle 1 after the collision is 15 m/s [left]. The velocity of particle 2
after the collision is 4.2 m/s [right].
!
!
m/s [ ight] ; vf = 172 m/s [right]
4. Given: m1 = 2.67 kg; m2 = 5.83 kg; vf [ =
2
! !
Required: vf ; vf
1
2
Analysis: Consider right to be positive, and let the direction of the chunks’ final motion be
&quot; m ! m2 % !
&quot; 2m2 % !
!
positive. Use the final velocity equations, vf = \$ 1
vi + \$
'
' vi and
1
# m1 + m2 &amp; 1 # m1 + m2 &amp; 2
&quot; m ! m1 % !
&quot; 2m1 % !
!
vf = \$ 2
vi + \$
'
' vi . Then, solve the resulting linear system.
2
# m1 + m2 &amp; 2 # m1 + m2 &amp; 1
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.4-5
Solution:
&quot; m ! m2 % !
&quot; 2m2 % !
!
vf = \$ 1
v
+
' i \$
' vi
1
# m1 + m2 &amp; 1 # m1 + m2 &amp; 2
&quot; 2.67 kg ! 5.83 kg % !
&quot;
%!
2(5.83 kg )
v
185 m/s = \$
+
' i1 \$
' vi 2
2.67
kg
+
5.83
kg
+
5.83
kg
2.67
kg
#
&amp;
#
&amp;
&quot; !3.16 % !
&quot; 11.66 % !
185 m/s = \$
+
v
v
# 8.50 '&amp; i 1 \$# 8.50 '&amp; i 2
!
!
1572.5 m/s = –3.16vi + 11.66vi
1
2
&quot; m ! m1 % !
&quot; 2m1 % !
!
v =\$ 2
vi + \$
'
' vi
2
# m1 + m2 &amp; 2 # m1 + m2 &amp; 1
&quot; 5.83 kg ! 2.67 kg % !
&quot;
%!
2(2.67 kg )
172 m/s = \$
' vi 2 + \$
' vi 1
# 2.67 kg + 5.83 kg &amp;
# 2.67 kg + 5.83 kg &amp;
&quot; 5.83 kg ! 2.67 kg % !
&quot; 2(2.67 kg ) % !
172 m/s = \$
' vi 2 + \$
' vi 1
8.50
8.50
#
&amp;
#
&amp;
!
!
m/s = 3.16vi +
vi
2
1
Solve the linear system.
!
!
!3.16vi + 11.66vi = 1572.5 m/s quation 1
1
2
!
!
5.34vi + 3.16vi = 1462 m/s Equation 2
1
2
534
Multiply Equation 1 by
316
!
!
!5.34vi + 19.704vi = 2657.3 m/s Equation 1
1
2
!
!
5.34vi + 3.16vi = 1462 m/s Equation 2
1
2
!
22.864vi = 4119.3
2
= 180.2 m/s [right] (one extra digit carried)
!
vi = 1.80 &quot; 102 m/s [right]
2
!
Substitute vi 2 = 180.2 into Equation 2.
!
!
5.34vi + 3.16vi = 1462 m/s
1
2
!
5.34vi + (3.16)(180.2 m/s) = 1462 m/s
1
!
5.34vi = 892.568 m/s
1
!
vi = 167 m/s [right]
1
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.4-6
Statement: The initial velocity of the more massive chunk is 1.80 &times; 102 m/s [right], and the
initial velocity of the less massive chunk is 167 m/s [right]. (Since all initial and final velocities
are in the same direction, the more massive chunk overtakes the less massive one and imparts
some of its momentum to it.)
!
!
5. (a) Given: m1 = 0.84 kg; vi = 4.2 m/s [iight] ; m2 = 0.48 kg; vi = 2.4 m/s [left] ;
1
2
3
k = 8.0 &times; 10 N/m
! !
Required: vf ; vf
1
2
&quot; m ! m2 % !
&quot; 2m2 % !
&quot; m ! m1 % !
&quot; 2m1 % !
!
!
Analysis: vf = \$ 1
vi + \$
vi ; vf = \$ 2
vi + \$
'
'
'
' vi
1
2
# m1 + m2 &amp; 1 # m1 + m2 &amp; 2
# m1 + m2 &amp; 2 # m1 + m2 &amp; 1
&quot; m1 ! m2 % !
&quot; 2m2 % !
!
vi + \$
Solution: vf 1 = \$
'
' vi
# m1 + m2 &amp; 1 # m1 + m2 &amp; 2
%
&quot; 0.84 kg ! 0.48 kg %
&quot;
2(0.48 kg )
=\$
' (4.2 m/s) + \$
' (!2.4 m/s)
# 0.84 kg + 0.48 kg &amp;
# 0.84 kg + 0.48 kg &amp;
!
vf = !0.60 m/s
1
&quot; m ! m1 % !
&quot; 2m1 % !
!
vf = \$ 2
v
+
' i
\$
' vi
2
# m1 + m2 &amp; 2 # m1 + m2 &amp; 1
%
&quot; 0.48 kg ! 0.84 kg %
&quot;
2(0.84 kg )
=\$
' (!2.4 m/s) + \$
' (4.2 m/s)
# 0.84 kg + 0.48 kg &amp;
# 0.84 kg + 0.48 kg &amp;
!
vf = 6.0 m/s
2
Statement: The velocity of cart 1 after the collision is 0.60 m/s [left]. The velocity of cart 2 after
the collision is 6.0 m/s [right].
!
!
(b) Given: m1 = 0.84 kg; vi = 4.2 m/s [iight] ; m2 = 0.48 kg; vi = 2.4 m/s [left] ;
1
2
!
3
vf = 3.0 m/s ght] ; k = 8.0 &times; 10 N/m
1
Analysis: Use the conservation of momentum to determine the velocity of cart 2 during the
collision, when cart 2 is moving 3.0 m/s [right]. Rearrange the conservation of momentum
equation to express the final velocity of cart 2 in terms of the other given values.
m1vi + m2 vi = m1vf + m2 vf
1
2
vf =
1
2
m1vi + m2 vi ! m1vf
1
2
1
m2
Then, use conservation of mechanical energy to determine the compression of the spring at this
particular moment during the collision. Consider right to be positive, and omit the vector
notation.
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.4-7
Solution: vf 2 =
m1vi + m2 vi ! m1vf
1
2
1
m2
(0.84 kg)(4.2 m/s) + (0.48 kg)(!2.4 m/s) ! (0.84 kg)(
0.48 kg
vf = !
m/s
=
/s)
2
Now use the law of conservation of mechanical energy to determine the compression of the
spring. Clear fractions first, and then isolate x.
1
1
1
1
m1vi2 + m2 vi2 = (m1vf2 + m2 vf2 ) + kx 2
1
2
1
2
2
2
2
2
m1vi2 + m2 vi2 ! (m1vf2 + m2 vf2 ) = kx 2
1
1
2
2
m v + m v ! (m v + m2 vf2 )
2
1 i1
x=
2
2 i2
2
1 f1
2
k
(
=
kg)(4.2 m/s)2 + (
kg)(!2.4 m/s)2 ! (
kg)(3.0 m/s)2 + (
&quot; 103 N/m
kg)(!0.30 m/s)2
x = 0.036 m
Statement: The compression of the spring when cart 1 is moving at 3.0 m/s [right] is
3.5 &times; 10−2 m.
!
!
(c) Given: m1 = 0.84 kg; vi = 4.2 m/s [iight] ; m2 = 0.48 kg; vi = 2.4 m/s [left] ;
1
2
k = 8.0 &times; 103 N/m
Required: x
Analysis: At the point of maximum compression of the spring, the two carts will have the same
!
velocity, vf . Use the conservation of momentum equation to determine this velocity.
m1vi + m2 vi = (m1 + m2 )vf
1
2
vf =
m1vi + m2 vi
1
2
m1 + m2
Then use the law of conservation of mechanical energy to determine the maximum compression
of the spring.
m1vi + m2 vi
1
2
Solution: vf =
m1 + m2
1
(0.84 kg)(4.2 m/s) + (0.48 kg)(!2.4 m/s)
0.84 kg + 0.48 kg
vf = 1.8 m/s
=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.4-8
1
1
1
1
m1vi2 + m2 vi2 = (m1 + m2 )vf2 + kx 2
1
2
2
2
2
2
m1vi2 + m2 vi2 = (m1 + m2 )vf2 + kx 2
1
2
kx 2 = m1vi2 + m2 vi2 ! (m1 + m2 )vf2
1
x =
2
x=
2
m v + m v ! (m1 + m2 )vf2
2
1 i1
2
2 i2
k
m v + m2 vi2 ! (m1 + m2 )vf2
2
1 i1
2
k
(0.84 kg)(4.2 m/s)2 + (0.48 kg)(!2.4 m/s)2 ! (0.84 kg + 0.48 kg)(1.8 m/s)2
=
8.0 &quot; 103 N/m
x = 0.041 m
Statement: The maximum compression of the spring is 4.1 &times; 10−2 m.
!
6. (a) Given: m1 = 2.0 kg; m2 = 4.0 kg; θ = 60.0&deg;; length of two strings = 3.0 m; vi 1 = 0 m/s
! !
Required: vf ; vf
1
2
Analysis: Draw a diagram of the situation. Let the y = 0 reference point be the vertical position
of ball 1 before the collision. Let ∆y be the vertical height of ball 2 above ball 1. Let l2 be the
vertical distance of ball 2 from the post.
Use conservation of energy to find the velocity of ball 2 just before the collision, at y = 0. Then,
use the equations for perfectly elastic collisions to find the velocities of the balls just after the
collision.
&quot; m ! m2 % !
&quot; 2m2 % !
&quot; m2 ! m1 % !
&quot; 2m1 % !
!
!
Ek = Eg; vf = \$ 1
;
v
+
v
v
=
v
+
' i \$
' i
\$ m + m ' i 2 \$ m + m ' vi 1
f2
1
# m1 + m2 &amp; 1 # m1 + m2 &amp; 2
# 1
# 1
2&amp;
2&amp;
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.4-9
Solution: From the diagram, l2 = (3.0 m)cos 60&deg; = 1.5 m. Thus, ∆y = 1.5 m.
Use conservation of energy.
Ek = Eg
1
m v 2 = m2 g!y
2 2 i2
vi = 2g!y
2
= 2(9.8 m/s 2 )(1.5 m)
vi = 5.42 m/s (one extra digit carried)
2
!
Now, use the equations for perfectly elastic collisions, using the fact that vi 1 = 0 m/s .
&quot; m ! m2 % !
&quot; 2m2 % !
!
vf = \$ 1
v
+
' i \$
' vi
1
# m1 + m2 &amp; 1 # m1 + m2 &amp; 2
&quot; 2m2 % !
=\$
' vi
# m1 + m2 &amp; 2
&quot;
2(4.0 kg ) %
=\$
' (5.42 m/s)
# 2.0 kg + 4.0 kg &amp;
!
vf = 7.23 m/s (one extra digit carried)
1
&quot; m ! m1 % !
!
vf = \$ 2
' vi
2
# m1 + m2 &amp; 2
&quot; m ! m1 % !
=\$ 2
' vi
# m1 + m2 &amp; 2
&quot; 4.0 kg ! 2.0 kg %
=\$
' (5.42 m/s)
# 2.0 kg + 4.0 kg &amp;
!
vf = 1.81 m/s (one extra digit carried)
2
Statement: The speed of ball 1 is 7.2 m/s, and the speed of ball 2 is 1.8 m/s.
!
!
(b) Given: m1 = 2.0 kg; m2 = 4.0 kg; vf = 7.23 m/s ; vf = 1.81 m/s
1
2
Required: hmax 1; hmax 2
Analysis: Use conservation of energy, Eg = Ek.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.4-10
Eg = Ek
Solution:
1
m1 ghm
1
=
hm
1
=
=
hm
Eg = Ek
2
m2 ghm
2
=
hm
2
=
=
1
1
1
m1 vf2
1
2
2
vf
1
2g
(
m/s ) 2
2(
m/ s 2 )
=
m
2
1
m v2
2 2 f2
vf2
2
2g
(1.81 m/s ) 2
2(
m/ s 2 )
hm 2 =
m
Statement: After the first collision, the maximum height of ball 1 is 2.7 m, and the maximum
height of ball 2 is 0.17 m.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.4-11
Section 5.5: Collisions in Two Dimensions: Glancing Collisions
Mini Investigation: Glancing Collisions, page 249
A. When one puck collides with a second puck at an angle, the speed of the second puck will be
less than the initial speed of the first puck, and as the angle of the collision increases, the speed
of the second puck will decrease.
B. There is less friction with pucks on an air table than with billiard balls or marbles, which will
affect the results. The results for pucks on an air table may be closer to those for an ideal,
frictionless system.
Tutorial 1 Practice, page 252
!
1. Given: Inelastic collision; m1 = 1.4 ! 104 kg; m2 = 1.5 ! 104 kg; vi = 45 km(h /N ;
1
!
vi = km(h /W
2
!
Required: vf
!
!
Analysis: According to the law of conservation of momentum, pT i = pT f . Since the initial
velocities are at right angles to each other, as shown in the figure below, you can calculate the
total velocity and momentum using the Pythagorean theorem and trigonometry:
p
p 2 = p12 + p22 , and tan ! = 2
p1
First, convert the velocities to metres per second.
km 1000 m
1h
!
v1 = 45
!
!
s
h
1 km
!
v1 = 12.5 m(s one e ra digit carried)
!
v2 =
!
v2 =
km 1000 m
1h
!
!
s
h
1 km
m(s two e ra digits carried)
Solution: Engine 1’s momentum is
!
!
p1 = m1v1
= 1.4 ! 104 kg) 12.5 m(s) /N6
!
p1 = 1.35 ! 105 kg ! m(s /N6 one e ra digit carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.5-1
Engine 2’s momentum is
!
!
p2 = m2 v2
= 1.5 ! 104 kg) 14.3 m(s) /W6
!
! 105 kg ! m(s /W6 two eMtra digits carried)
p2 =
Calculate the magnitude of the total momentum by applying the Pythagorean theorem:
p 2 = p12 + p22
p=
=
p12 + p22
1.35 &quot; 105 kg ! m(s)2 +
&quot; 105 kg ! m(s)2
p=
&quot; 105 kg ! m(s two eMtra digits carried)
Determine the direction by applying the tangent ratio:
p
tan ! = 2
p1
#p &amp;
! = tan &quot;1 % 2 (
\$ p1 '
#
) 105 kg * m(s &amp;
= tan %
(
5
\$ 1.35 ) 10 kg * m(s '
&quot;1
! = 52&deg;
The direction of the two engines is /N 52&deg; W6.
By conservation of momentum, the final total momentum of the engines must equal the initial
momentum. Since the collision is perfectly inelastic, both engines have the same final velocity:
!
!
!
pf = m1vf = m2 vf
!
!
pf = m1 = m2 )vf
!
pf
!
vf =
m1 = m2
=
&quot; 105 kg ! m(s /N 52&deg; W6
1.4 &quot; 104 kg = 1.5 &quot; 104 kg )
!
vf =
m(s /N 52&deg; W6
Statement: After the collision, the two engines are travelling together at a velocity of
9.3 m(s /N 52&deg; W6.
!
2. Given: m1 = 2 ! 10 kg; v1 = 2 ! 104 m(s /E6 ; m2 = 5 ! 10 kg;
!
v2 = ! 104 m(s /at right angle to star 16
!
Required: vf
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.5-2
Analysis: Assume that the direction of star 2 is north. According to the law of conservation of
!
!
momentum, pT = pT . Since the initial velocities are at right angles to each other, as shown in
i
f
the figure below, you can calculate the total velocity and momentum using the Pythagorean
theorem and trigonometry:
p
p 2 = p12 = p22 , and tan ! = 2
p1
Solution: Star 1’s momentum is
!
!
p1 = m1v1
= 2 ! 10 kg) 2 ! 104 m(s) /E6
!
p1 = 4 ! 10 kg ! m(s /E6
Star 2’s momentum is
!
!
p2 = m2 v2
= 5 &quot; 10 kg) &quot; 104 m(s) /N6
!
p2 = 1.5 &quot; 10 kg ! m(s /N6 one eMtra digit carried)
Calculate the magnitude of the total momentum by applying the Pythagorean theorem:
p 2 = p12 = p22
p=
=
p12 = p22
4 &quot; 10 kg ! m(s)2 = 1.5 &quot; 10 kg ! m(s)2
p=
&quot; 10 kg ! m(s one eMtra digit carried)
Determine the direction by applying the tangent ratio:
p
tan ! = 1
p2
# p &amp;
! = tan &quot;1 % 1 (
\$ p2 '
# 4 ) 10 kg * m(s &amp;
= tan &quot;1 %
(
\$ 1.5 ) 10 kg * m(s '
! = 14.9&deg; two eMtra digits carried)
The two stars’ direction is /N 10&deg; E6.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.5-
By conservation of momentum, the final total momentum of the stars must equal the initial
momentum. Since the collision is perfectly inelastic, both stars have the same final velocity:
!
!
!
pf = m1vf = m2 vf
!
!
pf = m1 = m2 )vf
!
pf
!
vf =
m1 = m2
! 10 kg ! m(s /N 10&deg; E6
2 &quot; 10 kg = 5 &quot; 10 kg)
=
!
vf = 2 &quot; 104 m(s /N 10&deg; E6
Statement: After the collision, the two stars are travelling together at a velocity of 2 ! 104 m(s,
at 10&ordm; to the initial path of the second star.
Section 5.5 Questions, page 253
!
!
!
1. Given: m1 = m2 = m; vi = 10.0 m(s /right6 ; vi = 0 m(s ; vf = 4.3 m(s ; θ
1
2
1
!
Required: vf
2
Analysis: Choose a coordinate system to identify directions: let positive x be to the right and
negative x be to the left. Let positive y be up and negative y be down.
Apply conservation of momentum independently in the x-direction and the y-direction to
determine the magnitude and direction of the final velocity of ball 2.
Solution: In the y-direction, the total momentum before and after the collision is zero:
pT = pT = 0
iy
fy
Therefore, after the collision:
mvf = f = 0
1y
2y
Divide both sides by m and substitute the vertical component of each velocity vector. The
vertical component of the velocity vector for ball 1 is vf sin ! . The vertical component of the
1
velocity vector for ball 2 is vf 2 sin !
vf sin &quot; = vf sin ! = 0
1
2
&deg;) = vf sin ! = 0
#4.3 m(s) sin
2
vf sin ! = 4.3 m(s) sin
2
&deg;)
In the x-direction, the total momentum before the collision is equal to the total momentum after
the collision. Only ball 1 has momentum in the x-direction before the collision, but both balls
have momentum in the x-direction after the collision.
mvi =
=
f
f
1
i1
1
=
f1
=
2
f2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.5-4
The horizontal component of the velocity vector of ball 1 after the collision is vf
1
horizontal component of the velocity vector of ball 2 after the collision is vf cos ! . The
2
horizontal component of the velocity vector of ball 1 before the collision is 10.0 m(s.
10.0 m(s = vf cos
&deg; = vf cos !
1
2
&quot; 1%
10.0 m(s = 4.3 m(s) \$ ' = vf cos φ
2
# 2&amp;
vf cos φ =
m(s
2
m(s
cos φ
Substitute this result into the previous result for vf :
vf =
2
2
vf sin ! = 4.3 m(s) sin
&deg;)
m(s)sin !
= 4.3 m(s) sin
cos!
&deg;)
2
sin !
4.3 m(s ) sin
=
cos!
m(s )
&deg;)
tan ! =
!= &deg;
Now, substitute to determine vf :
2
vf =
2
m(s
cos !
m(s
cos &deg;
vf =
m(s
=
2
Statement: The velocity of ball 2 after the collision is 8.3 m(s, 28&deg; above the horizontal
the initial path of ball 1).
!
!
2. Given: m1
m2 = 0.13 kg; vi = 2.0 m(s /E6 ; vi = 0 m(s ;
1
2
!
vf = 1.5 m(s /N &deg; E6 = 1.5 m(s /E 59&deg; N6
1
!
Required: vf
2
Analysis: Choose a coordinate system to identify directions: let positive x be to the right and
negative x be to the left. Let positive y be up and negative y be down.
Apply conservation of momentum independently in the x-direction and the y-direction to
determine the magnitude and direction of the final velocity of puck 2.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.5-5
Solution: In the y-direction, the total momentum before and after the collision is zero:
pT = pT = 0
iy
fy
Therefore, after the collision:
mvf = f = 0
1y
2y
Divide both sides by m and substitute the vertical component of each velocity vector. The
vertical component of the velocity vector for puck 1 is vf sin ! . The vertical component of the
1
velocity vector for puck 2 is vf sin φ .
2
vf sin &quot; = vf sin ! = 0
1
2
(1.5 m/s)(sin 59&deg;) = vf sin ! = 0
2
vf sin ! = (#1.5 m/s)(sin 59&deg;)
2
In the x-direction, the total momentum before the collision is equal to the total momentum after
the collision. Only puck 1 has momentum in the x-direction before the collision, but both pucks
have momentum in the x-direction after the collision.
mvi =
=
f
f
1
i1
1
=
f1
2
=
f2
The horizontal component of the velocity vector of puck 1 after the collision is vf cos 59&deg; . The
1
horizontal component of the velocity vector of puck 2 after the collision is vf cos φ . The
2
horizontal component of the velocity vector of puck 1 before the collision is 2.0 m(s.
2.0 m/s = vf cos 59&deg; = v cos φ
1
s 1cos 59&deg; = v cos φ
s=1
v cos φ =
v =
s
s
cos φ
1one extra digit carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.5-
Substitute this result into the previous result for vf :
2
vf 2 sin &quot; = (!1.5 m/s)(sin 59&deg;)
(1.23 / ) in &quot;
= (!1.9
c &quot;
/ )( in
in &quot; (!1.9 / )( in
=
c &quot;
(1.23 / )
&deg;)
&deg;)
tan &quot; = !1.09
&quot;=!
&deg; (one extra digit carried)
Now, substitute to determine vf :
2
vf =
2
1.23 /
c φ
1.23 /
c
&deg;
v =
/
=
2
Statement: The velocity of puck 2 after the collision is 1.8 m(s /S 44&deg; E6.
!
!
3. (a) Given: m1 = m2 = m; vi = 20.0 m/s 59&deg; ; vi = 15 m/s 59&deg;
1
2
!
vf = 10.0 m/s 59&deg;
1
!
Required: vf
;
2
Analysis: Choose a coordinate system to identify directions: let positive x be to the right and
negative x be to the left. Let positive y be up and negative y be down.
Apply conservation of momentum independently in the x-direction and the y-direction to
determine the magnitude and direction of the final velocity of puck 2.
Solution: In the y-direction, the total momentum before and after the collision is equal:
pT = pT
iy
fy
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.5-3
Therefore, after the collision:
mvi =
=
=
i
f
f
1y
2y
i1 y
=
i2 y
1y
=
f1 y
2y
=
f2 y
Substitute the vertical component of each velocity vector. The vertical component of the initial
velocity vector for puck 1 is − vi sin 45&deg;. The vertical component of the initial velocity vector
1
for puck 2 is − vi sin 45&deg;. The vertical component of the final velocity vector for puck 1 is
2
!vf 1 sin 59 . The vertical component of the final velocity vector for puck 2 is !vf sin &quot; .
2
!vi sin 59&deg; + (!vi
1
in 49&deg;) = !v
2
v
1
in 59&deg; ! v
in &quot;
2
in &quot; = in 59&deg;(vi + vi ! v )
2
1
v =
1
in 59&deg;(vi + vi ! v )
1
2
1
in &quot;
2
=
2
/ + 19
in &quot;
in 59&deg;(20.0
/ ! 10.0
/ )
/ ) in 59&deg;
2
in &quot;
In the x-direction, the total momentum before the collision is equal to the total momentum after
the collision.
mvi =
=
=
i
f
f
v =
1
2
i1
=
i2
1
=
f1
=
(29
2
f2
The horizontal component of the initial velocity vector of puck 1 is vi cos 45&deg;. The horizontal
1
component of the initial velocity vector of puck 2 is !vi cos 45&deg;. The horizontal component of
2
the final velocity vector of puck 1 is !vf cos 45&deg;. The horizontal component of the final
1
velocity vector of puck 2 is vf cos φ.
2
vi cof 59&deg; = (!vi cof 59&deg;) = !v cof 59&deg; = v cof &quot;
1
2
1
v =
2
(cof 59&deg;)(vi ! vi = v )
1
=
v =
2
2
1
cof &quot;
2
(cof 59&deg;)(20.0
(19
/f ! 19
cof &quot;
/f = 10.0
/f)
/f)(cof 59&deg;)
cof &quot;
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.5-8
Substitute this result into the previous result for vf :
2
(cof 59&deg;)(19
cofφ
/f)
=
(29
/f)(fin 59&deg;)
fin φ
2
2
( 19 /f) (29 /f)
2
2
=
cofφ
fin φ
fin φ 29 /f
=
cofφ 19 /f
9
tan φ =
3
φ= &deg;
Now, substitute to determine vf :
2
(15 m/s)(cof 59&deg;)
2
cof &deg;
v = 21 /f
vf =
2
Statement:
(b) The collision is non-perfectly inelastic, because kinetic energy is not conserved, but the
pucks do not move together after the collision.
!
!
4. Given: m1 = 1.4 ! 10 kg; m2
! 104 kg; vi = 32
; vi =
/
/
1
2
!
Required: vf
Analysis:
!
inelastic collision. Use the Pythagorean theorem and trigonometry to determine pf . Then, use
!
!
!
pf = mvf to determine vf .
First, convert the speeds to metres per second.
1000
1
!
!
v1 = 32
f
1
v1 =
/f (t o extra digitf carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.5-9
v2 =
!
1000
!
1
f
1
v2 = 13.3 /f (one extra digit carried)
Solution: pf =
p12 = p22
= (m1v1 )2 = (m2 v2 )2
=
(1.5 ! 103 sg)(
p =
s
= 1
! 105 sg)(13.33 /f) 2
000 sg ! /f (t o extra digitf carried)
pf = (m1 = m2 )vf
vf =
=
pf
m1 = m2
000 sg ! /f
1.5 &quot; 103 sg =
&quot; 105 sg
v = 13 /f
Convert the speed to kilometres per hour:
f
1
v1 = 13
!
!
1
f 1000
v1 =
/
tan ! =
=
=
p1
p2
m1v1
m2 v2
(1.5 ! 103 sg )(
1
! 10 sg )(13.3
5
s
/f )
tan ! =
! = 2.1&deg;
The angle is 2.1&deg; south of east, so it is 90&deg; – 2.1&deg; or 88&deg; east of south.
Statement: The velocity of the vehicles after the collision is 43 km(h /S 88&ordm; E6.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.5-10
5. (a)
!
(b) Given: θ = 25.5&deg;; φ = −45.9&deg;; vi =
1
Required: vf ; vf
1
!
/f ; vi = 0 m(s
2
2
Analysis: The momentum before and after the collision is equal. The momentum before the
collision consists of the momentum of ball 1 only, which is in the x-direction. The momentum
after the collision consists of the momentum of both balls, in both the x-direction and the ydirection. Consider the momentum in the x-direction and the y-direction separately. Use
trigonometry to determine the components.
Solution: Consider momentum in the x-direction first.
p =p
ix
fx
mvi
1x
= mvf
1x
vi
1x
= vf
= vf
1x
= mvf
2x
2x
/f = v cof ! = v cof&quot;
1
2
/f = v cof(#
&deg;) = v cof 29.9&deg;
1
2
Now, consider momentum in the y-direction. There is no momentum in the y-direction before the
collision.
p =p
iy
fy
0 = mvf
1y
0 = vf
= vf
1y
= mvf
2y
2y
0 = vf sin ! = vf sin &quot;
1
2
0 = vf sin(#
1
v =
2
v =
2
&deg;) = v fin 29.9&deg;
#v fin(#
1
2
&deg;)
fin 29.9&deg;
v fin
&deg;
1
fin 29.9&deg;
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.5-11
Substitute this eMpression into the other equation for vf and vf :
2
/f = v cof(!
&deg;) = v cof 29.9&deg;
/f = v cof(!
&quot; v fin
&deg;%
&deg;) = \$ 1
' cof 29.9&deg;
# fin 29.9&deg; &amp;
1
2
1
v =
1
/f
1
Substitute again to determine vf :
2
vf =
vf sin
1
&deg;
fin 29.9&deg;
/f)fin
(
=
fin 29.9&deg;
v =
/f
2
&deg;
2
Statement:
!
!
&quot;21
&quot;21
6. (a) Given: p 1 = 7.8 ! 10 kg # m/s [E] ; pp2 = 3.9 ! 10 sg # /f
Required: θ
Analysis: This is like a perfectly inelastic collision, with the particles as the objects that collide,
and the nucleus as the objects sticking together, eMcept that the nucleus fires off in the opposite
direction. Use trigonometry to determine θ.
Solution: tan ! =
p
p
1
2
#p &amp;
! = tan &quot;1 % 1 (
\$ p2'
# 7.8 ) 10&quot;21 kg * m/s &amp;
= tan &quot;1 %
(
&quot;21
\$ 3.9 ) 10 sg * /f '
!= &deg;
Statement: The direction of the nucleus is /W 24&deg; N6.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.5-12
!
!
&quot;21
&quot;21
(b) Given: pp1 = 7.8 ! 10 kg # m/s [E] ; pp2 = 3.9 ! 10 sg # /f
Analysis: Use the Pythagorean theorem to determine the final momentum of the nucleus.
2
2
2
Solution: pn = pp = pp
1
pn =
2
p = pp2
2
p1
2
= (7.8 ! 10&quot;21 kg # m/s)2 = (3.9 ! 10&quot;21 sg # /f)2
! 10&quot;21 sg # /f
pn =
Statement: The final momentum is 8.5 ! 10−21 kg&middot;m(s /W 24&deg; N6.
(c) Given: pn =
! 10&quot;21 sg # /f ; m = 2.3 ! 10&quot; sg
Analysis: p = mv
Solution: pn = mvn
vn =
=
vn =
pn
m
! 10&quot;21 sg # /f
2.3! 10&quot;
! 109
sg
/f
Statement:
! 105 m(s /W 24&deg; N6.
7. Given: m1 = 1.3 ! 10−23 kg; vi = 2.2 km(s = 2.2 ! 10 m(s; m2
! 10−23 kg; vi = 0 m(s;
1
vf
θ = 52&deg;
2
!
Required: vf
2
2
Analysis: The momentum before and after the collision is equal. The momentum before the
collision consists of the momentum of ball 1 only, which is in the x-direction. The momentum
after the collision consists of the momentum of both balls, in both the x-direction and the
y-direction. Consider the momentum in the x-direction and the y-direction separately. Use
trigonometry to determine the components.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.5-
Solution: Consider momentum in the x-direction first.
pi = pf
x
x
m1vi
(1.7 ! 10
&quot;27
kg)(2.2 ! 10
1x
= m1vf = m2 vf
1x
/f) = (
3
v
v
1x
1x
(
=
! 10
&quot;
2x
sg)v
! 10&quot;
1x
=(
! 10&quot; sg)(930
sg )(2.2 ! 103
(
=
! 10&quot;
/f) &quot; (
! 10
&quot;
/f)(cof 92&deg;)
sg )(930
/f)(cof 92&deg;)
sg )
/f (t o extra digitf carried)
Now, consider momentum in the y-direction. There is no momentum in the y-direction before the
collision.
pi = pf
y
y
0 = m1vf = m2 vf
1y
2y
0 = m1vf = m2 vf sin !
1y
2
0 = (1.7 &quot; 10
v
v
1y
1y
=
&quot;27
1y
&quot; 10&quot;
&quot;(
sg )(930
&quot; 10
=&quot;
&quot; 10&quot; sg)(930
kg)vf = (
&quot; 103
&quot;
/f)(fin 92&deg;)
/f)(fin 92&deg;)
sg
/f (t o extra digitf carried)
Use the Pythagorean theorem to determine the final speed of the neutron and trigonometry to
determine its direction.
vf2 = vf2 = vf2
1
1x
1y
vf = vf2 = vf2
1
1
1y
= (
/f)2 = (
=
! 103
v =
/f
1
tan ! =
v
v
! 103
/f)2
/f
1y
1
\$v
! = tan &quot;1 &amp;
&amp;% v
\$
= tan &amp;
%
&quot;1
1y
1
'
)
)(
# 103
/f '
)
/f (
!= &deg;
Statement: The final velocity of the neutron is 1.9 km(s,
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.5-14
8. The statement should be rewritten as: “For a head-on elastic collision between two objects of
equal mass, the after-collision velocities of the objects are at an 180&deg; angle to each other.”
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 5: Momentum and Collisions
5.5-15
Section 6.1: Newtonian Gravitation
Tutorial 1 Practice, page 293
1. Given: m1 = 1.0 &times; 1020 kg; m2 = 3.0 &times; 1020 kg; Fg = 2.2 &times; 109 N; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: r
Gm1m2
Analysis: Fg =
;
r2
Gm1m2
r2 =
Fg
r=
Gm1m2
Fg
Solution: r =
Gm1m2
Fg
2
\$
'
&quot;11 N # m
1.0 ! 1020 kg 3.0 ! 1020 kg
6.67
!
10
&amp;
2 )
kg (
%
(
=
)(
( 2.2 ! 10
9
)
N)
r = 3.0 ! 1010 m
Statement: The distance between the two asteroids is 3.0 &times; 1010 m.
2. Given: m = 1.9 &times; 1027 kg; r = 7.0 &times; 107 m; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: gJupiter
Gm1m2
, then use F = ma to substitute
r2
for Fg with the mass of an object on the surface, m2, and the acceleration of the object, which will
be the magnitude of the gravitational field strength on the surface of Jupiter, gJupiter.
Gm1m2
Fg =
r2
Gm1 m 2
m2 a =
r2
Gm
g Jupiter = 2 1
r
Gm
Solution: g Jupiter = 2 1
r
2 '
\$
27
&quot;11 N # m
&amp; 6.67 ! 10
) 1.9 ! 10 kg
2
%
kg (
=
(7.0 ! 107 m )2
(
)
g Jupiter = 26 m/s 2
Statement: The magnitude of the gravitational field strength on the surface of Jupiter is 26 m/s2.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.1-1
3. (a) Given: mA = 40.0 kg; mB = 60.0 kg; mC = 80.0 kg; rAB = 0.50 m; rBC = 0.75 m;
G = 6.67 &times; 10–11 N⋅m2/kg2
!
Required: Fnet
Gm1m2
r2 !
Solution: Determine FAB .
Analysis: Fg =
Gm1m2
r2
GmA mB
FAB =
2
rAB
Fg =
2 &amp;
#
&quot;11 N i m
6.67
!
10
%
( 40.0 kg 60.0 kg
2
kg
\$
'
(
=
)(
)
(0.50 m )2
FAB = 6.403! 10&quot;7 N (two extra digits carried)
!
Determine FBC .
Gm1m2
r2
GmB mC
FBC =
2
rBC
Fg =
2 '
\$
&quot;11 N # m
6.67
!
10
&amp;
) 60.0 kg 80.0 kg
2
kg
%
(
(
=
)(
)
(0.75 m )2
FBC = 5.692 ! 10&quot;7 N (two extra digits carried)
!
!
!
Fnet = FAB + FBC
= 6.403! 10&quot;7 N [left] + 5.692 ! 10&quot;7 N [right]
= 6.403! 10&quot;7 N [left] &quot; 5.692 ! 10&quot;7 N [left]
!
Fnet = 7.1! 10&quot;8 N [left]
Statement: The net force acting on B is 7.1 &times; 10–8 N [left].
(b) Given: mA = 40.0 kg; mB = 60.0 kg; mC = 80.0 kg; rAB = 0.50 m; rBC = 0.75 m;
G = 6.67 &times; 10–11 N⋅m2/kg2
!
Required: Fnet
Analysis: Fg =
Gm1m2
; determine the angle using the inverse tan function.
r2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.1-2
!
Solution: Determine FAB .
Gm1m2
r2
GmA mB
FAB =
2
rAB
Fg =
2 '
\$
&quot;11 N # m
6.67
!
10
&amp;
) 40.0 kg 60.0 kg
kg 2 (
%
(
=
)(
)
(0.50 m )2
FAB = 6.403! 10&quot;7 N (two extra digits carried)
!
Determine FBC .
Gm1m2
r2
GmB mC
FBC =
2
rBC
Fg =
2 &amp;
#
&quot;11 N i m
6.67
!
10
%
( 60.0 kg 80.0 kg
kg 2 '
\$
(
=
)(
)
(0.75 m )2
FBC = 5.692 ! 10&quot;7 N (two extra digits carried)
!
!
!
Fnet = FAB + FBC
2
2
Fnet = FAB
+ FBC
=
(6.403! 10&quot;7 N )2 + (5.692 ! 10&quot;7 N )2
Fnet = 8.6 ! 10&quot;8 N
#F &amp;
! = tan &quot;1 % BC (
\$ FAB '
# 5.692 ) 10&quot;7 N &amp;
= tan %
&quot;7
\$ 6.403) 10 N ('
&quot;1
! = 42&deg;
Statement: The net force acting on B is 8.6 &times; 10–8 N [W 42&deg; S].
Tutorial 2 Practice, page 295
1. Given: r = 7.0 &times; 106 m; mwhite dwarf = 1.2 &times; 1030 kg; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: gwhite dwarf
Gmwhite dwarf
Analysis: g white dwarf =
r2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.1-3
Solution: g white dwarf =
=
Gmwhite dwarf
r2
2 '
\$
30
&quot;11 N # m
6.67
!
10
&amp;
) 1.2 ! 10 kg
%
kg 2 (
(
(7.0 ! 10
6
m)
)
2
g white dwarf = 1.6 ! 106 N/kg
Statement: The surface gravitational field strength of the white dwarf is 1.6 &times; 106 N/kg, which is
over 100 000 times that of Earth.
2. Given: r2 = 2rSaturn
Required: g2
Gm
Analysis: g 2 = 2
r2
Solution: g 2 =
=
Gm
r22
Gm
( 2r )
2
Saturn
1 ! Gm \$
= # 2 &amp;
4 &quot; rSaturn %
1
g
4 Saturn
Statement: The surface gravitational field strength would be one quarter of the old surface
gravitational field strength.
g2 =
Research This: Gravitational Field Maps and Unmanned Underwater Vehicles,
page 295
A. Sample answers: A gravitational field map describes the strength of the gravitational field at
points across Earth. The map is created by using satellites to detect fine density differences in the
crust, which cause increases or decreases in the gravitational force. This information can be used
by a UUV to detect where it is on the planet based on the gravitational force.
B. Diagrams may vary depending on the type of UUV chosen. Students should highlight the key
feature of the UUV they choose, such as the propulsion system (a propeller is most common), the
power source (battery powered), the navigation system, and the sensors, which will vary with
purpose of the UUV, but may include depths sensors, sonar, or sensors to measure concentration
of compounds in the water.
C. Answers may vary. Students reports should explain the how UUVs use gravitational field
maps to compare with measurements collected by the UUV on about the direction, angle, and
strength of Earth’s magnetic field at its position. Students may also discuss the usefulness of
navigation by magnetic fields because UUVs travel to far for remote control and do not have
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.1-4
Section 6.1 Questions, page 296
1. For your weight to be one half your weight on the surface, the magnitude of the gravitational
acceleration must be one half of g.
Gm
g=
r
g=
g=
Gm
r
(
Gm
r
)
r , or about 1.41rE. Therefore, the altitude above Earth’s
The altitude from Earth’s centre is
surface is 1.41rE – 1rE = 0.41rE.
2. Given: r = 5.3 &times; 10–11 m; m1 = 1.67 &times; 10–27 kg; m2 = 9.11 &times; 10–31 kg;
G = 6.67 &times; 10–11 N⋅m2/kg2
Required: Fg
Gm1m2
Analysis: Fg =
r2
Gm1m2
Solution: Fg =
r2
2 '
\$
&quot;27
&quot;11 N # m
kg 9.11! 10&quot;31 kg
6.67
!
10
&amp;
) 1.67 ! 10
2
kg (
%
=
(5.3! 10&quot;11 m )2
(
)(
)
Fg = 3.6 ! 10&quot;47 N
Statement: The magnitude of the gravitational attraction between the proton and the electron is
3.6 &times; 10–47 N.
3. (a) The value for r is squared in the denominator, so as r increases, the gravitational force
decreases.
(b)
Gm1m2
Fg =
r2
Gm1m2
=
2
( 4r1 )
=
Fg =
Gm1m2
16r12
1 ! Gm1m2 \$
16 #&quot; r12 &amp;%
The gravitational force changes by a factor of
Copyright &copy; 2012 Nelson Education Ltd.
.
Chapter 6: Gravitational Fields
6.1-5
4. (a) Given: m1 = 225 kg; d = 8.62 &times; 106 m; m2 = 5.98 &times; 1024 kg; rE = 6.38 &times; 106 m;
G = 6.67 &times; 10–11 N⋅m2/kg2
Required: Fg
Analysis:
Gm1m2
Fg =
r2
Gm1mE
Fg =
2
( d + rE )
Solution: Fg =
Gm1mE
(d + r )
2
E
2 '
\$
&quot;11 N # m
24
&amp; 6.67 ! 10
) 225 kg 5.98 ! 10 kg
2
kg (
%
(
=
)(
)
(8.62 ! 106 m + 6.38 ! 106 m )2
Fg = 399 N
Statement: The gravitational force is 399 N toward Earth’s centre.
