Nelson SPH4U Textbook Solutions
G
(d) vav = ?
G
∆d
G
vav =
∆t
65 m [43° S of E]
=
120 s
G
vav = 0.54 m/s [43° S of E]
The average velocity is 0.54 m/s [43° S of E].
Applying Inquiry Skills
9. (a) Students can refer to the Learning Tip titled “The Image of a Tangent Line” on page 13 of the text to understand how
to use a plane mirror to check the accuracy of their tangents.
(b) One way to draw tangents accurately is to use the plane mirror method, as described in the Learning Tip. Another way
that is useful for a displacement-time graph of uniform acceleration is to draw the tangent parallel to an imaginary line
joining two points that are equal times away from the tangent time (e.g., at t = 3.5 s, draw the tangent parallel to the
line joining the points at t = 2.5 s and t = 4.5 s).
Making Connections
10. Use the equation d = vav∆t to complete the table.
Reaction Distance
Speed
17 m/s (60 km/h)
25 m/s (90 km/h)
33 m/s (120 km/h)
no alcohol
14 m
20 m
26 m
4 bottles
34 m
50 m
66 m
5 bottles
51 m
75 m
99 m
1.2 ACCELERATION IN ONE AND TWO DIMENSIONS
PRACTICE
(Page 20)
Understanding Concepts
1. All five examples could be units of acceleration.
2. (a) It is possible to have an eastward velocity with a westward acceleration. For example, a truck moving eastward whle
slowing down has a westward acceleration.
(b) It is possible to have acceleration when the velocity is zero. For example, at the instant that a ball tossed vertically
upward comes to a stop, its acceleration is still downward.
3. (a) When the flock’s acceleration is positive, the flock is moving south with increasing velocity.
(b) When the flock’s acceleration is negative, the flock is moving south with decreasing velocity.
(c) When the flock’s acceleration is zero, the flock is moving south with constant velocity.
G
4. vi = 0
G
vf = 9.3 m/s [fwd]
∆t = 3.9 s
G
aav = ?
G G
vf − vi
G
aav =
∆t
9.3 m/s [fwd] − 0
=
3.9 s
G
aav = 2.4 m/s 2 [fwd]
The runner’s average acceleration is 2.4 m/s2 [fwd].
16
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
5.
G
vi = 0
G
vf = 26.7 m/s [fwd]
G
a = 9.52 m/s2
(a) ∆t = ?
∆t =
=
G
G
vf − vi
G
aav
26.7 m/s − 0
9.52 m/s 2
∆t = 2.80 s
The Espace takes 2.80 s to achieve a speed of 26.7 m/s.
(b) v = ?
m 1 km 3600 s
v = 26.7 ×
×
s 1000 m
1h
v = 96.1 km/h
The speed is 96.1 km/h.
G G
? v −v
(c) ∆t = f G i
aav
L L
−
T
T
T=
 L 
 2
T 
?
? L  T2 
 
T = 

 T  L 
?
6.
7.
T=T
Thus, the equation is dimensionally correct.
G
aav =14 (km/h)/s [fwd]
∆t = 4.7 s
G
vi = 42 km/h [fwd]
G
vf = ?
G G
v − vi
G
aav = f
∆t
G G G
vf = vi + aav ∆t
= 42 km/h [fwd ] + (14 (km/h)/s [fwd ]) ( 4.7 s )
G
vf =108 km/h [fwd ]
Thus, the final velocity is 108 km/h [fwd].
G
aav = 1.37 × 103 m/s2
∆t = 3.12 × 10–3 s
G
vf =0 m/s
G
vi = ?
G G
G
v − vi
aav = f
∆t
G G G
vi = vf − aav ∆t
(
)(
= 0 m/s − 1.37 ×103 m/s 2 [W ] 3.12 × 10−2 s
G
vi = 42.8 m/s [E ]
)
The velocity of the arrow as it hits the target is 42.8 m/s [E].
Copyright © 2003 Nelson
Chapter 1 Kinematics
17
Try This Activity: Graphing Motion with Acceleration
(Page 23)
The required graphs are shown below, in which the position of zero displacement is located where the cart is near the bottom
of the ramp but is not experiencing a push.
PRACTICE
(Pages 23–24)
Understanding Concepts
8. (a) To determine the average acceleration from a velocity-time graph, determine the slope of the line if the acceleration is
constant.
(b) To determine the change in velocity from an acceleration-time graph, determine the area under the line.
9. (a) The motion starts with a westward velocity, but constant eastward acceleration. The motion then slows down to zero
velocity, then accelerates westward with increasing westward velocity. The magnitude of the westward acceleration is
somewhat less than the magnitude of the eastward acceleration.
(b) The motion is southward with northward acceleration slowing down to zero velocity.
(c) The motion is forward with constant acceleration forward. After a period of time, the motion increases to a higher
constant acceleration forward.
(d) The motion starts with northward acceleration then increases its northward acceleration. It starts to slow down
(southward acceleration), and then decreases its southward acceleration to zero.
10. (a)
18
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b)
11.
12. The car’s displacement is the area under the velocity-time graph. It is determined by adding the areas of rectangles and
triangles contained in each time segment. Referring to the figure in the text:
displacement = total area = A4 (0 to 3 s) + A5 (3 s to 5 s) +A6 (5 s to 9 s)
1
(12 m/s [S])(3.0s) = 18 m [S]
2
1
A5 = (12 m/s [S])(2.0s) + (18 m/s [S] − 12 m/s [S])(2.0s) = 30 m [S]
2
1
A6 = (18 m/s [S])(4.0s) + (24 m/s [S] − 18 m/s [S])(4.0s) = 84 m [S]
2
A4 =
displacement = A4 + A5 + A6 = 132 m [S]
The car’s displacement is 132 m [S].
Making Connections
13. (a) The word “idealized” means that the acceleration changes instantaneously from one value to another. In real situations,
changes from one acceleration value to another occur over a finite time interval.
(b) Calculations are much easier if idealized examples are used. For example, to find the change in velocity for an
idealized acceleration graph, we can find the area of a rectangle on the graph. That is much easier than finding the area
under a curved line.
(c)
14. The solution to this question depends on the software, calculator, or planimeter available. Each device is accompanied by
a set of instructions that students can follow to analyze graphs.
Copyright © 2003 Nelson
Chapter 1 Kinematics
19
PRACTICE
(Page 27)
Understanding Concepts
G
2
G G
a ( ∆t )
15. (a) ∆d = vi ∆t +
.
2
G 1 G G
(b) ∆d = (vi + vf ) ∆t .
2
2
2
16. vf = vi + 2a∆d
2
2
2
2
L ?L
 L 
 T  =  T  + 2  2  (L )
   
T 
2
L ?L
L
 T  = T  + 2 T 
   
 
Since the dimensions of each term are the same, the equation is dimensionally correct.
G 1 G G
17. (a) ∆d = (vi + vf ) ∆t
2
G
2 ∆d
∆t = G G
vi + vf
G 1 G G
∆d = (vi + vf ) ∆t
(b)
2
G
G G
2 ∆d
vi + vf =
∆t
G
G  2 ∆d  G
vf = 
−v
 ∆t  i