(b) Given: m1 = 225 kg; d = 8.62 &times; 106 m; mE = 5.98 &times; 1024 kg; rE = 6.38 &times; 106 m;
G = 6.67 &times; 10–11 N⋅m2/kg2
Required: g
Gm
Analysis: g = 2
r
Gm
Solution: g = 2
r
GmE
=
2
( d + rE )
=
2 '
\$
&quot;11 N # m
24
6.67
!
10
&amp;
) 5.98 ! 10 kg
%
kg 2 (
(
)
(8.62 ! 106 m + 6.38 ! 106 m )2
g = 1.77 m/s 2
Statement: The resulting acceleration is 1.77 m/s2 toward Earth’s centre.
5. Given: gTitan = 1.3 N/kg; m = 1.3 &times; 1023 kg; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: r
Gm
Analysis: g = 2
r
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.1-6
Solution: g Titan =
Gm
r2
r=
Gm
g Titan
2
\$
'
&quot;11 N # m
1.3! 1023 kg
6.67
!
10
&amp;
2 )
kg (
%
(
=
)
(1.3 N/ kg )
r = 2.6 ! 106 m
Statement: The radius of Titan is 2.6 &times; 106 m.
6. Given: gE = 9.8 N/kg; g2 = 3.20 N/kg
Required: r
Gm
Analysis: Use the equation g = 2 to determine the change in r given the change in the value
r
of g.
3.20 N/kg
g
Solution: 2 =
g E 9.8 N/kg
g 2 16
=
g E 49
16
g
49 E
16 ! Gm \$
= # 2 &amp;
49 &quot; rE %
g2 =
=
g2 =
Gm
49 2
r
16 E
Gm
!7 \$
#&quot; 4 rE &amp;%
2
Statement: The acceleration due to gravity is 3.20 N/kg at
r from Earth’s centre, or 0.75rE
above Earth’s surface,
7. Given: mSun = 2.0 &times; 1030 kg; r = 1.5 &times; 1011 m; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: g
Gm
Analysis: g = 2
r
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.1-7
Solution: g =
=
Gm
r2
2 &amp;
#
&quot;11 N i m
30
6.67
!
10
%
( 2.0 ! 10 kg
\$
kg 2 '
(
)
(1.5 ! 1011 m )2
g = 5.9 ! 10&quot;3 N/kg
Statement: The gravitational field strength of the Sun at a distance of 1.5 &times; 1011 m from its
centre is 5.9 &times; 10–3 N/kg.
8. Let m1 be the larger mass, and let x be the distance from m1 to the location of zero net force.
Set the two gravitational field strengths equal to each other, and develop a quadratic equation.
Solve for x.
Gm1
Gm2
2
x
(r ! x)2
G (r ! x)2
G m2 2
x
m1
(r ! x)2
m2 2
x
m1
r 2 ! 2rx = x 2
m2 2
x
m1
&quot; m2 % 2
2
\$ 1! m ' x ! 2rx = r
#
1&amp;
0
x=
!b &plusmn; b2 ! 4ac
2a
&quot; m %
!(!2r) &plusmn; (!2r)2 ! 4 \$ 1! 2 ' ( r 2 )
m1 &amp;
#
=
&quot; m %
2 \$ 1! 2 '
m&amp;
#
1
2r &plusmn; 4r 2 ! 4r 2 + 4
=
m2 2
r
m1
&quot; m %
2 \$ 1! 2 '
m1 &amp;
#
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.1-8
m
r
m
2r &plusmn;
=
&quot; m %
\$# ! m '&amp;
r&plusmn; r
=
&quot; m %
\$# ! m '&amp;
&plusmn;
x=r
m
m
m
m
&quot; m %
\$# ! m '&amp;
Since the greater value will not be between the two masses but will be the other side of m2 from
m1 :
!
x=r
m
m
&quot; m %
\$# ! m '&amp;
&quot;
m %
'
\$ !
m '
\$
x= r
\$ &quot; m %'
\$ \$ ! ''
\$# #
m &amp; '&amp;
!
x=r
&quot;
m %
'
\$ +
m '
\$
\$
m '
'
\$ +
\$#
m '&amp;
m
m
&quot; m %&quot;
m %
!
+
\$
'
\$#
m '&amp; #
m &amp;
r
x=
+
m
m
The location of zero force is
r
m
+
m
Copyright &copy; 2012 Nelson Education Ltd.
from the larger object, m1.
Chapter 6: Gravitational Fields
6.1-9
9. (a) Given: m1 = 537 kg; mE = 5.98 &times; 1024 kg; r = 2.5 &times; 107 m; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: g
Gm
Analysis: g = 2
r
Gm
Solution: g = 2
r
Gm
= 2E
r
2 '
\$
24
&quot;11 N # m
) 5.98 ! 10 kg
&amp; 6.67 ! 10
2
%
kg (
=
( 2.5 ! 107 m )2
(
)
g = 0.64 m/s 2
Statement: The resulting acceleration is 0.64 m/s2 toward Earth’s centre.
(b) Given: m1 = 537 kg; mE = 5.98 &times; 1024 kg; r = 2.5 &times; 107 m; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: Fg
Gm1m2
Analysis: g =
r2
Gm1m2
Solution: g =
r2
Gm1mE
=
r2
2 '
\$
&quot;11 N # m
24
6.67
!
10
&amp;
) 537 kg 5.98 ! 10 kg
2
kg (
%
=
( 2.5 ! 107 m )2
= 340 N
g
(
)(
)
Statement: The gravitational force is 340 N toward Earth’s centre.
10. Given: r = 2.44 &times; 106 m; mMercury = 3.28 &times; 1023 kg; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: gMercury
GmMercury
Analysis: g Mercury =
r2
GmMercury
Solution: g Mercury =
r2
2 '
\$
&quot;11 N # m
23
6.67
!
10
) 3.28 ! 10 kg
&amp;
2
%
kg (
=
2
2.44 ! 106 m
(
(
)
)
g Mercury = 3.67 N/kg
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.1-10
Statement: The surface gravitational field strength on Mercury is 3.67 N/kg. The value provided
in Table 2 is 3.7 N/kg, which is the same the value that I calculated to two significant digits.
11. (a) Given: g = 5.3 N/kg; mE = 5.98 &times; 1024 kg; rE = 6.38 &times; 106 m; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: r
Gm
Analysis: g = 2
r
Gm
Solution: g = 2
r
r=
Gm
g
2
\$
'
&quot;11 N # m
5.98 ! 1024 kg
6.67
!
10
&amp;
2 )
kg (
%
(
=
)
(5.3 N/ kg )
r = 8.675 ! 106 m (two extra digits carried)
Calculate the altitude above Earth’s surface:
8.675 &times; 106 m – 6.38 &times; 106 m = 2.3 &times; 106 m
Statement: The altitude of the satellite is 2.3 &times; 106 m.
(b) Given: m1 = 620 kg; mE = 5.98 &times; 1024 kg; rsatellite = 8.675 &times; 106 m;
G = 6.67 &times; 10–11 N⋅m2/kg2
Required: Fg
Gm1m2
Analysis: g =
r2
Gm1m2
Solution: g =
r2
2 '
\$
&quot;11 N # m
24
&amp; 6.67 ! 10
) 620 kg 5.98 ! 10 kg
2
kg (
%
=
(8.675 ! 106 m )2
(
g
)(
)
= 3.3! 103 N
Statement: The gravitational force on the satellite is 3.3 &times; 103 N toward Earth’s centre.
12. The motion of the Moon depends on Earth’s mass and G through the universal law of
gravitation. Using data on the mass and orbital radius of the Moon and G, we can determine
Gm1m2
Earth’s mass using the universal law of gravitation, g =
, and the equation for centripetal
r2
m
=
acceleration,
, since the two forces are equal.
r
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.1-11
Fg = Fc
Gm1m2 mv 2
=
r
r2
GmEarth mMoon
mMoon v 2
=
r
r2
GmEarth
= v2
r
rv 2
mEarth =
G
13. From question 8, the location of zero force is
r
m
+
m
from the larger object, m1. r is the
centre to centre distance between the Moon and Earth.
Since mE = 5.98 &times; 1024 kg and mMoon = 7.36 &times; 1022 kg:
r
r0 =
m2
1+
m1
r
=
1+
7.36 ! 1022
5.98 ! 1024
r
1.111
r0 = 0.9r
=
The mass should be 0.9r from the centre of the Earth, or
Copyright &copy; 2012 Nelson Education Ltd.
r
from the centre of the Moon.
10
Chapter 6: Gravitational Fields
6.1-12
Section 6.2: Orbits
Mini Investigation: Exploring Gravity and Orbits, page 298
A. When I increase the size of the Sun, Earth’s orbit changes: the orbit is closer to the Sun.
B. The Moon is pulled out of Earth’s orbit and orbits the Sun. When you increase the size of
Earth, the Moon orbits closer and faster.
C. The orbital radius and orbital period decrease.
D. Answers may vary. Students should describe a system using the satellite and planet simulation
that includes a gravity assist by having the satellite use the planet’s gravity to change direction.
The scale is too small to notice changes in speed.
Tutorial 1 Practice, page 302
1. Given: r = 5.34 &times; 1017 m; v = 7.5 &times; 105 m/s; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: m
Analysis: Rearrange the equation for speed to solve for mass:
Gm
v=
r
Gm
v2 =
r
rv 2
m=
G
rv 2
Solution: m =
G
2
(
5.34 ) 1017 m ) ( 7.5 ) 105 m/s )
=
kg =
6.67 ) 10
11
m
2
s
kg 2
= m2
m = 4.5 ) 1039 kg
Statement: The mass of the black hole is 4.5 &times; 1039 kg.
2. Given: r = 2.28 &times; 1011 m; m = 6.42 &times; 1023 kg; Fg = 1.63 &times; 1021 N; mSun = 1.99 &times; 1030 kg;
G = 6.67 &times; 10–11 N⋅m2/kg2
Required: v; T
Analysis: v =
2! r
Gm
;T=
v
r
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.2-1
Solution: Determine the orbital speed of Mars:
v=
GmSun
r
\$
m
2'
#
#
m
kg
2
&amp;
)
&quot;11
30
s
&amp; 6.67 ! 10
) 1.99 ! 10 kg
2
%
(
kg
=
11
2.28 ! 10 m
(
)
= 2.4128 ! 104 m/s (two extra digits carried)
v = 2.41! 104 m/s
Determine the period:
2! r
T=
v
2! ( 2.28 &quot; 1011 m )
=
m
2.4128 &quot; 104
s
1 min
1h
1d
1y
= 5.937 &quot; 107 s &quot;
&quot;
&quot;
&quot;
60 s
60 min 24 h 365 d
T = 1.90 y
Statement: The speed of Mars is 2.41 &times; 104 m/s, and its period is 1.90 Earth years.
3. Given: d = 600.0 km = 6.000 &times; 105 m; rE = 6.38 &times; 106 m; mE = 5.98 &times; 1024 kg;
G = 6.67 &times; 10–11 N⋅m2/kg2
Required: v; T
Analysis: Determine the orbital radius, then use the value for r to calculate the speed, v =
Then use the equation for period, T =
Gm
.
r
2! r
.
v
r = d + rE
= 6.000 ! 105 m + 6.38 ! 106 m
r = 6.98 ! 106 m
Solution:
v=
=
Gm
r
(6.67 ) 10&quot;11 N # m 2 /kg 2 )(5.98 ! 1024 kg )
6.98 ! 106 m
v = 7.559 ! 103 m/s (one extra digit carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.2-2
T=
=
2! r
v
2! (6.98 &quot; 106 m )
7.559 &quot; 103 m /s
1 min
= 5801.9 s &quot;
60 s
T = 97 min
Statement: The speed of the satellite is 7.56 &times; 103 m/s, and the period of the satellite is 97 min.
4. Given: d = 25 m; rMoon = 1.74 &times; 106 m; m = 7.36 &times; 1022 kg; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: v
Gm
Analysis: v =
r
r = d + rMoon
= 25 m + 1.74 ! 106 m
r = 1.740 ! 106 m
Determine the orbital speed:
v=
Gm
r
\$
'
m
kg # 2 # m 2 )
&amp;
22
&quot;11
s
&amp; 6.67 ! 10
) 7.36 ! 10 kg
2
kg
%
(
=
6
1.740 ! 10 m
(
)
v = 1.7 ! 103 m/s
Statement: The orbital speed of the satellite is 1.7 &times; 103 m/s.
Research This: Space Junk, page 302
A. Air resistance will slow a satellite and cause it to slip into a lower orbit.
B. The satellite may hit other satellites or burn up in the atmosphere.
C. Sample answer: Yes. Different styles of rockets and boosters are being considered for space
missions. Another way to reduce space junk is to equip satellites with small boosters that would
enable them to fall to Earth once they have become obsolete.
D. Sample answer: Space junk is regularly falling to Earth. To speed up the removal of space
junk, specific missions to remove space junk can be undertaken. A proposed technology that
could help reduce the amount of space junk is a “laser broom.” A laser broom is a ground-based
laser beam that would heat space junk enough to cause it to break apart into much smaller pieces
or change direction and fall to Earth.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.2-3
Hi Amelia;
I was researching the Internet and found out that space junk is actually many types of debris.
According to the NASA Orbital Debris Space Program Office space junk is
• abandoned spacecraft or spacecraft parts that no longer work—these items float around in
space circling Earth until they fall back down or collide with other space junk
• upper stages of launch vehicles—these are parts of space shuttles that are fired off or ejected in
stages, usually the upper parts of the rocket which get ejected last and are trapped in Earth’s orbit
• solid rocket fuel—some space shuttles use solid rocket fuel for propulsion and some can be left
over after launch in the container in which it was sent up
• tiny flecks of paint—when spacecraft enters space, heat or collisions with small particles chip
paint from the surface of the spacecraft
I find it interesting that paint flecks are considered space junk, and despite their size they can
actually do quite a bit of damage when they strike objects. Space junk can orbit Earth at a speed
of more than 3.5 ! 104 km/h. If a speck of paint travelling at that speed hits a space station, it
can create a 0.6 cm diameter hole in the window of the space station. Hard to believe something
that small can cause a lot of damage!
Section 6.2 Questions, page 303
1. Natural satellites are natural objects that revolve around another body due to gravitational
attraction, such as the Moon in the Earth–Moon system. Artificial satellites are objects that have
been manufactured and intentionally placed in orbit by humans, such as the International Space
Station.
2. Microgravity is a more accurate term than “zero gravity” to describe what astronauts
experience on the International Space Station. Microgravity is one millionth the value of g.
3. GPS satellites are a network of 24 satellites that coordinate several of their signals at once to
locate objects on Earth’s surface.
4. (a) A geosynchronous orbit is an orbit at a location above Earth’s surface such that the speed
of an object in a geosynchronous orbit matches the rate of the Earth’s rotation.
(b) A satellite in geosynchronous orbit appears to pass through the same position in the sky at the
same time every day to an observer on Earth.
(c) A satellite in a geostationary orbit appears to remain in the same position in the sky to an
observer on Earth.
5. Given: T = 24 h; mE = 5.98 &times; 1024 kg; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: r
2! r
Analysis: Use the equation for period to isolate v, T =
. Then set the value for v equal to the
v
equation for v to isolate and solve for r, v =
Copyright &copy; 2012 Nelson Education Ltd.
Gm
.
r
Chapter 6: Gravitational Fields
6.2-4
But first convert 24 h to seconds:
60 s
24 h 60 min
T=
!
!
1h
1d
1 min
T = 86 400 s
2! r
v
2! r
v=
T
T=
Gm 2! r
=
r
T
Gm 4! 2 r 2
=
r
T2
GmT 2
= r3
4! 2
GmT 2
= r3
Solution:
2
4!
%
(
m
kg \$ 2 \$ m 2 *
'
2
s
' 6.67 &quot; 10#11
* 5.98 &quot; 1024 kg (86400 s )
'&amp;
*)
kg 2
3
r =
4! 2
r = 4.2 &quot; 107 m
Statement: The orbital radius of a satellite in geosynchronous orbit is 4.2 &times; 107 m.
6. (a) Given: T = 164.5 y; r = 4.5 &times; 109 km = 4.5 &times; 1012 m
Required: v
2! r
Analysis: Use the equation for period to isolate and solve for v, T =
. But first convert the
v
period to seconds:
365 d 24 h 60 min
60 s
T = 164.5 y !
!
!
!
1h
1y
1d
1 min
(
)
T = 5.188 ! 109 s (two extra digits carried)
2! r
v
2! r
v=
T
T=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.2-5
Solution: Determine the orbital speed of Neptune:
2! r
v=
T
2! ( 4.5 &quot; 1012 m )
=
5.188 &quot; 109 s
= 5.450 &quot; 103 m/s (two extra digits carried)
v = 5.5 &quot; 103 m/s
Statement: The orbital speed of Neptune is 5.5 &times; 103 m/s.
(b) Given: r = 4.5 &times; 109 km = 4.5 &times; 1012 m; v = 5.450 &times; 103 m/s;
G = 6.67 &times; 10–11 N⋅m2/kg2
Required: m
Analysis: Use the equation for speed to isolate and solve for m,
=
G
:
r
G
r
G
2
=
r
r 2
=
G
=
Solution:
=
r 2
G
2
(
4.5 ! 1012 m ) (5.450 ! 103 m/s )
=
kg #
6.67 ! 10&quot;11
m
2
s
kg 2
# m2
= 2.0 ! 1030 kg
Statement: The mass of the Sun is 2.0 &times; 1030 kg.
7. Given: T = 29 y; v = 9.69 km/s = 9.69 &times; 103 m/s
Required: r
2! r
Analysis: T =
, but first convert the period to seconds:
v
365 d 24 h 60 min
60 s
T = 29 y !
!
!
!
1h
1y
1d
1 min
= 9.145 ! 108 s (two extra digits carried)
T = 9.1! 108 s
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.2-6
Solution: Determine the orbital radius of Saturn:
2! r
T=
v
Tv
r=
2!
(9.145 &quot; 108 s )(9.69 &quot; 103 m/s )
=
2!
12
r = 1.4 &quot; 10 m
Statement: The orbital radius of Saturn is 1.4 &times; 1012 m.
8. (a) Given: r = 5.03 &times; 1011 m; mSun = 1.99 &times; 1030 kg; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: v
G
Analysis: =
r
Solution:
=
=
G
r
\$
'
m
kg # 2 # m 2 )
&amp;
&quot;11
30
s
&amp; 6.67 ! 10
) 1.99 ! 10 kg
%
(
kg 2
(
)
5.03! 1011 m
= 1.6244 ! 104 m/s (two extra digits carried)
= 1.62 ! 104 m/s
Statement: The speed of the asteroid is 1.62 &times; 104 m/s.
(b) Given: r = 5.03 &times; 1011 m; v = 1.6244 &times; 104 m/s
Required: T
2! r
Analysis: T =
v
2! r
Solution: T =
v
2! (5.03&quot; 1011 m )
=
m
1.6244 &quot; 104
s
1 min
1h
1d
1y
= 1.9456 &quot; 108 s &quot;
&quot;
&quot;
&quot;
60 s
60 min 24 h 365 d
T = 6.17 y
Statement: The period of the asteroid is 6.17 y.
9. Given: m = 1.99 &times; 1030 kg; r = 4.05 &times; 1012 m; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: v
G
Analysis: =
r
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.2-7
Solution:
=
G
r
'
\$
m
kg # 2 # m 2 )
&amp;
30
&quot;11
s
) 1.99 ! 10 kg
&amp; 6.67 ! 10
2
kg
(
%
=
12
4.05 ! 10 m
(
= 5725.8
)
60 s
60 min
1 km
m
!
!
!
1000 m
s 1 min
1h
= 2.06 ! 104 km/h
Statement: The orbital speed of the exoplanet is 2.06 &times; 104 km/h.
10. Given: r = 4.03 &times; 1011 m; T = 1100 d; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: m
Analysis: Use the equation for T to isolate and then solve for v, T =
for speed to isolate and solve for m,
T = 1100 d !
=
2! r
. Then use the equation
v
G
. But first, convert the period to seconds:
r
24 h 60 min
60 s
!
!
1d
1h
1 min
T = 9.504 ! 107 s (two extra digits carried)
2! r
v
2! r
v=
T
G
=
r
G
2
=
r
r 2
=
G
Solution: Determine the orbital speed of the exoplanet:
2! r
v=
T
2! ( 4.03&quot; 1011 m )
=
9.504 &quot; 107 s
v = 2.664 &quot; 104 m/s (two extra digits carried)
T=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.2-8
Determine the mass of the star:
r
=
G
&quot;
m%
4.03! 10 m \$ 2.664 ! 104
s '&amp;
#
=
m
kg ) 2 ) m 2
s
6.67 ! 10(11
kg 2
(
11
)
2
= 4.3! 1030 kg
Statement: The mass of the star is 4.3 &times; 1030 kg.
11. Given: r = 9.38 &times; 106 m; m = 6.42 &times; 1023 kg; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: T
Analysis: Use the equation for period, T =
T=
=
T=
2! r
. For v in the equation for period, use
v
=
G
.
r
2! r
v
2! r
Gm
r
2! r 3
Gm
Solution: T =
=
2! r 3
Gm
2!
(9.38 &quot; 106 m )3
%
m
kg \$ 2 \$ m 2
'
#11
s
' 6.67 &quot; 10
kg 2
&amp;
= 27584 s &quot;
(
*
23
* 6.42 &quot; 10 kg
)
(
)
1 min
1h
1d
&quot;
&quot;
60 s
60 min 24 h
T = 0.319 d
Statement: The period of Phobos is 0.319 days.
12. Given: T = 24 h; mE = 5.98 &times; 1024 kg; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: v
Analysis: Using the equations for period and speed, isolate r in each: T =
2! r
Gm
, v=
. Then
v
r
set the two equations equal to each other and solve for v.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.2-9
But first convert 24 h to seconds:
60 s
24 h 60 min
T=
!
!
1h
1d
1 min
T = 86 400 s
2! r
v
Tv
r=
2!
Gm
r
Gm
v2 =
r
Gm
r= 2
v
T=
v=
Tv Gm
= 2
2!
v
2! Gm
v3 =
T
Solution:
Tv Gm
= 2
2!
v
2! Gm
v3 =
T
%
(
m
kg \$ 2 \$ m 2 *
'
24
s
2! ' 6.67 &quot; 10#11
* 5.98 &quot; 10 kg
2
kg
&amp;
)
=
86 400 s
(
= 3073
)
m
60 s
60 min
1 km
&quot;
&quot;
&quot;
s 1 min
1h
1000 m
v = 1.11&quot; 104 km/h
Statement: The orbital speed of a satellite in geosynchronous orbit is 1.11 &times; 104 km/h.
13. (a) Given: mSun = 1.99 &times; 1030 kg; rMercury = 5.79 &times; 1010 m; rVenus = 1.08 &times; 1011 m;
rEarth = 1.49 &times; 1011 m; rMars = 2.28 &times; 1011 m; G = 6.67 &times; 10–11 N⋅m2/kg2
Required: vMercury; vVenus; vEarth; vMars
Analysis: v =
Gm
r
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.2-10
Solution: vMercury =
GmSun
rMercury
\$
m
2'
kg
#
#
m
2
&amp;
)
30
&quot;11
s
&amp; 6.67 ! 10
) 1.99 ! 10 kg
2
kg
%
(
=
10
5.79 ! 10 m
(
)
vMercury = 4.79 ! 104 m/s
vVenus =
GmSun
rVenus
\$
'
m
kg # 2 # m 2 )
&amp;
&quot;11
30
s
&amp; 6.67 ! 10
) 1.99 ! 10 kg
2
kg
%
(
=
11
1.08 ! 10 m
(
)
vVenus = 3.51! 104 m/s
vEarth =
GmSun
rEarth
\$
'
m
kg # 2 # m 2 )
&amp;
&quot;11
30
s
&amp; 6.67 ! 10
) 1.99 ! 10 kg
2
kg
%
(
=
11
1.49 ! 10 m
(
)
vEarth = 2.98 ! 104 m/s
vMars =
GmSun
rMars
\$
'
m
kg # 2 # m 2 )
&amp;
&quot;11
30
s
&amp; 6.67 ! 10
) 1.99 ! 10 kg
2
kg
%
(
=
11
2.28 ! 10 m
(
)
vMars = 2.41! 104 m/s
Statement: The orbital speed of Mercury is 4.79 &times; 104 m/s.
The orbital speed of Venus is 3.51 &times; 104 m/s.
The orbital speed of Earth is 2.98 &times; 104 m/s.
The orbital speed of Mars is 2.41 &times; 104 m/s.
(b) The farther a planet is from the Sun, the slower its orbital speed.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.2-11
14. Given: d = 410 km = 4.1 &times; 105 m; rMoon = 1.74 &times; 106 m; m = 7.36 &times; 1022 kg;
G = 6.67 &times; 10–11 N⋅m2/kg2
Required: v; T
2! r
Gm
Analysis: v =
;T=
. First, calculate the orbital radius:
v
r
r = d + rMoon
= 4.1! 105 m + 1.74 ! 106 m
r = 2.15 ! 106 m (one extra digit carried)
Solution:
v=
Gm
r
\$
m
2'
#
#
m
kg
2
&amp;
)
&quot;11
22
s
&amp; 6.67 ! 10
) 7.36 ! 10 kg
2
kg
%
(
=
6
2.15 ! 10 m
(
)
= 1.511! 103 m/s (two extra digit carried)
v = 1.5 ! 103 m/s
Determine the period:
2! r
T=
v
2! 2.15 &quot; 106 m
=
m
1.511&quot; 103
s
T = 8.9 &quot; 103 s
Statement: The speed of the satellite is 1.5 &times; 103 m/s, and its period is 8.9 &times; 103 s.
(
)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 6: Gravitational Fields
6.2-12
Section 7.1: Properties of Electric Charge
Mini Investigation: Observing Electric Charge, page 325
A. Each charged object attracts the pieces of paper differently because different objects accept
different amounts of extra electrons. Therefore, they vary in their total charge. Objects with
greater charge have a stronger attraction to the pieces of paper.
B. Some pieces of paper fall off the charged object after a short while because the charged object
polarizes the nearby pieces of paper. The side of the paper attracted to the charged object has a
charge opposite to that of the charged object. When the two are in contact, electrons move from
the charged object to the paper, causing it to have a neutral charge. Without the attraction to the
charged object, the pieces of paper fall away from the object.
C. Paper in contact with the charged conductor can take on some of the excess charge of the
conductor. The conductor has enough charge to neutralize the opposite charge on the side of the
paper next to the conductor and to place extra charge onto the paper. The paper thus acquires the
same kind of charge as the conductor. The like charges repel each other, causing the paper to
“jump” from the conductor.
Section 7.1 Questions, page 326
1. To determine the sign of the charge on an unknown object using a glass rod and a piece of
silk, I would first rub the glass rod with the piece of silk, giving it a negative charge. Then I
would place the silk close to the charged object. If the silk is pulled toward the object, then the
object has a positive charge. If the silk is pushed away from the object, the object has a negative
charge.
2. The electric force between two like charges is repulsive, and the electric force between unlike
charges is attractive. The object that has a charge attracts unlike charges and repels like charges
in the object with zero charge. Induced charge separation creates a slight charge on the ends of
the object; because the unlike charge is closest to the inducing charge, the two objects are
attracted to one another.
3. The negative charge on the balloons repels the electrons in the material on the wall, so that the
electrons are repelled from the wall’s surface. This resulting positive charge in that area holds the
negative charge on the balloons, causing the balloons to attach to the wall.
4. When the pellets of foam plastic touch the rod, the pellets become negatively charged. The
negatively charged pellets are repelled by the negatively charged rod, causing the pellets to jump
away from the rod. When they jump, they spark.
5. When two objects, such as a glass rod and a silk cloth, are rubbed together, protons cannot
move from one object to the other. Protons cannot move from one object to another because they
are tightly bound to the nucleus of the atom.
6. (a) The charge on my body and the charge on the carpet are equal and opposite. After walking
across the carpet, my body now has an excess of electrons because the electrons moved from the
carpet to me. Therefore, the charge on my body is negative, and the charge on the carpet is
positive.
(b) Walking across a carpet on a cold, dry day illustrates charging by friction.
7. It is a good precaution to touch the metal casing before handling the memory chip because
sparks from charge build-up on the student can damage the circuitry of the chip. Touching the
casing acts as an electrical ground.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.1-1
8. To give a neutral object a positive charge using only a negatively charged object, I would use
the process of charging by induction. By bringing the negatively charged object near the neutral
object, the object becomes polarized. Using a ground connection on the far side of the neutral
object removes electrons, so that when the negatively charged object is pulled away, the formerly
neutral object has a net positive charge.
9. (a) When a glass rod is rubbed with a wool rag, electrons move from the glass to the wool.
The glass rod becomes negatively charged, and the wool rag becomes positively charged.
(b) When a plastic rod is rubbed with a silk scarf, electrons move from the scarf to the rod. The
plastic rod becomes positively charged, and the silk scarf becomes negatively charged.
(c) When a platinum rod with a negative charge is touched with a similar rod with a positive
charge, electrons move from the negatively charged rod to the other rod until the charge on one
rod is balanced. One rod will become neutral. The charge of the other rod will have the same
charge as the rod with the greater charge at the beginning.
(d) When a small metal rod touches a large positively charged metal sphere, electrons move from
the rod to the sphere, balancing only some positive charge on the sphere. Both the metal rod and
sphere are positively charged.
10. Friction between the father’s skin and the wool sweater he is wearing gave a net charge to the
father. When the student and his father shook hands, the charge transferred to the student.
11. (a) Clothes rubbing against each other in the dryer transfer electrons, so that some clothes
have negative charge and others have positive charge. These unlike charges attract each other,
causing the clothes to cling together.
(b) Compounds in a fabric softener sheet break into negative and positive ions, which interact
with the charges on the clothes. This process easily removes the built-up charge.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.1-2
Section 7.2: Coulomb’s Law
Tutorial 1 Practice, page 332
1. Given: q1 = 1.00 &times; 10–4 C; q2 = 1.00 &times; 10–5 C; r = 2.00 m; k = 8.99 &times; 109 N⋅m2/C2
Required: FE
kq q
Analysis: FE = 12 2
r
kq q
Solution: FE = 12 2
r
2 &amp;
#
9 N&quot; m
)5
)4
% 8.99 ! 10
( (1.00 ! 10 C )(1.00 ! 10 C )
2
C '
\$
=
(2.00 m)2
FE = 2.25 N
Statement: The magnitude of the electric force between the two charges is 2.25 N.
2. Given: q1 = q; q2 = –2q; r12 = 1.000 m; FE13 + FE23 = 0; k = 8.99 &times; 109 N⋅m2/C2
Required: r13
kq q
Analysis: Use FE = 12 2 to develop a quadratic equation to solve for r13.
r
Solution: FE13 + FE23 = 0
FE13 = ! FE23
k q1 q3
2
13
r
=!
k q2 q3
r232
q1
q
= ! 22
2
r13
r23
q
!2q
=!
2
r13
(1.000 + r
13
(1+ r )
13
2
)
2
= 2r132
1+ 2r13 + r132 = 2r132
0 = r132 ! 2r13 ! 1
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.2-1
0 = r132 ! 2r13 ! 1
!b &plusmn; b2 ! 4ac
x=
2a
! ( !2 ) &plusmn; ( !2 ) ! 4 (1)( !1)
r13 =
2 (1)
2
r13 =
2&plusmn; 4+4
2
r13 =
2&plusmn;2 2
2
r13 = 1&plusmn; 2
Only the positive distance is necessary:
r13 = 1+ 2 m
r13 = 2.414 m
Statement: The third charge is 2.414 m to the left of q.
3. Given: q1 = +2.0 &micro;C = +2.0 &times; 10–6 C; d1 = 0 m; q2 = –3.0 &micro;C = –3.0 &times; 10–6 C;
d2 = 40.0 cm = 0.40 m; q3 = –5.0 &micro;C = –5.0 &times; 10–6 C; d3 = 120.0 cm = 1.20 m;
k = 8.99 &times; 109 N⋅m2/C2
Required: FEnet at q2
kq q
Analysis: FE = 12 2
r
Solution:
Determine the electric force due to q1:
kq q
FE12 = 12 2
r
2 &amp;
#
9 N&quot; m
)6
)6
% 8.99 ! 10
( (2.0 ! 10 C )()3.0 ! 10 C )
2
\$
C '
=
(0.40 m )2
FE12 = )0.3371 N (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.2-2
Determine the electric force due to q3:
kq q
FE23 = 32 2
r
2 &amp;
#
)6
)6
9 N&quot; m
8.99
!
10
%
( ()5.0 ! 10 C )()3.0 ! 10 C )
2
\$
C '
=
(1.20 m ) 0.40 m )2
FE23 = 0.2107 N (two extra digits carried)
Determine
!
! the!net force:
FEnet = FE12 + FE23
= 0.3371 N [left] + 0.2107 N [left]
!
FEnet = 0.55 N [left]
Statement: The force on the –3.0 &micro;C charge is 0.55 N [left].
Section 7.2 Questions, page 333
1. Given: FE1 = 0.080 N; r2 = 3r1
Required: FE2
Analysis: Determine how the force changes when the distance is tripled, then substitute for the
kq q
original force, FE = 12 2 .
r
kq q
Solution: FE2 = 12 2
r2
=
kq1q2
(3r1 )2
=
kq1q2
9r12
1 ! kq q \$
= # 12 2 &amp;
9 &quot; r1 %
1
F
9 E1
1
= (0.080 N)
9
FE2 = 8.9 ' 10(3 N
=
Statement: The new force is 8.9 &times; 10–3 N.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.2-3
2. Given: FE1 = 0.080 N; r2 = 3r1; q1B = 3q1A
Required: FE2
Analysis: Determine how the force changes when the distance and the charge are tripled, then
kq q
substitute for the original force; FE = 12 2 .
r
kq1Bq2
Solution: FE2 =
r22
=
k(3q1A )q2
(3r1 )2
=
3kq1A q2
9r12
1 ! kq q \$
= # 1A2 2 &amp;
3 &quot; r1 %
1
= FE1
3
1
= (0.080 N)
3
FE2 = 2.7 ' 10(2 N
Statement: The new force is 2.7 &times; 10–2 N.
3. Given: q1 = 1.6 &times; 10–19 C; q2 = 1.6 &times; 10–19 C; r = 0.10 nm = 1.0 &times; 10–10 m;
k = 8.99 &times; 109 N⋅m2/C2
Required: FE
kq q
Analysis: FE = 12 2
r
kq q
Solution: FE = 12 2
r
2 &amp;
#
)19
9 N&quot; m
)19
8.99
!
10
( (1.6 ! 10 C )(1.6 ! 10 C )
%
2
\$
C '
=
(1.0 ! 10)10 m )2
FE = 2.3! 10)8 N
Statement: The magnitude of the electric force between the two electrons is 2.3 &times; 10–8 N.
1
4. Given: r2 =
r1
1.50
Required: FE2
kq q
Analysis: FE = 12 2
r
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.2-4
kq1q2
r22
Solution: FE2 =
=
kq1q2
! r1 \$
#&quot;
&amp;
1.50 %
2
! kq q \$
FE2 = 2.25 # 12 2 &amp;
&quot; r1 %
FE2 = 2.25FE1
Statement: The magnitude of the electric force will increase by a factor of 2.25.
5. Given: q1 = 1.00 &micro;C = 1.00 &times; 10–6 C; q2 = 1.00 &micro;C = 1.00 &times; 10–6 C; m = 1.00 kg;
g = 9.8 m/s2; FE = Fg; k = 8.99 &times; 109 N⋅m2/C2
Required: Fg; r
kq q
Analysis: Rearrange the equation FE = 12 2 to solve for r. Then determine Fg.
r
kq q
FE = 12 2
r
r=
kq1q2
FE
Fg = mg
= (1.00 kg)(9.8 m/s 2 )
Fg = 9.8 N
Solution:
r=
=
kq0q2
F
#
% 8.99 !
\$
9
N &quot; m2 &amp;
((
2
'
!
)
)(
!
)
)
9.8 N
r=
m
Statement: The distance between the charges is 0.030 m.
6. (a) Given: m1 = 9.11 &times; 10–31 kg; m2 = 1.67 &times; 10–27 kg; r = 1.0 nm = 1.0 &times; 10–9 m;
G = 6.67 &times; 10–11 N⋅m2/kg2
Required: Fg
Gm1m2
Analysis: Fg =
r2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.2-5
Solution: Fg =
=
Gm1m2
r2
2
\$
&quot;11 N # m
6.67
!
10
&amp;
kg 2
%
'
&quot;27
&quot;31
kg )
) (9.11! 10 kg )(1.67 ! 10
(
(1.0 ! 10&quot;9 m )2
Fg = 1.0 ! 10&quot;49 N
Statement: The magnitude of the gravitational force between the electron and the proton is
1.0 &times; 10–49 N.
(b) Given: q1 = 1.6 &times; 10–19 C; q2 = 1.6 &times; 10–19 C; r = 1.0 &times; 10–9 m; k = 8.99 &times; 109 N⋅m2/C2
Required: FE
kq q
Analysis: FE = 12 2
r
Solution:
kq q
F = 02 2
r
2 &amp;
#
9 N&quot; m
)09
)( ! )09 )
(( !
% 8.99 !
2
'
\$
=
( ! )9 m )2
F =
! ) N
Statement: The magnitude of the electric force between the electron and proton is 2.3 &times; 10–10 N.
(c) If the distance were increased to 1.0 m, there would be no change because the ratios of the
forces are independent of the separation distance.
7. Given: q1 = q; q2 = 3q; r1 = 50; r2 = –40; FE13 = FE23
Required: r13
kq q
Analysis: Use FE = 12 2 to develop a quadratic equation to solve for r13. First determine r12:
r
r12 = 50– (–40) = 90
Solution:
FE13 = FE23
kq1 q3
r132
=
kq2 q3
r232
q1 q2
=
r132 r232
q
=
r132 (
3q
!r
)
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.2-6
(90 ! r )
2
13
= 3r132
8100 ! 180r13 + r132 = 3r132
0 = 2r132 + 180r13 ! 8100
0 = r132 + 90r13 ! 4050
!b &plusmn; b2 ! 4ac
2a
2
0 = r13 + 90r13 ! 4050
x=
! ( 90 ) &plusmn; ( 90 ) ! 4 (1)( !4050 )
r13 =
2 (1)
2
=
!90 &plusmn; 24300
2
=
!90 &plusmn; 90 3
2
r13 = !45 &plusmn; 45 3
Only the positive distance is necessary:
r13 = 45 ! 45 3
r13 = 33
x = –40 + 33
x = –7
Statement: The third charge is at x = –7.
8. Given: q1 = 2.0 &times; 10–6 C; q2 = –1.0 &times; 10–6 C; r12 = 10 cm = 0.10 m; FE13 + FE23 = 0
Required: r13
kq q
Analysis: Use FE = 12 2 to develop a quadratic equation to solve for r13.
r
Solution: FE13 + FE23 = 0
FE13 = ! FE23
kq1 q3
r132
=!
kq2 q3
r232
q
q1
= ! 22
2
r13
r23
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.2-7
2.0 ! 10&quot;6 C
&quot;1.0 ! 10&quot;6 C
=
&quot;
2
r132
(0.10 + r )
13
2
1
=
2
r13 ( 0.10 + r )2
13
2 ( 0.10 + r13 ) = r132
2
0.01+ 0.2r13 + 2r132 = r132
r132 + 0.2r13 &quot; 0.01 = 0
!b &plusmn; b2 ! 4ac
2a
2
r13 + 0.2r13 ! 0.01 = 0
x=
r13 =
r13 =
! ( !0.2 ) &plusmn;
( !0.2)2 ! 4(1)( !0.01)
2 (1)
0.2 &plusmn;
+
r13 =
2
&plusmn;
2
2
r13 =
&plusmn; 2
Only the positive distance is necessary:
r13 = 0.1 m + 0.1 2 m
r13 = 0.24 m
Statement: The third charge is 0.24 m, or 24 cm, beyond the smaller charge.
9. (a) Given: q = 7.5 &times; 10–6 C; L = 25 cm = 0.25 m; k = 8.99 &times; 109 N⋅m2/C2
Required: FEnet at q3
Analysis: Determine the distance between particles using the Pythagorean theorem. Then use
kq q
FE = 12 2 to determine the magnitude of the electric force between two particles.
r
Solution: Determine the distance:
r = (0.25 m)2 + (0.25 m)2
= 0.125 m 2
r = 0.3536 m (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.2-8
Determine the magnitude of the electric force:
kq q
FE = 12 2
r
2 &amp;
#
)6
)6
9 N&quot; m
8.99
!
10
%
( (7.5 ! 10 C )(7.5 ! 10 C )
2
\$
C '
=
(0.3536 m )2
FE = 4.044 N (two extra digits carried)
The x-components of the forces will add to zero, so calculate the y-components of the forces.