18. Start with the defining equation for constant acceleration and the equation for displacement in terms of average velocity:
G
G G
G ∆v
a=
∆d = vav ∆t
∆t
G G
G G
G (vi + vf )
G (vf − vi )
∆d =
∆t
a=
2
∆t
(a) To derive the constant acceleration equation in which the final velocity has been eliminated: first solve acceleration
G
equation for vf , then substitute into the equation for displacement.
G G
G (vi + vf )
∆d =
∆t
2
G G G
G (vi + vi + a ∆t )
G G
G vf − vi
∆
=
∆t
d
a=
2
∆t
G
G
G G G
2vi ∆t + a ∆t 2
vf = vi + a ∆t
=
2
G G
1G
∆d = vi ∆t + a ∆t 2
2
20
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b) To derive the constant acceleration equation in which the initial velocity has been eliminated: first solve the
G
acceleration equation for vi , then substitute into the equation for displacement.
G G
G (vi + vf )
∆d =
∆t
2
G G
G
G (vf − a ∆t + vf )
∆d =
∆t
G G G
2
vi = vf − a ∆t
G
G
2vf ∆t − a ∆t 2
=
2
G G
1G
∆d = vf ∆t − a ∆t 2
2
G
2
19. a =18 m/s [E]
∆t = 1.6 s
G
vi =73 m/s [W]
G
vf = ?
G G
G v − vi
a= f
∆t
G G G
vf = vi + a ∆t
= 73m/s[W] + 18 m/s 2 [E](1.6s)
G
vf = 44 m/s[W]
The velocity of the birdie is 44 m/s [W].
G
20. vi = 41 m/s [S]
G
vf = 47 m/s [N]
∆t = 1.9 ms = 1.9 × 10–3 s
G
a =?
G G
G v −v
a= f i
∆t
47 m/s[N] − 41m/s[S]
=
1.9 × 10−3 s
G
a = 4.6 × 104 m/s 2 [N]
The acceleration is 4.6 × 104 m/s2 [N].
21. Note: As was described in the text, pages 24 and 25, and applied in Sample Problem 6, page 26, we can omit the vector
G
G
notation when vf 2 or vi 2 terms are involved. However, in the solution shown here as well as the solution for question 24,
we have kept the vector notation in order to show what the final direction is.
G
vi = 0
G
a = 2.3 m/s2 [fwd]
∆t = 3.6 s
G
(a) ∆d = ?
G G
1G
2
∆d = vi ∆t + a ( ∆t )
2
(2.3m/s [fwd ]) (3.6 s)
= 0.0 m +
2
G
∆d = 15 m [fwd]
The sprinter’s displacement is 15 m [fwd].
Copyright © 2003 Nelson
2
2
Chapter 1 Kinematics
21
G
(b) vf = ?
G G
G vf − vi
a=
∆t
G G G
vf = vi + a ∆t
(
)
= 0 m/s + 2.3 m/s 2 [fwd] (3.6 s)
G
vf = 8.3 m/s [fwd]
The sprinter’s final velocity is 8.3 m/s [fwd].
G
22. vi = 7.72 × 107 m/s [E]
G
vf = 2.46 × 107 m/s [E]
G
∆d = 0.478 m [E]
G
(a) a = ?
G
G
G G
vf 2 = vi 2 + 2a ∆d
G
G
G v 2 −v2
a = f Gi
2 ∆d
(2.46 ×10 m/s [E]) − (7.72 ×10 m/s [E])
=
2
7
7
2
2(0.478 m [E])
G
a = −5.60 × 1015 m/s 2 [E]
G
a = 5.60 × 1015 m/s 2 [W]
The electron’s acceleration is 5.60 × 1015 m/s2 [W].
(b) ∆t = ?
G 1 G G
∆d = (vi + vf ) ∆t
2
G
2∆d
t
∆ = G G
vi + vf
=
2 ( 0.478 m [E])
(7.72 ×10 m/s [E]) + (2.46 ×10 m/s [E])
7
7
∆t = 9.39 × 10−9 s
The acceleration occurs over 9.39 × 10–9 s.
Applying Inquiry Skills
23. The experiment can be simple. Besides the book and the desk, the only equipment required is a metre stick and a
stopwatch. Slide the book along the desk at a constant speed for a predetermined distance that is long enough that the time
interval to cover the distance is at least 2.0 s (e.g., slide the book at a constant speed of about 0.50 m/s for 2.0 s). Remove
the pushing force from the book and determine the displacement from that instant to the stopping position. The
acceleration can be found by applying the equation vf 2 = vi 2 + 2a∆d , where vf = 0, vi is the measured value of the speed
while the book is being pushed, and ∆d is the distance the book slides after it is no longer pushed.
Making Connections
G
 1000 m  1 h 
24. vi = 75.0 km/h [N] = (75.0 km/h [N]) 

 = 20.8 m/s [N]
 km  3600 s 
G
a = 4.80 m/s2 [S]
reaction time = ?
22
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
First we must calculate the change in displacement:
G 2 G2
G G
vf = vi + 2a ∆d
G vGf 2 − vGi 2
∆d =
G
2a
=
(0 m/s )2 − ( 20.8 m/s[N])2
2(4.80 m/s 2 [S])
G
∆d = 45.2 m[N]
Calculate reaction distance (distance before stopping) = 48.0 m – 45.2 m = 2.8 m
reaction distance
reaction time =
vi
=
2.8 m
20.8 m/s
reaction time = 0.13s
Thus, the reaction time is 0.13 s.
PRACTICE
(Page 29)
Understanding Concepts
G
25. vi = 25 m/s [E]
G
vf =25 m/s [S]
∆t = 15 s
G
aav = ?
Using components, with +x eastward and +y southward:
∆v x = vfx + ( − vix )
∆v y = vfy + ( − viy )
∆v x = −25 m/s
∆v y = 25 m/s
G
∆v = ∆v x 2 + ∆v y 2
=
( 25 m/s )2 + ( 25 m/s )2
G
∆v = 35.3 m/s
tanθ =
∆v y
∆v x
 25 m/s 
θ = tan −1 