FEnet = 2FE sin 45&deg;
= 2(4.044 N)sin 45&deg;
FEnet = 5.7 N [down]
Statement: The net force on the charge at the bottom is 5.7 N [down].
(b) Given: q = 7.5 &times; 10–6 C; L = 0.25 m; k = 8.99 &times; 109 N⋅m2/C2; FE = 4.044 N
Required: FEnet at q2
kq q
Analysis: FE = 12 2
r
Solution: Determine the magnitude of the force between the two particles on the x-axis:
kq q
FE = 12 2
r
2 &amp;
#
9 N&quot; m
)6
)6
8.99
!
10
%
( (7.5 ! 10 C )(7.5 ! 10 C )
2
\$
C '
=
(0.25 m + 0.25 m )2
= 2.023 N (two extra digits carried)
FE = 2.0 N
Determine the x- and y-components of the diagonal force:
F x = F s 45&deg;
Fy=F
45&deg;
= (4.044 N)s 45&deg;
FEx = 2.860 N
= (4.044 N)
45&deg;
FEy = 2.860 N
2.860 N + 2.023 N = 4.883 N [to the right]
The vertical force is 2.860 N [up].
Determine the magnitude of the net force:
FEnet = (4.883 N)2 + (2.860 N)2
= 32.02 N 2
= 5.659 N (two extra digits carried)
FEnet = 5.7 N
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.2-9
Determine the direction of the net force:
2.860 N
tan ! =
4.883 N
# 2.860 N &amp;
! = tan &quot;1 %
\$ 4.883 N ('
! = 30&deg;
Statement: The net force on the charge on the right is 5.7 N [E 30&deg; N].
(c) Given: q = 7.5 &times; 10–6 C; L = 0.25 m; qe = 1.6 &times; 10–19 C; k = 8.99 &times; 109 N⋅m2/C2
Required: FEnet at origin
kq q
Analysis: FE = 12 2 ; the two charges on the x-axis have a net force of zero, so the only force is
r
an attractive force from the charge at the bottom.
kq q
Solution: FE = 12 2
r
2 &amp;
#
)6
)19
9 N&quot; m
8.99
!
10
%
( (7.5 ! 10 C )(1.6 ! 10 C )
2
C '
\$
=
(0.25 m )2
= 1.726 ! 10)13 N (two extra digits carried)
FE = 1.7 ! 10)13 N
Statement: The net force on the electron is 1.7 &times; 10–13 N [down].
10. Given: m = 5.00 g = 5.00 &times; 10–3 kg; L = 1.00 m; θ = 30.0&deg;; g = 9.8 m/s2;
k = 8.99 &times; 109 N⋅m2/C2
Required: q
Analysis: The electric, gravitational, and tension forces on the pith balls give a net force of zero.
Since the electric force is entirely horizontal and the gravitational force is entirely vertical, first
determine the gravitational force, Fg = mg. Then use trigonometry to determine the tension force
kq q
and the electric force. Use FE = 12 2 to determine the charge on each pith ball. Draw a sketch of
r
the situation.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.2-10
Solution: Determine the gravitational force on one pith ball:
Fg = mg
= (5.00 ! 10&quot;3 kg)(9.8 m/s 2 )
Fg = 0.490 N
Determine the force of tension on one pith ball:
! 30.0&deg; \$
Fg = FT cos #
&quot; 2 &amp;%
FT =
Fg
cos15.0&deg;
0.490 N
=
cos15.0&deg;
FT = 0.5073 N
Determine the electric force on one pith ball:
! 30.0&deg; \$
FE = FT sin #
&quot; 2 &amp;%
= (0.5073 N)sin15.0&deg;
FE = 1.313' 10(2 N
Determine the distance between pith balls:
! 30.0&deg; \$
=2 s #
&quot; 2 &amp;%
= 2(1.00 m)s 15.0&deg;
= 0.5176 m
Determine the charge on the pith balls:
kq q
FE = 12 2
r
kq 2
FE = 2
r
F r2
q2 = E
k
q=
FE r 2
k
(1.313! 10&quot;2 N )(0.5176 m )2
=
2 '
\$
9 N # m
)
&amp;% 8.99 ! 10
C2 (
q = 6.26 ! 10&quot;7 C
Statement: The charge on each pith ball is 6.26 &times; 10–7 C.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.2-11
Section 7.3: Electric Fields
Tutorial 1 Practice,
page 337
!
1. (a) Given: FE = 2.5 N [left]; q = –5.0 C
!
Required: !
!
!
Analysis: FE = q!
!
!
Solution: FE = q!
!
! FE
!=
q
2.5 N [left]
–5.0 C
= &quot;0.50 N,C [left]
!
! = 0.50 N,C [right]
Statement: The
! electric field in which the charge is located is 0.50 N,C [toward the right].
(b) Given: FE = 2.5 N [left]; q = –0.75 C
!
Required: !
!
!
Analysis: FE = q!
!
!
Solution: FE = q!
!
! FE
!=
q
=
2.5 N [left]
–0.75 C
= &quot;3.3 N,C [left]
=
!
! = 3.3 N,C [right]
Statement: The electric field in which the charge is located is 3.3 N,C [toward the right].
–6
9
2 2
2. Given: r =
! 2.50 m; q = 6.25 &quot; 10 C; k = 8.99 &quot; 10 N!m ,C
Required: !
kq
Analysis: ! = 2
r
kq
Solution: ! = 2
r
2 &amp;
#
9 N) m
*6
%\$ 8.99 &quot; 10
(' +6.25 &quot; 10 C
2
C
=
+2.50 m 2
! = 8.99 &quot; 103 N,C
Statement: The electric field at the point is 8.99 &quot; 103 N,C [toward the right].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.3-1
3. Given: r1 = 0.668 m; r2 = 0.332 m; q1 = 5.56 &quot; 10–9 C; q2 = –1.23 &quot; 10–9 C;
k = 8.99 &quot; 109 N!m2,C2
!
Required: ! net
Analysis: The direction of the electric
kq
magnitude of the electric field at point Z; ! = 2 ;
te r:
r
r = r 1 r2
r = 1.000 m
Solution:
kq
!1 = 21
r
2 &amp;
#
9 N) m
*9
(' +5.56 &quot; 10 C
%\$ 8.99 &quot; 10
2
C
=
+1.000 m 2
!1 =
!2 =
N,C +two e ra digits carrie
kq2
r22
2 &amp;
#
9 N) m
*9
8.99
&quot;
10
%\$
(' +–1.23&quot; 10 C
2
C
=
+0.332 m 2
! 2 = *1.0032 &quot; 102 N,C +two e ra digits carrie
! !
!
! net = !1 + ! 2
=(
N,C [right]) + ( &quot;1.0032 &quot; 102 N,C [right])
= *50.336 N,C [right]
!
! net = 50.3 N,C [left]
Statement: The electric field at the point Z is –
50.3 N,C [toward the left].
Research This: Fish and Electric Fields, page 344
A. Answers may vary. Sample answers: I chose electric eels. Electric eels produce electric fields
for self-defence; they also use electric fields to stun prey.
B. Answers may vary. Sample answers: Fish that stun prey with electric fields are typically
freshwater species because salt water conducts electricity. Therefore
strong enough field in salt water to stun prey can also injure itself in the salt water.
C. Answers may vary. Sample answers: Low light levels and poor visibility in certain rivers
would make it difficult for fish to detect prey visually. Since the transmission of electric fields is
not affected by the amount o
is a
beneficial adaptation for fish living in these rivers.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.3-2
Section 7.3 Questions, page 345
1. For a proton and an electron placed in a uniform electric field the magnitudes of the forces
will be equal.
2. Given: r = 1.5 m; q = 3.5 C; k = 8.99 &quot; 109 N!m2,C2
Required: !
kq
Analysis: ! = 2
r
kq
Solution: ! = 2
r
2 '
\$
9 N# m
&amp;% 8.99 &quot; 10
)( +3.5 C
2
C
=
+1.5 m 2
!=
&quot; 1010 N,C
Statement: The magnitude of the electric field at the point is 1.4 &quot; 1010 N,C.
!
3. Given: r1 = 10 cm = 0.10 m; r2 = 25 cm = 0.25 m; q1 = 4.5 &quot; 10–6 C; ! net = 0 N,C;
k = 8.99 &quot; 109 N!m2,C2
Required: q2
! 2 r22
kq1
kq2
Analysis: !1 = 2 ; ! 2 = 2 ; q2 =
k
r1
r2
Solution: Determine the magnitude of the electric field from q1 at the origin.
kq
!1 = 21
r1
2 &amp;
#
9 N) m
*6
8.99
&quot;
10
%\$
(' +4.5 &quot; 10 C
2
C
=
+0.10 m 2
= 4.046 &quot; 106 N,C +two e ra digits carrie
The second charge’s electric field must have the same magnitude at the origin:
! 2 r22
q2 =
k
#
6 N &amp;
2
%\$ 4.046 &quot; 10 C (' +0.25 m
=
2 &amp;
#
9 N ) m
%\$ 8.99 &quot; 10
(
C2 '
q2 = 2.8 &quot; 10*5 C
Statement: The charge on q2 is 2.8 &quot; 10–5 C.
4. T
so
opposite charge on the other side of the ring. Th
0 N,C.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.3-3
5.
he number of lines originating from a charge is determined
by the relative strength of the charge from which the lines originate to the other charges in the
sketch.
!
3
6. Given: r1 = ; r2 = ; ! net = 0 N,C
4
4
Required: q1:q2
Analysis:
kq
Use ! = 2 to determine the ratio.
r
Solution:
!1 = ! 2
k q1 k q2
= 2
r12
r2
q1
&quot; %
\$# 4 '&amp;
2
q1
=
q2
&quot;3 %
\$# 4 '&amp;
2
q2
9 2
16 16
q1 q2
=
1 9
q1 1
=
q2 9
Statement: The ratio between the charges q1:q2 is 1:9.
7. Given: q = 7.5 C; r = 2.3 m; k = 8.99 &times; 109 N!m2,C2
!
Required: ! net
Analysis: The point charges on the x-axis result in a zero electric field because they have equal
2
=
will only contribute a vertical component. The direction of the electric field will be down.
kq
Determine the magnitude of the electric field from a point charge at the origin ! = 2 . Then
r
determine the total electric field at the origin.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.3-4
Solution: The magnitude of the electric field from a point charge at the origin:
kq
!= 2
r
2 '
\$
9 N# m
8.99
&quot;
10
)( +7.5 C
&amp;%
2
C
=
+2.3 m 2
! = 1.275 &quot; 1010 N,C +two extra digits carrie
The total electric field at the origin:
! net = ! + ! sin 45&deg; + ! sin 45&deg;
=!+
2!
2
= ! + 2!
= 1.275 &quot; 1010 N,C + 2+1.275 &quot; 1010 N,C
! net = 3.1&quot; 1010 N,C [down]
Statement: The electric field at the origin is 3.1 &times; 1010 N,C [down].
8.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.3-5
Section 7.4: Potential Difference and Electric Potential
Tutorial 1 Practice, page 349
!
1. (a) Given: ! = 145 N/C [right]; q = –1.6 &times; 10–19 C; di = 1.5 m; df = 4.6 m
Required: ∆EE
Analysis: ! EE = !q&quot; ! d
Solution: ! EE = !q&quot; ! d
= !q&quot; ( df ! di )
\$
N'
= ! !1.6 # 10!19 C &amp; 145 ) ( 4.6 m ! 1.5 m )
C(
%
\$
N'
= 1.6 # 10!19 C &amp; 145 ) ( 3.1 m )
C(
%
!17
= 7.192 # 10 N * m (two extra digits carried)
! EE = 7.2 # 10!17 J
(
(
)
)
Statement: The change in the electric potential energy of the electron is 7.2 &times; 10–17 J.
(b) Given: vi = 1.7 &times; 107 m/s; ΔEE = 7.192 &times; 10–17 J; m = –9.11 &times; 10–31 kg
Required: vf
1
1
Analysis: ! EE + ! Ek = 0 ; ! Ek = mvf 2 ! mvi 2
2
2
Solution:
! EE + ! Ek = 0
&quot;1
%
1
! EE + \$ mvf2 ! mvi2 ' = 0
2
#2
&amp;
1 2 1 2
mv = mv ! ! EE
2 f 2 i
vf =
%
2&quot;1 2
mvi ! ! EE '
\$
m# 2
&amp;
= vi2 !
2 ! EE
m
=
(1.7 ( 10
=
(1.7 ( 10
7
7
) ((9.11 ( 10
2
m/s !
)
2
m/s !
2 7.192 ( 10!17 J
!31
kg
)
)
m
)m
s2
kg
1.4384 ( 10!16 kg )
9.11 ( 10!31
vf = 1.1 ( 107 m/s
Statement: The final speed of the electron is 1.1 &times; 107 m/s.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.4-1
2. Given: q = 1.6 &times; 10–19 C; ∆d = 0.75 m; ε = 23 N/C
Required: W
Analysis: W = q! ! d
Solution: W = q! ! d
\$
N'
= 1.6 &quot; 10#19 C &amp; 23 ) ( 0.75 m )
C(
%
#18
= 2.76 &quot; 10 N * m
W = 2.8 &quot; 10#18 J
Statement: The work done in moving the proton 0.75!m is 2.8 &times; 10–18 J.
3. Given: q = –1.6 &times; 10–19 C; ∆Ek = +4.2 &times; 10–16 J; ! d = 0.18 m [right]
!
Required: !
Analysis: ! EE + ! Ek = 0 ; ! EE = !q&quot; ! d
(
Solution:
)
! EE + ! Ek = 0
(q) ! d + ! Ek = 0
! Ek = q) ! d
)=
=
! Ek
q!d
4.2 # 10!16 N \$ m
!1.6 # 10!19 C ) 0.18 m )
) = (1.5 # 104 N/C
Since the magnitude of the electric field is negative, the direction of the electric field is in the
opposite direction of the displacement.
Statement: The magnitude and direction of the electric field are as follows:
1.5 &times; 104 N/C [toward the left].
Tutorial 2 Practice, page 353
1. (a) Given: ∆V = 1.6 &times; 104 V; ∆d = 12 cm = 0.12 m; q = –1.6 &times; 10–19 C; m = 9.11 &times; 10–31 kg
Required: vf
1
Analysis: ! EE + ! Ek = 0 ; ! EE = q !V ; ! Ek = mvf 2
2
! EE + ! Ek = 0
!
\$
( q !V ) + #&quot; 12 mv &amp;% = 0
2
f
1 2
mv = 'q !V
2 f
vf =
'2q !V
m
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.4-2
Solution:
!2q !V
vf =
m
=
=
!2 ( !1.6 &quot; 10!19 C ) (1.6 &quot; 104 V )
9.11&quot; 10!31 kg
(3.2 &quot; 10
!19
\$
m m'
C &amp; 1.6 &quot; 104 kg # 2 # )
%
s C(
!31
9.11&quot; 10 kg
)
vf = 7.5 &quot; 107 m/s
Statement: The electrons strike the screen at a speed of 7.5 &times; 107 m/s.
(b) Given: ∆V = 1.6 &times; 104 V; ∆d = 0.12 m
Required: ε
!V
Analysis: ! =
!d
Solution:
!V
!=
!d
m
1.6 &quot; 104 N #
C
=
0.12 m
! = 1.3 &quot; 105 N/C
Statement: The magnitude of the electric field is 1.3 &times; 105 N/C.
2. (a) Given: ∆dXW = 6.0 cm = 0.060 m; vi = 0 m/s; ∆V = 4.0 &times; 102 V; q = –1.6 &times; 10–19 C;
m = 9.11 &times; 10–31 kg
Required: vf
Analysis: Determine the final speed of the electron using the equation vf 2 = vi 2 + 2 ! d . But
first calculate the acceleration of the electron using the equations a =
FE
m
=
!V
q!
and ! =
:
!d
m
q!
m
q &quot; !V %
a= \$
m # ! d '&amp;
a=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.4-3
Solution:
q ! !V \$
a= #
m &quot; ! d &amp;%
=
(1.6 ' 10
(19
!
m m\$
C # 4.0 ' 102 kg ) 2 ) &amp;
&quot;
s C%
)
(9.11' 10
(31
)
kg ( 0.060 m )
a = 1.171' 1015 m/s 2 (two extra digits carried)
Determine the final speed of the electron as it reaches hole W:
vf = vi2 + 2 ! d
&quot;
m%
= 02 + 2 \$ 1.171! 1015 2 ' ( 0.060 m)
#
s &amp;
= 1.185 ! 107 m/s (two extra digits carried)
f
= 1.2 ! 107 m/s
Solution: The speed of the electron at hole W is 1.2 &times; 107 m/s.
(b) Given: ∆dYZ = 6.0 cm = 0.060 m; vi = 0 m/s; vf = 1.185 &times; 107 m/s; ∆V = 7.0 &times; 103 V;
q = –1.6 &times; 10–19 C; m = 9.11 &times; 10–31 kg
Required: ∆dZ0, the distance from Z at which the speed of the electron is 0 m/s
q ! !V \$
Analysis: a = #
;
m &quot; ! d &amp;%
vf2 = vi2 + 2 ! d
!d =
2
f
!
2
i
2
Solution:
q ! !V \$
a= #
m &quot; ! d &amp;%
!
\$
(1.6 ' 10 C ) #&quot; 7.0 ' 10 kg ) sm ) mC &amp;%
=
(9.11' 10 kg )(0.060 m )
(19
3
2
(31
a = 2.049 ' 1016 m/s 2 (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.4-4
Determine the distance the electron travels from Y before its speed becomes 0:
vf2 ! vi2
!d =
2
(1.185 &quot; 10
=
7
(
m/ s
)
2
! 02
2 2.049 &quot; 1016 m/ s 2
)
! d = 0.0034 m
Determine the distance the from Z:
∆dZ0 = 0.060 m – 0.0034 m
∆dZ0 = 0.057 m, or 5.7 cm
Statement: The electron changes direction 5.7 cm to the left of Z.
!
3. Given: L = 8.0 cm = 0.080 m; ∆d = 4.0 cm = 0.040 m; vi = 6.0 &times; 107 m/s [E];
∆V = 6.0 &times; 102 V; q = –1.6 &times; 10–19 C; m = 9.11 &times; 10–31 kg
!
Required: v
Analysis: There is an upward force on the electron because the negative plate is below the
!V
negatively charged electron. First, determine the magnitude of the electric field, ! =
. Then
!d
calculate the amount of time it takes the electron to pass through the plates,
!d
L
L
v=
; vi =
; !t = . Then calculate the resultant velocity using the equation
!t
!t
vi
vf = v f 2 +
2
f
, where the components of the final velocity are the initial velocity for vxf and
vyf = ay∆t, where a =
E
m
=
q!
. Finally, calculate the angle using the inverse tangent function.
m
Solution:
!V
!=
!d
=
6.0 &quot; 102 N #
m
C
0.040 m
! = 1.500 &quot; 104 N/C (two extra digits carried)
Determine the amount of time it takes the electron to pass through the plates:
! =
i
=
(0.080 m )
&quot;
7 m %
\$# 6.0 ! 10
'
s &amp;
! = 1.333! 10 –9 s (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.4-5
Determine the upward velocity of the electron:
q!
v yf =
!t
m
(1.6 &quot; 10#19 C ) \$&amp;% 1.500 &quot; 104 N ')(
C (
=
1.333&quot; 10#9 s )
#31
9.11&quot; 10 kg
m
2.400 &quot; 10#15 kg * 2
s (1.333&quot; 10#9 s )
=
#31
9.11&quot; 10 kg
v yf = 3.504 &quot; 106 m/s (two extra digits carried)
Determine the magnitude of the final velocity:
vf = v 2f +
=
2
f
(6.0 ! 107 m/s )2 + (3.504 ! 106 m/s )2
= 6.0 ! 107 m/s
Determine the direction of the final velocity (the angle north of east):
&quot;v %
tan ! = \$ f '
# f&amp;
f
%
'
f &amp;
&quot;
! = tan (1 \$
#
f
&quot; 3.504 ) 106 m/s %
= tan (1 \$
'
7
# 6.0 ) 10 m/s &amp;
! = 3.3&deg;
Statement: The final velocity of the electron is 6.0 &times; 107 m/s [E 3.3&deg; N].
Section 7.4 Questions, page 354
1. (a) Given: q = –1.6 &times; 10–19 C; Vi = 30 V; Vf = 150 V
Required: ∆EE
! E
Analysis: ! =
q
!
E
= q!
!
E
= q(
f
! i)
Solution: ! EE = q (Vf ! Vi )
= ( !1.6 &quot; 10!19 C )(150 V ! 30 V )
= !1.92 &quot; 10!17 J (one extra digit carried)
! EE = !1.9 &quot; 10!17 J
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.4-6
Statement: The change in the electron’s potential energy is a decrease of 1.9 &times; 10–17 J.
(b) Given: q = –1.6 &times; 10–19 C; ∆d = 10 cm = 0.10 m; ∆EE = –1.92 &times; 10–17 J
!
Required: !
! EE
Analysis: ! EE = !q&quot; ! d ; = =
. Since electrons travel from regions of low potential to
q!d
regions of high potential, and electrons move against the direction of an electric field, the
direction of the field will be opposite the direction of the electron.
! EE
Solution: ! = &quot;
q!
=&quot;
&quot;1.92 # 10&quot;17 N \$ m
(1.6 # 10&quot;19 C) (0.10 m )
! = 1.2 # 103 N/C
Statement: The average electric field along the electron’s path is –1.2 &times; 103 N/C.
2. Given: ∆d = 3.0 mm = 3.0 &times; 10–3 m; ε = 250 V/m
Required: ∆V
!V
Analysis: ! = &quot;
; !V = !&quot; ! d
!d
Solution: !V = !&quot; ! d
#
V&amp;
3.0 ) 10!3 m
= ! % 250
(
m'
\$
(
)
!V = !0.75 V
Statement: The magnitude of the electric potential difference is 0.75 V.
3. (a) Given: q = 1.6 &times; 10–19 C; Vi = 75.0 V; Vf = –20.0 V
Required: ∆Ek
! ! Ek
Analysis: !V =
q
! Ek = !q !V
! Ek = !q (Vf ! Vi )
Solution: ! Ek = !q (
f
!
i
)
= ! (1.6 &quot; 10!19 C )( –20.0 V ! 75.0 V )
!
k
= 1.52 &quot; 10!17 J
Statement: The change in the proton’s kinetic energy is 1.52 &times; 10–17 J.
(b) Given: q = –1.6 &times; 10–19 C; Vi = 75.0 V; Vf = –20.0 V
Required: ∆Ek
Analysis: ! Ek = !q Vf ! Vi
(
)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.4-7
Solution: ! Ek = !q (
f
!
i
)
= ! ( !1.6 &quot; 10!19 C )( –20.0 V ! 75.0 V )
!
k
= !1.52 &quot; 10!17 J
Statement: The change in the electron’s kinetic energy is –1.52 &times; 10–17 J.
4. (a) Given: q = –1.6 &times; 10–19 C; ∆V = 45 V
Required: W
Analysis:
! EE
!V =
q
! EE = q !V
W = ! ! EE
W = !q&quot;V
Solution: W = !q !V
= ! ( !1.6 &quot; 10!19 C )( 45 V )
= 7.2 &quot; 10!18 J
Statement: The work done to push the electron is 7.2 &times; 10–18 J against the electric field.
(b) The electric field is doing the work.
5. (a) Electrons move from a region of low potential energy to a region of high potential energy.
(b) Given: q = –1.6 &times; 10–19 C; ∆V = 2.5 &times; 104 V
Required: ∆Ek
Analysis: ∆Ek + ∆EE = 0; ∆EE = –∆Ek
!
! =
q
!!
! =
q
! = !q !
Solution:
! Ek = !q !V
= ! ( !1.6 &quot; 10!19 C ) ( 2.5 &quot; 104 V )
! Ek = 4.0 &quot; 10!15 J
Statement: The change in one of the electron’s kinetic energy is 4.0 &times; 10–15 J.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.4-8
(c) Given: vi = 0 m/s; ΔEk = 4.0 &times; 10–15 J; m = 9.11 &times; 10–31 kg
Required: vf
Analysis:
1 2 1 2
! Ek =
!
2 f 2 i
1 2
=
!0
2 f
f
=
2 ! Ek
Solution:
vf =
=
2 ! Ek
m
\$
m '
2 &amp; 4.0 ! 10&quot;15 kg # 2 # m )
%
(
s
(9.11 ! 10
&quot;31
kg
)
vf = 9.4 ! 107 m/s
Solution: The final speed of the electron is 9.4 &times; 107 m/s.
6. Given: ε = 2.26 &times; 105 N/C; di = 2.55 m; df = 4.55 m
Required: ∆V
Analysis:
!V
=
!d
&quot;V = &quot;d
Solution: !V = ! ! d
= ! ( df &quot; di )
= ( 2.26 # 105 N/C )( 4.55 m &quot; 2.55 m )
!V = 4.52 # 105 V
Statement: The change in the electric potential between the points is 4.52 &times; 105 V.
7. (a) Given: ε = 150 N/C; L = 6.0 cm = 0.060 m; vi = 4.0 &times; 106 m/s; q = –1.6 &times; 10–19 C;
m = 9.11 &times; 10–31 kg
!
Required: v yf
Analysis: There is an upward force on the electron because the negative plate is below the
negatively charged electron. First, calculate the amount of time the electron takes to pass through
!d
L
L
the plates, v =
; vi =
; !t = . Then determine the vertical component of the final
!t
!t
vi
velocity using vyf = ay∆t, where a y =
Copyright &copy; 2012 Nelson Education Ltd.
FE
m
=
q!
.
m
Chapter 7: Electric Fields
7.4-9
Solution: Determine the amount of time it takes the electron to pass through the plates:
! =
i
=
(0.060 m )
&quot;
6 m %
'
\$# 4.0 ! 10
s &amp;
! = 1.5 ! 10 –8 s
Determine the upward velocity of the electron:
q!
v yf
!t
m
\$
N'
1.6 &quot; 10#19 C &amp; 150
C )(
%
1.5 &quot; 10#8 s
#31
9.11 &quot; 10 kg
)
)
2.400 &quot; 10#17 kg *
9.11 &quot; 10#31 kg
m
s 2 1.5 &quot; 10#8 s
)
3.952 &quot; 105 m/s (two extra digits carried)
v yf
4.0 &quot; 105 m/s
Statement: The vertical component of the electron’s final velocity is 4.0 &times; 105 m/s [up].
(b) Given: ε = 150 N/C; L = 0.060 m; vi = 4.0 &times; 106 m/s; vyf = 3.952 &times; 105 m/s;
q = –1.6 &times; 10–19 C; m = 9.11 &times; 10–31 kg
Required: v
Analysis: Determine the magnitude of the final velocity using the equation
f
=
2
f
+
2
f
. Then
use the inverse tangent ratio to determine the angle.
Solution: vf = v 2f +
2
f
( 4.0 ! 106 m/s )2 + (
=
! 105
)2
= 4.0 ! 10
&quot;
tan ! = \$
#
%
'
f &amp;
f
&quot;
! = tan (1 \$
#
%
'
f &amp;
f
&quot; 3.952 ) 105 m/s %
= tan (1 \$
'
6
# 4.0 ) 10 m/s &amp;
! = 5.6&deg;
Statement: The final velocity of the electron is 4.0 &times; 106 m/s [5.6&deg; from the x-axis].
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.4-10
8. (a) Given: ε = 20 N/C; dA = 0 m; dB = 4 m
Required: ∆V
!V
!V = = ! d
Analysis: = =
!d
Solution: !V = !&quot; ! d
(
)
V&amp;
(4 m ! 0 m )
m ('
!V = !&quot; d B ! dA
#
= ! % 20
\$
!V = !80 V
Statement: The potential difference is –80 V.
(b) Given: ε = 20 N/C; dA = 4 m; dB = 6 m
Required: ∆V
Analysis: !V = !&quot; ! d
Solution: !V = !&quot; ! d
(
)
V&amp;
(6 m ! 4 m )
m ('
!V = !&quot; d B ! dA
#
= ! % 20
\$
!V = !40 V
Statement: The potential difference is –40 V.
9. The net work done is 0 J because the points are at the same potential.
10. Given: V = 20 V; q = 0.5 C
Required: W
Analysis:
!W
! =
q
W = !q
Solution: W = !q !
= ! ( 0.5 C )( 20 V )
W = !10 J
Statement: The amount of work done in bringing the charge to the point was 10 J.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.4-11
Section 7.5: Electric Potential and Electric Potential Energy Due to
Point Charges
Tutorial 1 Practice, page 360
1. (a) Given: q1 = +6.0 &times; 10–6 C; q2 = –3.0 &times; 10–6 C; q3 = –3.0 &times; 10–6 C; r = 3.0 m;
k = 8.99 &times; 109 N⋅m2/C2
Required: V1; V2; V3
Analysis: The potential at each midpoint is the sum of the potentials due to the three charges.
kq
Use V =
to calculate the potential due to each charge. The distance from a midpoint to an
r
endpoint is 0.5r, and the distance from a midpoint to an opposite vertex is 0.5 3r .
Solution: Calculate the potential between q1 and q2, V1:
kq
kq3
kq
V1 = 1 + 2 +
0.5r 0.5r 0.5 3r
=
=
k!
2 \$
2q1 + 2q2 +
q3 &amp;
#
r&quot;
3 %
J( m
C 2 ! 2 6.0 ' 10)6 C + 2 )3.0 ' 10)6 C + 2 )3.0 ' 10)6 C \$
#&quot;
&amp;%
3.0 m
3
8.99 ' 109
(
) (
)
(
)
V1 = 7.6 ' 103 J/C
Calculate the potential between q1 and q3, V2:
kq2
kq
kq
V2 = 1 +
+ 3
0.5r 0.5 3r 0.5r
\$
k!
2
2q
+
q
+
2q
1
2
2
&amp;%
r #&quot;
3
J( m
8.99 ' 109
C 2 ! 2 6.0 ' 10)6 C + 2 )3.0 ' 10)6 C + 2 )3.0 ' 10)6 C \$
=
#&quot;
&amp;%
3.0 m
3
=
(
)
(
) (
)
V2 = 7.6 ' 103 J/C
Calculate the potential between q2 and q3, V3:
kq
kq
kq1
V3 =
+ 2 + 3
0.5 3r 0.5r 0.5r
\$
k! 2
q1 + 2q2 + 2q3 &amp;
#
r&quot; 3
%
J( m
8.99 ' 109
C 2 ! 2 6.0 ' 10)6 C + 2 )3.0 ' 10)6 C + 2 )3.0 ' 10)6 C \$
=
#&quot;
&amp;%
3.0 m
3
=
(
) (
) (
)
V3 = )1.5 ' 104 J/C
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.5-1
Statement: The potential between q1 and q2, V1, is 7.6 &times; 103 J/C. The potential between q1 and
q3, V2, is 7.6 &times; 103 J/C. The potential between q2 and q3, V3, is –1.5 &times; 104 J/C.
(b) Given: q1 = +6.0 &times; 10–6 C; q2 = –3.0 &times; 10–6 C; q3 = –3.0 &times; 10–6 C; r = 3.0 m;
k = 8.99 &times; 109 N⋅m2/C2
Required: EE
Analysis: The total electric potential energy is the sum of the three electric potential energies of
kq q
a pair of charges. Use EE = 1 2 to calculate the electric potential energy for each pair of
r
charges.
Solution: Calculate the electric potential energy between q1 and q2, EE1:
kq q
EE1 = 1 2
r
#
&amp;
9 J&quot; m
6.0 ! 10)6 C )3.0 ! 10)6 C
% 8.99 ! 10
2 (
\$
C '
=
3.0 m
EE1 = )5.394 ! 10)2 J (two extra digits carried)
The electric potential energy between q1 and q3, EE2, is equal to EE1 because q2 = q3.
Calculate the electric potential energy between q2 and q3, EE3:
kq q
EE3 = 2 3
r
#
&amp;
9 J&quot; m
)3.0 ! 10)6 C )3.0 ! 10)6 C
% 8.99 ! 10
2 (
\$
C '
=
3.0 m
)2
EE3 = 2.697 ! 10 J (two extra digits carried)
Calculate the total electric potential energy, EE:
EE = EE1 + EE2 + EE3
(
(
)(
)(
)
)
= (!5.394 &quot; 10!2 J) + (!5.394 &quot; 10!2 J) + (2.697 &quot; 10!2 J)
EE = !8.1&quot; 10!2 J
Statement: The total electric potential energy of the group of charges is –8.1 &times; 10–2 J.
2. Given: q = 4.5 &times; 10–6 C; s = 1.5 m; k = 8.99 &times; 109 N⋅m2/C2
Required: V at the centre of the square
Analysis: The potential at the centre is the sum of the potentials due to the four charges. Use
kq
V=
to calculate the potential due to each charge. The distance from a vertex to the centre is
r
1
2s , or
.
2
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.5-2
kq
r
4kq
=
s
Solution: V = 4
2
#
J&quot; m &amp;
4 2 % 8.99 ! 109
(4.5 ! 10)6 C )
\$
C 2 ('
=
1.5 m
V = 1.5 ! 105 J/C
Statement: The electric potential at the centre of the square is 1.5 &times; 105 J/C.
3. Given: q = –1.6 &times; 10–19 C; ri = 5.0 &times; 10–12 m; m = 9.11 &times; 10–31 kg; k = 8.99 &times; 109 N⋅m2/C2
Required: vf
kq q
Analysis: Determine the initial electric potential energy using the equation EEi = 1 2 . The final
ri
1 2
mv . The initial kinetic energy and final potential energy are
2 f
both 0. Use the conservation of energy to determine the final speed of each electron.
EEi + Eki = EEf + Ekf
kinetic energy of each electron is
kq1q2
1
1
+ 0 = 0 + mvf2 + mvf2
ri
2
2
kq 2
= mvf2
ri
vf =
kq 2
mri
Solution: vf =
=
kq 2
mri
#
&amp;
m
kg &quot; 2 &quot; m 2 (
2
%
)19
9
s
(
)
)1.6
!
10
C
% 8.99 ! 10
(
\$
'
C2
(9.11! 10
)31
)
kg (5.0 ! 10)12 m )
vf = 7.1! 106 m/s
Statement: The final speed of each electron is 7.1 &times; 106 m/s.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.5-3
4. Given: q = 1.6 &times; 10–19 C; v1 = 2.3 &times; 106 m/s; v2 = 1.2 &times; 106 m/s; m = 1.673 &times; 10–27 kg;
k = 8.99 &times; 109 N⋅m2/C2
Required: rf
kq q
Analysis: Calculate the final electric potential energy using the equation EEi = 1 2 . The initial
rf
1 2
mv , and the initial potential energy and final kinetic energy
2
are both 0. Use the conservation of energy to determine the final separation of the protons.
EEi + Eki = EEf + Ekf
kinetic energy of each proton is
kq q
1
1
0 + m1vi12 + m2 vi22 = 1 2 + 0
2
2
rf
2kq 2
mv + mv =
rf
2
i1
2
i2
rf =
2kq 2
mvi12 + mvi22
Solution:
2kq 2
rf =
mvi12 + mvi22
=
#
m
&quot;
&quot; m2
kg
%
2
s
2 % 8.99 ! 109
%\$
C2
(1.673! 10
)27
)
2
(
&amp;
(
( 1.6 ! 10)19 C
('
(
)
)
2
#
#
m&amp;
m&amp;
kg % 2.3! 106
+ 1.673! 10)27 kg % 1.2 ! 106
(
s '
s ('
\$
\$
2
rf = 4.1! 10)14 m
Statement: The separation of the protons when they are closest to each other is 4.1 &times; 10–14 m.
Section 7.5 Questions, page 361
1. (a) The proton moves to a region of lower potential energy and lower electric potential.
(b) The electron moves to a region of lower potential energy and higher electric potential.
2. Two particles that are at locations where the electric potential is the same do not necessarily
have the same electric potential energy. The potential energy equals the product of the charge
and the electric potential. If the particles have different charges, then they have different
potential energies.
3. No work, or 0 J, is required to move a charge from one spot to another with the same electric
potential. The work done equals the change in kinetic energy, and the change in kinetic energy
equals the negative change in potential energy if energy is conserved. If the electric potential
does not change, then the electric potential energy does not change. Therefore, the kinetic energy
does not change, and no work is done.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.5-4
4. Given: q1 = 4.5 &times; 10–5 C; q2 = 8.5 &times; 10–5 C; EE = 40.0 J; k = 8.99 &times; 109 N⋅m2/C2
Required: r
kq q
Analysis: EE = 1 2
r
kq q
r= 1 2
EE
Solution: r =
kq1q2
EE
&quot;
N %
9
4.5 ! 10(5 C 8.5 ! 10(5 C
8.99
!
10
\$
2 '
#
C &amp;
=
40.0 N ) m
= 0.86 m
(
)(
)
r = 86 cm
Statement: The distance between the charges is 86 cm.
5. Given: q1 = 4.5 &times; 10–5 C; q2 = 8.5 &times; 10–5 C; ri = 2.5 m; rf = 1.5 m; k = 8.99 &times; 109 N⋅m2/C2
Required: ∆EE
Analysis:
kq q kq q
! EE = 1 2 ! 1 2
rf
ri
&quot; 1 1%
! EE = kq1q2 \$ ! '
# rf ri &amp;
&quot; 1 1%
Solution: ! EE = kq1q2 \$ ! '
# rf ri &amp;
&quot;
&quot; 1
J) m %
1 %
4.5 ( 10!5 C 8.5 ( 10!5 C \$
= \$ 8.99 ( 109
!
'
# 1.5 m 2.5 m '&amp;
#
C2 &amp;
(
)(
)
! EE = 9.2 J
Statement: The electric potential energy increases by +9.2 J.
6. Given: q1 = 3.5 &times; 10–6 C; q2 = 7.5 &times; 10–6 C; ri → ∞; rf = 2.5 m; k = 8.99 &times; 109 N⋅m2/C2
Required: W
Analysis:
kq q kq q
= 1 2! 1 2
ri
rf
=
kq1q2
!0
rf
=
kq1q2
rf
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.5-5
Solution:
=
=
kq1q2
r
#
%
\$
! 10
&quot;m&amp;
(
C2 '
(
)(
! 10)6 C 9.5 ! 10)6 C
)
2.5 m
)2
=
! 10
Statement: The work required to bring the point charges together is 9.4 &times; 10–2 J.
7. Given: q1 = –1.6 &times; 10–19 C; q2 = 1.6 &times; 10–19 C; ri = 5.00 &times; 10–11 m; rf → ∞;
k = 8.99 &times; 109 N⋅m2/C2
Required: W
Analysis:
kq q kq q
= 1 2! 1 2
rf
ri
= 0!
=!
kq1q2
ri
kq1q2
ri
Solution:
=!
kq1q2
ri
\$
'
9 J# m
–1.6 &quot; 10!19 C 1.6 &quot; 10!19 C
&amp; 8.99 &quot; 10
2 )
%
C (
=!
5.00 &quot; 10!11 m
= 4.6 &quot; 10!18 J
Statement: The work required to separate the electron and the proton is 4.6 &times; 10–18 J.
8. (a) Given: rf = 15 cm = 0.15 m; V = –8.5 &times; 104 V; k = 8.99 &times; 109 N⋅m2/C2
Required: q
kq
Vr
Analysis: V =
; q=
r
k
Vr
Solution: q =
k
#
4 J &amp;
%\$ !8.5 &quot; 10 C (' ( 0.15 m )
=
#
9 J ) m &amp;
%\$ 8.99 &quot; 10 C 2 ('
(
)(
)
q = !1.4 &quot; 10!6 C
Statement: The charge on the sphere is –1.4 &times; 10–6 C.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.5-6
(b) Given: rf = 0.15 m; V = –8.5 &times; 104 V
Required: ε
Analysis:
kq
!= 2
r
V
!=
r
V
Solution: ! =
r
%
4 N\$ m (
'&amp; &quot;8.5 # 10
C *)
=
(0.15 m )
! = &quot;5.7 # 105 N/C
Statement: The magnitude of the electric field near the surface of the sphere is –5.7 &times; 105 N/C.
(c) By convention, the electric field points radially inward from the surface of a negatively
charged sphere.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.5-7
Section 7.6: The Millikan Oil Drop Experiment
Tutorial 1 Practice, page 364
1. Given: r = 110 cm = 1.10 m; N = 1.2 &times; 108; e = 1.602 &times; 10–19 C; k = 8.99 &times; 109 N⋅m2/C2
Required: FE
Analysis: Determine the magnitude of the charge on each sphere using q = Ne. The charges will
be positive because the spheres have a deficit of electrons. Then calculate the force of repulsion,
kq q
FE = 12 2 .
r
Solution:
q = Ne
= (1.2 ! 108 ) (1.602 ! 10&quot;19 C )
q = 1.92 ! 10&quot;11 C (one extra digit carried)
FE =
kq1q2
r2
2 &amp;
#
9 N&quot; m
)11
)11
8.99
!
10
%
( 1.92 ! 10 C 1.92 ! 10 C
2
C '
\$
(
=
)(
(1.10 m )
FE = 2.7 ! 10)12 N
)
2
Statement: The force of repulsion between the two plastic spheres is 2.7 &times; 10–12 N.
2. Given: m = 2.48 &times; 10–15 kg; ∆d = 1.7 cm = 0.017 m; ∆Vb = 260 V; e = 1.602 &times; 10–19 C;
g = 9.8 m/s2
Required: q; N
mg ! d
Analysis: Determine the charge on the oil drop, q =
. Then calculate the number of
!Vb
q
. The top plate is positively charged, so the field between the
e
plates points downward. However, the electric force is balancing the gravitational force, so the
particle is moving against the electric field. Therefore the charge will be negative.
Solution: Determine the charge on the oil drop:
mg ! d
q=
!Vb
#
m&amp;
2.48 ! 10&quot;15 kg % 9.8
( 0.017 m
\$
s2 '
=
m m
260 kg )
)
s2 C
q = 1.589 ! 10&quot;18 C (two extra digits carried)
q = 1.6 ! 10&quot;18 C
The charge on the oil drop is –1.6 &times; 10–18 C.
excess electrons, q = Ne; N =
(
)
(
Copyright &copy; 2012 Nelson Education Ltd.
)
Chapter 7: Electric Fields
7.6-1
Determine the excess of electrons:
q
N=
e
1.589 ! 10&quot;18 C
=
1.602 ! 10&quot;19 C
N = 10
Statement: The charge on the oil drop is –1.6 &times; 10–18 C. The oil drop has an excess of 10
electrons, or –10e.
3. Given: ε = 1.0 &times; 102 N/C; m = 2.4 &times; 10–15 kg; e = 1.602 &times; 10–19 C; g = 9.8 m/s2
Required: q
mg
Analysis: q =
. The ionosphere is positively charged, so Earth’s electric field points toward
!
Earth’s surface. The electric force is balancing the gravitational force, so the particle is moving
against the electric field. Therefore the charge will be negative.
Solution: Determine the charge on the object:
mg
q=
!
\$
m'
2.4 &quot; 10#15 kg &amp; 9.8
)
%
s2 (
=
\$
2 N '
&amp;% 1.0 &quot; 10 C )(
= 2.352 &quot; 10#16 C (two extra digits carried)
q = 2.4 &quot; 10#16 C
Determine the charge as a multiple of the elementary charge:
qe
q=
e
(
2.352 ! 10&quot;16 C ) e
=
(1.602 ! 10&quot;19 C )
q = 1.5 ! 103 e
Statement: The oil drop has a charge of –2.4 &times; 10–16 C, or –1.5 &times; 103e.
(
)
Section 7.6 Questions, page 365
1. Given: q = 3.8 &times; 10–14 C; e = 1.602 &times; 10–19 C
Required: N
Analysis:
q = Ne
q
N=
e
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.6-2
q
e
3.8 ! 10&quot;14 C
=
1.602 ! 10&quot;19 C
N = 2.4 ! 105
Statement: To give the object a positive charge of 3.8 &times; 10–14 C, 2.4 &times; 105 C electrons must be
removed.
2. Given: r = 0.35 m; N = 6.1 &times; 106 C; e = –1.602 &times; 10–19 C; k = 8.99 &times; 109 N⋅m2/C2
Required: ε; V
Analysis: Determine the charge on the object using q = Ne. Then calculate the magnitude of the
kq
electric field using ! = 2 and the magnitude of the electric potential using V = !&quot; d .
r
Solution: Determine the charge on the object:
q = Ne
= (6.1 ! 106 )(&quot;1.602 ! 10&quot;19 C)
q = &quot;9.77 ! 10&quot;13 C (one extra digit carried)
Determine the magnitude of the electric field:
kq
!= 2
r
2 '
\$
*13
9 N# m
&amp; 8.99 &quot; 10
) *9.77 &quot; 10 C
2
C
%
(
=
2
0.35 m
Solution: N =
(
*2
(
)
)
= *7.172 &quot; 10 N/C (two extra digits carried)
! = *7.2 &quot; 10*2 N/C
Determine the magnitude of the electric potential:
V = !&quot; d
= !(!7.172 # 10!2 N/C)(0.35 m)
V = 2.5 # 10!2 V
Statement: At a distance of 0.35 m, the magnitude of the electric field is 7.2 &times; 10–2 N/C and the
magnitude of the electric potential is 2.5 &times; 10–2 V.
3. Given: ∆d = 2.00 mm = 2.00 &times; 10–3 m; ∆V = 240 V; m = 5.88 &times; 10–10 kg; e = 1.602 &times; 10–19 C;
g = 9.8 m/s2
Required: q; N
mg ! d
Analysis: Determine the charge on the oil droplet using q =
. Then determine the excess
!Vb
number of electrons using q = Ne;
=
q
. The top plate is positively charged, so the field
between the plates points downward. However, the electric force is balancing the gravitational
force, so the particle is moving against the electric field. Therefore the charge will be negative.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.6-3
Solution: Determine the charge on the oil droplet:
mg ! d
q=
!Vb
#
m&amp;
&quot;3
5.88 ! 10&quot;10 kg % 9.8
( 2.00 ! 10 m
2
\$
s '
=
m m
240 kg )
)
s2 C
= 4.802 ! 10&quot;14 C (two extra digits carried)
q = 4.8 ! 10&quot;14 C
Determine the excess of electrons:
q
=
(
)
(
)
4.802 ! 10&quot;14 C
1.602 ! 10&quot;19 C
= 3.0 ! 105
Statement: The charge on the oil droplet is –4.8 &times; 10–14 C. The oil droplet has an excess of
3.0 &times; 105 electrons.
4. (a) The charge on the drop is positive because the force needed to suspend the drop is in the
same direction as the field.
(b) Given: m = 3.3 &times; 10–7 kg; ε = 8.4 &times; 103 N/C; e = 1.602 &times; 10–19 C; g = 9.8 m/s2
Required: N
mg
q
=
Analysis: q =
; q = Ne;
!
Solution: Determine the charge on the object:
mg
q=
!
\$
m'
3.3 &quot; 10# kg &amp; 9.8
)
%
s2 (
=
\$
3 N '
&amp;% 8.4 &quot; 10 C )(
q = 3.85 &quot; 10#10 C (one extra digit carried)
Determine the deficit of electrons:
q
=
=
(
)
3.85 ! 10&quot;10 C
1.602 ! 10&quot;19 C
= 2.4 ! 109
Statement: The drop of water has a deficit of 2.4 &times; 109 electrons.
=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.6-4
5. (a) Given: m = 5.2 &times; 10–15 kg; ∆d = 0.21 cm = 2.1 &times; 10–3 m; ∆V = 220 V; g = 9.8 m/s2
Required: q
mg ! d
Analysis: q =
. The bottom plate is positively charged, so the field between the plates
!Vb
points upward. However, the electric force is balancing the gravitational force, so the particle is
moving against the electric field. Therefore the charge will be positive.
mg ! d
Solution: q =
!Vb
#
m&amp;
&quot;3
5.2 ! 10&quot;15 kg % 9.8
( 2.1 ! 10 m
2
\$
s '
=
m m
220 kg )
)
s2 C
= 4.864 ! 10&quot;19 C (two extra digits carried)
q = 4.9 ! 10&quot;19 C
Statement: The charge on the oil drop is 4.9 &times; 10–19 C.
(b) Given: q = 4.864 &times; 10–19 C; e = 1.602 &times; 10–19 C
Required: N
q
=
Analysis: q = Ne;
(
Solution:
=
)
(
)
q
4.864 ! 10&quot;19 C
=
1.602 ! 10&quot;19 C
=3
Statement: The oil drop has a deficit of 3 electrons.
6. Given: mA = 4.2 &times; 10–2 kg; NA = 1.2 &times; 1012; NB = 3.5 &times; 1012; r = 0.23 m; e = 1.602 &times; 10–19 C;
k = 8.99 &times; 109 N⋅m2/C2; g = 9.8 m/s2
Required: θ
Analysis: The angle between the wall and the thread is also the angle of the tension force on
sphere A. The only other forces on sphere A are due to gravity and electric attraction. Use q = Ne
kq q
to determine the charges, FE = 12 2 to determine the electric force, and Fg = mg to determine
r
the gravitational force. Use the tangent ratio to determine the angle of the tension force.
Solution: Determine the charge on each sphere:
qA = A
qB = B
12
&quot;19
= (1.2 ! 10 )(1.602 ! 10 C)
= (3.5 ! 1012 )(1.602 ! 10&quot;19 C)
qA = 1.92 ! 10&quot;7 C (one extra digit carried)
qB = 5.61 ! 10&quot;7 C (one extra digit carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.6-5
Determine the electric force:
kq q
FN = 12 2
r
#
&quot; m2 &amp;
9
)7
)7
8.99
!
10
%
( 1.92 ! 10 C 4.51 ! 10 C
2
C '
\$
=
2
0.23 m
(
)2
(
)(
)
)
FN = 1.830 ! 10
(t o extra digit carried)
Determine the force of gravity:
= A
g
g
= ( 4.2 ! 10&quot;2 kg ) ( 9.8 m/s 2 )
= 4.116 ! 10&quot;1 N (two extra digits carried)
Use the tangent ratio to determine the angle of the tension force:
&quot;F %
tan ! = \$ E '
# Fg &amp;
&quot;F %
! = tan (1 \$ E '
# Fg &amp;
(2
&quot;
N%
(1 1.830 ) 10
= tan \$
'
(1
# 4.116 ) 10 N &amp;
= 2.544&deg; (two extra digits carried)
! = 2.5&deg;
Statement: The thread is at an angle of 2.5&deg; from the wall.
(b) Given: θ = 2.544&deg;; Fg = 4.116 &times; 10–1 N
Required: FT
Fg
Analysis: cos! =
FT
Fg
FT =
cos!
Fg
Solution: FT =
cos!
4.116 &quot; 10#1 N
=
cos 2.544&deg;
FT = 0.41 N
Statement: The tension in the thread is 0.41 N [up the thread].
7. (a) Both Earth’s electric field and gravitational field point in the same direction and have
approximately the same shape. The change in Earth’s electric field and gravitational field is the
same as the altitude increases, because both follow the inverse-square law.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.6-6
(b) Given: ε = 1.0 &times; 102 N/C; q = 1.602 &times; 10–19 C; g = 9.8 m/s2
Required: m
mg
Analysis: q =
!
q!
m=
g
q!
Solution: m =
g
%
kg m (
1.602 &quot; 10#19 C ' 1.0 &quot; 102
\$ *
C s2 )
&amp;
=
%
m(
9.8
'
*
&amp;
s2 )
m = 1.6 &quot; 10#18 kg
Statement: The mass of a particle with an elementary charge on it that can be suspended by
Earth’s electric field is 1.6 &times; 10–18 kg.
8. The tiny dust particles observed in a beam of sunlight may have become charged by friction
and are now suspended by Earth’s electric field or some other nearby electric field. I could test
the answer by charging another object, such as a balloon or a piece of fabric, and testing whether
the particles are attracted or repelled by the object without coming into contact. If so, then the
particles have a charge.
(
)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 7: Electric Fields
7.6-7
Section 8.1: Magnets and Electromagnets
Section 8.1 Questions, page 385
1. A permanent magnet is a magnet that always behaves as a magnet, unlike some materials that
only act as magnets in response to an applied magnetic field.
2. (a) Earth’s magnetic field is directed slightly off from Earth’s axis. The south magnetic pole
attracts the north pole of a bar magnet. This means that, from a physical standpoint, the north
pole of Earth’s magnetic field is located near Earth’s south geographic pole.
(b) Charged particles from the Sun are directed downward toward Earth’s surface by Earth’s
magnetic field. They are deflected near the equator and channelled along magnetic field lines
toward the poles. These particles energize gas atoms in the upper atmosphere, causing the gas
atoms to release the extra energy as rays of light, producing the beautiful colours of the auroras.
(c) Magnetic compasses align with Earth’s magnetic field. A hiker can use a compass to identify
a northerly direction.
3. (a) I would hold the conductor in my right hand with my thumb pointing along its length. The
direction of the magnetic field is in the direction of my curled fingers.
(b) I would hold the loop of wire in my right hand with my thumb pointing along the loop. The
direction of the magnetic field is in the direction of my curled fingers.
(c) I would make a fist and hold my hand so that my fingers curl in the direction of the electric
current. My thumb then points in the direction of the magnetic field lines in the core.
4. The field lines inside the coils are straight and directed along the length of the solenoid. The
field lines outside the coils curve around from one end to the other.
5. The doorbell contains an electromagnet, which is a solenoid with a core of a magnetic metal.
When the switch is closed, current flows through the coils of the electromagnet, producing a
magnetic field according to the principle of electromagnetism. By the right-hand rule for a
solenoid, the magnetic field attracts the metal plate away from the contact. When contact is
broken, the hammer hits the doorbell.
6. To demonstrate the principle of electromagnetism, I would follow these steps:
Make an electromagnet from an iron rod in a solenoid. Have the wires of the solenoid connected
to a circuit with a battery. Using various types of batteries, experiment with how the
electromagnet can attract paper clips as the voltage changes.
7. Answers may vary. Students’ paragraphs should note that each crystal, called a magnetosome,
is composed of a single magnetic domain. Bacteria use the magnetic areas like a compass to
direct them downward, away from oxygen.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.1-1
Section 8.2: Magnetic Force on Moving Charges
Tutorial 1 Practice, page 390
1. (a) Given: q = 1.60 &times; 10–19 C; v = 9.4 &times; 104 m/s; B = 1.8 T; θ = 90&deg;
Required: FM
Analysis: FM = qvB sin θ; by the right-hand rule, the direction of the electric force is south.
Solution: FM = qvBsin !
m' \$
kg '
\$
= (1.60 &quot; 10 #19 C ) &amp; 9.4 &quot; 10 4
)( &amp;% 1.8
) sin 90&deg;
%
C *s (
s
= 2.707 &quot; 10 #14 kg * m/s 2
= 2.707 &quot; 10 #14 N (two extra digits carried)
FM = 2.7 &quot; 10 #14 N
Statement: The magnetic force on the proton is 2.7 &times; 10–14 N [S].
(b) Given: m = 1.67 &times; 10–27 kg; g = 9.8 m/s2
Required: Fg
Analysis: Fg = mg
Solution: Fg = mg
= (1.67 ! 10 &quot;27 kg)(9.8 m/s 2 )
= 1.637 ! 10 &quot;26 N (two extra digits carried)
Fg = 1.6 ! 10 &quot;26 N
Statement: The gravitational force on the proton is 1.6 &times; 10–26 N.
(c) Determine the ratio of the two forces on the proton:
Fg 1.637 ! 10 &quot;26 N
=
FM 2.707 ! 10 &quot;14 N
Fg 6.0 ! 10 &quot;13
=
FM
1
The gravitational force on the proton is 6.0 &times; 10–13 times the magnetic force on the proton.
2. Given: q = –1.60 &times; 10–19 C; v = 3.5 &times; 105 m/s; FM = 7.5 &times; 10–14 N; θ = 90&deg;
Required: B
Analysis: by the right-hand rule, the direction of the electric field is into the page;
FM = qvBsin !
B=
FM
qvsin !
Solution: B =
FM
qvsin !
m'
%
7.5 &quot; 10 #14 kg \$ 2
&amp;
s (
=
( #1.60 &quot; 10 #19 C )%) 3.5 &quot; 10 5 m '* sin 90&deg;
&amp;
s (
B = 1.3 T
Statement: The magnitude of the electric field is 1.3 T and it is directed into the page.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.2-1
3. (a) Given:! q = 1.60 &times; 10–19 C; v = 2.24 &times; 108 m/s; B = 0.56 T; θ = 90&deg;
Required: FM
Analysis: FM = qvB sin θ; by the right-hand rule, the direction of the force is outward from the
spiral.
Solution: FM = qvBsin !
kg (
%
= 1.60 &quot; 10 #19 C 2.24 &quot; 10 8 m/s ' 0.56
* sin 90&deg;
&amp;
C \$s )
FM = 2.0 &quot; 10 #11 N
Statement: The magnetic force on the electron is 2.0 &times; 10–11 N, outward from the spiral.
(b) Given: q = 1.60 &times; 10–19 C; v = 2.24 &times; 108 m/s; B = 5.5 &times; 10–5 T; θ = 90&deg;
Required: FM
Analysis: FM = qvB sin θ
Solution: FM = qvBsin !
kg (
%
= 1.60 &quot; 10 #19 C 2.24 &quot; 10 8 m/s ' 5.5 &quot; 10 #5
* sin 90&deg;
&amp;
C \$s )
FM = 2.0 &quot; 10 #15 N
Statement: The magnetic force on the electron after it leaves
the spiral is 2.0 &times; 10–15 N.
!
!
4. (a) Given:! q = –1.60 &times; 10–19 C; v = 6.7 &times; 106 m/s [E]; B = 2.3 T; θ = 47&deg;
Required: FM
Analysis: FM = qvB sin θ; by the right-hand rule, the direction of the electric field is north.
Solution: FM = qvBsin !
kg (
%
= ( &quot;1.60 # 10 &quot;19 C ) ( 6.7 # 10 6 m/s ) ' 2.3
* sin 47&deg;
&amp;
C \$s )
= &quot;1.803 # 10 &quot;12 N (two extra digits carried)
FM = &quot;1.8 # 10 &quot;12 N
Statement: The magnetic force on the electron is 1.8 &times; 10–12 N [N].
(b) Given: m = 9.11 &times; 10–31 kg; FM = 1.803 &times; 1012 N
Required: a
Analysis: FM = ma
F
a= M
m
FM
Solution: a =
m
m&amp;
\$
1.803 ! 10 &quot;12 kg # 2
%
s '
=
&quot;31
9.11! 10
kg
(
)(
)
(
)(
)
(
)
a = 2.0 ! 10 m/s
Statement: The acceleration of the electron is 2.0 &times; 1018 m/s2.
18
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.2-2
(c) Given: m = 1.67 &times; 10–27 kg; FM = 1.803 &times; 1012 N
Required: a
Analysis: FM = ma
F
a= M
m
F
Solution: a = M
m
m&amp;
\$
1.803 ! 10 &quot;12 kg # 2
%
s '
=
&quot;27
kg
1.67 ! 10
(
)
a = 1.1! 10 m/s
Statement: The acceleration of the proton is 1.1 &times; 1015 m/s2.
15
2
Section 8.2 Questions, page 391
1. The right-hand rule for a straight conductor and the right-hand rule for a solenoid both
describe how to determine the direction of the magnetic field if you know the direction of a
current. The right-hand rule for a moving charge in a magnetic field allows you to determine the
direction of the resulting magnetic force.
2. The particle has a positive charge, since it acts in the same direction as that determined by the
right-hand rule.
3. The particle has a negative charge according to the right-hand rule. If the charge tripled while
the velocity was halved, the magnitude of the force would be 1.5 that of the original situation:
!1 \$
FM = ( 3q ) # v&amp; Bsin '
&quot;2 %
3
qvBsin '
2
4. (a) Given: q = 1.60 &times; 10–19 C; v = 1.4 &times; 103 m/s; B = 0.85 T; θ = 90&deg;
Required: FM
Analysis: FM = qvB sin θ
Solution: FM = qvBsin !
kg (
%
= (1.60 &quot; 10 #19 C ) (1.4 &quot; 10 3 m/s ) ' 0.85
* sin 90&deg;
&amp;
C \$s )
FM = 1.9 &quot; 10 #16 N
Statement: The magnetic force on the proton is 1.9 &times; 10–16 N.
(b) The magnitude of the magnetic force on the electron is also 1.9 &times; 10–16 N because the proton
and electron have the same magnitude of charge.
FM =
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.2-3
5. Given: q = –1.60 &times; 10–19 C; v = 235 m/s; B = 2.8 T; FM = 5.7 &times; 10–17 C
Required: θ
Analysis: FM = qvBsin !
F
sin ! = M
qvB
F
Solution: sin ! = M
qvB
m(
%
#17
kg \$ 2 *
'&amp; 5.7 &quot; 10
s )
=
%
kg (
( #1.60 &quot; 10#19 C ) %'&amp; 235 ms (*) '&amp; 2.8 C \$ s *)
= #0.5414
! = sin #1 (#0.5414)
! = #33&deg;
Statement: The angle between the path of the electron and the electric field is 33&deg;.
6. By the right-hand rule, the particle is deflected downward on the plane of the page.
7. (a) Given: q = 6.4 &micro;C = 6.4 &times; 10–6 C; θ = 27&deg;; v = 170 m/s; B = 0.85 T
Required: FM
Analysis: FM = qvB sin θ
Solution: FM = qvBsin !
kg (
%
= 6.4 &quot; 10 #6 C (170 m/s ) ' 0.85
* sin &deg;
&amp;
C \$s )
FM = 4.2 &quot; 10 #4 N
Statement: The magnitude of the magnetic force on the particle is 4.2 &times; 10–4 N.
(b) By the right-hand rule, the magnetic force is in the –z direction.
(c) 0 N; there would be no force because the angle between the velocity and the magnetic field
is 0&deg;.
8. By the right-hand rule, the magnetic force is in the +z direction.
–6
9. Given: q =
! –7.9 &micro;C = –7.9 &times; 10 C; v = 580 m/s; θ = 55&deg;; B = 1.3 T [+y direction]
Required: FM
Analysis: FM = qvB sin θ; by the right-hand rule, the magnetic force is in the –z direction; the
given angle is with respect to the x-axis, so subtract it from 90&deg; to get the angle between the
velocity and the magnetic field.
Solution: FM = qvBsin !
kg (
%
= &quot;7.9 # 10 &quot;6 C ( 580 m/s ) ' 1.3
* sin ( 90&deg; &quot; 55&deg; )
&amp;
C \$s )
FM = &quot;3.4 # 10 &quot;3 N
Statement: The magnitude of the magnetic force on the particle is 3.4 &times; 10–3 N [–z direction].
(
(
)
)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.2-4
10. (a) Given: m = 6.644 &times; 10–27 kg; a = 2.4 &times; 103 m/s2
Required: FM
Analysis: FM = ma
Solution: FM = ma
= (6.644 ! 10 &quot;27 kg)(2.4 ! 10 3 m/s 2 )
= 1.595 ! 10 &quot;23 N (two extra digits carried)
FM = 1.6 ! 10 &quot;23 N
Statement: The magnitude of the magnetic force on the alpha particle is 1.6 &times; 10–23 N.
(b) Given: q = 2(1.60 &times; 10–19 C); θ = 90&deg;; B = 1.4 T; FM = 1.595 &times; 10–23 N
Required: v
Analysis: FM = qvB sin !
FM
qB sin !
FM
Solution: v =
qB sin !
v=
\$
m'
–23
&amp;% 1.595 &quot; 10 kg # s 2 )(
=
\$
kg '
2(1.60 &quot; 10 –19 C) &amp; 1.4
) sin 90&deg;
C#s(
%
v = 3.6 &quot; 10 –5 m/s
Statement: The speed of the alpha particle is 3.6 &times; 10–5 m/s.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.2-5
Section 8.3: Magnetic Force on a Current-Carrying Conductor
Tutorial 1 Practice, page 395
1. Given: L = 155 mm = 0.155 m; I = 3.2 A; B = 1.8 T; θ = 90&deg;
Required: Fon wire
Analysis: Fon wire = ILB sin θ
Solution: Fon wire = ILBsin !
C%
kg %
&quot;
&quot;
0.155 m ) \$ 1.8
= \$ 3.2
(
' sin 90&deg;
'
#
#
s&amp;
C (s &amp;
Fon wire = 0.89 N
Statement: The magnitude of the force on the wire is 0.89 N.
(b) By the right-hand rule, the magnetic force is in the –x direction.
2. Given: Fon wire = 0.75 N; I = 15 A; θ = 90&deg;; B = 0.20 T
Required: L
Analysis: Fon wire = ILBsin !
F
L = on wire
IBsin !
F
Solution: L = on wire
IBsin !
m&amp;
#
%\$ 0.75 kg &quot; s 2 ('
=
kg &amp;
C&amp;#
#
0.20
15
%\$
(%
( sin 90&deg;
s '\$
C&quot;s'
= 0.25 m
L = 25 cm
Statement: The length of the wire is 25 cm.
3. Given: Fon wire = 1.4 &times; 10–5 N; L = 0.045 m; θ = 18&deg;; B = 5.3 &times; 10–5 T
Required: I
Analysis: Fon wire = ILBsin !
F
I = on wire
LBsin !
Fon wire
Solution: I =
LBsin !
m'
%
1.4 &quot; 10 #5 kg \$ 2
&amp;
s (
=
%
kg '
( 0.045 m ) )&amp; 5.3 &quot; 10 #5
* sin18&deg;
C\$ s (
I = 19 A
Statement: The current in the wire is 19 A.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.3-1
4. Given: I = 1.5 A; L = 5.7 cm = 0.057 m; θ = 90&deg;; Fon wire = 5.7 &times; 10–6 N
Required: B
Analysis: Fon wire = ILBsin !
F
B = on wire
IL sin !
F
Solution: B = on wire
IL sin !
m'
%
5.7 &quot; 10 #6 kg \$ 2
&amp;
s (
=
C'
%
)&amp; 3.2 *( ( 0.057 m ) sin 90&deg;
s
B = 6.7 &quot; 10 #5 T
Statement: The magnitude of Earth’s magnetic field around the lamp is 6.7 &times; 10–5 T.
Section 8.3 Questions, page 396
1. (a) Given: B = 1.4 T; L = 2.3 m; Fon wire = 1.8 N; θ = 90&deg;
Required: I
Analysis: Fon wire = ILBsin !
F
I = on wire
LBsin !
F
Solution: I = on wire
LBsin !
m%
#
1.8 kg &quot; 2
\$
s &amp;
=
#
kg %
( 2.3 m ) '\$ 1.4
( sin 90&deg;
C&quot; s &amp;
I = 0.56 A
Statement: The current in the conductor is 0.56 A.
(b) When the magnetic force is a maximum, the angle is 90&deg; because that is when sin θ is a
maximum.
2. (a) Given: L = 120 mm = 0.120 m; θ = 56&deg;; B = 0.40 T; I = 2.3 A
Required: Fon wire
Analysis: Fon wire = ILB sin θ
Solution: Fon wire = ILBsin !
kg %
C%
&quot;
&quot;
= \$ 2.3 ' ( 0.120 m ) \$ 0.40
' sin 45&deg;
#
#
C (s &amp;
s&amp;
Fon wire = 7.8 ) 10 *2 N
Statement: The magnitude of the force on the wire is 7.8 &times; 10–2 N.
(b) By the right-hand rule, the direction of the magnetic force is upward.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.3-2
3. (a) Given:! L = 2.6 m; I = 2.5 A; B = 5.0 &times; 10–5 T; θ = 90&deg;
Required: Fon wire
Analysis: Fon wire = ILB sin θ; by the right-hand rule, the magnetic force is downward.
Solution: Fon wire = ILBsin !
C%
&quot;
= \$ 2.5
(
#
s '&amp;
kg %
&quot;
m ) \$ 5.0 ( 10 )5
' sin
#
C *s &amp;
&deg;
Fon wire = 3.2 ( 10 )4 N
Statement: The force on the wire is 3.2 &times; 10–4 N [down].
(b) Given: L = 2.6 m; I = 2.5 A; B = 5.0 &times; 10–5 T; θ = 72&deg;
Required: Fon wire
Analysis: Fon wire = ILB sin θ
Solution: Fon wire = ILBsin !
kg %
C%
&quot;
&quot;
= \$ 2.5
2.6 m ) \$ 5.0 ( 10 )5
(
' sin 72&deg;
'
#
#
C *s &amp;
s&amp;
Fon wire = 3.1( 10 )4 N
Statement: The magnitude of the force on the wire is 3.1 &times; 10–4 N.
4. Given: L = 1.4 m; I = 3.5 A; B = 1.5 T; θ = 90&deg;
Required: Fon wire
Analysis: Fon wire = ILB sin θ
Solution: Fon wire = ILBsin !
C%
kg %
&quot;
&quot;
1.4 m ) \$ 1.5
= \$ 3.5
(
' sin 90&deg;
'
#
#
s&amp;
C (s &amp;
Fon wire = 7.4 N
Statement: The magnitude of the force on the wire is 7.4 N.
5. (a) The magnetic field is in the same direction as the wire, so the force is 0 N.
(b) Write the length of each segment in terms of L and θ:
L
cos! =
Lhypotenuse
Lhypotenuse =
L
cos!
Lopposite
L
= L tan !
tan ! =
Lopposite
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.3-3
Write the magnetic force on each segment in terms of L and θ:
Fon wire = ILBsin !
Fhypotenuse = ILhypotenuse Bsin !
&quot; L %
Bsin !
= I\$
# cos! '&amp;
Fhypotenuse = ILB tan !
Fon wire = ILBsin !
Fopposite = ILopposite Bsin 90&deg;
= I ( L tan ! ) B
Fopposite = ILB tan !
The magnitudes of the two forces are equal. By the right-hand rule, the force on the hypotenuse
is into the page and the force on the opposite side is out of the page. That means that the sum of
the forces is 0 N.
(c) The magnetic force on a closed loop in a uniform magnetic field is zero.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.3-4
Section 8.4: Motion of Charged Particles in Magnetic Fields
Tutorial 1 Practice, page 401
1. Given: q = 3.2 ! 10–19 C; m = 6.7 ! 10–27 kg; B = 2.4 T; v = 1.5 ! 107 m/s
Required: r
mv
Analysis: r =
qB
mv
Solution: r =
qB
( 6.7 ! 10
)
m&amp;
#
kg % 1.5 ! 10 7
(
\$
s'
=
#
kg &amp;
( 3.2 ! 10 &quot;19 C ) % 2.4
(
\$
C)s'
r = 0.13 m
Statement: The radius of the ion’s path is 0.13 m.
2. Given: q = 1.60 ! 10–19 C; m = 1.67 ! 10–27 kg; B = 1.5 T; r = 8.0 cm = 0.080 m
Required: v
mv
Analysis: r =
qB
rqB
v=
m
rqB
Solution: v =
m
#
kg &amp;
( 0.080 m )(1.60 ! 10 &quot;19 C ) %\$ 1.5
(
C )s '
=
1.67 ! 10 &quot;27 kg
&quot;27
(
)
v = 1.1! 10 7 m/s
Statement: The speed of the proton is 1.1 ! 107 m/s.
3. Given: q = 1.60 ! 10–19 C; v = 6.0 ! 105 m/s; m1 = 1.67 ! 10–27 kg;
m2 = 2(1.67 ! 10–27 kg) = 3.34 ! 10–27 kg; !d = 1.5 mm = 0.0015 m
Required: B
mv
Analysis: r =
qB
In a mass spectrometer, the difference between the entry point and the ion detector is 2r. The
greater the mass of an ion, the greater the radius. So, the deuterium ion is detected at 2r + 0.0015
m from the entry point, where r is the radius of the path of the hydrogen ion:
! d = 2rdeuterium &quot; 2rhydrogen
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.4-1
Solution: ! d = 2rdeuterium &quot; 2rhydrogen
=
=
B=
2mdeuterium v 2mhydrogen v
&quot;
qB
qB
2v ( 2mhydrogen &quot; mhydrogen )
qB
2vmhydrogen
q! d
m%
#
2 6.0 ! 10 5
(1.67 &quot; 10 !27 kg )
\$
&amp;
s
=
!19
1.60
&quot;
10
C ) ( 0.0015 m )
(
B = 8.4 T
Statement: The magnitude of the magnetic field is 8.4 T.
4. (a) Since the electric force is up, the balancing magnetic force must be down. By the righthand rule, the magnetic field should be directed out of the page.
(b) The magnetic force is FM = qvB since the angle is 90&deg;. The electric force is FE = !q. These
forces are equal when the speed is proper:
F =F
! q = qvB
! = vB
!
v=
B
!
The proper velocity is v = .
B
(c) Since speed only affects the magnetic force, an ion moving too fast will experience a greater
magnetic force and be pushed downward. An ion moving too slowly will experience a greater
electric force and move upward.
Mini Investigation: Simulating a Mass Spectrometer, page 401
A. The ball bearings experience a magnetic force and deflect by different amounts, depending on
their masses. This effect is similar to what happens in a mass spectrometer.
B. This activity does not quite model the function of a mass spectrometer because the bearings
do not experience a uniform magnetic force. The force gets stronger at the bottom of the ramp,
and gravity will have a more significant effect in this simulation than on particles in a mass
spectrometer.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.4-2
Section 8.4 Questions, page 404
1. The mass spectrometer makes use of the magnetic force on a moving charged particle. Atoms
are converted into ions and then accelerated into a finely focused beam. The force deflects a
particle by an amount depending on its mass and its charge. Electric detectors identify how far
the ion travelled in the mass spectrometer.
2. Given: q = 3(1.60 ! 10–19 C) = 4.80 ! 10–19 C; mU238 = 3.952 ! 10–25 kg;
mU235 = 3.903 ! 10–25 kg; B = 9.5 T; !d = 2.2 mm = 0.0022 m
Required: v
mv
Analysis: r =
qB
In a mass spectrometer, the difference between the entry point and the ion detector is 2r. The
greater the mass of an ion, the greater the radius. So, the U-238 ion is detected at 2r + 0.0022 m
from the entry point where r is the radius of the path of the U-235 ion: ! d = 2rU238 &quot; 2rU235
! d = 2rU238 &quot; 2rU235
2mU238 v 2mU235 v
=
&quot;
qB
qB
2v ( mU238 &quot; mU235 )
qB
qB! d
v=
2 ( mU238 &quot; mU235 )
qB! d
Solution: v =
2 ( mU238 &quot; mU235 )
=
#
kg &amp;
3(1.60 ! 10 &quot;19 C ) % 9.5
( ( 0.0022 m )
\$
C )s '
=
2 3.952 ! 10 &quot;25 kg &quot; 3.903 ! 10 &quot;25 kg
(
)
v = 1.0 ! 10 6 m/s
Statement: The initial speed of the ions is 1.0 ! 106 m/s.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.4-3
3. Given: q = –1.60 ! 10–19 C; m = 9.11 ! 10–31 kg; B = 0.424 T; Ek = 2.203 ! 10–19 J
Required: r
mv
1
Analysis: Use Ek = mv 2 to determine the speed of the electron; then use r =
to determine
qB
2
Solution: Determine the speed of the electron:
1
Ek = mv 2
2
2Ek
v=
m
m &amp;
\$
2 2.203 ! 10 &quot;19 kg # 2 # m
%
'
s
=
&quot;31
9.11! 10
kg
(
)
v = 6.958 ! 10 5 m/s
o extra digits carried)
Determine the radius of the path:
mv
r=
qB
=
( 4(..!
(
!
)
#
&amp;
9g % 3(421 ! 2
(
\$
s'
#
9g &amp;
&quot;.4
5 )%
(
\$
5)s'
&quot;0.
r = 4(01 ! &quot;3
–6
Statement: The radius of the electron’s path is 9.34 ! 10
m.
!
!
!
–9
3
4. Given: q = 4 ! 10 C; v. = 3 ! 10 m/s [E 45&deg; N]; F. is upward; v2 = 2 ! 104 m/s [up];
!
F2 = 4 ! 10–5 N [W]
Required: B
Analysis: The upward force in the first situation means that, by the right-hand rule, the direction
of the magnetic field must be in the x–y plane, and somewhere within 180&deg; counterclockwise of
E 45&deg; N. The westward force in the second situation means that, by the right-hand rule, the
direction of the magnetic field must be in the y–z plane. The only possible direction that fits both
scenarios is north. Use this information and FM = qvB sin &quot; to solve for the magnitude of the
field.
F = qvBsi !
B=
F
qvsi !
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.4-4
Solution: B =
=
F
qvsi !
#
1)
\$
(1 )
&quot;4
&quot;2
9g *
#
5) % / )
\$
1
&amp;
s/ '
&amp;
( si
s '
&deg;
B=
Statement: The magnetic field is 0.5 T [N].
5. Given: q = –1.60 ! 10–19 C; m = 9.11 ! 10–31 kg; !V = 100.0 V; B = 0.0400 T
Required: r
Analysis: Use the law of the conservation of energy, !EE + !Ek = 0, along with the equations
E
.
to determine the speed of the electron. Then calculate the radius using
E9 = mv / and ! V =
q
/
mv
r=
.
qB
Solution: Determine the speed of the electron:
! E + ! E9 =
.
q! V + mv / =
/
. /
mv = &quot;q! V
/
&quot;/q! V
v=
m
&quot;/ ( &quot;
=
#
&amp;
5 )%
9g ) / ) (
\$
s 5'
&quot;0.
4(..!
9g
&quot;.4
!
(
)
v = 2(4/36 ! 3 s t o extra digits carried)
Determine the radius of the path:
mv
r=
qB
=
( 4(..!
(
!
)
#
&amp;
9g % 2(4/36 ! 3
(
\$
s'
#
9g &amp;
&quot;.4
5 )%
(
\$
5)s'
&quot;0.
r=
! &quot;1
Statement: The radius of the path described by the electron is 8.44 ! 10–4 m = 0.844 mm.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.4-5
6. (a) Given: &quot; = 90&deg;; v = 5.0 ! 102 m/s; B = 0.050 T
Required: !
Analysis: FE = FM; FM = qvB sin &quot;; FE = !q
F =F
! q = qvBsi &quot;
Solution: ! = vBsi
\$
=&amp;
%
&deg;
&quot;
/
'\$
)&amp;
s (%
9g '
)
5 #s (
! = /2 5
Statement: The strength of the electric field is 25 N/C.
(b) Given: q = 1.60 ! 10–19 C; m = 1.67 ! 10–27 kg; B = 0.050 T; v = 5.0 ! 102 m/s
Required: r
mv
Analysis: r =
qB
Solution: r =
mv
qB
&quot;
m%
(1.67 ! 10 –27 kg ) \$ 5.0 ! 102
s '&amp;
#
=
&quot;
kg %
(1.60 ! 10 –19 C ) \$ 0.050
'
C(s&amp;
#
r = 1.0 ! 10 –4 m
Statement: The radius of the proton’s path to point P is 1.0 ! 10–4 m.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.4-6
Section 8.5: Applications of Electric and Magnetic Fields
Research This: Privacy Concerns Associated with RFID Technology, page 406
A. Student answers may vary depending on the application they choose, but should reflect
careful consideration of the issues. For example, some people are concerned that buying items
with RFID tags will allow the government or corporations to track their movements, which is
considered an invasion of privacy. In the future, issues regarding privacy will arise because each
RFID is associated with an object and collecting data about the object can mean collecting data
about the person purchasing the object. If a person makes a purchase using a credit or debit card,
your personal information is now associated with the object. Purchasing habits can be registered
B. Student answers may vary, but should show evidence of research and thoughtful
consideration. For example, if there were a way to destroy the RFID tags when an item is
purchased, that would assuage those people who are concerned about being tracked.
C. RFID tags are currently being used for automated vehicle identification and registration,
product tracking and theft control, public transit payments, toll roads, and electronic payment
through cell phones.
D. Some privacy issues associated with RFID tags include the release of personal information
such as phone numbers, addresses, and medical information. Insurance companies, employers,
government institutions, and others could make decisions based on the personal information they
obtain, and these decisions could impact individuals.
E. To ensure the privacy of personal information, businesses should establish privacy policies
and procedures related to the use of RFID tags, conduct audits on the security of RFID
technology, enforce compliance with policies and procedures, and explore technology
alternatives that do not require collecting personal information.
Section 8.5 Questions, page 409
1. Answers may vary. Sample answer: RFID tags act as transponders to communicate data
through the use of electromagnetic waves. The tag detects a specific radio signal sent by an
identification code back to the transceiver.
2. (a) (i) Fewer cashiers will be needed, since computers can read the signals all at once.
(ii) There will be less shoplifting, since detectors can read the RFID chips leaving the store.
(iii) RFID technology will improve convenience for shoppers, since check-out time will
decrease.
(iv) RFID technology will improve the efficiency of inventory tracking, because grocery store
staff will not need to count stock items manually.
(b) Answers may vary. Sample answer: Some negative side effects of RFID tags in stores are
customers feeling their privacy is being invaded (since they are going home with radio
transmitters) and an increase in theft—customers could switch RFID tags on products before
checking out. Retail businesses may view the RFID tags as security against theft. Job losses
could result due to a reduced need for cashiers and security personnel in the store.
3. Answers may vary. Sample answer: Some of the challenges are standardization, cost factor,
and defects in manufacture and detection of tags. Stores will have to analyze the benefits of
buying, programming, and installing thousands of RFID tags, especially when they will still need
employees to make sure RFID tags are on the correct product and all tags are being read.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.5-1
4. Answers may vary. Sample answer: Under normal conditions, a magnetorheological (MR)
fluid is solid, but it changes to a liquid in response to a magnetic field.
5. (a) Sample answers: Magnetorheological fluid is used as a fluid damper in buildings, in car
shock absorbers, in washing machines, in prosthetics, and in exercise equipment.
(b) Student answers will vary based upon their selection, but should reflect an understanding of
the application. Sample answer: Magnetic dampers can be used in shock absorbers and
appliances like washing machines to decrease noise and vibration, which, in turn, saves energy.
6. A smart structure is a building constructed with MR fluids.
7. Sample answer: In the 1980s, some studies showed a link between magnetic field strength and
an increased risk of cancer. Since that time, however, scientists have reviewed over two decades
of research involving people exposed to electric and magnetic fields. To date, they have found no
clear evidence linking high exposures with cancers and only a weak association between
exposure to electric and magnetic fields and childhood leukemia.