 25 m/s 
θ = 45°
G
So, ∆v = 35m/s [45° S of W ]. Therefore,
G
∆v
G
aav =
∆t
35 m/s [45° S of W ]
=
15 s
G
2
aav = 2.4 m/s [45° S of W ]
Thus, the car’s average acceleration is 2.4 m/s2 [45° S of W].
Copyright © 2003 Nelson
Chapter 1 Kinematics
23
G
26. vi = 6.4 m/s [E]
G
aav = 2.0 m/s2 [S]
∆t = 2.5 s
G
vf = ?
G G
v − vi
G
aav = f
∆t
G G G
vf = vi + aav ∆ t
(
)
= 6.4 m/s [E] + 2.0 m/s 2 [S] ( 2.5s )
G
vf = 6.4 m/s [E] + 5.0 m/s [S]
G
vf = vfx 2 + vfy 2
(6.4 m/s )2 + (5.0 m/s )2
=
G
vf = 8.1m/s
tan θ =
vfy
vfx
 5.0 m/s 
θ = tan −1 

 6.4 m/s 
θ = 38°
The final velocity of the watercraft is 8.1 m/s [38° S of E].
G
27. vi = 26 m/s [22° S of E]
G
vf = 21 m/s [22° N of E]
G
aav = ?
Using +x eastward and +y northward:
∆vx = ( 21m/s ) cos 22° − ( 26 m/s ) cos 22°
∆v y = ( 21m/s ) sin 22° − ( −26 m/s ) sin 22°
∆v y = 17.6 m/s
∆vx = −4.6 m/s
∆v x
∆t
−4.6 m/s
=
2.5 × 10−3 s
= −1.9 ×103 m/s 2
aav,x =
aav,y =
aav,x
aav,y
∆v y
∆t
17.6 m/s
=
2.5 × 10−3 s
= 7.0 ×103 m/s 2
G
aav = aavx 2 + aavy 2
=
(−1.9 ×10 m/s ) + (7.0 ×10 m/s )
3
2 2
3
2 2
G
aav = 7.3 × 103 m/s 2
tan θ =
aavy
aavx
 7.0 × 103 m/s 2 
θ = tan −1 
3
2 
 1.9 × 10 m/s 
θ = 75°
Thus, the average acceleration of the puck is 7.3 × 103 m/s2 [75° N of W].
24
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
G
28. aav =9.8 m/s2 [down]
∆t = 2.0 s
G
vf =24 m/s [45° below horizontal]
G
vi = ?
G G
v − vi
G
aav = f
∆t
G G G
vi = vf − aav ∆t
(
)
= 24 m/s [45° below the horizontal] − 9.8 m/s 2 [down ] ( 2.0s )
G
vi = 24 m/s [45° below the horizontal] − 19.6 m/s [down ]
vix = ( 24 m/s ) cos 45°
viy = ( −24 m/s ) sin 45° − ( −20 m/s )
vix = 17 m/s
viy = 2.6 m/s
G
vi = vix 2 + viy 2
=
(17 m/s )2 + ( 2.6 m/s )2
G
vi = 17 m/s
tanθ =
viy
vix
 2.6 m/s 
θ = tan −1 

 17 m/s 
θ = 10°
The ball’s initial velocity is 17 m/s [10° above the horizontal].
 82.0 km   1000 m   1 h 
29. vi = 82.0 km/h = 


 = 22.8 m/s
 1 h   1 km   3600 s 
vi = vf = 22.8 m/s
 60 s 
3
∆t = 15 min = 15 min 
 = 9.00 × 10 s
 min 
As stated in the question, +x east and +y north.
∆v
aav,x = x
∆t
22.8 m/s (cos12.7° ) − 22.8 m/s (sin 38.2° )
=
9.00 × 103 s
aav,x = 9.0 × 10 −3 m/s 2
aav,y =
=
∆v y
∆t
−22.8 m/s (sin12.7° ) − 22.8 m/s (cos 38.2° )
9.00 × 103 s
aav,y = −2.5 × 10 −2 m/s2
Thus, the x-component of the average acceleration is 9.0 × 10–3 m/s2 and the y-component is –2.5 × 10–2 m/s2.
Section 1.2 Questions
(Pages 30–31)
Understanding Concepts
1.
2.
Instantaneous acceleration equals average acceleration during motion of constant acceleration.
It is possible to have a northward velocity with westward acceleration if there is a change in direction. For example, if a
truck is initially moving northward at 50 km/h and changes direction to obtain a final velocity of 50 km/h [45° W of N],
the change in velocity and, thus, the acceleration just at the instant of the initial velocity, is westward.
Copyright © 2003 Nelson
Chapter 1 Kinematics
25
3.
G
vi = 1.65 × 103 km/h [W]
G
vf = 1.12 × 103 km/h [W]
∆t = 345 s
G
(a) a = ?
G G
G vf − vi
a=
∆t
1.12 × 103 km/h [W] − 1.65 × 103 km/h [W ]
=
345 s
G
a = 1.54 (km/h)/s [E ]
The average acceleration of the aircraft is 1.54 (km/h)/s [E].
G
 1000 m  1 h 
(b) a = 1.54 (km/h)/s 


 km  3600 s 
G
a = 0.427 m/s 2 [E]
The average acceleration of the aircraft is 0.427 m/s2 [E].
4. (a)
(b) To determine the instantaneous acceleration at t = 2.0 s, calculate the slope of the tangent to the curve indicated on the
graph.
G G
5. (a) Students can determine the data for the velocity-time graph by using the constant acceleration equation ∆d = vav ∆t
(applied at the times indicated), or by drawing the position-time graph and finding the tangents to the curve. The table
below gives the data. The velocity-time graph is a straight line, and its slope indicates the acceleration.
t (s)
G
v (m/s [W])
26
0
0.2
0.4
0.6
0.8
0
2.6
5.2
7.8
10.4
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
G
(b) a = ?
G
2
G G
a ( ∆t )
∆d = vi ∆t +
2
G
G 2 ∆d
a=
2
( ∆t )
=
G
(where vi = 0)
2(4.16 m [W])
(0.80 s )
2
G
a = 13 m/s 2
6. (a) The motion starts at a specific location at a high velocity with a negative acceleration, reaching zero velocity midway
through the motion. The motion then undergoes increasing velocity in the opposite direction until reaching the initial
position.
(b) This motion is increasing velocity in the negative direction, followed by a decreasing velocity in the same direction,
eventually reaching zero velocity. This is followed by increasing velocity in the positive direction. The magnitudes of
the negative and positive accelerations are equal.
(c) The motion in this graph starts with a constant velocity, then accelerates at a high rate for a short time before slowing