8. Magnetic resonance imaging (MRI) devices use a superconducting magnet to create a large,
stable magnetic field that interacts with hydrogen atoms in the human body to produce precise
three-dimensional images of internal body systems.
9. (a) There are two types of magnets in a magnetic resonance imaging (MRI) device. The
superconducting electromagnet generates the primary magnetic field, which interacts with all the
hydrogen atoms in the body. Then, three gradient magnets are activated, which are smaller and
but also more precise. By altering these magnets, the magnetic field can be specifically focused
on a selected area. The MRI can then gather information about the particular location.
(b) The hydrogen atoms will align with the magnetic field, either in the same direction or the
opposite direction. The difference in the number of hydrogen atoms pointing one way or the
other depends on the type of material (such as skin, bone, or organs) and whether it is healthy or
abnormal and diseased. To obtain observable data, the hydrogen atoms in a specific area are
given potential energy by a radio frequency pulse. After the pulse, those atoms aligned against
the magnetic field release the energy, and the MRI device detects the energy release. The amount
of energy tells the MRI device what type of tissue is at the location, so the MRI can create a
three-dimensional map of the person in the machine.
10. (a) Sample answer: Magnetophoretic technology can be used to detect and isolate cancer
cells in bone marrow transplants.
(b) Sample answer: Magnetophoretic technology can be used to separate cells for study or stem
cell research.
11. Answer may vary. Students should explain that superconduction refers to conduction with
almost no loss of energy. Superconduction traditionally requires very cold conditions, but some
research is being conducted into exploring moderate temperatures so superconductors can be just
as effective at room temperature. Students should list applications or benefits of
superconductivity, including environmental impacts due to reduced waste of energy, which also
saves money.
12. Answer may vary. Students should report that many migratory animals, such as birds and
loggerhead turtles, use geomagnetism to navigate. Scientists have exposed these animals to
magnetic fields that simulate various locations on Earth and noted that the animals travel in the
direction they would at the simulated location.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 8: Magnetic Fields
8.5-2
Section 9.1: Properties of Waves and Light
Section 9.1 Questions, page 443
1. The frequency of a wave is determined by the frequency of the wave’s source.
2. The speed of a wave is determined by the medium in which it travels.
3. The amplitude of a wave is determined partly by its source and partly by the conditions of the
medium in which it travels.
4. The wavelength of a wave is determined by both the wave speed and the frequency. This
mathematical relationship is called the universal wave equation.
5. (a) Given: incident ray makes an angle of 10&deg; with the surface
Required: angle of incidence, !i
Analysis: The angle of incidence is measured with respect to the normal. Therefore
!i = 90&deg; &quot; 10&deg; .
Solution: !i = 90&deg; &quot; 10&deg;
!i =
&deg;
Statement: The angle of incidence is 80&deg;.
(b) The angle of reflection, !r, equals the angle of incidence. Therefore !r = 80&deg;.
(c)
6. Given: sketch of a wave; f = 40 Hz
Required: v
Analysis: The wavelength is the length of one complete wave. Measure the length of several
complete waves on the sketch, and calculate an average value for !. Then use the universal wave
equation v = f! to determine v.
Solution: The wave first crosses the x-axis at approximately 0.1 cm. Three cycles later it crosses
the x-axis at 4.0 cm.
4.0 cm &quot; 0.1 cm
!=
/
! = 1./ cm
v = f!
&quot;
= \$ 40
#
1%
1./ cm
s '&amp;
v=
cm s
Statement:
7. Given: !d = 0./ m; !t = /.5 s; f = 4.6 Hz
Required: !
Analysis: Use the distance and time information to calculate the wave speed, v =
rearrange the universal wave equation, v = f!, to isolate and solve for wavelength.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
!d
. Then
!t
9.1-1
v = f!
!=
v
f
!d
!t
0./ m
=
/.5 s
v = 0.0853 m s two extra digits carrie
Solution: v =
!=
=
v
f
0.0853 m s
1
4.6
s
! = 2 &quot; 10#2 m
Statement: The wavelength is 2 &quot; 10–2 m.
8. Given: T = 0.05 s
Required: f
Analysis: Frequency is the inverse of period, f =
Solution: f =
1
.
T
1
T
1
0.05 s
= 20 Hz
Statement: The frequency is 20 Hz.
9. Given: v = /.0 &quot; 108
f = 5.0 &quot; 1014 Hz
Required: !
Analysis: Rearrange the universal wave equation, v = f!, to isolate and solve for wavelength.
v = f!
=
!=
v
f
Solution: ! =
v
f
/.0 &quot; 108 m s
=
1
5.0 &quot; 1014
s
! = 6.0 &quot; 10#3 m
Statement: The wavelength of the light is 6.0 &quot; 10–3 m.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.1-2
10. Given: v = /.0 &quot; 108
! = 350 nm = 3.5 &quot; 10–9 m
Required: f
Analysis: Rearrange the universal wave equation, v = f!, to isolate and solve for frequency.
v = f!
v
f =
!
v
Solution: f =
!
/.0 &quot; 108 m s
=
3.5 &quot; 10#3 m
f = 4.0 &quot; 1014 Hz
Statement: The frequency of the red light waves is 4.0 &quot; 1014 Hz.
; f = 6.0 &quot; 1014 Hz
11. Given: c = /.0 &quot; 108
Required: !
Analysis: Rearrange the universal wave equation, v = f!, to isolate and solve for wavelength.
v = f!
!=
v
f
Solution: ! =
=
v
f
/.0 &quot; 108 m s
1
6.0 &quot; 1014
s
! = 5.0 &quot; 10#3 m
Statement: The wavelength of the violet light is 5.0 &quot; 10–3 m.
12. Given: distance to mirror = 2.5 m;
distance between source and reflected ray at source wall = 1.2 m
Required: !i
Analysis: !i = ! r ; sketch the situation. The normal at the point of incidence divides the triangle
into two congruent right triangles. Use the tangent ratio to determine !i .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.1-/
# 0.6 m &amp;
Solution: ! i = tan &quot;1 %
\$ 2.5 m ('
! i = 1/&deg;
Statement: The angle of incidence is 1/&deg;.
13. Given: v = 1.5 ! 10/ m s; f = 4.4 ! 102 Hz
Required: !
Analysis: Rearrange the universal wave equation, v = f!, to isolate and solve for wavelength.
v = f!
!=
v
f
Solution: ! =
v
f
1.5 &quot; 10/ m s
=
1
4.4 &quot; 102
s
! = /.4 m
Statement: The wavelength of this frequency of sound in water is /.4 m.
! = 2.0 m
14. Given: v
Required: f
Analysis: Rearrange the universal wave equation, v = f!, to isolate and solve for frequency.
v = f!
v
f =
!
v
Solution: f =
!
20.0 m s
=
2.0 m
f = 10 Hz
Statement: The frequency of the wave is 10 Hz.
15. Given: f = /.1 kHz = /.1 &quot; 10/ Hz; ! = 0.1/ m
Required: v
Analysis: v = f!
Solution: v = f !
#
1&amp;
= % /.1 &quot; 10/ ( 0.1/ m
s'
\$
v = 4.0 &quot; 102 m s
Statement: The speed of the wave is 4.0 &quot; 102
16. Given: f = 3.9 &quot; 1014 Hz; v = /.0 &quot; 108
Required: !
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.1-4
Analysis: Rearrange the universal wave equation, v = f!, to isolate and solve for wavelength.
v = f!
!=
v
f
Solution: ! =
=
v
f
/.0 &quot; 108 m s
1
3.9 &quot; 1014
s
! = /.8 &quot; 10#3 m
Statement: The wavelength of the radiation is /.8 &quot; 10–3 m.
17. Given: f = /10 MHz = /.1 &quot; 108 Hz; v = /.0 &quot; 108
Required: !
Analysis: Rearrange the universal wave equation, v = f!, to isolate and solve for wavelength.
v = f!
v
!=
f
v
Solution: ! =
f
/.0 &quot; 108 m s
/.1&quot; 108 Hz
! = 0.93 m
Statement: The wavelength of the microwaves is 0.93 m.
18. Sample answer: Mirrors can reflect images because they have a smooth reflecting surface.
When several incident light rays strike the mirror, they are reflected in the same direction, which
creates a clear image to an observer. This is called specular reflection.
Method 1: Using proportional reasoning. Frequency and wavelength are related by the universal
wave equation v = f!. For fixed wave speed, wavelength is inversely proportional to frequency.
If one frequency is a factor of three larger than another, its corresponding wavelength is onethird of the other wavelength.
Method 2: Using algebra. v = f1!1 and v = f2!2; f2 = /f1. Set the two values for v equal to each
other.
=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.1-5
f 2 !2 = f1!1
!2 f1
=
!1 f 2
=
f1
/ f1
!2 1
=
!1 /
!1
/
The ratio of the second wavelength to the first wavelength is / : 1.
20. Given: f1 = 0.1/ Hz; !1 = 0.56 m; f2 = 0.45 Hz
Required: !2
Analysis: Use the universal wave equation, v = f!, to calculate the wave speed. Then use the
wave speed and f2 to determine !2.
v
!2 =
Solution: v = f !
f2
= 0.8/ Hz 0.56 m
0.4648 m s
=
v = 0.4648 m s two extra digits carrie
1
0.45
s
!2 = 1.0 m
Statement: When the frequency is 0.45 Hz, the new wavelength is 1.0 m.
21. (a) Sample answer: A flat mirror causes specular reflection because its surface is smooth and
regular and reflects the rays of a parallel beam of light in one direction.
(b) Sample answer: A piece of notebook paper causes diffuse reflection because the paper fibres
have many orientations and reflect the rays of a parallel beam of light in many different
directions.
(c) Sample answer: The surface of a puddle on a calm day causes specular reflection because it
is smooth and regular and reflects the rays of a parallel beam of light in one direction.
(d) Sample answer: The surface of a lake on a windy day causes diffuse reflection because the
rough waves reflect the rays of a parallel beam of light in many directions.
!2 =
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.1-6
Section 9.2: Refraction and Total Internal Reflection
Tutorial 1 Practice, page 449
1. The angle of incidence is 65&deg;. The fact that the experiment takes place in water does not
change the angle of incidence.
2. Given: ! i = 47.5&deg;; ! R = 34.0&deg;; nair = 1.0003
Required: n2
Analysis: Index of refraction is a physical property that can be used to identify a substance. Use
Snell’s law, n1 sin θ1 = n2 sin θ 2 , to calculate the index of refraction of the medium. Then match it
to a substance in Table 1.
Solution: n1 sin !1 = n2 sin ! 2
n2 =
n1 sin !1
sin ! 2
(1.0003)sin 47.5&deg;
sin 34.0&deg;
n2 = 1.32
According to Table 1, the index of refraction lies between that of ice and liquid water but is
closer to water.
Statement: The medium is probably water.
3. Given: !1 = 35&deg;; ! R = 25&deg;; nair = 1.0003
=
Required: n2
Analysis: n1 sin θ1 = n2 sin θ 2
Solution: n1 sin !1 = n2 sin ! 2
n2 =
n1 sin !1
sin ! 2
(1.0003)sin 35&deg;
sin 25&deg;
n2 = 1.36
Statement: The index of refraction of the water is 1.36.
4. Given: n = 2.42; c = 3.0 ! 108 m/s
Required: v
=
v
Analysis: Use the definition of index of refraction, n = , to solve for the speed v.
c
c
n=
v
c
v=
n
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.2-1
c
n
3.0 ! 108 m/s
=
2.42
v = 1.2 ! 108 m/s
Statement: The speed of light in diamond is 1.2 ! 108 m/s .
Solution: v =
5. Given: n = 1.46; !1 = 5.6 &quot; 10#7 m
Required: λ2
Analysis: Use the alternative definition of index of refraction, n =
n=
!1
!2
!2 =
!1
n
λ1
, to solve for λ2 .
λ2
!1
n
5.6 &quot; 10#7 m
=
1.46
!2 = 3.8 &quot; 10#7 m
Solution: !2 =
Statement: The wavelength of light in quartz is 3.8 ! 10&quot;7 m .
6. Given: n = 1.45; !1 = 450 nm = 4.5 &quot; 10#7 m
Required: f 2
Analysis: The frequency of light does not change when light passes from one medium into
another. The frequency of the light inside the glass is the same as in vacuum. Rearrange the
universal wave equation, v = fλ, to solve for f.
v = f!
v
f =
!
v
Solution: f =
!
3.0 &quot; 108 m /s
=
4.5 &quot; 10#7 m
f = 6.7 &quot; 1014 Hz
Statement: The frequency of the light is 6.7 ! 1014 Hz .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.2-2
Tutorial 2 Practice, page 452
1. (a) Given: θ1 = 40.0&deg;; n1 = 1.0003; n2 = 1.465
Required: θ 2
Analysis: n1 sin θ1 = n2 sin θ 2
Solution: n1 sin !1 = n2 sin ! 2
sin ! 2 =
n1 sin !1
n2
# n sin !1 &amp;
! 2 = sin &quot;1 % 1
(
\$ n2 '
# (1.0003)sin 40.0&deg; &amp;
= sin &quot;1 %
('
1.465
\$
= 26.033&deg; (two extra digits carried)
! 2 = 26.0&deg;
Statement: The angle of refraction at the left boundary of the prism is 26.0&deg;.
(b) Given: θ2 = 26.033&deg; ; n3 = 1.465
Required: θ 4
Analysis: From the geometry of the prism, the angle of incidence at the right boundary, θ3, is
θ3 = 60.0&deg; − θ2 . Determine θ 2 , then use Snell’s law, n3 sin θ3 = n4 sin θ4 , to calculate the angle of
refraction, θ 4 .
Solution: ! 3 = 60.0&deg; &quot; ! 2
= 60.0&deg; &quot; 26.033&deg;
! 3 = 33.967&deg; (two extra digits carried)
n3 sin ! 3 = n4 sin ! 4
sin ! 4 =
n3 sin ! 3
n4
# n sin ! 3 &amp;
! 4 = sin &quot;1 % 3
(
\$ n4 '
# (1.465)sin 33.967&deg; &amp;
= sin &quot;1 %
('
1.0003
\$
! 4 = 54.9&deg;
Statement: The angle of refraction of the exiting light is 54.9&deg;.
2. Sample answer: The light entered and exited the prism on faces that were not parallel. You
would only see the exit angle equal to the incident angle if the faces were parallel, as in a sheet
of glass.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.2-3
3. Given: θ1 = 55&deg;; n = 1.60
Required: θ, the angle of the outgoing ray as measured with the horizontal
Analysis: Calculate the first angle of refraction, θ 2 , using Snell’s law, n1 sin θ1 = n2 sin θ 2 . Then
determine the second angle of incidence, θ3 , using θ3 = 60.0&deg; − θ 2 . Use Snell’s law to calculate
the second angle of refraction, θ 4 . Determine the exit angle, θ, with respect to the horizontal.
Solution:
The first angle of refraction is θ 2 .
n1 sin !1 = n2 sin ! 2
sin ! 2 =
n1 sin !1
n2
# n sin !1 &amp;
! 2 = sin &quot;1 % 1
(
\$ n2 '
# (1.0003)sin55&deg; &amp;
= sin &quot;1 %
('
1.60
\$
! 2 = 30.81&deg; (two extra digits carried)
The second angle of incidence is θ3 .
! 3 = 60.0&deg; &quot; ! 2
= 60.0&deg; &quot; 30.81&deg;
! 3 = 29.19&deg; (two extra digits carried)
The second angle of refraction is θ 4 .
n3 sin ! 3 = n4 sin ! 4
sin ! 4 =
n3 sin ! 3
n4
# n sin ! 3 &amp;
! 4 = sin &quot;1 % 3
(
\$ n4 '
# (1.60)sin
= sin &quot;1 %
1.0003
\$
&deg;&amp;
('
! 4 = 51&deg;
The normal on the right side of the prism is directed at 30&deg; above the horizontal, so the exit angle
is θ = θ 4 − 30&deg;
= 51&deg; − 30&deg;
θ = 21&deg;
Statement: The light exits at 21&deg; below the horizontal.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.2-4
Research This: Using Spectroscopy to Determine Whether Extra-Solar Planets
Can Support Life, page 452
A. Answers may vary. Sample answers: Light reflected from other planets can be seen and
analyzed on Earth. When the light passes through a spectrometer, it is dispersed (broken up) into
its component colours and makes a spectrum similar to the one shown in the text. Scientists use
the dark lines in the spectrum to identify the atom or molecule that absorbed the missing colours.
This atom or molecule had to be on the planet where the light was reflected.
B. Answers may vary. Sample answers: Astrophysicists and astrobiologists look for oxygen,
carbon, nitrogen, and hydrogen. On Earth, these are the main elements involved in biological
processes. Finding these elements elsewhere could indicate the right conditions for
extraterrestrial life.
C. Answers may vary. Sample answers: Scientists think of light as a wave when using a
diffraction grating in a spectrometer. But they also think of light as particle when it is absorbed
by or emitted from an atom.
Tutorial 3 Practice, page 457
1. Answers may vary. Sample answer: I will use a liquid with n = 1.20 for my comparison.
Given: n1 = 1.20; n2 = 1.0003
Required: θ c
⎛n ⎞
Analysis: θc = sin −1 ⎜ 2 ⎟
⎝ n1 ⎠
⎛n ⎞
Solution: θ c = sin −1 ⎜ 2 ⎟
⎝ n1 ⎠
⎛ 1.0003 ⎞
= sin −1 ⎜
⎟
⎝ 1.20 ⎠
θ c = 56.4&deg;
Statement: If the index of refraction of the liquid is decreased to 1.20, then the critical angle
increases to 56.4&deg;. As the index of refraction decreases, the critical angle increases.
2. Given: n1 = 1.50; n2 = 1.33
Required: θ c
⎛n ⎞
Analysis: θc = sin −1 ⎜ 2 ⎟
⎝ n1 ⎠
⎛n ⎞
Solution: θ c = sin −1 ⎜ 2 ⎟
⎝ n1 ⎠
⎛ 1.33 ⎞
= sin −1 ⎜
⎟
⎝ 1.50 ⎠
θc = 62.5&deg;
Statement: The critical angle for light at the benzene–water boundary is 62.5&deg;.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.2-5
3. Given: n1 = 1.40; n2 = 1.0003
Required: θ c
⎛n ⎞
Analysis: θc = sin −1 ⎜ 2 ⎟
⎝ n1 ⎠
⎛n ⎞
Solution: θ c = sin −1 ⎜ 2 ⎟
⎝ n1 ⎠
⎛ 1.0003 ⎞
= sin −1 ⎜
⎟
⎝ 1.40 ⎠
θ c = 45.6&deg;
Statement: The critical angle for light on the glass–air boundary is 45.6&deg;.
4. Given: nd = 2.42; ng = 1.52; nz = 1.92; nair = 1.0003
Required: θc; θg; θz
⎛n ⎞
Analysis: θc = sin −1 ⎜ air ⎟
⎝ nmed ⎠
Solution:
Diamond:
#n &amp;
! = sin &quot;1 % i (
\$ n '
# 1.0003 &amp;
= sin &quot;1 %
('
\$
!
=
&deg;
Crown glass:
#n &amp;
! c,g = sin &quot;1 % air (
\$ ng '
# 1.0003 &amp;
= sin &quot;1 %
\$ 1.52 ('
! c,g = 41.2&deg;
Zircon:
#n &amp;
! c,z = sin &quot;1 % air (
\$ nz '
# 1.0003 &amp;
= sin &quot;1 %
\$ 1.92 ('
! c,z = 31.4&deg;
Statement: The critical angle for diamond is 24.4&deg;. The critical angle for zircon is 31.4&deg;. The
critical angle for crown glass is 41.2&deg;.
Diamond has a smaller critical angle than crown glass and zircon, so a light ray passing through
diamond is more likely to reflect off the surface. If the light passes into the diamond from an
angle that is less than the normal angle of 90&deg; (most probable), then the refraction will be more
likely to disperse the spectrum than a material such as glass, which has a far lower index of
refraction. The diamond appears to glitter.
Additional information: Light rays that pass through a piece of material like diamond may reflect
off the surface several times before finally passing out of the material in a different direction than
when they entered. This effect gives a viewer the impression that light sources inside the material
produced the light, even if the light came from a source outside the material.
Section 9.2 Questions, page 458
1. Answers may vary. Sample answer: When light travels from one medium to another, its
direction of propagation changes. This change in direction during refraction makes the light ray
appear to “bend.”
2. When light is reflected or refracted, its direction changes. The change in angle between the
incident ray and the outgoing ray is the angle of deviation.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.2-6
3. Given: n =
Required: λ2
; !1 =
nm = 6.3 &quot; 10#7 m
Analysis: Rearrange the equation for index of refraction, n =
n=
!1
!2
!2 =
!1
n
λ1
, to solve for wavelength.
λ2
!1
n
6.3 &quot; 10#7 m
=
1.33
! =
&quot; 10#7 m
Solution: ! =
Statement: In water, the laser light has a wavelength of 4.7 &times; 10–7 m, or 470 nm.
4. Given: v = 3.0 &times; 108 m/s; c = 3.0 &times; 108 m/s
Required: n
v
Analysis: n =
c
v
Solution: n =
c
3.0 ! 108 m/s
=
3.0 ! 108 m/s
n = 1.0
Statement: The index of refraction of the medium is 1.0.
5. Given: θ1 = 30.0&deg;; n1 = 1.47; n2 = 1.33; n3 = 1.0003
Required: θ3
Analysis: One method is to use Snell’s law, n1 sin θ1 = n2 sin θ 2 , to determine the angle of
refraction in the water film. This angle is the incident angle for the second refraction into air. Use
Snell’s law to determine the angle of refraction in air. A second method recognizes that the film
of water does not matter because its surfaces are parallel. We could use Snell’s law to go directly
from glass to air.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.2-7
Solution:
First method:
n1 sin !1 = n2 sin ! 2
sin ! 2 =
n1 sin !1
n2
# n sin !1 &amp;
! 2 = sin &quot;1 % 1
(
\$ n2 '
# (1.47)sin 30.0&deg; &amp;
= sin &quot;1 %
('
1.33
\$
! 2 = 33.548&deg; (two extra digits carried)
n sin ! = n3 sin ! 3
sin ! 3 =
n sin !
n3
# n sin ! &amp;
! 3 = sin &quot;1 %
(
\$ n3 '
# (1.33)sin
= sin &quot;1 %
1.0003
\$
&deg;&amp;
('
! 3 = 47.3&deg;
Second method:
n1 sin !1 = n3 sin ! 3
sin ! 3 =
n1 sin !1
n3
# n sin !1 &amp;
! 3 = sin &quot;1 % 1
(
\$ n3 '
# (1.47)sin 30.0&deg; &amp;
= sin &quot;1 %
('
1.0003
\$
! 3 = 47.3&deg;
Statement: The angle of refraction of the final outgoing ray is 47.3&deg;.
6. Given: θ1 = 30.0&deg;; n1 = 1.44; n2 = 1.0003
Required: n2
Analysis: n1 sin θ1 = n2 sin θ 2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.2-8
Solution: n1 sin !1 = n2 sin ! 2
sin ! 2 =
n1 sin !1
n2
# n sin !1 &amp;
! 2 = sin &quot;1 % 1
(
\$ n2 '
# (1.44)sin 30.0&deg; &amp;
= sin &quot;1 %
('
1.0003
\$
! 2 = 46.0&deg;
Statement: The angle of refraction is 46.0&deg;.
7. Given: θ1 = 50.0&deg;; n1 = 1.33; n2 = 1.0003
Required: n2
Analysis: n1 sin θ1 = n2 sin θ 2
Solution: n1 sin !1 = n2 sin ! 2
sin ! 2 =
n1 sin !1
n2
# n sin !1 &amp;
! 2 = sin &quot;1 % 1
(
\$ n2 '
# (1.33)sin50.0&deg; &amp;
= sin &quot;1 %
('
1.0003
\$
(
)
! 2 = sin &quot;1 1.0185
There is no solution for ! 2 .
Statement: The incident angle of the laser beam in water is greater than the critical angle in
water. The laser beam undergoes total internal reflection.
8. (a) Given: n1 = 1.65; n2 = 1.33
Required: θ c
⎛n ⎞
Analysis: θc = sin −1 ⎜ 2 ⎟
⎝ n1 ⎠
⎛n ⎞
Solution: θ c = sin −1 ⎜ 2 ⎟
⎝ n1 ⎠
⎛ 1.33 ⎞
= sin −1 ⎜
⎟
⎝ 1.65 ⎠
θc = 53.7&deg;
Statement: The critical angle for light at a glass–water boundary is 53.7&deg;.
(b) The light starts in the medium with the higher index of refraction, which is the glass. There
can be no total internal reflection if the light starts in the medium with the lower index of
refraction.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.2-9
9. (a) Given: θ2 = 45&deg;; n1 = 1.0003; n2 = 1.30
Required: θi = θ1
Analysis: n1 sin θ1 = n2 sin θ 2
Solution:
n1 sin θ1 = n2 sin θ 2
sin θ1 =
n2 sin θ 2
n1
⎛ n2 sin θ 2 ⎞
⎟
⎝ n1 ⎠
θ1 = sin −1 ⎜
⎛ 1.30sin 45&deg; ⎞
= sin −1 ⎜
⎟
⎝ 1.0003 ⎠
θ1 = 67&deg;
Statement: The angle of incidence is 67&deg;.
(b) Given: n1 = 1.30; n2 = 1.0003
Required: θ c
⎛n ⎞
Analysis: θc = sin −1 ⎜ 2 ⎟
⎝ n1 ⎠
⎛n ⎞
Solution: θ c = sin −1 ⎜ 2 ⎟
⎝ n1 ⎠
⎛ 1.0003 ⎞
= sin −1 ⎜
⎟
⎝ 1.30 ⎠
θ c = 50.3&deg;
Statement: The critical angle for light at the transparent material–air boundary is 50.3&deg;.
10. (a) Given: θ1 = 30.0&deg;; n1 = 1.33; n2 = 1.63
Required: θR = θ 2
Analysis: n1 sin θ1 = n2 sin θ 2
Solution: n1 sin !1 = n2 sin ! 2
sin ! 2 =
n1 sin !1
n2
# n sin !1 &amp;
! 2 = sin &quot;1 % 1
(
\$ n2 '
# (1.33)sin 30.0&deg; &amp;
= sin &quot;1 %
('
1.63
\$
! 2 = 24.1&deg;
Statement: The angle of refraction is 24.1&deg;.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.2-10
(b) Light incident on a water–carbon disulfide boundary cannot undergo total internal reflection
because the index of refraction of carbon disulfide is greater than the index of refraction of
water.
11. Answers may vary. Sample answer: Fibre optics, which use total internal reflection, are used
in medicine to view inside various parts of the body. One example of an instrument that uses
fibre optics is the endoscope. Doctors use an endoscope to examine a patient’s internal tissues
and organs.
Additional information: An angioscope is a coated fibre optic cable with a fish-eye lens that can
be inserted into a blood vessel to diagnose constrictions, blockages, or weaknesses. A
gastroscope is another variation of fibre optic cable that is swallowed and is used to view the
esophagus, stomach, and some of the small intestine. Most medical fibre optic scopes also have
mechanisms for taking tissue samples or for removing diseased tissue.
12. (a) Answers may vary. Sample answer: Hibernia Atlantic and Emerald Express are two
international companies with plans for new fibre optic transatlantic cables.
(b) The biggest advantage of submarine cables is that the time for transmission and reception of
the signal is significantly shorter than when using satellite communication. This may not seem
like a major advantage for conversations, but most transatlantic communication involves
investment trading, when every millisecond counts. The newest cables are aiming for round-trip
transit times of 60 ms. The biggest disadvantage of submarine cables is cost. This technology
contributes to the escalating prices for communications.
13. Answers may vary. Sample answer: Signal reduction, usually called attenuation, in optical
fibres occurs for a number of reasons. One reason is that impurities in the fibre may absorb the
signal. More significantly, there are losses due to reflection from the core or cladding, and losses
due to splicing of the cables. These losses occur because the signal reflects off these surfaces at
an angle that will allow transmission out of the fibre.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.2-11
Section 9.3: Diffraction and Interference of Water Waves
Tutorial 1 Practice, page 461
1. Given: λ = 1.0 m; w = 0.5 m
Required:
λ
w
Analysis: Diffraction should be noticeable if
Solution:
λ 1.0 m
=
w 0.5 m
λ
w
λ
w
≥ 1, so solve for
λ
w
.
=2
Statement: Yes, the diffraction should be noticeable because
2. Given: λ = 630 nm = 6.3 &times; 10–7 m
Required: maximum width w for noticeable diffraction
Analysis: Use the condition that
Solution:
λ
≥
w
λ≥w
λ
w
λ
w
is greater than 1.
≥ .
&times; 0\$m ≥ w
Statement: The maximum slit width for significant diffraction to be produced is 6.3 &times; 10–7 m.
Mini Investigation: Interference from Two Speakers, page 464
A. Answers may vary. Sample answer: The distances to the two speakers should differ by zero or
a whole number of wavelengths to get constructive interference.
B. Answers may vary. Sample answers: The distances to the two speakers should differ by a
half-whole number of wavelengths to get destructive interference.
C. Answers may vary. Sample answers: If a sound of known frequency and wavelength is
played, students can compare their estimates with the known values.
Tutorial 2 Practice, page 468
1. Given: two-source interference; λ = 2.1 m
Required: d, smallest path difference for a node
⎛
⎞
Analysis: Use PnS − PnS2 = ⎜ n − ⎟ λ with n = 1.
2⎠
⎝
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.3-1
⎛
⎞
Solution: PnS − PnS2 = ⎜ n − ⎟ λ
2⎠
⎝
⎛
⎞
= ⎜ − ⎟ 2.1 m
⎝ 2⎠
P S − P S2 = .2 m
Statement: The smallest path difference for a node is 1.2 m.
2. (a) Given: n = 3; P3S1 = 35 cm; P3S2 = 42 cm
Required: λ
⎛
⎞
Analysis: PnS − PnS2 = ⎜ n − ⎟ λ
2⎠
⎝
P S − PnS2
λ= n
1
n−
2
P3S1 − P3S2
Solution: λ =
1
n−
2
35 cm − 42 cm
=
2.1
λ = 2.8 cm
Statement: The wavelength of the waves is 2.8 cm.
(b) Given: f = 0.1 P λ = 2.8 cm
Required: v
Analysis: v = f λ
Solution: v = f λ
= 54
1 2.8 cm1
v = 2 cm3
Statement: The speed of the waves is 29 cm/s.
3. (a) Given: n = 2S P2S = 29.5 cm; P2S2 = 25.0 cm; v = 7.5 cm/s
Required: λ
⎛
⎞
Analysis: PnS − PnS2 = ⎜ n − ⎟ λ
2⎠
⎝
P S − PnS2
λ= n
n−
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.3-2
Solution: λ =
=
P2S − P2S2
n−
2
29.5 cm − 25.0 cm
λ=
cm
Statement: The wavelength of the waves is 3.0 cm.
(b) Given: v = m.1 cm3s; λ = 1.0 cm
Required: f
Analysis: v = f λ
f =
Solution: f =
v
λ
v
λ
cm
s
=
1.0 cm
f = 2.5 Hz
Statement: The frequency at which the sources are vibrating is 2.5 Hz.
m.1
Section 9.3 Questions, page 469
1. Diffraction of waves through a slit is maximized when the wavelength is comparable to or
somewhat greater than the slit width.
2. Answers may vary. Sample answer: When the waves reach my friend in phase, there is
constructive interference and he hears a loud sound. When the phase of one speaker is changed
by 180&deg;, the waves reach my friend out of phase and he hears a sound with decreased volume.
3. (a) Given: λ = 7.3 &times;20−5
Required: maximum width w for noticeable diffraction
Analysis: Use the condition that
Solution:
λ
w
λ
w
≥ .
≥2
λ≥w
−5
7.3 &times;20
≥w
Statement: The maximum slit width for noticeable diffraction is 6.3 &times; 10–4 m.
(b) Sample answer: If the slit is wider than 6.3 &times; 10–4 m, there may still be some diffraction. The
wider the slit is, the less diffraction will be noticeable.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.3-3
4. Given: d = .0
Required: !
Analysis: d
λ = 0.
⎛
⎝
n =2
⎞
⎠
θn = ⎜ n − ⎟ λ
2⎞
⎛
⎜n − ⎟λ
⎠
θn = ⎝
d
Solution:
⎛
2⎞
⎜⎝ n − ⎟⎠ λ
m θn =
d
⎛ ⎛ 2⎞
⎜ ⎜⎝ 2− ⎟⎠
θ2 = m −2 ⎜
⎝
2.0 s
θ2 = &deg;
⎞
s ⎟
⎟
⎠
Statement: The angle of the first nodal line is 7.2&deg;.
5. Sample answer: For the interference pattern from a two-point source to be stable, the phase
between the sources must not change.
6. (a)
(b) Given: d = 1.0 cmS n = 5S x = 45 cm − 35 cm = 0 cmS L = 10 cmS f = 5.0 Hz
Required: λ
Analysis: The distance from the nodal points to the midpoint between the sources is close to the
⎛
⎞ Lλ
distance L between the line joining the sources and the metre stick. Rearrange xn = ⎜ n − ⎟
2⎠ d
⎝
xn d
.
to determine the wavelength, λ =
⎛
⎞
⎜n− ⎟L
2⎠
⎝
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.3-4
Solution: λ =
xn d
⎛
⎞
⎜n− ⎟L
2⎠
⎝
(10 cm)(5.0 cm )
⎛ 1⎞
⎜1 − ⎟ (50 cm )
⎝ 2⎠
λ = 2 cm
Statement: The wavelength of the waves is 2 cm.
(c) Given: λ = 2 cmS f = 5.0 Hz
Required: v
Analysis: v = f λ
Solution: v = f λ
= (5.0 Hz)(2 cm)
v = 12 cm/s
Statement: The speed of the waves is 12 cm/s.
7. (a) Given: n = 3S x3 = 15 cm; L = 77 cm; d = 5.0 cmS !3 = 21&deg; ;
0
c s = 4.2 cm
Required: Calculate λ using three different methods.
⎛
⎞ Lλ
, with n = 3. For the second method, use
Analysis: For the first method, use xn = ⎜ n − ⎟
2⎠ d
⎝
⎛
⎞
d θ n = ⎜ n − ⎟ λ , with n = 3. For the third method, use the measurement between wave crests
2⎠
⎝
to determine the wavelength directly.
Solution:
First method:
2 ⎞ Lλ
⎛
xn = ⎜ n − ⎟
2⎠ d
⎝
x3d
λ=
⎛
⎞
⎜3− ⎟ L
2⎠
⎝
=
(15 cm1(7.0 cm 1
5⎞
⎛
cm )
⎜3− ⎟(
2⎠
⎝
=
cm
t n
λ = a.a cm
=
cn
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.3-5
Second method:
⎛
⎞
d θn = ⎜ n − ⎟ λ
2⎠
⎝
d θn
λ=
1
n−
2
(5.0 cm1
=
1
−
2
= .0 cm (
&deg;
n
cn t
cm
λ=
Third method:
The distance between five consecutive crests corresponds to four whole wavelengths:
5λ = 5.2 cm
5.2 cm
5
= 2.01 cm
c0 6)
λ = 1.0 cm
Each of the three methods of analysis resulted in a wavelength between 1.0 cm and 1.1 cm, with
an average of 1.0 cm.
Statement: The wavelength is 1.0 cm.
(b) Answers may vary. Sample answer: The three methods are based on data with two significant
digits. The three results differed only by one unit in the last digit. I think that these results are
consistent and that no particular measurement stands out as being incorrect.
λ=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.3-6
Section 9.4: Light: Wave or Particle?
Research This: Very Long Baseline Interferometry, page 475
A. The Dominion Radio Astrophysical Observatory in Penticton, British Columbia, uses Very
Long Baseline Interferometry (VLBI).
B. Answers may vary. Sample answer: Usually, data at each telescope are recorded digitally onto
computer hard drives. The data from each telescope are combined to form an image.
C. Answers may vary. Sample answer: The signal is sampled at each telescope, then stored and
shipped to a central location for later processing with data from other telescopes.
D. Answers may vary. Sample answer: Many scientists store the data on disc and ship the discs.
Newer methods use Internet transfer to send the data to a central location.
E. Answers may vary. Sample answer: When the data are played back, an atomic clock is used to
synchronize the data with the time at different telescopes.
F. Answers may vary. Sample answers: The word “interferometry” is from the word
“interferometer.” An interferometer uses interference patterns from electromagnetic radiation to
form images. Interference is a wave-like property of light. The data from all the telescopes in a
baseline array are combined to form interference patterns, which reveal information about the
object under study. Another wave-like property used in interferometry is reflection. The
electromagnetic radiation from the object reaches the large, parabolic radio telescope dishes, and
the radio waves are reflected to a focus and then to a receiver. From the receiver, the signals are
transmitted for image processing.
Additional information: Radio waves reveal different types of information in the object under
study than visible light radiation. For example, quasars are massive, high-energy objects at the
edges of the known universe. In visible light, through an optical telescope, for example, quasars
appear as points of light. Images compiled from radio waves reveal the massive quantities of
energy emanating from quasars, as well as their enormous size.
Section 9.4 Questions, page 476
1. (a)
(b)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.4-1
(c)
2. Answers may vary. Sample answer: Based on what I have learned in this section about the
wave and particle properties of light, I think that the wave model explains properties of light
better than the particle model. The wave model explains reflection, refraction, dispersion, and
interference better than the particle model. The particle model partially explains rectilinear
propagation.
Additional information: Newton’s corpuscular model predicts that light does not need a medium
in which to travel, which is true.
3. Sample answer: Double-slit interference patterns provide strong evidence that light is a wave.
4. Answers may vary. Sample answer: The frequency and wavelength of light do not change
when light is reflected. So, according to the universal wave equation, v = fλ, the speed must stay
the same.
5. Answers may vary. Sample answer: Newton did not detect any pressure from light, so he
reasoned that the mass of a light particle is very low.
6. Answers may vary. Sample answer: Newton’s theory of light was dominant for so long
because Newton had a very high reputation in the physics community because of his successes in
other studies, especially gravity. Another reason that Newton’s theory took so long to refute was
that the technology of Young’s experiment did not exist until much later.
Additional information: Newton was also the powerful head of the British Royal Society and
could influence the opinions of others.
7. Sample answer: Huygens’ principle applies to all waves, including water and sound waves.
the wave and particle models of light.
(a) Light travels from the Sun to Earth in a straight line, in a vacuum. Rectilinear propagation is
explained by a particle model of light.
(b) The energy travelling for TV, radio, and X-ray technologies is best understood if light is
considered to be a wave because waves carry energy.
9. (a) Answers may vary. Sample answer: The laser light passing through an open window
shows no diffraction because the wavelength of the light is much smaller than the width of the
window.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.4-2
window, the wavelength of the radiation must be about the width of the window or somewhat
greater. Based on my research for the Research This activity on baseline interferometry, I know
waves can probably also diffract, although it may not be enough to be noticeable. So I would use
longer-wavelength and consider decreasing the size of the window, if that was an option.
10. Answers may vary. Sample answer: Young’s double-slit experiment demonstrated that light
waves can interfere, confirming that light is a wave. Therefore, Young’s experiment contradicted
Newton’s corpuscular theory of light. Newton’s corpuscular theory of light cannot account for
the observed interference pattern.
11. (a) Answers may vary. Sample answer: Light waves cause the electrons to vibrate. These
vibrations are called surface plasmons. A dielectric material is a material that does not conduct
direct current. When a dielectric material is placed against a metallic surface, travelling waves of
electron vibrations can be trapped in the interface between the materials. The travelling waves
are called surface plasmon polaritons.
(b) Answers may vary. Sample answer: Surface plasmon polaritons have multiple applications in
nanotechnology, for example, optical data storage.
12. Answers may vary. Sample answer: Newton believed that light was composed of particles,
which he called corpuscles, and that corpuscles were subject to gravitational attraction. He
therefore believed that these particles had mass. In explaining refraction, the attraction of a
massive body (water or glass) caused the light to bend and speed up. Newton tried to disprove
Grimaldi’s beliefs about diffraction of light. Newton argued that Grimaldi’s observations of light
diffraction were a result of collisions between light particles at the edges of the slit, and not a
result of waves of light spreading out. By the time Newton wrote Opticks, his book on light, he
explained diffraction as a kind of refraction. This is since understood to be incorrect. Newton’s
greatest contribution to optics was the demonstration that light could be broken down into its
spectral colours by a prism, and that after passing through a second prism the light would appear
white. His theory could explain this at a time when other theories could not. However, in 1850,
Foucault showed that light travels more slowly in water, not faster, and Newton’s theory was put
to rest. His entire theory of light has been demonstrated to be incorrect, except for the prediction
that light does not need a medium in which to travel.