down with negative acceleration at a lower rate to zero velocity.
G
7. vi =26 m/s [E]
G
a = 5.5 m/s2 [W] = −5.5 m/s2 [E]
∆t = 2.6 s
G
vf = ?
G G
G vf − vi
a=
∆t
G G G
vf = vi + a ∆t
= 26 m/s[E] + ( −5.5 m/s 2 [E])(2.6s)
8.
9.
G
vf = 12 m/s[E]
The car’s velocity is 12 m/s [E].
G
a = −9.7 m/s2 [fwd]
∆t = 2.9 s
G
vi = ?
G G
G vf − vi
a=
∆t
G
G
vi = vf − a ∆t
= 0 − ( −9.7 m/s2 [fwd])(2.9 s)
G
vi = 28 m/s[fwd]
The car’s initial speed is 28 m/s.
The data points for the position-time graph can be found by finding the total area on the velocity-time graph up the each
second. The results are shown in the table.
t (s)
G
d (m [W])
Copyright © 2003 Nelson
0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
0
2.5
10.0
22.5
37.5
52.5
67.5
78.8
82.5
78.8
67.5
56.2
52.5
Chapter 1 Kinematics
27
The acceleration-time graph is generated by calculating the slopes of the line segments on the velocity-time graph.
10.
G
a = 4.4 m/s2 [fwd]
∆t = 3.4 s
G
vi = 0
G
(a) vf = ?
G G
G v − vi
a= f
∆t
G
G G
vf = vi + a ∆t
(
)
= 0 + 4.4 m/s 2 [fwd ] (3.4 s)
G
vf = 15 m/s [fwd]
The jumper’s final velocity is 15 m/s [fwd].
G
(b) ∆d = ?
G G
1G
2
∆d = vi ∆t + a ( ∆t )
2
1
= 0 + 4.4 m/s 2 [fwd] (3.4 s)2
2
G
∆d = 25 m [fwd]
The jumper’s displacement is 25 m [fwd].
G
11. vi = 0
G
vf = 2.0 × 107 m/s [E]
G
∆d = 0.10 m [E]
G
(a) a = ?
G
G
G G
vf 2 = vi 2 + 2a ∆d
G
G v2
a= f G
2 ∆d
(2.0 × 107 m/s[E])2
=
2(0.10 m [E])
G
a = 2.0 × 1015 m/s 2 [E]
(
)
The acceleration of the electron is 2.0 × 1015 m/s2 [E].
28
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b) ∆t = ?
G G
G v − vi
a= f
∆t
G G
vf − vi
∆t = G
a
2.0 × 107 m/s [E] − 0
=
2.0 ×1015 m/s 2 [E]
∆t = 1.0 × 10−8 s
The electron takes 1.0 × 10–8 s to reach its final velocity.
G
12. vi = 204 m/s [fwd]
G
vf = 508 m/s [fwd]
∆t = 29.4 s
G
∆d = ?
G 1 G G
∆d = ( vi + vf ) ∆t
2
( 204 m/s [fwd] + 508 m/s [fwd])
=
(29.4 s )
2
G
∆d = 1.05 × 104 m [fwd]
The displacement of the rocket is 1.05 × 104 m [fwd].
G
13. vi = 0
G
vf = 4.2 × 102 m/s [fwd]
G
∆d = 0.56 m [fwd]
G
(a) vav = ?
G G
vi + vf
G
vav =
2
0 + 4.2 ×10 2 m/s [fwd]
=
2
G
2
vav = 2.1 ×10 m/s [fwd]
The average velocity of the bullet is 2.1 × 102 m/s [fwd].
(b) ∆t = ?
G G
∆d = vav ∆t
G
∆d
∆t = G
vav
=
0.56 m [fwd]
2.1× 10 2 m/s [fwd]
∆t = 2.7 × 10−3 s
The uniform acceleration occurs over 2.7 × 10–3 s.
14. (a) After 45 s, the car and the van have the same velocity (from the graph).
(b) Let the subscript V represent the van and the subscript C represent the car. The displacements of the two vehicles are
equal at some time t, and can be found by determining the areas under the lines on the graphs.
G
G
∆d V = Atriangle,V + Arectangle,V
∆d C = Atriangle,C + Arectangle,C
G
G
G
G
= (vV1 )av ∆tV1 + vV2 ∆tV2
= (vC1 )av ∆tC1 + vC2 ∆tC2
 20 m/s + 0 m/s 
=
 60s + ( 20 m/s )(t − 60s )
2