13. In his book Micrographia, published in 1665, Robert Hooke described light as vibrations,
and compared the movement of light to the movement of water waves. Hooke suggested that the
vibrations of light are perpendicular to the direction of travel. He also proposed that light was not
made up of particles as Newton suggested.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.4-3
Section 9.5: Interference of Light Waves: Young’s Double-Slit
Experiment
Tutorial 1 Practice, page 482
1. Given: double-slit interference; n = 5; θ5 = 3.8&deg;; d = 0.042 mm = 4.2 &times; 105 m
Required: λ
1⎞
⎛
Analysis: Rearrange the equation d sin θ n = ⎜ n − ⎟ λ to solve for wavelength;
2⎠
⎝
#
1&amp;
d sin ! n = % n &quot; ( )
2'
\$
)=
d sin ! n
1
n&quot;
2
d sin θ n
1
n−
2
(4.2 &times; 10−5 m) sin 3.8&deg;
=
1
5−
2
−7
λ = 6.2 &times; 10 m
Statement: The wavelength of the monochromatic light is 6.2 &times; 10–7 m.
2. Given: double-slit interference; d = 0.050 mm = 5.0 &times; 10–5 m; λ = 650 nm = 6.5 &times; 10–7 m;
L = 2.6 m
Required: Δx , at the centre of the interference pattern
Lλ
Analysis: Δx =
d
Solution:
Lλ
Δx =
d
(2.6 m)(6.5 &times;10 −7 m)
=
5.0 &times; 10−5 m
Solution:
λ=
= 3.4 &times; 10−2 m
Δx = 3.4 cm
Statement: The fringe separation at the centre of the interference pattern is 3.4 cm.
3. Given: double-slit interference; n = 3; λ = 652 nm = 6.52 &times; 10–7 m; d = 6.3 &times; 10–6 m
Required: θ3
1⎞
⎛
Analysis: Rearrange the equation d sin θ n = ⎜ n − ⎟ λ to solve for the angle of the fringe;
2⎠
⎝
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.5-1
#
1&amp;
d sin ! n = % n &quot; ( )
\$
2'
##
1&amp; &amp;
%\$ n &quot; (' ) (
%
2
! n = sin &quot;1 %
('
\$
d
Solution:
##
1&amp; &amp;
%\$ n &quot; (' ) (
%
2
! n = sin &quot;1 %
('
\$
d
##
&amp;
1&amp;
&quot;7
%\$ 3&quot; (' (6.52 * 10 m ) (
%
2
= sin &quot;1 %
(
\$
'
6.3* 10&quot;6 m
! 3 = 15&deg;
Statement: The fringe is observed at the angle 15&deg;.
Mini Investigation: Wavelengths of Light, page 483
A. Both interference patterns consisted of a horizontal band of light filled with coloured and dark
vertical fringes. The fringes were closer together with the green filter than with the red filter.
B. Table 1: Number of Lines between Two Sliders
Number of lines found
Colour of light
between two sliders
red
9
green
11
C. Table 2: Calculating the Wavelength of Light
Red light
L (m)
1.0
d (m)
1.76 &times;10−4
n
6
x (m)
0.015
Δx (m)
2.5 &times;10−3
λ (m)
4.4 &times;10−7
Sample calculation:
Green light
1.0
1.76 &times;10−4
7
0.015
2.1&times;10−3
3.9 &times;10−7
Δx λ
=
L d
d Δx
λ=
L
(1.76 &times;10−4 m )(2.5 &times;10−3 m)
=
1.0 m
λ = 4.4 &times;10−7 m
The wavelength is 4.4 &times; 10–7 m.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.5-2
Section 9.5 Questions, page 484
1. Answers may vary. Sample answer: When all other factors are kept constant, the fringes for
the red light, λ = 650 nm, are more widely spaced than the fringes for the blue light, λ = 470 nm.
2. Given: double-slit interference; d = 0.20 mm = 2.0 &times;10−4 m; L = 3.5 m; n = 1 dark fringe;
x1 = 4.6 mm = 4.6 &times; 10–3 m
Required: λ
1 ⎞ Lλ
⎛
Analysis: Rearrange the equation xn = ⎜ n − ⎟
to solve for wavelength;
2⎠ d
⎝
1 ⎞ Lλ
⎛
xn = ⎜ n − ⎟
2⎠ d
⎝
xn d
λ=
1⎞
⎛
⎜n− ⎟L
2⎠
⎝
Solution:
xn d
!=
#
1&amp;
%\$ n &quot; 2 (' L
=
(4.6 ) 10&quot;3 m)(2.0 ) 10&quot;4 m)
# 1&amp;
%\$ 1&quot; 2 (' (3.5 m)
= 5.3) 10&quot;7 m
! = 530 nm
Statement: The wavelength of the light is 530 nm.
3. Answers may vary. Sample answer: An underwater double-slit experiment would have
different results than a double-slit experiment in air because, in water, the speed of light is slower
and the wavelength is shorter. The spacing between the resulting fringes would be closer
together underwater than in air.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.5-3
4. Given: double-slit interference; d = 0.30 mm = 3.0 &times;10−4 m; n = 5 dark fringe;
x5 = 12.8 &times;10−2 m; λ = 4.5 &times;10−7 m
Required: L
1 ⎞ Lλ
⎛
Analysis: Rearrange the equation xn = ⎜ n − ⎟
to solve for distance to the screen;
2⎠ d
⎝
&quot;
1 % L(
xn = \$ n ! '
2&amp; d
#
L=
xn d
&quot;
1%
\$# n ! 2 '&amp; (
Solution: L =
=
xn d
&quot;
1%
\$# n ! 2 '&amp; (
(12.8 ) 10!2 m)(3.0 ) 10!4 m)
&quot;
1%
!7
\$# 5 ! 2 '&amp; (4.5 ) 10 m)
L = 19 m
Statement: The distance at which the screen is placed is 19 m.
5. (a) Given: double-slit interference; d = 0.15 mm = 1.5 &times; 10–4 m; L = 2.0 m;
Δx = 0.56 cm = 5.6 &times;10−3 m
Required: λ
Lλ
Analysis: Rearrange the equation Δx =
to solve for wavelength
d
d Δx
Solution: λ =
L
(1.5 &times;10−4 m)(5.6 &times;10−3 m)
=
(2.0 m)
λ = 4.2 &times;10−7 m
Statement: The wavelength of the source is 4.2 &times; 10–7 m, or 420 nm.
(b) Given: double-slit interference; d = 0.15 mm = 1.5 &times;10−4 m; L = 2.0 m;
λ = 600 nm = 6 &times;10−7 m
Required: Δx , of the dark fringes
Lλ
Analysis: Δx =
d
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.5-4
L&quot;
d
(2.0 m)(6 # 10\$7 m)
=
(1.5 # 10\$4 m)
Solution: !x =
!x = 8 # 10\$3 m
Statement: The spacing of the dark fringes is 8 &times; 10–3 m, or 0.8 cm.
6. Given: double-slit interference; n = 2 dark fringes; ! 2 = 5.4&deg;
d
Required:
λ
d
1⎞
⎛
Analysis: Rearrange the equation d sin θ n = ⎜ n − ⎟ λ to solve for ;
λ
2⎠
⎝
1⎞
⎛
d sin θ n = ⎜ n − ⎟ λ
2⎠
⎝
1
n−
d
2
=
λ sin θ n
Solution:
1
n−
d
2
=
λ sin θ n
1
2
=
sin 5.4&deg;
2−
d
λ
= 16
Statement: The ratio of the separation of the slits to the wavelength is 16:1.
7. (a) Given: double-slit interference; d = 0.80 mm = 8.0 &times;10−4 m; L = 49 cm = 0.49 m;
Δx = 0.33 mm = 3.3 &times;10−4 m
Required: λ
Analysis: Rearrange the equation Δx =
Lλ
d
d Δx
λ=
L
Lλ
to solve for wavelength;
d
Δx =
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.5-5
Solution:
d Δx
Δλ =
L
(8.0 &times;10−4 m)(3.3 &times;10−4 m)
=
(0.49 m )
λ = 5.39 &times;10−7 m (one extra digit carried)
λ = 5.4 &times;10−7 m
Statement: The wavelength of the monochromatic light is 5.4 &times; 10–7 m, or 540 nm.
(b) Given: double-slit interference; d = 0.60 mm = 6.0 &times;10−4 m; L = 0.49 m; λ = 5.39 &times; 10–7 m
Required: Δx , of the dark fringes
Lλ
Analysis: Δx =
d
Solution:
Lλ
Δx =
d
(0.49 m)(5.39 &times;10−7 m)
=
(6.0 &times;10−4 m)
Δx = 4.4 &times;10−4 m
Statement: The spacing of the dark fringes is 4.4 &times; 10–4 m, or 0.44 mm.
8. (a) Given: double-slit interference; L = 2.5 m; λ = 5.1&times;10−7 m; Δx = 12 mm = 1.2 &times;10−2 m
Required: d
Lλ
Analysis: Rearrange the equation Δx =
to solve for distance between the slits;
d
Lλ
Δx =
d
Lλ
d=
Δx
Solution:
Lλ
d=
Δx
(2.5 m)(5.1&times;10−7 m)
=
(1.2 &times;10−2 m)
d = 1.1 &times; 10−4 m
Statement: The slit spacing is 1.1 &times; 10–4 m, or 0.11 mm.
Lλ
(b) Solutions may vary. Sample solution: From Δx =
, the fringe separation is inversely
d
proportional to the slit spacing. If the slit spacing is reduced by a factor of three, the fringe
separation will increase by a factor of three. The fringe separation will change to
3 &times; 12 mm = 36 mm, or 3.6 cm.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.5-6
9. Answers may vary. Student answers should include some of the following points. Newton
developed a particle theory of light that explained rectilinear propagation and reflection well. His
reputation put the work of others at the time in the shadows. Grimaldi was working on a theory
for the diffraction and dispersion of light, and Hooke was looking a wave theory. Later,
Newton’s theory that light travelled faster in a medium than in a vacuum was shown
experimentally to be incorrect. With this failure, Newton’s explanations of refraction, colour, and
dispersion were also refuted. Huygens’ wave ideas and Young’s double-slit interference
experiment showed that light is definitely a wave. By 1900, all known properties of light could
be explained by a wave model.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 9: Waves and Light
9.5-7
Section 10.1: Interference in Thin Films
Tutorial 1 Practice, page 507
1. The second soap film is thicker. The longer wavelength of the second film means the
film at that point must be thicker for constructive interference to occur.
4! =
nm =
&quot; 10#3 m
2. Given: nsoap film =
Required: t
Analysis: Phase changes occur at both reflections, so use the formula for destructive
interference4 use m = 0 4
!
1\$
1\$
!
m
+
λ
m
+
#&quot;
&amp;% λ
#
&amp;
2
2
% 4t=
2t = &quot;
2nsoap film
nsoap film
m
Solution:
!
1\$
#&quot; m + 2 &amp;% !
t=
2nsoap film
=
!
1\$
0
+
#&quot;
2 &amp;%
2
&quot; 10#3 m
(
)
#3
&quot; 10 m
mt =
Statement: The smallest thickness of soap film capable of producing reflective
&quot; 10&quot;3 m .
4! =
nm =
&quot; 10#3 m
3. Given: noil =
m
Required: t
Analysis: The yellow light undergoes a phase change at the air–oil reflection interface,
but not at the oil–water interface. Since we do not want to see the yellow light, use the
nλ
nλ
4t=
formula for destructive interference4 use n = 1 4 2t =
2noil
noil
m
Solution:
n!
t=
2noil
=
1
2
(
&quot; 10#3 m
)
#3
mt = 1.30 &quot; 10 m
Statement: The oil slick needs to be 1.30 &quot; 10–3 m thick for the yellow light be invisible.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.1-1
4!=
nm =
&quot; 10#3 m
4. Given: ncoating =
m
Required: t
Analysis: The red light undergoes a phase change at both the air–coating reflection
interface and the coating–lens interface. Since we do not want to see the red light, use the
formula for destructive interference4 use m = 04
!
!
1\$
1\$
m
m
λ
λ
#&quot;
#&quot;
2 &amp;%
2 &amp;%
4t=
2t =
ncoating
2ncoating
m
m
!
1\$
#&quot; m + 2 &amp;% !
Solution: t =
2ncoating
=
5.1&quot; 10#3 m
2(
)
#3
mt = 1.1&quot; 10 m
Statement: The thickness of the magnesium fluoride anti-reflection coating needs to be
1.1 &quot; 10–3 m.
Mini Investigation: Observing a Thin Film on Water, page 507
A. Sample answer: The dark areas represent areas where destructive interference occurs.
B. Sample answer: The patterns were caused by different thicknesses of the film and the
different wavelengths of light.
C. Sample answer: The pattern changed when the colour of the light changed because the
thicknesses of oil cause destructive interference for some colours.
Tutorial 2 Practice, page 510
1. Given: mt = 0.012 cm = 1.2 &quot; 10&quot; m4 L =
cm =
&quot; 10&quot;1 m4
&quot; 10– m
Required: λ
Analysis: Calculate &quot;x, the separation between the fringes. Then rearrange the equation
L!
2t&quot;
&quot; =
to solve for wavelength4 ! =
.
2t
L
m
m
Solution: Calculate &quot;x:
&quot; 10# m
! =
3
&quot; 10# m one eItra digit carrie
m! =
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.1-2
!=
=
2t&quot;
L
(
2 1.2 # 10\$ m
)(
# 10\$ m
)
# 10\$1 m
\$3
m! = 3.5 # 10 m
Statement: The wavelength of the light is 3.5 &quot; 10–3 m.
2. Given: L = 5.0 cm = 5.0 &quot; 10–2 m4 λ
&quot; 10–3 m4 there are 52 cycles of
alternating light patterns in L
Required: t
L
5.0 &quot; 10#2 m
4 &quot; =
Analysis: &quot;x, the separation between the fringes, is given by
.
52
m52 m
L!
L!
to solve for thickness4 t =
.
Rearrange the equation &quot; =
2t
m
m 2&quot;
L!
Solution: t =
2&quot;
(5.0 # 10
=
\$2
m
)(
# 10\$3 m
% 5.0 # 10
2'
52
&amp;
mt =
Statement:
\$2
m(
*
)
# 10−3 m
)
&times; 10– m.
Research This: Thin Films and Cellphones, page 510
A. Answers may vary. Sample answer: Cellphones have several layers of thin films. The
infrared light incident on the screen enters the top layer, and the other layers turn this
energy into visible light. The multiple layers amplify the light energy so that it is visible.
B. Answers may vary. Sample answer: As of late 2010, applying night vision technology
to glasses and cellphones is still in the research phase. Given adequate financing, this
technology could be implemented in cellphones within a year or two.
C. Answers may vary. Sample answer: For this technology to become usable, more
research needs for the thin film technology to be practical to be added to glasses or
cellphone screens. There may be a cost barrier as well, which is typical of new
inventions. However, this barrier would be overcome as the application became more
based on the research and could include images, schematics, and equations.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.1-
Section 10.1 Questions, page 511
1. Given: ncoating = 1. 4 nlens = 1. 4 t = 133. nm = 1.
&times;10−3 m
Required: λ
Analysis: Phase changes occur at the air–coating interface but not at the coating–lens
interface. For minimal reflection, use the formula for destructive interference,
2t ncoating
nλ
, to solve for wavelength4 λ =
use n = 1 .
2t =
n
ncoating
m
m
Solution:
(
!=
=
(
2t ncoating
2
(
)
)
n
&quot; 10#3 m
#3
)( )
1
m
=
&quot; 10
nm
m! =
Statement: The wavelength of the light t
2. Given: nfilm =
4 ! = 3.00 &quot; 10#3 m4 n = 1
m
Required: t
Analysis: Phase changes occur at the air–film interface and at the film–water interface.
nλ
4
For maIimum reflection, use the formula for constructive interference4 2t =
nfilm
m
nλ
.
t=
2nfilm
m
Solution:
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.1-
t=
=
n!
2nfilm
1 3.00 &quot; 10#3 m
2(
)
#3
mt = 2.31&quot; 10 m
Statement: The thickness of the film is 2.31 &times; 10–3 m.
4 nglass =
4 !red = 5.00 &quot; 10#3 m
3. Given: nfilm =
m
Required: t
Analysis: Phase changes occur at the air–film interface and at the film–glass interface.
nλ
For maIimum reflection, use the formula for constructive interference, 2t =
4
nfilm
m
nλ
. Use n = 1.
t=
2nfilm
m
Solution:
n!
t=
2nfilm
=
1 5.00 &quot; 10#3 m
2
(
)
#3
mt = 2.22 &quot; 10 m
Statement: The thickness of the soapy water film is 2.22 &times; 10–3 m.
4 nglass = 1.104 ! = 5.00 &quot; 10#3 m
4. Given: nfilm =
m
Required: t
Analysis: Phase changes occur at the air–film interface but not at the film–glass
interface. For maIimum reflection, use the formula for constructive interference,
!
!
1\$
1\$
#&quot; m + 2 &amp;% λ
#&quot; m + 2 &amp;% λ
4t=
. Use m = 0.
2t =
nfilm
2nfilm
m
m
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.1-
!
1\$
#&quot; m + 2 &amp;% λ
Solution: t =
2nfilm
m
!
1\$
0
+
!
#&quot;
2 &amp;%
t=
2nfilm
( )(5.00 &quot; 10
=
2(
)
#3
m
)
#3
mt = 1.11 &quot; 10 m
Statement: When the indeI of refraction of the glass plate changes to 1.10, the thickness
of the soapy water film becomes 1.11 &times; 10–3 m.
5. The interference between reflections on the top and bottom of the soap film produces
the colours. No, the bubble is not of uniform thickness. t is thinnest in the blue bands
because this is the condition for which there is constructive interference with the smallest
thickness of film. The film is thickest in the red areas.
6. An incident ray of light reflects from both the top surface of a soap film and the bottom
surface of the soap film. More light is reflected than passes through the film. This gives
the appearance of brightness on the top of the soap film and less bright, or darker, on the
bottom of the soap film.
4 nwater =
4 !1 =
&quot; 10#3 m4 !2 =
&quot; 10#3 m
7. Given: nglass =
m
Required: t
Analysis: Phase changes occur at the air–film interface but not at the film–glass
interface. For maIimum reflection, use the equation for constructive interference4
!
!
1\$
1\$
#&quot; m + 2 &amp;% λ
#&quot; m + 2 &amp;% λ
4 t=
. Use m = 0, 1, 2
2t =
, ... for each wavelength to see
nfilm
2nfilm
m
m
which value produces a thickness that results in both lights being reflected.
!
1\$
#&quot; m + 2 &amp;% λ
Solution: t =
2nglass
m
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.1-5
For λ1, there are a number of thicknesses that will reflect the light.
For m = 0:
For m = 2:
!
1\$
!
1\$
&quot;3
m
+
m
+
m
&times;
10
#&quot;
#&quot;
2 &amp;%
2 &amp;%
t=
t=
2
2
(
(
mt =
(
)
)
&times; 10&quot;3 m
For m = 1:
!
1\$
m
+
#&quot;
2 &amp;%
t=
(
2
&times; 10&quot;3 m
(
)
)
&quot;3
mt = 2.310 &times; 10 m
&times; 10&quot;3 m
(
&quot;3
)
mt =
&times; 10
For m
!
#&quot; m +
t=
1\$
2 &amp;%
mt =
&times; 10&quot;3 m
m
(
2
&times; 10&quot;3 m
(
(
)
(
&times; 10&quot;3 m
mt =
For m = 1:
!
1\$
&times; 10&quot;3 m
#&quot; m + 2 &amp;%
t=
2
(
(
(
)
&times; 10
mt =
For m
!
1\$
#&quot; m + 2 &amp;%
t=
2
)
(
)
&times; 10&quot;3 m
mt =
The smallest common thickness is
Statement:
Copyright &copy; 2012 Nelson Education Ltd.
&quot;3
&times; 10&quot;3 m .
mt =
)
)
For λ2, there are also a number of thicknesses that will reflect the light.
For m = 0:
For m = 2:
!
1\$
!
1\$
&quot;3
m
+
&times; 10&quot;3 m
m
+
m
&times;
10
#
&amp;
#&quot;
&amp;
2%
&quot;
2%
t=
t=
2
2
(
)
)
)
m
&times; 10&quot;3 m
(
)
)
&times; 10&quot;3 m
&times; 10–3 m.
Chapter 10: Applications of the Wave Nature of Light
10.1-3
Section 10.2: Single-Slit Diffraction
Tutorial 1 Practice, page 516
1. When light travels from air to a medium that is denser than air, such as water, the light is
w&quot;y
refracted and the wavelength of the light shortens. According to the equation ! =
, λ is
L
proportional to Δy, so when λ is reduced, the central maximum will also be reduced. So the
central maximum would be narrower if the equipment were submerged in water.
2. Given: ! = 7.328 &quot; 10#7 m; w = 43 &micro;m = 4.3 &times; 10–5 m; L = 3.0 m
Required: Δy
w&quot;y
Analysis: Rearrange the equation ! =
to solve for the distance between adjacent minima;
L
L&quot;
!y =
w
Solution:
L&quot;
!y =
w
(3.0 m)(7.328 # 10\$7 # 10\$7 m )
=
4.3 # 10\$5 m
= 5.1 # 10\$2 m
!y = 5.1 cm
Statement: The separation of adjacent minima is 5.1 cm.
3. Given: w = 3.00 ! 10&quot;6 m; #1 = 25.0&deg;
Required: λ
Analysis: The angle between the first dark fringes is equal to the angle for the width of the
central maximum, which is twice the angle for the first dark fringe, given by w sin θ n = λ . In this
case, n = 1 and 2! = 25.0&deg; ; ! = wsin &quot; n .
Solution:
2θ = 25.0&deg;
θ = 12.5&deg;
! = wsin &quot; n
= (3.0 # 10\$6 m)sin12.5&deg;
! = 6.49 # 10\$7 m
Statement: The wavelength of the light is 6.49 &times; 10–7 m.
4. Given: θa = 56&deg;; θb = 34&deg;
w
Required: a
wb
Analysis: The formula that relates the angles, the common wavelength, and the slit sizes is
w
w sin θn = λ ; wa sin &quot; a = &quot; and wb sin &quot; b = &quot; . Divide the equations to determine the ratio a .
wb
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.2-1
Solution:
wa sin &quot; a &quot;
=
wb sin &quot; b &quot;
wa sin &quot; b
=
wb sin &quot; a
=
sin34&deg;
sin56&deg;
wa
= 0.67
wb
Statement: The ratio of the slit widths,
wa
wb
, is 0.67.
Section 10.2 Questions, page 519
1. Given: single-slit diffraction; ! = 794 nm = 7.94 &quot; 10#7 m; L = 1.0 m
n = 9, y9 = 6.48 cm = 0.0648 m
Required: w
y
&quot;L
Analysis: !y = 9 ; !y =
n
w
&quot;L
w=
!y
y
Solution: !y = 9
n
0.0648 m
=
9
!y = 7.2 &quot; 10 –3m
!L
w=
&quot;y
(7.94 # 10\$7 m)(1.0 m)
7.2 # 10 –3 m
w = 1.1 # 10 –3 m
Statement: The width of the slit is 1.1&quot; 10&quot;4 m .
2. Given: single-slit diffraction;
600 nm 6.00 &quot; 10#7 m; \$1 6.9&deg;
Required: w
Analysis: The first dark fringe is located where w sin θ n = λ . Rearrange the equation
=
w sin θn = λ to solve for slit width; w =
Copyright &copy; 2012 Nelson Education Ltd.
!
.
sin &quot; n
Chapter 10: Applications of the Wave Nature of Light
10.2-2
Solution: w =
!
sin &quot;1
6.00 # 10\$7 m
sin6.9&deg;
w = 5.0 # 10\$6 m
Statement: The width of the slit is 5.0 &times; 10–6 m.
3. Given: single-slit diffraction; λ = 450 nm = 4.50 &times; 10–7 m; L = 10.0 m;
w = 0.15 mm = 1.5 &quot; 10&quot;4 m
Required: y, the distance between the first and third dark fringes
λL
Analysis: Δy =
; y = 2!y
w
&quot;L
Solution: !y =
w
(4.50 # 10\$7 m)(10.0 m )
=
1.50 # 10\$4 m
=
= 3.0 # 10\$2 m
y = 2!y
= 6.0 # 10\$2 m
y = 6.0 cm
Statement: The distance between the first and third dark fringes is 6.0 cm.
4. Given: single-slit diffraction; λ = 550 nm = 5.50 &times; 10–7 m; L = 2.0 m;
y1 = 5.5 mm = 5.5 &quot; 10&quot;3 m
Required: w
!L
λL
Analysis: Rearrange the equation y1 =
to solve for slit width; w =
y
w
Solution: w =
=
!L
y1
(5.50 &quot; 10#7 m)(2.0 m )
5.50 &quot; 10#3 m
= 2.0 &quot; 10#4 m
w = 0.20 mm
Statement: The width of the slit is 0.20 mm.
5. Given: single-slit diffraction; λ = 630 nm = 6.30 &times; 10–7 m; L = 3.0 m;
w = 0.25 mm = 2.5 ! 10&quot;4 m
Required: 2Δy , the width of the central maximum
&quot;L
Analysis: Multiply the equation !y =
by 2 to obtain 2Δy .
w
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.2-3
Solution: 2!y =
=
2&quot; L
w
(
2 4.10 #
\$5
2.30 #
)(
m 1.0 m
\$
)
m
=
# \$2 m
2!y=
m
Statement: The width of the central maximum is 1.5 cm.
6. Given: Δy = 0.120 cm = 1.20 &times;10−3 m; w = 0.0295 cm = 2.95 &times;10−4 m; L = 60.0 cm = 0.60 m
Required: λ
&quot;L
w&quot;y
Analysis: Rearrange the equation !y =
to solve for wavelength; ! =
w
L
w&quot;y
Solution: ! =
L
(2.95 # 10\$4 m)(1.20 # 10\$3 m )
=
6.0 # 10\$1 m
! = 5.90 # 10\$7 m
Statement: The wavelength of the yellow light is 5.90 &times;10−7 m .
7. (a) The distance between successive maxima in single-slit diffraction is given by Δy =
λL
. If
w
I double the wavelength, λ , then the distance Δy will also double. The angles of the maxima
and the minima would be approximately doubled.
λL
(b) If I multiplied both the wavelength, λ , and the slit width, w, in the equation Δy =
by 2,
w
the 2s will cancel each other out. Therefore, there will be no effect on Δy . The interference
pattern will be the same.
8. Blue light has an average wavelength of 475 nm, and green light has an average wavelength of
510 nm. If I replaced the blue light with the green light, then I would be increasing the
wavelength. Therefore, spacing of the intensity maxima would be greater.
9. Given: single-slit diffraction
Required: θ10
Analysis: Assume the width of a typical doorway is w = 0.92 m. Assume the visible light has a
n&quot;
wavelength of 500 nm. Rearrange the equation sin &quot; n =
to solve for the angle;
w
\$ n# '
! n = sin &quot;1 &amp; )
% w(
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.2-4
\$ n# '
Solution: ! n = sin &quot;1 &amp; )
% w(
(
)
\$ (10) 5.00 * 10&quot;7 m '
)
= sin &amp;
&amp;%
)(
0.92 m
&quot;1
(
= sin &quot;1 5.43* 10&quot;6
)
!10 = 3.1* 10 &deg;
Statement: The angle of the tenth minimum for a doorway that is 0.92 m wide is 3.1 &times; 10–4&deg;.
10. To improve the resolution of a digital image, I could use more pixels per square centimetre.
Or, I could use a wider aperture (size of slit) to increase the resolution. However, the wider
aperture would reduce the depth of field (range of the focus).
11. I would be able to resolve the double stars in Mizar with the telescope because, in addition to
enlarging the image, the telescope’s aperture is wider than the aperture in my eye. The wider
aperture increases the resolution, allowing me to see the two stars.
12. In a double-slit interference pattern, there are more intensity maxima than in a single-slit
interference pattern. In the double-slit interference pattern, there is less space between fringes
because the second slit causes additional destructive interference.
&quot;4
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.2-5
Section 10.3: The Diffraction Grating
Tutorial 1 Practice, page 523
1
. As N increases, w
N
decreases. The diffraction grating with 10 000 lines/cm has more lines per centimetre than the
second diffraction grating, so the separation between adjacent principal maxima in the first
grating would have to be smaller.
2. Given: ! = 660 nm = 6.60 &quot; 10#7 m; N = 8500 lines/cm ; m = 1
Required: θ, the angular separation between successive maxima
1
Analysis: Use the equation w =
to calculate the slit separation. Then use the equation
N
m&quot;
.
mλ = w sin θm to locate the maximum for m; sin ! m =
w
Solution:
1
w=
N
1
1m
=
!
8500 lines/ cm 100 cm
1. Slit separation and number of lines are related by the equation w =
w = 1.176 ! 10&quot;6 m (two extra digits carried)
m&quot;
w
(1)(6.60 # 10\$7 m )
sin !1 =
1.176 # 10\$6 m
!1 = 34&deg;
Statement: The angular separation of successive maxima is 34&deg; .
3. Given: ! = 694.3 nm = 6.943 &quot; 10#7 m; m = 3; \$ 3 = 22.0&deg;
Required: N
sin ! m =
Analysis: Use the equation mλ = w sin θm to calculate the slit separation; w =
the equation w =
Solution:
m!
w=
sin &quot; m
m!
. Then use
sin &quot; m
1
1
to determine the number of grating lines; N = .
w
N
(3)(6.943 # 10\$7 m)
=
sin 22.0&deg;
w = 5.5602 # 10\$6 m (one extra digit carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.3-1
N=
=
1
w
1
&quot;6
!
1 m
100 cm
5.5602 ! 10 m
N = 1798 lines/cm
Statement: The grating has 1798 lines per centimetre.
Research This: Blu-ray Technology, page 524
A. Answers may vary. Sample answer: The technology is called Blu-ray because it uses a blue
laser instead of the red laser used in DVDs.
B. Answers may vary. Sample answer: In Blu-ray technology, the data are placed on the top of a
disc coated with a polycarbonate layer. There are pits in the disc. Each pit contains a signal that
is interpreted as a zero or a one, much like a computer. The laser reads the pits.
C. Answers may vary. Sample answer: Blu-ray technology can hold more data than CDs and
DVDs. The quantity of data is at least five times greater than the quantity that a DVD can store,
which means that images can be much more detailed on Blu-ray discs. Blu-ray is also able to
play much faster than CDs and DVDs.
D. Answers may vary. Sample answer: The improvements are possible through the use of the
blue laser, which has a much smaller wavelength than the red laser used with CD and DVD
players. Blu-ray technology can also store information on as many as 20 layers within the disc.
vary based on presentation format, but students should show at least three images to demonstrate
the representative features of the three technologies. Hazards associated with using high-power
lasers such as the lasers used in Blu-ray technology should be included. For example, if the laser
is pointed at a person’s eyes, that person’s eyesight could be damaged.
Section 10.3 Questions, page 525
1. The surface of a CD has many closely spaced parallel lines, like a diffraction grating.
Consequently, when white light reflects from the surface of a CD, we see a rainbow-like pattern
because the surface acts like a diffraction grating.
2. Given: N = 2800 lines/cm
1
Analysis: w =
N
Solution:
1
w=
N
1
=
2800 lines/cm
= 3.6 ! 10&quot;4 cm
w = 3.6 ! 10&quot;6 m
Statement: The distance between two lines in the diffraction grating is 3.6 &times;10−6 m .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.3-2
3. Given: N = 10 000 lines/cm; θ1 = 31.2&deg;; θ2 = 36.4&deg;; θ3 = 47.5&deg;
Required: λ1 , λ2 , λ3
1
Analysis: Use the equation w =
to calculate the slit separation. Then use the equation
N
wsin &quot; m
.
mλ = w sin θm to determine the wavelength for each value of m; ! =
m
Solution:
1
w=
N
1
w=
10 000 lines/cm
= 1.0 ! 10&quot;4 cm
w = 1.0 ! 10&quot;6 m
For m = 1:
wsin &quot; m
!=
m
(1.0 # 10\$6 m)sin31.2&deg;
=
1
\$7
= 5.18 # 10 m
! = 518 nm
For m = 3:
wsin &quot; m
!=
m
(1.0 # 10\$6 m)sin47.5&deg;
=
3
\$7
= 2.46 # 10 m
! = 246 nm
For m = 2:
wsin &quot; m
!=
m
= (1.0 # 10\$6 m)sin36.4&deg;
2
\$7
=
# 10 m
!=
nm
Statement: The wavelengths that produce these maxima are 518 nm, 297 nm, and 246 nm.
4. Given: N = 6000 lines/cm; w = 2.0 cm; ! = 450 nm = 4.50 &quot; 10#7 m; m = 1
Required: θ
Analysis: Divide the number of slits, N, by the length of the slit:
6000 lines/cm
1
= 3000 lines/cm . Then use w =
to calculate the slit separation, and rearrange
2
N
m&quot;
the equation mλ = w sin θm to calculate the angle; sin ! m =
.
w
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.3-3
Solution:
1
w=
N
w=
1
3000 lines/ cm
!
1m
100 cm
w = 3.333 ! 10&quot;6 m (two extra digits carried)
sin &quot; m =
=
m&quot;
w
(1)(4.50 # 10\$7 m )
3.333 # 10\$6 m
!= &deg;
Statement: Blue light produces the first intensity maximum at 7.8&deg; .
5. Given: λ = 600.0 nm = 6.000 &times; 10–7 m; w = 25 &micro;m = 2.5 &times; 10–5 m; m = 1
Required: θ
m&quot;
Analysis: mλ = w sin θm ; sin ! m =
w
Solution:
m&quot;
sin ! m =
w
(1)(6.000 # 10\$7 m )
=
2.5 # 10\$5 m
! = 1.4&deg;
Statement: The first-order maximum in intensity is at the angle 1.4&deg; .
6. Given: ! = 780 nm = 7.80 &quot; 10#7 m; L = 10 m; \$y = 0.50 m; m = 1
Required: w
Analysis: Use the sine ratio to calculate !1 . Then rearrange the equation mλ = w sin θm to
calculate the slit separation; w =
m!
.
sin &quot; m
Solution:
0.50 m
sin ! =
10 m
sin ! = 0.05
m!
w=
sin &quot;1
(1)(7.80 # 10\$7 m)
=
0.05
w = 1.6 # 10\$5 m
Statement: The spacing between the lines in the diffraction grating is 1.6 &times;10−5 m .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.3-4
7. Given: N = 300 i s m L =
m !y = 3.6 m = 0.036 m m = 3
Required: λ
1
Analysis: Use w =
to calculate the slit separation. Use the tan ratio to calculate θ3 . Then
N
wsin &quot; m
rearrange the equation mλ = w sin θm to calculate the wavelength; ! =
.
m
Solution:
1
w=
N
1
1m
=
!
300 lines/ cm 100 cm
w = 3.333 ! 10&quot;5 m (two extra digits carried)
tan ! =
0.036 m
0.84 m
! = 2.454&deg;
(two extra digits carried)
wsin &quot; m
m
(3.333 # 10\$5 )sin 2.454&deg;
=
3
\$7
= 4.76 # 10 m
! = 480 nm
Statement: The wavelength of the light is 480 nm.
8. Given: N = 3000 lines/cm; λ = 5.4 &times;10−7 m
Required: maximum value of m
1
Analysis: Use w =
to calculate the slit separation. Then use the equation mλ = w sin θm to
N
m&quot;
determine the angle in terms of m; sin ! m =
.
w
Solution:
1
w=
N
1
1m
=
!
3000 lines/ cm 100 cm
!=
w = 3.333 ! 10&quot;6 m (two extra digits carried)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.3-5
For a maximum m, sin ! m &lt; 1.0 :
m&quot;
w
(m)(5.4 # 10\$7 m)
=
3.333 # 10\$6 m
s n ! m = 0.1620m
We need
0.1620m &lt; 1.0
sin ! m =
1.0
0.1620
m &lt; 6.17
m=6
Statement: The maximum order number possible is the 6th order.
9. (a) Given: w = 0.50 nm = 5.0 ! 10&quot;10 m; # = 0.050 nm = 5.0 ! 10&quot;11 m
Required: θ1 , θ2 , θ3
m&lt;
Analysis: Use the equation mλ = w sin θm to calculate the angles; sin ! m =
Solution:
For m = 1:
m&quot;
sin ! m =
w
(1)(5.0 # 10\$11 m)
=
5.0 # 10\$10 m
! = 5.7&deg;
m&quot;
.
w
For m = 3:
&quot;
sin ! =
(3)(5.0 # 10\$11 m)
5.0 # 10\$10 m
! = 17&deg;
=
For m = 2:
m&quot;
sin ! m =
w
(2)(5.0 # 10\$11 m)
=
5.0 # 10\$10 m
! = 12&deg;
Statement: The angles for the first three maxima are 5.7&deg;, 12&deg;, and 17&deg; .
(b) w = 5.0 ! 10&quot;10 m; # = 600 nm = 6.0 ! 107 m; m = 1
Required: θ1
Analysis: mλ = w sin θm ; sin ! =
Copyright &copy; 2012 Nelson Education Ltd.
&quot;
Chapter 10: Applications of the Wave Nature of Light
10.3-6
Solution: sin ! m =
=
m&quot;
w
(1)(6.0 # 10\$9 m )
5.0 # 10\$10 m
= 1.2
!1 = no angle possible
Statement: The angle for the first bright fringe does not exist.
(c) The wavelength 600 nm is within the range of visible light, but no fringe angle was possible
in part (b). Visible light is not usually diffracted by crystal lattices. It may be possible to get a
fringe but only if the wavelength of the light is sufficiently short.
10. Given: !i = 5.00 &quot; 10 sm 5.00 &quot; 10#7 m \$ i =
&deg; \$ = 18.0&deg;; m = 1
Required: natmosphere
m!
Analysis: Use the equation mλ = w sin θm to calculate w; w =
. Then use the same
sin &quot; m
equation with the value of w and the new angle, θB, to calculate the wavelength in the planet’s
atmosphere. Take the ratio of the wavelengths to determine the index of refraction for the
!
planet’s atmosphere, sm phere = A .
!B
Solution: w =
m!
sin &quot; m
(1)(5.0 # 10\$ m)
si 1/./&deg;
w = 0.361 # 0/\$6 m (
wsin &quot; m
=
!B =
extra digit carried)
m
(1.462 # 10\$6 m)sin18.0&deg;
=
1
\$7
!B = 4.518 # 10 m (one extra digit carried)
atmosphere
=
=
atmosphere
!A
!B
5.00 &quot; 10#7 m
4.518 &quot; 10#7 m
= 1.11
Statement: The index of refraction of the planet’s atmosphere is 1.11.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.3-7
Tutorial 1 Practice, page 530
1. Given: f = 107.1 MHz = 1.071 ! 108 Hz; c = 3.0 ! 108 m/s
Required: λ
c
Analysis: c = λ f ; λ =
f
Solution: ! =
=
c
f
3.0 &quot; 108 m/ s
1.071&quot; 108 Hz
! = 2.8 m
Statement: The wavelength of the signal is 2.8 m.
2. Given: f = 3.0 ! 1017 Hz; c = 3.0 ! 108 m/s
Required: λ
c
Analysis: c = λ f ; λ =
f
Solution: ! =
=
c
f
3.0 &quot; 108 m/ s
3.0 &quot; 1017 Hz
= 1.0 &quot; 10#9 m
! = 1.0 &quot; 10#7 cm
Statement: The wavelength of the X-rays is 1.0 ! 10–7 cm.
3. Given: ! = 638 nm = 6.38 &quot; 10#7 m; c = 3.0 ! 108 m/s
Required: T
Analysis: λ f = c
c
f =
λ
1
T=
f
λ
T=
c
!
Solution: T =
c
6.38 &quot; 10#7 m
=
3.0 &quot; 108 m /s
T = 2.1&quot; 10#15 s
Statement: The period of the wave is 2.1 ! 10&quot;15 s .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.4-1
4. Given: f = 60 Hz; x = 5.0 ! 103 km = 5.0 ! 106 m; c = 3.0 ! 108 m/s
Required: n, number of wavelengths
Analysis: Calculate the wavelength of the electrical transmission using the universal wave
c
equation, c = λf; λ = . Then divide the wavelength by the distance across North America;
f
λ
= .
Solution: ! =
c
f
=
!
5.0 &quot; 106 m
3.0 &quot; 108 m/s
=
=
5.0 &quot; 106 m
60 Hz
=1
! = 5.0 &quot; 106 m
Statement: The number of wavelengths of the electrical transmission required to cross North
America is 1.
Section 10.4 Questions, page 531
1. Given: f = 5.0 ! 1014 Hz; c = 3.0 ! 108 m/s
Required: λ
c
Analysis: c = λ f ; λ =
f
Solution: ! =
=
c
f
3.0 &quot; 108 m/ s
5.0 &quot; 1014 Hz
= 6.0 &quot; 10#7 m
! = 6.0 &quot; 102 nm
Statement: The wavelength of the light in the CD player is 6.0 ! 102 nm.
2. Given: ! = 550 nm = 5.50 &quot; 10#7 m; c = 3.0 ! 108 m/s
Required: f
Analysis: λ f = c
c
f =
λ
c
Solution: f =
!
3.0 &quot; 108 m /s
=
5.50 &quot; 10#7 m
f = 5.5 &quot; 1014 Hz
Statement: The frequency of the light that is most sensitive to the human eye is 5.5 ! 1014 Hz.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.4-2
3. Given: f1 = 88 MHz = 8.8 ! 107 Hz; f 2 = 108 MHz = 1.08 ! 108 Hz; c = 3.0 ! 108 m/s
Required: λ1 , λ2
Analysis: λ f = c
λ=
Solution: !1 =
=
c
f
c
f1
3.0 &quot; 108 m/ s
!2 =
c
f2
3.0 &quot; 108 m /s
=
1.08 &quot; 108 Hz
8.8 &quot; 107 Hz
!1 = 3.4 m
!2 = 2.8 m
Statement: The wavelengths of the FM radio stations range from 3.4 m to 2.8 m.
4. Given: ! = 0.10 nm = 1.0 &quot; 10#10 m; c = 3.0 ! 108 m/s
Required: f
c
Analysis: c = λ f ; f =
λ
c
Solution: f =
!
3.0 &quot; 108 m /s
=
1.0 &quot; 10#10 m
f = 3.0 &quot; 1018 Hz
Statement: The frequency of the X-ray is 3.0 ! 1018 Hz.
5. (a) Given: ! =
cm = 1.224 &quot; 10#1 m; c = 3.0 ! 108 m/s
Required: f
Analysis: λ f = c
c
f =
λ
c
Solution: f =
!
3.0 &quot; 108 m /s
=
1.224 &quot; 10#1 m
f = 2.5 &quot; 109 Hz
Statement: The frequency of the X-ray wave is 2.5 ! 109 Hz.
(b) Most microwave ovens contain rotating carousels to heat the food evenly. By rotating the
food, the nodes cannot heat the same spot all the time, so the food cooks more evenly. With a
microwave wavelength of 12 cm, the nodes are located within the oven itself.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.4-3
6. Given: f =
Required: λ
Analysis: c = λ f
λ=
c
f
Solution: ! =
c
f
=
Hz
! 103 Hz; c = 3.0 ! 108 m/s
3.0 &quot; 108 m/ s
2.4 &quot; 109 Hz
= 1.25 &quot; 10#1 m
! = 12 cm
Statement: The wavelength of the radio waves in the cordless phone is 12 cm.
7. Given: f = 680 kHz = 6.80 ! 105 Hz; c = 3.0 ! 108 m/s
Required: λ
Analysis: λ f = c
λ=
Solution: ! =
=
c
f
c
f
3.0 &quot; 108 m/ s
6.80 &quot; 105 Hz
= 4.4 &quot; 102 m
! = 440 m
Statement: The wavelength of the broadcasting frequency used by 680 News is 440 m.
8. Given: = 6.0 cm = 6.0 ! 10&quot;2 m; = 7.5 GHz = 7.5 ! 109 Hz; c = 3.0 ! 108 m/s;
m=1
Required: θ
Analysis: First calculate the wavelength using the universal wave equation, c = λf;
c
&quot;
λ = . Then use the equation mλ = w sin θm to calculate the angle; sin ! =
.
f
Solution: ! =
=
c
f
3.0 &quot; 108 m /s
7.5 &quot; 109 Hz
! = 4.0 &quot; 10#2 m
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.4-4
sHs ! =
zHs ! =
&quot;
\$
/ 4.1 #
m/
\$
m
8.1 #
!= &deg;
Statement: The angle from the central maximum to the first diffraction minimum is 42&deg; .
9. Some of the radiation in the electromagnetic spectrum is only detectable in deep space because
Earth’s atmosphere absorbs the radiation at several different wavelengths, so it does not pass
through the atmosphere to the surface. To be able to detect these portions of the electromagnetic
spectrum, the detectors have to be in space, above Earth’s atmosphere. Some examples of
electromagnetic radiation from space that does not reach Earth’s surface are some wavelengths
of infrared radiation from distant objects, X-rays, and gamma rays.
Additional information: All objects emit infrared radiation. To avoid interfering with very faint
astronomical objects emitting infrared radiation, the detectors (telescopes) need to be kept
extremely cold. That is only possible in deep space.
10. Television correspondents in a distant part of the world are so far away that their responses
are delayed due to the travel time of the signal. There is also the time required to process the
signal.
11. (a) Given: f = 75 MHz = 75 ! 106 Hz; d = 134 m; c = 3.0 ! 108 m/s
Required: type of interference
c
Analysis: Calculate the wavelength of the signal, c = λf; λ = . Divide the distance by
f
wavelength.
c
!=
Solution:
f
=
3.0 &quot; 108 m/ s
75 &quot; 106 Hz
!=4m
1
134 m
= 33 wavelengths
2
4m
The interference is constructive interference because the path difference is a half-whole-number
multiple of the wavelength, and there is a 180&deg; phase change from the reflection.
Statement: The kind of interference that results is constructive interference.
(b) The signal is now being reflected from 134 m – 42 m = 92 m, which is exactly 23
wavelengths. This is now destructive interference because the path difference is a whole-number
multiple of wavelengths, and there is a 180&deg; phase change from the reflection.
12. Answers may vary. The report should explain how an antenna converts electrical currents
into electromagnetic radiation and that an antenna can be either a transmitter or a receiver. The
size of the antenna will depend on the type of signals being transmitted. For example, the
wavelengths of FM signals are 2 m to 4 m. Generally, the length of the antenna should be about
half the wavelength of the radio waves you are trying to send or receive.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.4-5
Section 10.5: Polarization of Light
Mini Investigation: Observing Polarization from Reflection, page 534
A. Sample answer: When the second filter is rotated, the bright light is blocked out completely.
B. Answers may vary. Sample answer: Glare on a surface is reduced when I use a Polaroid filter
because the polarizer blocks some of the light travelling through the filter. I see less of the light
reflecting from the surface, therefore reducing the glare.
C. Answers may vary. Sample answer: Variations in light seen from changing the location of a
Polaroid filter against various regions of the sky occur because polarized light in different parts
of the sky reaches me at different angles. Depending on the angle, light rays either pass through
the filter or are blocked by the filter.
Research This: Holography, page 536
A. Answers may vary. Sample answer: Holography is the process of making three-dimensional
images, called holograms, on a single film. To produce a hologram, a laser beam is split into two
halves by a beam splitter. One half of the laser beam reflects off a mirror to the object that will
be made into a hologram, and reflects onto a photographic plate. The hologram is created on the
photographic plate. The other half of the laser beam, called the reference beam, does not come
into contact with the object. The reference beam travels to the photographic plate. On the
photographic plate, both beams intersect and interfere with each other, producing the hologram.
B. Sample answer: To make a hologram, the materials required are a laser, a beam splitter, a
mirror, a photographic plate, and the object that will be the hologram.
C. Answers may vary. Sample answer: For holography to work, the laser beam must be
polarized. The beams are reflected at Brewster’s angle, so they are perfectly polarized.
Polarization results in a clearer hologram.
D. Answers may vary. Sample answer: Holography is used for security. For example, holograms
are used on money and credit cards because holograms are difficult to counterfeit (although it is
getting easier to duplicate holograms). Holography is also used in art and interactive graphics.
Novel applications of holography are data storage, museum tour guides who are holograms, and
holograms as jewellery.
E. Answers may vary. Sample answer: Presentations will include the key points from the
student’s research, from questions A to D, and could include an actual hologram, if a kit is
available. The presentation could also include images from the Internet of a hologram and an
illustration of the setup and equipment required.
Section 10.5 Questions, page 537
1. The light waves in polarized light vibrate in a single plane, whereas the light waves in
unpolarized light vibrate in several different planes.
2. To test whether a pair of sunglasses has polarizing lenses or simply darkened plastic lenses, I
would hold two of the same kind of lenses together and rotate them relative to each other to look
for changes in transmission intensity. I could also test the lenses by rotating one of the lenses in
front of my eyes to see how reflections from different objects change. With either method, if
light intensity changes depending on the angles of incidence and refraction, the lenses are
polarized.
3. Selective absorption polarizes light with a polarizing material. This polarizing material
transmits only a certain polarization of light and absorbs the other polarizations. Light that leaves
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.5-1
the polarizing material is always linearly polarized along the direction of the axis of the
polarizing material.
Polarization by reflection polarizes reflected light by the motion of electric charges
within the reflecting material. The charges can vibrate more easily in a direction that leads to
them emitting light polarized parallel to the surface.
Scattering polarizes light because the light scatters from air molecules with a polarization
that depends on the direction of the scatter. Observers in a particular direction only see light with
a certain polarization.
4. (a) If I place a sheet of Polaroid, whose transmission axis is parallel to the transmission axis of
the polarizer, between a polarizer and an analyzer, the light will not change. This is because the
sheet of Polaroid and the polarizer have parallel axes, so the light leaving the polarizer is
unchanged after passing through the Polaroid filter.
(b) If I place a sheet of Polaroid, whose transmission axis is parallel to the transmission axis of
the analyzer, between a polarizer and an analyzer, the light will not change. This is because the
sheet of Polaroid and the analyzer have parallel axes, so the light leaving the analyzer is
unchanged after passing through the Polaroid filter.
(c) If I place a sheet of Polaroid, whose transmission axis is at 45&deg;, between a polarizer and an
analyzer, more light will get through because light leaving the intermediate Polaroid is now at
45&deg; polarization with respect to the final polarization axis.
5. The sky often looks very different when viewed wearing polarizing sunglasses than otherwise
because of the different polarizations of light from different directions in the sky.
6. The light reflected from the surface of a still pond is horizontally polarized. When light is
reflected from a surface, the reflected light is completely polarized parallel to the surface, which,
in the case of the still pond, is horizontally. If the axis of the polarizer is set vertically, the
reflected light will be completely absorbed, thus showing that the reflected light was horizontally
polarized.
7. The sky appears blue because light from the Sun encounters small particles in the atmosphere,
which scatter the waves. Shorter waves are scattered more than longer waves. Since the blue
portion of the visible light spectrum has shorter wavelengths, the sky appears blue.
8. Given: Iin = 250 candelas; Iout = 17 % of Iin = (0.17)(250 candelas)
Required: !
I
Analysis: I out = I in cos 2 ! ; cos 2 ! = out
I in
Solution: cos 2 ! =
=
I out
I in
(0.17)(250 candelas )
250 candelas
= 0.17
cos! = 0.412 (one extra digit carried)
! = 66&deg;
Statement: The polarization angle of the incident light with respect to the polarization angle of
the filter is 66&deg;.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.5-2
9. Given: n1 = 1.00; n2 = 1.54
Required: ! B
Analysis: tan ! B =
n2
n1
Solution: tan ! B =
n2
n1
1.54
1.00
! B = 57&deg;
Statement: The angle of incidence that results in perfectly polarized light is 57&deg;. This angle is
called Brewster’s angle.
10. Answers may vary. Sample answer: Materials are able to reflect polarized light when the
light source is unpolarized because electrons in the material’s surface emit light with a
polarization perpendicular to the electron’s direction of vibration. Electrons on the material
surface can vibrate much more easily parallel to the surface than perpendicular to the surface.
The light reflected by the surface is light emitted by the surface electrons, so the electrons’
parallel vibrations produce mostly perpendicularly polarized light.
11. Polarized sunglasses can make it easier to see through the window on a sunny day because
the window partially polarizes sunlight through reflection. This light can then be filtered out
using glasses with a different polarization.
12. Yes, it is possible for light intensity to be unaffected by a polarization filter if all the
incoming light has the same polarization angle as the filter.
13. Answers may vary. Sample answer: 3D movies show two images overlaid on each other, one
for each eye of the viewer. One method of projecting the images is to project the two images
with different polarizations. The viewer wears glasses with filters in each eyepiece, which are
also polarized differently to match the two images. This way, the light from only one of the
images reaches each eye, and each eye sees a different image. The differences in the images’
perspective create the impression of viewing a three-dimensional object.
14. Answers may vary. Sample answers: Each lens of the glasses has a different polarizing filter
so only the light from the movie that is similarly polarized is seen through that lens and the other
light is blocked. Therefore, each eye sees a different image, which produces the threedimensional effect.
tan ! B =
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 10: Applications of the Wave Nature of Light
10.5-3
Section 11.1: The Special Theory of Relativity
Mini Investigation: Understanding Frames of Reference, page 576
A. The speed of the ball is greater when the person tossing the ball walks forward. The
speed appears to be the sum of the speed at which the ball is tossed and the speed of the
person tossing the ball.
B. The speed of the ball was even faster when the student throwing the ball walked
forward more quickly. This supports what we inferred in Question A.
C. When the student throwing the ball walked slowly backward, the speed decreased. The
decrease occurred because the person tossing the ball was moving away from the catcher.
D. Yes, if you know the speed of an object in one inertial reference frame, you can
determine its speed in another inertial reference frame, at least at the slow speeds used in
this investigation. We can determine the speed of the ball if we know the speed in the
catcher’s inertial frame and the speed of the pitcher relative to the catcher. When we
change this relative speed between the two inertial frames, we can determine the speed of
the ball in the new inertial frame.
Section 11.1 Questions, page 579
(a) The three most natural reference frames to use would be: 1) a frame moving alongside
the skater at the same velocity; 2) a frame fixed on the deck of the boat; and, 3) a frame
fixed on the shoreline.
(b) For reference frame 1 in part (a), the student on skates would not be moving along,
just moving in place. But the boat would be moving at constant speed, as would the
shoreline (unless the skater’s velocity relative to the boat was equal but opposite to the
velocity of the boat relative to the shoreline). For reference frame 2, the skater would be
moving along, the shoreline would be moving past, but the boat would appear fixed in
place. For reference frame 3, the skater would appear to be moving with velocity equal to
the vector sum of the boat’s velocity relative to the shoreline and the skater’s velocity
relative to the boat. In this inertial frame, the boat would appear to be moving past the
shoreline, but the shoreline would appear fixed.
2. (a) An inertial frame of reference moves at constant velocity, whereas a non-inertial
frame accelerates (i.e., its velocity changes).
Two examples of an inertial frame of reference being at rest on the ground and being in
an airplane cruising at constant altitude, with a fixed direction and fixed speed.
Two examples of a non-inertial frame of reference are being in a car accelerating from a
traffic light and being on a merry-go-round at a carnival.
3. (a) According to special relativity, the astronaut would measure the speed of light
to be c.
(b) The speed of the light measured by a person on Earth would equal c.
4. (a) Lutaaq would see Gabor’s ball follow a parabolic arc and Gabor moving along
underneath the ball. The ball would be seen to fall directly into Gabor’s hand.
(b) Gabor would see the same thing that Lutaaq had seen, that is, Lutaaq’s ball following
a parabolic arc and Lutaaq moving along underneath the ball, having the ball land in her
hand.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.1-1
5. The feature of Einstein’s coil and magnet thought experiment that Einstein found
troubling was that in the inertial frame of the coil, an electric field causes a current, but in
the inertial frame of the magnet, a magnetic field produces a current. The explanation
depended on the inertial frame of the explainer.
6. The two postulates of the special theory of relativity are
1. The laws of physics are the same in all inertial reference frames.
2. For an observer in at least one inertial reference frame, the speed of light in a vacuum
is independent of the motion of the light source.
7. The conclusion that results from the combination of Einstein’s two postulates is that all
inertial observers, regardless of their motion, will measure the same speed of light in a
vacuum, regardless of the motion of the light source.
8. A thought experiment is an imagined experiment that may be possible to do but
impractical. The experiments are generally used to test an hypothesis or show a problem
with an idea. As an example of the former, if one wonders if a laser beam from Earth
could be used to deflect an incoming comet, then one could imagine the experiment,
using current knowledge about lasers and comets. The hypothesis might be that the
vaporized material causes the comet to change course. In Einstein’s thought experiment
concerning the magnet and coil, he was demonstrating a problem with the then-current
ideas of electrodynamics.
9. To determine whether your ship is in an inertial frame of reference, you must show that
the ship is not accelerating. Qualitatively, if you were fixed to your seat, you would feel
any acceleration in your body. For a quantitative measure, the simplest experiment would
be to hold a ball out in front of you such that it is not moving relative to you or the ship.
Then, carefully let go of the ball without giving it any push or pull. Does the ball move?
If so, your frame is non-inertial and you can measure the acceleration. (Unless your
spacecraft is as massive as a small planet and you are not at its centre, then the ball
should float, free of the influence of gravity.)
10. (a) The ball rolls forward but then suddenly slows down when the train car suddenly
accelerates forward. You would have felt this acceleration too. It means your reference
frame suddenly became non-inertial.
(b) You pushed the ball straight ahead (forward), but it curved to the right because your
train car is accelerating to the left. You would sense this acceleration too. It means that
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.1-2
Section 11.2: Time Dilation
Tutorial 1 Practice, page 585
c
1. Given: !ts = 1.00 s v =
Required: !t( &quot; !ts
Analysis: Use
!t(
=
!ts
1
.
v2
1&quot; 2
c
# !t
&amp;
Solution: !t( &quot; !ts = !ts % ( &quot; 1(
\$ !ts
'
#
&amp;
1
&quot; 1(
= !ts %
2
% 1&quot; v
(
\$
'
c2
&amp;
#
1
= !ts %
&quot; 1(
2
c)
(
% 1&quot;
2
('
%\$
c
#
= !ts %
\$
1
&amp;
&quot; 1(
'
= 1.00 s)
!t( &quot; !ts 3
Statement: The proper ti(e interval of 1.00 s of the clock appears to increase by 0.25 s
when the clock (oves with a speed of 0.60c relative to the observer.
2. Given: !t( =
&quot; 10#6 s4 v =
&quot; 10 ( s4 c = /.0 &quot; 10 ( s
Required: !ts
Analysis:
!t(
=
!ts
1
v2
c2
# !t &amp;
!ts = !t( % s (
\$ !t( '
1&quot;
# !t &amp;
= !t( % s (
\$ !t( '
!ts = !t( 1&quot;
v2
c2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.2-1
v2
1&quot; 2
c
Solution: !ts = !t(
= !t( 1&quot;
&quot; 10 ( s )2
/.0 &quot; 10 ( s )2
= !t(
# &amp;
1# % (
\$ 5'
= !t(
25 ! 16
25
2
# /&amp;
) 10!6 s) % (
\$ 5'
=
!ts = 2.2 ) 10!6 s
Statement: At rest, the particle’s lifeti(e is 2.2 &quot; 10&quot;6 s , which is less than the lifeti(e
of the sa(e particles in a fast-(oving bea(.
s4 !t( = 10.0 s4 c = /.0 &quot; 10 ( s
3. Given: !ts =
Required: v
Analysis: Use
!t(
=
!ts
!t(
=
!ts
1
1&quot;
2
v
c2
to solve for v.
1
1&quot;
v2
c2
!ts
v2
= 1&quot; 2
!t(
c
2
# !ts &amp;
v2
=
1&quot;
% !t (
c2
\$ ('
# !t &amp;
v2
= 1&quot; % s (
2
c
\$ !t( '
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.2-2
# !t &amp;
v2
Solution: 2 = 1&quot; % s (
c
\$ !t( '
# &amp;
= 1! % (
\$ 5'
2
2
=
25
v /
=
c 5
/
v= c
5
# /&amp;
= % ( /.0 ) 10 ( s)
\$ 5'
v=
) 10 ( s
Statement: The spacecraft is (oving at
4. (a) Given: !t( = /0.0 h4 v =
c
! 10 ( s relative to Earth.
Required: !ts
Analysis:
!t(
=
!ts
1
v2
c2
# !t &amp;
!ts = !t( % s (
\$ !t( '
1&quot;
# !t &amp;
Solution: !ts = !t( % s (
\$ !t( '
= !t( 1&quot;
= !t( 1&quot;
v2
c2
c )2
c2
= /0.0 h)
=
!ts =
Statement:
(b) Given: !ts =
h two eHtra digits carried)
h
h4 v =
c
Required: !t(
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.2-/
Analysis:
!t(
=
!ts
!t( =
1
v2
c2
1&quot;
!ts
v2
c2
1&quot;
Solution: !t( =
!ts
v2
1&quot; 2
c
h
=
c )2
1&quot;
!t( =
c2
h
!t(
for both speeds. The !ts factor is dropped out.
!ts
Statement: By going faster, the crew (easured the ti(e between events to increase fro(
/0.0 h to 68.6 h.
5. (a) Given: v = 1.1 ! 10 ( s4 c = /.0 ! 108 ( s
!t
Required: (
!ts
ratios
Analysis: Use
Solution:
!t(
=
!ts
!t(
=
!ts
1
v2
1&quot; 2
c
.
1
1&quot;
v2
c2
1
=
1&quot;
= 1+
1.1 &quot; 10 )2
/.0 &quot; 108 )2
&quot; 10#
!t(
= 1.000 000 001
!ts
Statement: The ti(e dilation factor is 1.000 000 001.
(b) For objects going (uch slower than the speed of light, (easuring the effects of ti(e
dilation requires eHtre(ely high accuracy.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.2-
Section 11.2 Questions, page 587
1. For observer 2 to (easure the sa(e ti(e for the light pulse on the light clock that
observer 1 (easures, she would have to be in the sa(e inertial fra(e as observer 1. That
railway car.
2. (a) The process will always take longer for the observer who is (oving relative to the
process.
(b) The observer who is at rest with respect to the process (easures the proper ti(e of
the process.
-Keating
3. (a)
locks on aircraft), the clock that orbits Earth will be observed by a
person on Earth to run slow. So when the clock returns to Earth, it will have recorded less
elapsed ti(e. That is, the clocks will no longer be synchronized.
(b) Although the clock that orbited will have less elapsed ti(e after returning to Earth, it
will thereafter run at the sa(e rate as other clocks that are stationary on Earth. It will not
run slow.
. The
(c) As described in part
ti(es will be different.
(d) We are assu(ing that the clocks are ideal4 they run as designed, keeping perfect ti(e.
So, the stationary clock did not have the wrong ti(e. Any differences between the
stationary and the orbiting clock are due to the nature of ti(e as revealed by special
relativity.
(e) The orbiting clock was also assu(ed to be ideal. Its ti(e is correct, and all differences
with the stationary clock were due to the nature of ti(e, not the clock. It did not have the
wrong ti(e.
4. Notice that 1) the aircraft is travelling in the opposite direction to that of the ground on
the spinning Earth, and 2) the aircraft took eHactly one day to travel around the world,
60 s
60 (in
h
! 10 s !
!
!
3 1 day
1 h
1 day
1 (in
Therefore, for an observer at rest with respect to the centre of Earth, the plane was
-Keating eHperi(ent, the airport clock would show less elapsed ti(e. The
airport clock would have run slower.
5. The accuracy of a GPS syste( depends on correcting satellite clocks for special
relativity because the GPS satellites, which send the signals, are (oving rapidly with
respect to the GPS receivers, which are stationary on Earth. Thus, according to the
receiver, the clocks on the satellites run slow. Even if the ti(e dilation effect is s(all,
differences in elapsed ti(e continue to build. To ensure that that both the receiver and
satellite agree on the elapsed ti(e, the GPS syste( should take into account ti(e dilation
fro( special relativity.
6. (a) Roger, not Mia, (oves in Roger’s inertial reference fra(e. Thus, Roger, not Mia,
(easures Roger’s proper ti(e.
(b) Given: !ts = /0 s4 v = 0.85c
Required: !t(
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.2-5
Analysis:
!t(
=
!ts
!t( =
1
1&quot;
!ts
v2
c2
1&quot;
Solution: !t( =
=
v2
c2
!ts
v2
1&quot; 2
c
/0 s
0.85c ) 2
1&quot;
c2
!t( =
s
Statement: Roger observes that over a period of /0 s in his reference fra(e, Mia’s watch
has elapsed 57 s.
c
7. Given: !ts = 1.0 s4 v =
Required: !t(
!t
Analysis: ( =
!ts
!t( =
1
v2
1&quot; 2
c
!ts
v2
c2
1&quot;
Solution: !t( =
!ts
1&quot;
=
1&quot;
v2
c2
1.0 s
c )2
c2
!t( = /.2 s
Statement: The observer on Earth finds that the signals arrive every /.2 s.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.2-6
Section 11.3: Length Contraction, Simultaneity, and Relativistic
Momentum
Tutorial 1 Practice, page 591
1. Given: Ls = 5.0 m; Lm = 4.5 m; c = 3.0 &times; 108 m/s
Required: v
Analysis:
Lm
v2
= 1! 2
Ls
c
2
&quot; Lm %
v2
=
1
!
\$ L '
c2
# s&amp;
&quot;L %
v2
= 1! \$ m '
2
c
# Ls &amp;
2
&quot;L %
v = c 1! \$ m '
# Ls &amp;
&quot;L %
Solution: v = c 1 ! \$ m '
# Ls &amp;
2
2
= (3.0 ( 108 m/s) 1 !
(4.5 m )2
(5.0 m )2
v = 1.3 ( 108 m/s
Statement: To have had a relativistic contraction to 4.5 m, the 5.0 m long object must
have moved at 1.3 &times; 108 m/s.
2. Given: Ls = 120 m; v = 0.80c
Required: Lm
Lm
v2
Analysis:
= 1! 2
Ls
c
v2
Lm = Ls 1 ! 2
c
Solution: Lm = Ls 1 !
v2
c2
= (120 m) 1 !
(0.80 c )2
c2
Lm = 72 m
Statement: The relativistic contraction reduces the length of the spacecraft to 72 m.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.3-1
3. (a) Given: Ls = 2.5 m; Lm = 2.2 m; c = 3.0 &times; 108 m/s
Required: v
Analysis:
Lm
v2
= 1! 2
Ls
c
2
&quot; Lm %
v2
=
1
!
\$ L '
c2
# s&amp;
&quot; Lm %
v2
=
1
!
\$ L '
c2
# s&amp;
2
&quot;L %
v = c 1! \$ m '
# Ls &amp;
&quot;L %
Solution: v = c 1 ! \$ m '
# Ls &amp;
2
2
= (3.0 ( 108 m/s) 1 !
(2.2 m )2
(2.5 m )2
v = 1.4 ( 108 m/s
Statement: To have contracted from 2.5 m to 2.2 m, the car must have moved at
1.4 &times; 108 m/s.
(b) Given: Ls = 33 m; Lm = 26 m; c = 3.0 &times; 108 m/s
Required: v
2
&quot;L %
Analysis: Same as in part (a) above, v = c 1 ! \$ m ' .
# Ls &amp;
&quot;L %
Solution: v = c 1 ! \$ m '
# Ls &amp;
2
= (3.0 ( 108 m/s) 1 !
(26 m )2
(33 m )2
v = 1.8 ( 108 m/s
Statement: To have contracted from 33 m to 26 m, the rocket must have moved at
1.8 &times; 108 m/s.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.3-2
Tutorial 2 Practice, page 596
1. (a) Given: m = 1.67 ! 10&quot;27 kg; v = 0.85c; c = 3.0 ! 108 m/s
Required: pclassical
Analysis:
classical
= mv
Solution: pclassical = mv
= m (0.85c)
= (1.67 ! 10&quot;27 kg)(0.85)(3.0 ! 108 m/s)
= 4.259 ! 10&quot;19 kg # m/s (two extra digits carried)
pclassical =4.3 ! 10&quot;19 kg # m/s
Statement: The proton’s classical momentum is 4.3 ! 10&quot;19 kg # m/s .
(b) Given: m = 1.67 ! 10&quot;27 kg; v = 0.85c; pclassical = 4.259 ! 10&quot;19 kg # m/s
Required: prelativistic
Analysis: prelativistic =
Solution:
relativistic
=
mv
2
v
1! 2
c
and
classical
= mv , so
relativistic
=
classical
v2
1! 2
c
.
classical
v2
c2
4.259 &quot; 10!19 kg # m/s
1!
=
1!
relativistic
(0.85c )2
c2
= 8.1 &quot; 10!19 kg # m/s
Statement: The proton’s relativistic momentum in the lab frame of reference is
8.1 ! 10&quot;19 kg # m/s , about twice the classical value.
2. Given: m = 0.1 kg; v = 0.30c; c = 3.0 ! 108 m/s
Required: prelativistic
Analysis: prelativistic =
mv
1!
v2
c2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.3-3
Solution: prelativistic =
mv
2
v
c2
m(0.30c)
1!
=
1!
=
(0.30 c )2
c2
(0.1 kg)(0.30)(3.0 &quot; 108 m/s)
0.91
prelativistic = 9.4 &quot; 10 kg # m/s
Statement: The projectile’s relativistic momentum with respect to Earth is
9.4 ! 10 kg &quot; m/s .
6
3. Given: m = 1.67 ! 10&quot;27 kg; v = 0.750c; c = 3.0 ! 108 m/s
Required: prelativistic
Analysis: prelativistic =
Solution: prelativistic =
=
mv
v2
1! 2
c
mv
2
v
1! 2
c
m(0.750c)
1!
=
(0.750 c )2
c2
(1.67 &quot; 10 –27 kg)(0.750)(3.0 &quot; 108 m/s)
0.4375
prelativistic = 5.68 &quot; 10 kg # m/s
Statement: The proton’s relativistic momentum in the lab frame of reference is
5.68 ! 10&quot;19 kg # m/s .
4. (a) The motion affects only the component of length along the direction of motion. So,
only direction y is affected.
(b) Given: Lxs = 0.100 m; Lys = 0.100 m; Lzs = 0.100 m (proper lengths of the cube);
!19
v = 0.950c (speed along -axis); c = 3.0 ! 108 m/s
Required: relativistic volume of the cube,
m
Analysis: Vm = Lxm Lym Lzm . We know that x- and z-directions are unaffected by the
motion, so L m =
and L m =
.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.3-4
Lm
v2
Use
= 1 ! 2 , rearranged to solve for Lm .
Ls
c
We only need to calculate Lym = Lys
v2
1! 2 .
c
Solution: Vm = Lxm Lym Lzm
= Lxs Lys Lzs 1 !
v2
c2
= (0.100 m)(0.100 m)(0.100 m) 1 !
(0.950 c )2
c2
Vm = 3.12 &quot; 10!4 m 3
Vm = Lxs Lys Lzs
1 ! (v / c)2
= (0.100 m)(0.100 m)(0.100 m) 1 ! (0.950)2
Vm = 3.12 &quot; 10!4 m 3
Statement: The cube contracts along its direction of motion, resulting in a relativistic
volume of 3.12 ! 10&quot;4 m 3 . This is less than the proper volume Vs = 1.00 &times; 10–3 m3
( = Lxs Lys Lzs ).
(c) Given: Vs = 1.00 ! 10&quot;3 m 3 ; density = 2.26 ! 104 kg/m 3 ; v = 0.950c; c = 3.0 ! 108 m/s
Required: prelativistic
Analysis: First, use the proper volume and density to calculate the rest mass, m. Then,
mv
use prelativistic =
.
v2
1! 2
c
mv
Solution: prelativistic =
2
v
1! 2
c
(V &quot;density)(0.950c)
= s
(0.950 c ) 2
1!
c2
(1.00 # 10!3 m 3 )(2.26 # 104 kg/m 3 )(0.950)(3.0 # 108 m/s)
0.3122
10
= 2.06 # 10 kg &quot; m/s
=
prelativistic
Statement: The cube’s relativistic momentum is 2.06 ! 1010 kg &quot; m/s .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.3-5
Section 11.3 Questions, page 597
1. Given: Lm = 475 m; v = 0.755c
Required: Ls
Analysis:
Lm
v2
= 1 ! 2 , so Ls =
Ls
c
Solution: Ls =
=
Lm
v2
1! 2
c
.
Lm
v2
1! 2
c
475 m
1!
(0.755c )2
c2
Ls = 724 m
Statement: The proper length of spacecraft 2 is 724 m.
2. Given: Lm1 =
v1 =
c v =
c
Required: Lm2
Analysis: Apply the length contraction formula to each astronaut (the proper lengths are
the same).
Lm1
v2 L
v2
= 1 ! 12 ; m2 = 1 ! 22
Ls
Ls
c
c
Divide the relation for astronaut 2 by that for astronaut 1.
Lm2
=
Lm1
1!
1!
v22
c 2 , so L = L
m2
m1
v12
c2
1!
1!
v22
c2
v12
c2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.3-6
1!
Solution: Lm2 = L
1
1!
v22
c2
v12
c2
1!
= 8.0 ly)
1!
(0.85c )2
c2
(0.55c )2
c2
Lm2 = 5.0 ly
Statement: The second astronaut, who travels at 0.85c, finds the distance to the star to
be 5.0 ly.
3. Fran&ccedil;ois does not observe the explosions to occur simultaneously. Assume Fran&ccedil;ois is
travelling east relative to Soledad. According to him, Soledad is moving west at a speed
of 0.95c. Now Soledad has arranged the explosions in her moving frame to occur such
that the light from each one reaches her at the same time in her inertial frame of
reference. However, once the light leaves the explosion sites, they travel at speed c. So, if
Fran&ccedil;ois had seen the explosions as simultaneous, then Soledad would have detected the
western explosion first, because she was moving toward that explosion. Since she instead
observed the explosions simultaneously, the western explosion must have occurred later
than the eastern explosion. In other words, Fran&ccedil;ois sees the explosion next to the front of
his railway car first.
4. (a) Given: m = 1.67 ! 10&quot;27 kg; v = 0.99c; c = 3.0 ! 108 m/s
Required: pclassical
Analysis: classical = mv
Solution: pclassical = mv
= m(0.99c)
= (1.67 ! 10&quot;27 kg)(0.99)(3.0 ! 108 m/s)
= 4.960 ! 10&quot;19 kg # m/s (two extra digits carried)
pclassical = 5.0 ! 10&quot;19 kg # m/s
Statement: This proton’s momentum, according to Newton’s definition, is
8.0 ! 10&quot;19 kg # m/s .
(b) Given:
classical
= 4.960 ! 10&quot;19 kg # m/s; v = 0.99c
Required: prelativistic
Analysis: prelativistic =
mv
2
and
classical
= mv , so
relativistic
=
classical
v
v2
1! 2
1! 2
c
c
relativistic factor separately because it will be used in part (c) below.
Copyright &copy; 2012 Nelson Education Ltd.
. Calculate the
Chapter 11: Relativity
11.3-7
Solution:
relativistic
=
classical
v2
c2
4.960 &quot; 10!19 kg # m/s
1!
=
1!
(0.99 c )2
c2
4.960 &quot; 10!19 kg # m/s
(two extra digits carried)
7.088
= 3.52 &quot; 10!18 kg # m/s
=
relativistic
Statement: The proton’s relativistic momentum is
! 0.&quot; 95 # m)s .
(c) From the intermediate step in part (b) above, the relativistic momentum exceeds that
of the classical value by a ratio of 7 : 1.
5. Given: prelativistic / pclassical = 5
Required: v
mv
Analysis: prelativistic =
relativistic
1
=
classical
1!
relativistic
=
classical
&quot;
\$
#
v
c2
1
1!
2
%
relativistic
' =
classical &amp;
1!
2
and
2
v
1! 2
c
classical
= mv , so
relativistic
. Rewrite this equation in terms of
=
classical
v2
1! 2
c
, or
v
.
c
v2
c2
1
v2
1! 2
c
1
v2
=
c2 &quot;
\$
#
%
relativistic
'
classical &amp;
v2
= 1!
c2
&quot;
\$
#
v
= 1!
c
&quot;
\$
#
2
1
relativistic
classical
%
'
&amp;
2
1
%
relativistic
'
classical &amp;
2
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.3-8
Solution:
v
= 1!
c
&quot;
\$
#
= 1!
1
%
relativistic
'
classical &amp;
2
1
25
v
= 0.980
c
v = 0.980c
Statement: To have increased its momentum by a relativistic factor of 5, the speed of the
particle must be 0.98c.
6. Given: melectron = 9.11 ! 10&quot;31 kg; velectron = 0.999c; mship = 4.38 ! 107 kg;
= 3.0 ! 108 m/s
Required: ship
Analysis: The speed of the ship will be much less than that of the electron, so its
relativistic momentum will be very close to its classical momentum. This is not the case
for the electron, so we will use the relativistic expression for just the electron.
m
v
= mship vship , pelectron = electron electron , and pship = pelectron , so
ship
(velectron )2
1!
c2
m 2 kon velectron
ms67g vs67g =
(velectron )2
1!
c2
melectron velectron
vship =
(velectron )2
mship 1 !
c2
melectron velectron
Solution: vship =
(v
)2
mship 1! electron
c2
=
(9.11&quot; 10!31 kg )(0.999)(3.0 &quot; 108 m/s)
(4.38 &quot; 107 kg ) 1! (0.999)2
vship = 1.39 &quot; 10!28 m/s
Statement: If the ship has the same momentum as the electron, its speed is
1.39 ! 10&quot;28 m/s . (This speed is so low that the ship would have to travel for nearly a
million million years before travelling the distance of one hydrogen atom.)
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.3-9
7. No, if you were travelling on a spacecraft at 0.99c relative to Earth, you would not feel
compressed in the direction of travel. According to the principle of relativity, experiments
cannot tell us if we are at rest or moving at constant velocity. If we could get the feeling
of being compressed, then we could then find some experiment that would determine this
compression. As no such experiment exists, we cannot feel any changes in our bodies,
including compression.
8. We do not notice the effects of length contraction in our everyday lives because the
speeds that we experience are much less than the speed of light, so we do not notice the
effects of relativistic length contraction. For example, formula-one race cars can slightly
exceed 300 km/h, which is 1/1000 of the speed of light. The length contraction factor
even at this high speed is 1 ! 10!6 , which is about 0.999 999 5. A car that is 5 m long at
rest would shrink by only 5 &micro;m at this speed, a distance we could only detect using a
microscope. This is why cars do not appear shorter when they drive past us at high
speeds.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.3-10
Section 11.4: Mass–Energy Equivalence
Tutorial 1 Practice, page 602
! 10 3 c = /.0 ! 10 6)s
1. Given: Erest =
Required: m
Analysis: Rearrange the rest-6ass equation by dividing both sides by c2: m =
Solution: m =
=
Erest
c2
Erest
c2
! 10 7g ! 6 2 )s 2
(/.0 &quot; 10
2
6 2 )s 2
m=
7g
Statement: The cellphone’s rest 6ass is 0.25 7g.
2. (a) Given: m = 1 &quot; 10&quot; 7g3 =
c3 c = /.0 &quot; 10 6)s
e
Required: Etotal in units
Analysis: First calculate the rest energy of the proton, Erest = mc 2 , and convert it to units
of MeV. Then, use the equation Etotal =
Erest
1&quot;
2
.
c2
Solution: Erest = mc 2
= (1
!
&quot; 10&quot; 7g (/.0 &quot; 10 6)s 2 #
&quot;
Erest =
MeV (t o e ra digits carrie
MeV
Etotal =
1&quot;
Etotal =
1 MeV \$
&quot; 10&quot;1/ &amp;%
(
c
2
c2
&times; 10/ MeV
Statement: The given proton
Copyright &copy; 2012 Nelson Education Ltd.
! 10/ MeV.
Chapter 11: Relativity
11.4-1
(b) Given: Erest =
MeV3
=
c
Required: E7 in units of MeV
Analysis: Use the equations Etotal =
Erest
1&quot;
E7 =
Erest
2
2
and E7 = Etotal
Erest to obtain
c2
&quot; Erest
1&quot;
c2
&quot;
%
1
E7 = Erest \$
&quot; 1'
2
\$ 1&quot;
'
#
&amp;
c2
v
3
4
Use = 0.800 = , 6a7ing the deno6inator .
5
c
5
!
\$
&quot; &amp;
Solution: E = E #
# &quot;
&amp;
&quot;
%
c
! 1
\$
E = 939.38 MeV) #
&quot; 1&amp;
! 3\$
## &amp; &amp;
&quot; &quot; 5% %
Ek = 6.26 &times; 102 MeV
Statement: The proton
3. (a) Given: m =
c = 3.0 &quot; 108
Required: Erest
Analysis: Use the rest-6ass equation, Erest = mc2.
Solution: E = mc
=
(3.0 &quot; 108
&quot; 102 MeV.
r)2
E ers = 2.1 &quot; 1018
Statement: The typical CANDU fuel bundle has a rest-6ass energy of 2.1 &quot; 1018 J.
(b)
&quot; 1010 J)day
years.
gives 5.8 &quot; 10 day
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-2
4. Given: m = 2500 kg; Ek = 1.5 &quot; 1020 J; c = 3.0 &quot; 108 m/s
Required: v
Analysis: Rearrange the equation for E7.
1
= mc (
&quot; 1)
2
1&quot;
k
2
mc
1&quot;
1
+1=
1&quot;
2
c
2
k
2
c
2
c
=
c2
+1
1
&quot; k
%
+
1
\$ mc 2
'
#
&amp;
= 1&quot;
Solution:
2
1
=
mc 2
1&quot;
c2
1
&quot; k
%
\$ mc 2 + 1'
#
&amp;
v
=
c
&quot;
=
&quot;
= 1&quot;
c
2
!
#&quot; mc
2
\$
+ &amp;
%
!
\$
1.5 &times; 1020
#&quot; (2.5 &times; 103 )(3.0 &times; 108 )2 + 1&amp;%
2
1
!2
\$
#&quot; 3 + 1&amp;%
2
=
5
= 0.80
= 0.80c
Statement: The asteroid has a speed of 0.80c.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-/
Section 11.4 Questions, page 603
1. Given: m = 1 kg; = 0.95c
Required: m e
Analysis: m e
s
s
rs
rs
=
m
1&quot;
Solution: m e
s
rs
=
2
c2
m
2
1&quot;
=
1&quot;
c2
1
(0.95c ) 2
c2
m e s rs = 3.2
Statement: The relativistic 6ass of the 1 7g object is /.2 7g.
2. The fa6ous equation E = mc2 tells us that the rest 6ass of an object and its energy are
equivalent, or Erest = mc 2 . This equation shows that even at rest, and with no other
sources of potential energy, a 6ass nevertheless has energy. It suggests the possibility,
now well established, that 6ass can be converted to energy and vice versa, and thus the
conservation of relativistic energy is really a conservation of 6ass and energy. In
particular, the rest 6ass m can be converted to another for6 of energy in the a6ount of
Erest = mc2.
&quot; 106
c = 3.0 &quot; 108 r
3. Given: E ers =
Required: m
Analysis: Erest = mc 2 , so m =
Solution: m =
=
Erest
c2
.
E ers
c2
&quot; 106
(3.0 ! 108 )2
#
2
2
r2
r2
m=
! 10&quot;11
Statement: A rest 6ass of about 4.
TNT explosive.
Copyright &copy; 2012 Nelson Education Ltd.
&quot; 10-11 7g has as 6uch rest energy as 1 7g of
Chapter 11: Relativity
11.4-4
4. Given: m = 1 &times; 10&quot; 7g3 c = /.0 &times; 108 6)s
Required: Erest for two 6asses m that annihilate
Analysis: Erest = mc 2 . Use the rest 6ass energy of one proton and 6ultiply by two.
Solution: E ers = 2mc 2
= 2(1
&times; 10&quot;
)(3.0 &times; 108
r)2
E ers = 3.01 &times; 10&quot;10
Statement: When a proton and anti-proton annihilate each other, the energy released is
the su6 of their rest-6ass energies, which equals /.01 &quot; 10–10 J.
5. (a) Given: E ers = 1.28 MeV c = 3.0 &quot; 108 r
Required: m
Analysis: Erest = mc 2 , so m =
Erest
c2
. Change the energy units to joules.
Eters
c2
! 1.60 &times; 10&quot;13 J \$
1.28 MeV
=
&amp;
(3.0 &times; 108 )2 m 2 /s 2 #&quot; 1 MeV
%
Solution: m =
m = 2.28 &times; 10&quot;30 kg
Statement: The subato6ic particle’s rest 6ass is 2.28 &quot; 10–/0 7g.
(b) Given: Eters = 1.28 MeV; Esotal = 1.72 MeV
Required: E7
Analysis: The total energy is the su6 of the rest energy and the 7inetic energy. Thus, the
7inetic energy is the difference between the two given energies.
Solution: Ek = Esotal &quot; E e t
= 1.72 MeV &quot;
MeV
E =
MeV
Statement: The relativistic 7inetic energy of the subato6ic particle is 0.44 MeV.
&quot;
(c) Given: Eters = 1.28 MeV; Esotal = 1.72 MeV c =
Required: v
Analysis: Use the expression for the relativistic total energy and solve for v)c.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-5
Erest
Etotal =
1&quot;
2
&quot; Etotal %
\$E ' =
# rest &amp;
1&quot;
2
c2
2
c
2
c
=
c2
1
1&quot;
2
c2
1
&quot; Etotal %
\$E '
# rest &amp;
= 1&quot;
c
&quot; Etotal %
\$E '
# rest &amp;
&quot; Etotal %
\$E '
# rest &amp;
= 1&quot;
=
2
1
= 1&quot;
c
2
1
= 1&quot;
Solution:
2
2
1
! Esotal \$
#E &amp;
&quot; et %
2
1
! 1.72 MeV \$
#
&amp;
MeV %
&quot;
t oe
2
a
t sa e
= 2.88 &times; 18
Statement: The particle’s speed in the laboratory is 2.00 &quot; 108 6)s (about 2)/ the speed
of light .
6. Given: m =
&quot; 1822 kg; = 1.02 &quot; 103 m/s; c = 3.0 &quot; 108 m/s
Required: 6ass that, when converted to energy, would produce the observed 7inetic
energy of the Moon
Analysis: The speed of the Moon is 6ore than 5 orders of 6agnitude less than the speed
of light. Hence, the 7inetic energy of the Moon equals the classical 7inetic energy to at
least 10 deci6al places. So, we calculate the classical 7inetic energy of the Moon and
c2.
deter6ine the 6ass equivalent using E7-classical
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-
Solution:
k-classical
= (mass equivalent)c 2
mass equivalent =
k-classical
2
c
1 2
m
2
=
c2
1
(7.35 &quot; 1022 kg)(1.02 &quot; 103 m/s )2
= 2
(3.0 &quot; 108 m/s )2
mass equivalent = 4.25 &quot; 1011 kg
Statement: A conversion of 4.25 &quot; 1011 7g to energy would be enough to accelerate the
Moon fro6 rest to its present orbital speed.
7. Given: Eters r = 2809.4 MeV; Erest P = 938.3 MeV; Erest N = 939.6 MeV
Required: energy released, Ereleased, when two neutrons co6bine with one proton to for6
one tritiu6 nucleus
Etotal
Analysis: Use conservation of 6ass–energy, Etotal
the neutrons and protons are initially at rest. After the nuclear reaction, we have only the
rest energy of the tritiu6 plus Ereleased, where Ereleased 6ay include any allowed
Solution:
e = Et
ter
Et
Erest P + 2Erest N = Erest + Ere e se
938.3 MeV + 2 939.6 MeV = 2809.4 MeV + Ere e se
Ere e se =
MeV
Statement: The energy released when the proton co6bines with the two neutrons to for6
a tritiu6 nucleus is 8.1 MeV (which 6ay contribute to the tritiu6’s 7inetic energy and
8. Given: E7 = Erest
Required: v
mc 2
Analysis: Substitute Erest = mc 2 into t =
and then substitute into
&quot;
E7 = Etotal – Erest to obtain E7 =
Erest
1&quot;
Copyright &copy; 2012 Nelson Education Ltd.
2
2
c2
&quot; Erest . Set E = Erest and solve for v.
c2
Chapter 11: Relativity
11.4-
E =
Erest
&quot;
2
c2
&quot;
E = Erest \$
\$
#
E
+ =
Erest
&quot;
&quot;
&quot;
%
'
'
&amp;
2
c2
2
c2
Set E = Erest , so
2=
&quot; Erest
E
= .
Erest
v2
&quot; 2
c
v2
=
c2 4
v2 3
=
c2 4
3
v
=
2
c
v=
c 3
2
c 3
2
= 3.0 &quot;
Solution: v =
8
s
v = 2.60 &quot;
s
Statement: The particle’s speed is 0 &quot; 108 6)s.
9. Given: m =
kg; Ek = 2.0 &quot; 10 ; c = 3.0 &quot;
Required: v
Analysis: Rearrange the equation for E7.
8
Copyright &copy; 2012 Nelson Education Ltd.
8
s
Chapter 11: Relativity
11.4-8
&quot;
= mc 2 \$
\$
#
+ =
mc 2
&quot;
2
&quot;
c2
2
c2
c
Solution:
=
c2
+
&quot;
%
\$# 2 + '&amp;
mc
=
v
=
c
&quot;
=
&quot;
c
2
=
mc 2
&quot;
%
&quot; '
2
'
&quot; 2
&amp;
c
&quot;
2
%
&quot;
\$# 2 + '&amp;
mc
!
\$
#&quot; 2 + &amp;%
mc
!
#&quot;
2
2
2.0 &times;
&times; 3 3.0 &times;
8 2
\$
+ &amp;
%
2
= 0.40
= 0.40c
Statement: The spacecraft travels at 0.40 ti6es the speed of light.
10. Given: Erest = 512 7eV3 the classical 7inetic energy E7-class = /0.0 7eV
Required: the relativistic 7inetic energy, E7
1
Analysis: Rewrite the classical equation for 7inetic energy, k-class = m 2 , as
2
2
2 k-class
= 2 . Then, substitute that into the relativistic 7inetic energy equation. Use
mc 2
c
2
Erest = mc to calculate.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-9
&quot;
= mc 2 \$
\$
#
&quot;
= mc 2 \$
\$
#
%
&quot; '
2
'
&quot; 2
&amp;
c
&quot;
2
ss
mc 2
&quot;
Solution: E = mc 2 \$
\$
#
%
&quot; '
'
&amp;
&quot;
2E Tlass
mc 2
%
&quot; 1'
'
&amp;
&quot;
%
1
&quot; 1'
= (512 keV) \$
\$ 1&quot; 2(30.0 keV ) '
'&amp;
\$#
512 keV
Ek = 32.9 keV
Statement: The actual 7inetic energy of the electron is /2.9 7eV.
11. Given: mfuel = 100.0 6g3 c = /.0 &quot; 108 6)s3 car 6ileage = /0 76 fro6 1.0 &quot; 108 J
of fuel
Required: nu6ber of 7ilo6etres travelled on mfuel , d
Analysis: First, using Erest = mc 2 , calculate the rest energy in 100.0 6g of fuel, then
6ultiply by the 6ileage ratio, /.0 &quot; 10– 76)J.
Solution: rest = mc 2
= (1.000 &times; 10&quot;4 kg)(3.0 &times; 108 m/s)2
rest
= 9.0 &times; 1012 J
30.0 km
1.0 &times; 108 J
30.0 km
= 9.0 &times; 1012 J &times;
1.0 &times; 108 J
=
rest
&times;
= 2.7 &times; 106 km
Statement: The car could hypothetically drive
in the rest 6ass of 100.0 6g of fuel.
Copyright &copy; 2012 Nelson Education Ltd.
&quot; 10 76 using the energy contained
Chapter 11: Relativity
11.4-10
12. Given: reduction in uraniu6 6ass, !m = 100.000 7g – 99./12 7g =
&quot; 108 6)s
c=
Required: Erest f om !mc 2
Analysis: Erest = !mc 2
Solution: E
= !mc 2
=
E
=
e
&quot;
m e2
&quot;
Statement: The a6ount of energy released during the four years
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 11: Relativity
&quot; 10 J.
11.4-11
Section 12.1: Introducing Quantum Theory
Section 12.1 Questions, page 619
1. In classical physics, energy can be transferred by collisions between particles and larger
objects, including near-collisions when objects interact at a distance (e.g., charged particles,
astronomical objects) and by waves, including water waves, sound, and light.
2. Classical particles differ from classical waves in that a particle occupies a small region of
space with some mass density that abruptly changes outside the particle. In contrast, a wave
spreads out over a large region and oscillates. Also, for transferring energy, particles tend to
transfer energy in discrete parcels (lumps), one at a time, whereas waves continuously transfer
energy and can interfere destructively.
3. (a) The double-slit experiment with electrons shows that the energy arrives in discrete lumps,
which indicates that the electrons are particles.
(b) The double-slit experiment with electrons also shows destructive interference, which
indicates that the electrons are waves.
4. In the early twentieth century, classical physics could describe nearly everything physical.
However, there were a few puzzling things related to microscopic phenomena, particularly the
behaviour of atoms and electrons, that disagreed with predictions of classical physics. Atoms and
electrons behave differently than large objects such as baseballs, rockets, and planets. In
particular, microscopic objects can exhibit wave-like behaviour. Quantum theory was developed
to describe both the wave-like and particle-like behaviour of microscopic things such as atoms
and electrons.
5. The electron double-slit experiment showed that electrons can interfere like waves, which is a
non-classical behaviour. The result could not be explained by classical physics and thus gave a
major push to the development of quantum theory.
6. Quantum theory is more complete than Maxwell’s theory because it can describe both the
wave-nature and the particle-nature of light. Maxwell’s theory describes only the wave-nature of
light.
7. Golf balls hit toward a wall with two narrow slits would leave a distribution of marks on a
wall behind the slits. The distribution would show two dense areas directly behind each slit.
Many balls would not even hit the slits, but assuming that a high fraction of them did, the
distribution would look like the one below.
8. (a) The distribution of baseballs behind the two open slits would look similar to (b).
(b) The distribution of electrons behind the two open slits would look similar to (a). (The slits
would have to be narrower in accordance with the smaller size of the electrons.)
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Chapter 12: Quantum Mechanics
12.1-1
9. Some of our intuitions gained through our experiences in the macroscopic world of large
objects do not apply to the quantum world because objects in the quantum world exhibit both
wave-like and particle-like behaviour. Things in the macroworld exhibit only wave-like or
particle-like behaviour.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 12: Quantum Mechanics
12.1-2
Section 12.2: Photons and the Quantum Theory of Light
Tutorial 1 Practice, page 624
1. Given: W = 4.60 ! 10&quot;19 J; h = 6.63 ! 10&quot;34 J # s
Required: Ephoton
Analysis: Ephoton = W
Solution: Ephoton = W
Ephoton = 4.60 ! 10&quot;19 J
Statement: The lowest photon energy that can cause emission of electrons from calcium
is 4.60 &times;10−19 J , the work function of calcium.
2. Given: !0 = 268 nm = 2.68 &quot; 10#7 m; h = 6.63&quot; 10#34 J \$s; c = 3.0 ! 108 m/s
Required: the work function, W
Analysis: The lowest photon energy must satisfy the equation Ephoton = W . Use
Einstein’s equation for the photon energy, written in terms of the wavelength, and
substitute the threshold wavelength λ0.
hc
Ephoton =
λ0
Solution: Ephoton =
=
hc
!0
(6.63&quot; 10#34 J \$ s ) (3.0 &quot; 108 m/s )
(2.68 &quot; 10 –7 m )
Ephoton = 7.42 &quot; 10#19 J
Statement: The minimum photon energy to release an electron from the material is
7.42 ! 10&quot;19 J , which is much closer to silver’s value ( 7.43 ! 10&quot;19 J ) than to lead’s
( 6.81 ! 10&quot;19 J ). So, the material is silver.
Tutorial 2 Practice, page 626
1. Given: ! = 450 nm = 4.50 &quot; 10 –7 m; h = 6.63&quot; 10#34 J \$s
Required:
photon
, the momentum of the photon
Analysis: Use the relation
Solution:
photon
photon
photon
=
!
.
=
!
6.63&quot; 10#34 J \$s
=
4.50 &quot; 10 –7 m
= 1.5 &quot; 10#27 kg \$ m / s
Statement: The photon’s momentum is 1.5 ! 10&quot;27 kg # m / s .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 12: Quantum Mechanics
12.2-1
2. Given: ! = 630 nm = 6.30 &quot; 10 –7 m; h = 6.63&quot; 10#34 J \$s; = 3.0 &quot; 108 m/s
Required: Ephoton , the energy of the photon
Analysis: Use the relation Ephoton =
Solution: Ephoton =
hc
.
!0
hc
!0
(6.63&quot; 10#34 J \$ s )(3.0 &quot; 108 m/s )
6.30 &quot; 10 –7 m
= 3.2 &quot; 10#19 J
=
Ephoton
Statement: The photon’s energy is 3.2 &times;10−19 J .
3. Given: Ephoton = 2.2 &times;10−14 J; c = 3.0 &times;108 m/s
Required:
photon
, the momentum of the gamma photon
Analysis: Ephoton = hf and pphoton =
Solution:
hf
, so
c
photon
=
Ephoton
.
Ephoton
photon
=
photon
2.2 ! 10&quot;14 J
=
3.0 ! 108 m/s
= 7.3! 10&quot;23 kg # m / s
Statement: The momentum of the gamma ray is 7.3! 10&quot;23 kg # m / s .
Tutorial 3 Practice, page 629
1. Given: T = 5100 K
Required: !
Analysis: Use Wien’s law, !max =
Solution: !max =
2.90 &quot; 10#3 m \$ K
.
2.90 &quot; 10#3 m \$ K
2.90 &quot; 10#3 m \$ K
5100 K
= 5.7 &quot; 10#7 m
=
!max
Statement: The maximum wavelength of the 5100 K blackbody is 5.7 &times;10−7 m , which
gives a yellow colour.
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Chapter 12: Quantum Mechanics
12.2-2
2. Given: !max = 510 nm = 5.10 &quot; 10#7 m
Required: T
Analysis: Use Wien’s law, rearranged to obtain T.
2.90 &quot; 10#3 m \$ K
!max =
2.90 &quot; 10#3 m \$ K
=
!max
Solution:
=
2.90 ! 10&quot;3 m # K
\$max
2.90 ! 10&quot;3 m # K
5.10 ! 10&quot;7 m
= 5700 K
Statement: The temperature of the aqua-coloured blackbody is 5700 K.
3. Given: T = 37 ˚C = 310 K
Required: !max
=
Analysis: Use Wien’s law, !max =
Solution: !max =
2.90 &quot; 10#3 m \$ K
.
2.90 &quot; 10#3 m \$ K
2.90 &quot; 10#3 m \$ K
3.1&quot; 102 K
!max = 9.4 &quot; 10#6 m
Statement: The wavelength of maximum intensity for a human body is 9.4 &times; 10–6 m, or
9.4 &micro;m. This is infrared light, which can be imaged with an infrared camera, but not the
human eye.
=
Research This: Exploring Photonics, page 630
Answers may vary. Students will choose a technology or process that uses the particle
nature of light. The sample answers for A to C below are for the photomultiplier tube:
A. A photomultiplier tube relies upon the photoelectric effect, and thus relies upon the
particle nature of light. The tube also relies upon secondary emission, another process
that is similar to the photoelectric effect but occurs for massive, charged particles such as
electrons. Photomultiplier tubes consist of a photocathode and a series of electrodes,
called dynodes, in an evacuated glass enclosure. The electrodes are each maintained at a
more positive potential. A photon that strikes the photoemissive cathode releases an
electron from the photocathode (material with very low work function). The released
electron gets focused and accelerated by a magnetic electron lens and strikes the first
dynode. The collision releases several electrons in a process called secondary emission.
The electrons strike the next dynode, releasing even more electrons, and the process
cascades along many more dynodes. This cascading effect creates 105 to 107 electrons for
each photon hitting the first cathode, depending on the number of dynodes and the
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Chapter 12: Quantum Mechanics
12.2-3
accelerating voltage. The resulting avalanche of electrons hits the output (anode), where
the amplified signal can be measured.
B. The photomultiplier tube counts photons, which would not be possible to interpret
without the quantum theory of light.
C. The avalanche photodiode is the solid-state equivalent to the photomultiplier tube.
D. As of 2009, there were over 5000 Canadian companies in the Canadian Photonics
Consortium, and these companies employed nearly 300 000 people.
Section 12.2 Questions, page 631
1. Given: W = 5.0 eV = 8.0 ! 10–19 J; h = 6.63 ! 10&quot;34 J # s
Required: f0, the minimum photon frequency that can eject an electron
Analysis: Use Einstein’s relation for the photon energy, Ephoton = hf0 , rearranged for f0,
and then substitute the photon energy with the work function, Ephoton = W .
Ephoton = hf0
f0 =
Ephoton
h
Now, substitute W for Ephoton .
f0 =
h
Solution: f0 =
h
8.0 ! 10&quot;19 J
=
6.63! 10&quot;34 J #s
f0 = 1.2 ! 1015 Hz
Statement: The minimum frequency to eject an electron is 1.2 ! 1015 Hz , which is
ultraviolet, according to Table 3.
2. (a) The lower-energy material, cesium, is a better choice than aluminum for the metal
piece in the photocell because it will capture all the photons that the aluminum will plus
those with energy between 1.95 eV and 4.20 eV. Judging from Table 3, we would
probably need a work function less than half the 5.0 eV of the aluminum to capture the
lower-frequency visible light.
(b) Given: W = 1.95 eV = 3.12 ! 10–19 J; h = 6.63 ! 10&quot;34 J # s
Required: f0, the minimum photon frequency that can eject an electron
Analysis: Use Einstein’s relation for the photon energy, Ephoton = hf0 , rearranged for f0,
and then substitute the photon energy with the work function, Ephoton = W .
Ephoton = hf0
f0 =
Ephoton
h
Now, substitute W for Ephoton .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 12: Quantum Mechanics
12.2-4
f0 =
h
Solution: f0 =
h
3.12 ! 10&quot;19 J
=
6.63! 10&quot;34 J #s
f0 = 4.71! 1014 Hz
Statement: The lowest photon frequency that can be measured with the photocell is
4.71! 1014 Hz .
(c) A frequency of 4.71! 1014 Hz is near the lower boundary of the visible range of the
electromagnetic spectrum ( 4.3! 1014 Hz ).
3. The higher-intensity red light will not be able to eject electrons according to the
photoelectric effect. The ability of light to eject electrons depends only on the frequency,
not the intensity. So, if the low-intensity red light could not eject electrons, the higherintensity red light cannot either.
4. (a) Given: f = 100 MHz = 1.00 ! 108 Hz; h = 6.63! 10&quot;34 J #s; = 3.0 ! 108 m/s
Required: the energy, E, and momentum, p, of the photon
hf
E
Analysis: Combine the equations E = hf and p =
to obtain = . Since we can
c
use E to calculate p, first calculate E.
Solution: E = hf
= (6.63! 10&quot;34 J # s )(1.00 ! 108 Hz )
E = 6.63! 10&quot;26 J (two extra digits carried)
=
E
6.63! 10&quot;26 J
=
3.0 ! 108 m/s
= 2 ! 10&quot;34 kg # m / s
Statement: The energy of an FM radio station with a frequency of 100 MHz is
7 ! 10&quot;26 J , and the momentum is 2 ! 10&quot;34 kg # m / s .
(b) Given: ! = 633 nm = 6.33&quot; 10 –7 m; h = 6.63&quot; 10#34 J \$s; = 3.0 &quot; 108 m/s
Required: the energy, E, and momentum, p, of the photon
hc
Analysis: Combine the equations E = hf and ! =
to obtain E =
, and use
!
from part (a). Since we can use E to calculate p, first calculate E.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 12: Quantum Mechanics
=
E
12.2-5
hc
!
(6.63&quot; 10#34 J \$ s )(3.0 &quot; 108 m/s)
=
6.33&quot; 10 –7 m
= 3.1422 &quot; 10#19 J (two extra digits carried)
Solution: E =
E = 3.14 &quot; 10#19 J
=
E
3.1422 ! 10&quot;19 J
=
3.0 ! 108 m/s
= 1.05 ! 10&quot;27 kg # m / s
Statement: The energy of a red light with a wavelength of 633 nm is 3.14 ! 10&quot;19 J and
the momentum is 1.05 ! 10&quot;27 kg # m / s .
(c) Given: ! = 0.070 nm = 7.0 &quot; 10 –11 m; h = 6.63&quot; 10#34 J \$s; = 3.0 &quot; 108 m/s
Required: the energy, E, and momentum, p, of the photon
Analysis: The same as in part (b) above.
hc
Solution: E =
!
(6.63&quot; 10#34 J \$ s )(3.0 &quot; 108 m/s)
=
7.0 &quot; 10 –11 m
= 2.841&quot; 10#15 J (two extra digits carried)
E = 2.8 &quot; 10#15 J
=
E
2.841! 10&quot;15 J
3.0 ! 108 m/s
= 9.5 ! 10&quot;24 kg # m / s
Statement: The energy of an X-ray photon of wavelength 0.070 nm is 2.8 &times;10−15 J , and
the momentum is 9.5 ! 10&quot;24 kg # m / s .
5. An X-ray photon has greater energy than an ultraviolet photon because it has a greater
frequency.
6. (a) There are 1.60 &times; 10–19 J in one electron-volt, so we multiply the 13.6 eV by
1.60 &times; 10–19 J to get 2.176 &times; 10–18 J (one extra digit carried). There are 2.18 &times; 10–18 J
in 13.6 eV.
=
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 12: Quantum Mechanics
12.2-6
(b) Given: E = 2.176 &times; 10–18 J; h =
! 10&quot;
#s =
! 109 m/s
Required: f ; λ
Analysis: Use Einstein’s relation for the energy of a photon, E = hf , rearranged to
isolate f: f =
E
. Then, use the relation between wavelength and frequency, ! =
h
, to
calculate λ.
E
h
2.176 ! 10&quot;18 J
=
6.63! 10&quot;34 J #s
Solution: f =
= 3.2821! 1015 Hz (two extra digits carried)
f = 3.28 ! 1015 Hz
!=
=
3.0 &quot; 108 m/s
3.2821&quot; 1016
! = J.14 &quot; 10#8 m
Statement: The frequency of the highest-energy photons emitted by a hydrogen atom is
3.28 ! 1015 s &quot;1 , and the wavelength is 9.14 ! 10&quot;8 m , which correspond to ultraviolet
photons.
7. The working of a solar cell relies on quantum mechanics through the photoelectric
effect and the quantum theory of semiconductors. The photoelectric effect removes an
electron from a special region within the semiconductor wafer, and the electron flows
through a circuit, producing electrical power.
The special region inside the semiconductor is similar to that in an LED. One side has
been made to have electron charge carriers, and the other side to have missing electrons,
which act as “holes” that carry positive charge according to the quantum theory of
semiconductors. At the boundary, the electrons and holes combine, forming the special
region called the depletion region. The depletion region does not have many charge
carriers. The combining of the electrons and holes creates an electric field and thus a
voltage. To generate power, the depletion region needs charges that can flow. These
charges are produced by the photovoltaic effect, which is similar to the photoelectric
effect; briefly, an incident photo travels into the depletion region and causes an electron
to move up to a higher energy level in which it can conduct. The electron then gets
pushed by the electric field through the circuit, generating electricity.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 12: Quantum Mechanics
12.2-7
Section 12.3: Wave Properties of Classical Particles
Tutorial 1 Practice, page 634
1. Given: p = 1.8 ! 10&quot;25 kg # m/s; h = 6.63 ! 10&quot;34 J # s
Required: λ
h
Analysis: Use the de Broglie relation, λ = .
p
h
Solution: ! =
p
=
6.63 &quot; 10#34 kg \$ m 2 /s
1.8 &quot; 10#25 kg \$ m/s
! = 3.7 &quot; 10#9 m
Statement: The de Broglie wavelength of the electron is 3.7 ! 10&quot;9 m , or 3.7 nm.
2. Given: m = 1.7 ! 10&quot;27 kg8 v = 3.4 ! 105 m/s8 h = 6.63 ! 10&quot;34 J # s
Required: λ
Analysis: The speed of the proton is much less than light speed, so we can use the
h
h
classical momentum p = mv. Thus, the de Broglie relation, λ = , becomes λ =
.
p
mv
h
Solution: ! =
mv
6.63 &quot; 10#34 kg \$ m 2 /s
=
(1.7 &quot; 10#27 kg )(3.4 &quot; 105 m/s )
! = 1.1 &quot; 10#12 m
Statement: The proton’s de Broglie wavelength is 1.1&times;10−12 m .
3. Given: m = 140 g = 1.4 ! 10 –1 kg; v = 140 km/h; h = 6.63 ! 10&quot;34 J # s
Required: λ
Analysis: The speed of the proton is much less than light speed, so we can use the
h
h
classical momentum p = mv. Thus, the de Broglie relation, λ = , becomes λ =
.
p
mv
First, convert kilometres per hour to metres per second.
km 103 m
1h
140
!
!
= 3.89 ! 101 m/s (one extra digit carried)
3
h
1 km 3.6 ! 10 s
Solution: ! =
h
mv
6.63 &quot; 10#34 J
=
(1.40 &quot; 10 –1 kg)(3.89 &quot; 101 m/s)
! = 1.2 &quot; 10#34 m
Statement: The de Broglie wavelength of the baseball is 1.2 &times;10−34 m .
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 12: Quantum Mechanics
12.3-1
4. The de Broglie wavelength of the baseball is 19 orders of magnitude smaller than the
diameter of a proton; therefore, we could never expect to see any wave-like behavior of a
macroscopic object like a baseball.
Research This: Exploring Quantum Computers, page 638
A. Quantum computers differ fundamentally from digital computers in the basic unit of
information. For a digital computer, the basic unit is the bit, an element that can be in
only one of two states, a “0” and a “1”. For a quantum computer, the basic unit is a
quantum bit or “qubit.” A qubit can be in any superposition of two states, just like the
electron trapped in a box (Figure 4 on page 636 of the Student Book) can be in a
superposition of state 1 and state 2. Moreover, reading the state of a qubit is much
different than reading the state of a bit. The reading of the state destroys the quantum
superposition.
B. Several problems presently stand in the way of building practical quantum computers.
One is the difficulty of making a computer with many qubits. Another problem is the
fragility of the quantum superposition state; it is relatively easy to disturb the system, so
that the superposition state gets destroyed. Another difficulty is finding a way to easily
C. A quantum computer’s design should allow it to perform very quickly at some
computations that are very difficult for digital computers, so some possible applications
of quantum computing include the factoring of large numbers, database searching, and
the simulation of quantum mechanical systems.
Section 12.3 Questions, page 639
1. Given: m = 9.11 ! 10&quot;31 kg; # = 150 nm = 1.5 ! 10 –7 m; h = 6.63 ! 10&quot;34 J \$ s
Required: speed of the electron, v
Analysis: Notice that the wavelength here is larger than that in the solution to Sample
Problem 1 of Tutorial 1, and in that case the electron’s speed is much less than that of
light. A larger wavelength means that the speed is slower, so use the classical momentum
in the de Broglie relation and solve for v.
h
h
!=
, so v =
.
mv
m!
h
Solution: v =
m!
6.63 &quot; 10#34 J \$ s
=
(9.11 &quot; 10#31 kg)(1.5 &quot; 10 –7 m)
v = 4.9 &quot; 103 m/s
Statement: The electron’s speed is 4.9 ! 103 m/s, a non-relativistic speed.
2. Given: mproton/melectron = 1800; λproton = λelectron
Required: Eproton/Eelectron
Analysis: Assume that the two particles are non-relativistic (otherwise, we would need to
know if the energy is the total energy or just the kinetic energy). In addition to using the
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Chapter 12: Quantum Mechanics
12.3-2
de Broglie relation, use the classical relation between E and v, as well as that between p
and v.
h
h
!electron =
and !proton =
pelectron
pproton
But, as λproton = λelectron, then it follows that pproton = pelectron. The classical kinetic energy
can be written in terms of the classical momentum mv.
1
Eproton = mproton v 2 proton
2
2
1 (mproton vproton )
=
2
mproton
2
1 p proton
Eproton =
2 mproton
The same relation holds for the electron.
1 p 2 electron
Eelectron =
2 melectron
Thus,
1 p 2 electron
Eelectron 2 melectron
=
2
Eproton
1 p proton
2 mproton
2
Eelectron p electron mproton
=
Eproton melectron p 2 proton
2
Eelectron p electron mproton
Solution:
=
Eproton melectron p 2 proton
=
mproton
melectron
Eelectron 1800
=
1
Eproton
Statement: When the proton’s wavelength equals that of the electron, then they both
have the same momentum. And when they have the same momentum and have nonrelativistic speeds, then the ratio of their classical kinetic energies is 1800:1, with the
electron having the higher energy because it is lighter.
Copyright &copy; 2012 Nelson Education Ltd.
Chapter 12: Quantum Mechanics
12.3-3
3. (a) Given: m = 1000.0 kg; v = 100.0 km/h; h = 6.63 ! 10&quot;34 J # s
Required: λ
Analysis: Use the de Broglie relation, assuming non-relativistic speed: λ =
First, convert kilometres per hour to metres per second.
km 103 m
1h
100.0
!
!
= 2.778 ! 101 m/s (one extra digit carried)
3
h
3.6
!
10
s
1 km
Solution: ! =
=
h
.
mv
h
mv
6.63&quot; 10#34 J \$s
(1000.0 kg)(
&quot; 101 m/s)
!=
&quot; 10# m
Statement: The de Broglie wavelength of the car travelling at 100.0 km/h is
2.39 ! 10&quot;38 m .
(b) Given: m = 1000.0 kg; v = 10.0 ! 103 km/h = 1.0 ! 104 km/h; h = 6.63! 10&quot;34 J #s
Required: λ
Analysis: The car’s speed, though fast, is still non-relativistic (~2800 m/s), so we can use
h
the same relation as in (a), λ =
. First, convert kilometres per hour to metres per
mv
second.
103 m
1h
4 km
1.0 ! 10
!
!
= 2.778 ! 103 m/s (one extra digit carried)
3
h
1 km 3.6 ! 10 s
Solution: ! =
=
h
mv
6.63&quot; 10#34 J
&quot; 103 m/s)
(1000.0 kg)(
!=
&quot; 10#40 m
Statement: The de Broglie wavelength of the car travelling at 1.0 ! 104 km/h is
2.39 ! 10&quot; m .
(c) The de Broglie wavelength of the car at rest is undefined. As the speed decreases, the
wavelength increases, and at zero speed, the wavelength blows up. (This result seems
impossible, because we always see parked cars as solid objects and not spread out.
However, consider how small the speed needs to be for the wavelength of the car to
exceed 1 &micro;m. The observer would have to establish that the speed of the car was less
than about 6 &times; 10–31 m/s. Such a determination would be impossible, so we do not see
parked cars spread out like a wave.)
4. In classical physics, particles occupy a definite position in space, and we can calculate
exactly how a particle’s position changes with time. Moreover, we can determine both
the particle’s position and the particle’s velocity at each instant of time with arbitrary
precision. In quantum mechanics, we do not know what happens to the particle between
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Chapter 12: Quantum Mechanics
12.3-4
measurements. Moreover, a measurement cannot determine the particle’s position and
velocity with arbitrary precision. Instead, quantum mechanics gives us the probabilities
for obtaining various outcomes of the measurement.
(a) An example of experimental evidence for wave-like properties of matter is the
Davisson–Germer experiment with electrons diffracting from a crystal. Other
experiments have shown diffraction of larger particles.
(b) An example of experimental evidence for particle-like properties of electromagnetic
radiation is the early experiments by Heinrich Hertz on the photoelectric effect. (Other
experiments include that of the photovoltaic effect (e.g., solar cells)).
6. I think wave functions are real. Wave functions cannot be observed directly, so one
might conclude that they are not real. However, we can say the same thing about atoms,
and yet atoms seem to be quite real; we can touch objects and we can feel the wind.
Similarly, we can infer the existence of wave functions through their influence on
measurements. For example, the probability distribution of electrons striking the wall
behind a pair of slits is a result of the electron’s wave function.
7. Presently, all interpretations of quantum mechanics are consistent with the same
observable results that we measure and experience. Yet quantum mechanics describes
things that we cannot observe directly, such as the wave function. This indeterminacy of
various aspects of quantum mechanics makes it possible for several views to be
consistent with what we observe. Thus, different interpretations of quantum mechanics
exist.
I think the Copenhagen interpretation is most likely because I am comfortable
with the idea that there are things we simply cannot know. I do not like the pilot-wave
interpretation, as it seems to imply that future events are predetermined. Future events
might be predetermined, but I am not comfortable with the idea. Similarly, I do not like
the many-worlds interpretation because it is hard for me to picture the universe
continually splitting in two. The collapse interpretation is not so objectionable, but I
prefer the Copenhagen interpretation.
8. According to the Heisenberg uncertainty principle, one cannot take exact
measurements of an electron (or any other object) when it is at rest. If the electron is at
rest, then Δp = 0 and Δx, the uncertainty in the electron’s position, blows up. Thus, we
could not determine where the electron was.
9. Willard Boyle earned a PhD in physics from McGill University. He worked at Bell
Labs in New Jersey, then left for a job providing NASA with technological support for
the Apollo space program, and then returned to Bell Labs in 1964, where he worked on
developing electronic devices, including the charge-coupled device. Charge-coupled
devices (CCDs) are designed around the photoelectric effect and the quantum mechanics
of semiconductors. Other physical aspects of their operation are the motion of charges
under an applied voltage, and for CCDs used for imaging, their operation depends on
optics. They were originally designed to be used in several applications, including use as
a memory device and shift register, but their most common application is for imaging.
They were immediately useful in astronomy because, as imaging sensors, they could
detect far fainter objects than those detected using film. They are now used in nearly all
digital cameras.
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Chapter 12: Quantum Mechanics
12.3-5
Section 12.6: The Standard Model of Elementary Particles
Tutorial 1 Practice, page 647
1. The orbits in Figure 5 have energy levels 4 and 7. So, energy levels 5 and 6 are
allowed between these values.
2. The first three energy levels of a Bohr atom are sketched below.
Section 12.6 Questions, page 653
1. The Rutherford planetary model of the atom is inconsistent with classical physics
because classical electrodynamics shows that an accelerating charged particle radiates
energy. If an atom had orbiting electrons, as in the planetary model, then classically they
would radiate and fall into the nucleus. Such a collapse would occur quickly. As we
know that atoms are stable (otherwise, how could we exist?), the planetary model is
inconsistent with classical physics.
2. The progression of atomic models involved the Rutherford model, the Bohr model, the
later quantum model (Schr&ouml;dinger and others), and the standard model. Rutherford’s
model was classical and did not work well, as explained in Question 1 above. The Bohr
model, proposed by Niels Bohr and later strengthened by Louis de Broglie, was the first
quantum model. It explained the hydrogen atom very well, but could not explain atoms
with many electrons. Later quantum models, based on the Schr&ouml;dinger equation, did well
at explaining the behaviour of larger atoms and molecules, but did not explain the
nucleus. The standard model is a complete model of the entire atom, describing not only
the electrons, but also the nucleus.
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Chapter 12: Quantum Mechanics
12.6-1
3. (a) Sketches of the constituent parts of the proton and neutron are below. The smaller
circles with “u” and “d” are the quarks, and the wiggly lines connecting them represent
the gluons, the force-mediating particles that hold the nucleus together.
(b) Answers may vary. Sample answer: I decided to change one of the d quarks in the
neutron to a c quark.
(c) In looking up the above particle in a list of baryons, I found it labelled the “charmed
lambda.” Because the charm quark has the same charge as the up quark, the charmed
lambda has a charge of +1. It is unstable, with a lifetime of about 2.00 ! 10–13 s, and it is
more than twice as heavy as the proton, at 2286.46 MeV/c2.
4. Bosons have an important role in the standard model because they mediate the forces
between particles. For example, the photon is the boson that mediates the electrodynamic
force. In addition, the Higgs boson gives particles their mass.
5. (a) Fermions and bosons together are the building blocks of matter. The fermions
include the leptons and quarks and have can have charge. The bosons mediate the forces
and provide mass.
(b) Mesons and baryons are hadrons, in that they are composed of quarks. Whereas the
baryons have three quarks and include the particles of the nucleus, the proton and
neutron, the mesons have just two quarks and are all unstable.
(c) Leptons and hadrons are both particles, but leptons do not contain quarks. Leptons are
the family of particles that include the electron, its neutrino, and its antiparticle. Unlike
the leptons, hadrons are composite particles, being made out of quarks and including
mesons and baryons.
6. The standard model describes how matter is composed and how interactions occur
through the electromagnetic force and both the weak and the strong nuclear force. The
model describes matter as being made up of quarks and leptons, with their forces
mediated by bosons. However, the standard model does not describe gravity.
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Chapter 12: Quantum Mechanics
12.6-2
7. The only significant missing piece in the standard model is the Higgs boson. The
standard model says the Higgs exists, but it has never been detected. So, the discovery of
the Higgs boson would help to verify the standard model and help us understand the
origin of mass.
8. Scientist may think that there must be a theory of everything for a few reasons. One,
there is the general idea that the universe can be understood. Without a theory of
everything, we do not understand why the microscopic realm is described by one theory,
the standard model, whereas the large-scale structure is described by another theory,
general relativity. In addition, the trend in science is to unify. First Newton unified the
mechanics on Earth with the motions of the planets. Later Maxwell unified electricity and
magnetism. And more recently, the standard model unified electromagnetism with the
nuclear forces. Next on the list is to unify all forces in one grand theory. With a grand
theory of everything, scientists may better understand the origin of the universe as well as
many other mysteries, and also discover new phenomena.
9. Newton unified physics in the sense that the same gravitational force that objects on
Earth respond to also acts on celestial objects. Moreover, Newton proposed that all matter
attracts other matter gravitationally; that is, gravity is a universal force between all
masses.
10. Answer may vary. Various pieces of the standard model have come from many
scientists. Students will research the history of the standard model and choose a scientist
who made an important contribution to the standard model. Scientists fitting this
description would have to include many Nobel Prize winners—Murray Gell-Mann,
Jerome I. Friedman, Henry W. Kendall, Richard E. Taylor, Yoichiro Nambu, Makoto
Kobayashi, Toshihide Maskawa, Emilio Gino Segr&egrave;, Owen Chamberlain, Sin-Itiro
Tomonaga, Julian Schwinger, Richard P. Feynman, Sheldon Lee Glashow, Abdus Salam,
Steven Weinberg, Gerardus ’t Hooft, Martinus J.G. Veltman, David Politzer, David
Gross, Frank Wilczek, Burton Richter, Samuel Ting, Martin Perl, Frederick Reines, Leon
M. Lederman, Melvin Schwartz, Jack Steinberger, Chen Ning Yang, Tsung-Dao (T.D.)
Lee, Carlo Rubbia, and Simon Van der Meer—and many others, including Gerson
Goldhaber, Francois Pierre, James Bjorken, John Iliopoulos, Luciano Maiani, Donald
Perkins, Charles Prescott, Harald Fritzsch, George Zweig, O.W. Greenberg, M.Y. Han,
Harald Fritzsch, and Chien-Shiung Wu.
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Chapter 12: Quantum Mechanics
12.6-3
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