G
∆d V = 600 m+ ( 20 m/s )(t − 60s )
Copyright © 2003 Nelson
 15 m/s + 0 m/s 
=
 30s + (15 m/s )(t − 30s )
2


G
∆d C = 225 m+ (15 m/s )(t − 30s )
Chapter 1 Kinematics
29
G
G
Set ∆d V = ∆dC and solve for t:
600 m+ ( 20 m/s )(t − 60s ) = 225 m+ (15 m/s )(t − 30s )
(20 m/s ) t − (15 m/s ) t = 225 m − (15 m/s )(30 s ) + ( 20 m/s )(60 s ) − 600 m
t=
375 m
5 m/s
t = 75s
Thus, the van V overtakes the car C at 75 s.
G
G
(c) Using t = 75 s, substitute into the equation for ∆d V or ∆d C .
G
G
G
∆d V = (vV1 )av ∆tV1 + vV2 ∆tV2
 20 m/s + 0 m/s 
=
 60s + ( 20 m/s )( 75s − 60s )
2


= 600 m+ 300 m
G
∆d V = 900 m
The displacement from the intersection when V overtakes C is 900 m.
G
15. vA = 4.4 m/s [31° S of E]
G
vB = 7.8 m/s [25° N of E]
∆t = 8.5 s
G
a =?
Using +x east and +y north, we find the components of the velocities and then the accelerations.
vAx = 4.4 m/s (cos 31° )
vBx = 7.8 m/s (cos 25° )
vAx = 3.8 m/s
vBx = 7.1m/s
vAy = −4.4 m/s (sin 31° )
vBy = 7.8 m/s (sin 25° )
vAy = −2.3 m/s
vBy = 3.3m/s
vBx − vAx
∆t
7.1m/s − 3.8 m/s
=
8.5s
ay =
ax =
a x = 0.39 m/s
=
2
vBy − vAy
∆t
3.3 m/s − ( −2.3m/s )
8.5s
a y = 0.66 m/s 2
G
a = ax 2 + a y 2
=
(0.39 m/s ) + (0.66 m/s )
2 2
2 2
G
a = 0.76 m/s 2
tan θ =
ay
ax
 0.66 m/s 2 
θ = tan −1 
2 
 0.39 m/s 
θ = 31°
The bird’s average acceleration is 0.76 m/s2 [31° E of N].
30
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
G
16. vi = 155 km/h [E]
G
vf = 118 km/h [S]
∆t = 56.5 s
G
a =?
Using components with +x east and +y south:
vfy − viy
v −v
ay =
a x = fx ix
∆t
∆t
0 km/h − 155 km/h
118 km/h − 0 km/h
=
=
56.5s
56.5s
a x = −2.74 (km/h)/s
a y = 2.09 (km/h)/s
G
aav = a x 2 + a y 2
( −2.74 (km/h)/s )2 + ( 2.09 (km/h)/s )2
=
G
aav = 3.45 (km/h)/s
tan θ =
ay
ax
 2.74 (km/h)/s 
θ = tan −1 

 2.09 (km/h)/s 
θ = 52.7°
Thus, the helicopter’s average acceleration is 3.45 (km/h)/s [52.7° W of S].
Applying Inquiry Skills
17. In both cases, the variable most difficult to measure is the time interval for the acceleration. Since the final speed is zero
and the initial speed is given, it remains to find the displacement the object undergoes during the acceleration. Then the
variables can be used in the equation vf 2 = vi 2 + 2a∆d . Estimates of the acceleration will vary if they are just guesses, but
should be fairly close if they are determined by a quick calculation with estimated quantities.
(a) The displacement of the bullet during stopping can be found by measuring the penetration of the bullet plus the
distance the wood moves. Assuming the value is 15 cm [fwd], the acceleration is:
G 2 G2
v −v
G
aav = f G i
2 ∆d
=
0 − (175 m/s [fwd])
2
2(0.15 m [fwd])
G
5
2
aav = −6.8 × 10 m/s [fwd]
The average acceleration of the bullet is –6.8 ¯ 105 m/s2 [fwd].
(b) The stopping distance can be measured by adding the total crunch distance of the car plus any change of position of the
barrels. Assuming the value is 1.5 m [fwd], and changing the initial speed to 24 m/s, the acceleration is:
G 2 G2
vf − vi
G
aav =
G
2 ∆d
=
0 − (24 m/s) [fwd])
2
2(1.5 m [fwd])
G
2
2
aav = −1.9 × 10 m/s [fwd]
The average acceleration of the test car is –1.9 ¯ 102 m/s2 [fwd].
Copyright © 2003 Nelson
Chapter 1 Kinematics
31
Making Connections
18. In the relay, the second, third, and fourth runners have already accelerated to vf before they handoff to the next runner.
Thus, the time is shorter than if you include acceleration time for all runners.
1.3 ACCELERATION DUE TO GRAVITY
PRACTICE
(Pages 32–33)
Understanding Concepts
1.
2.
The skydiver’s velocity is much greater than the diver’s velocity. Air resistance increases with velocity and cannot be
neglected for the skydiver.
The disadvantage is that what might seem logical or reasonable does not agree with what actually happens. One reason for
this is some events occur too rapidly for our senses to be able to observe slight differences. Aristotle’s reasoning that
heavy objects fell faster than lighter ones provides an example of the disadvantage of not using experimentation to
determine the dependency of one variable on another, in this case the dependency of the acceleration of a falling body on
the mass of the body.
Applying Inquiry Skills
3.
The experimental setup would require a vacuum chamber in which the coin and the feather are released simultaneously.
(This device is available commercially from scientific supply companies.)
Making Connections
4.
Since there is no atmosphere on the Moon, falling objects do not experience air resistance.
Case Study: Predicting Earthquake Accelerations
(Page 34)
(a) The map stretches from Northern California (40° north latitude) to a few kilometres north of Vancouver (50° north
latitude), and from the middle of Vancouver Island (125° west longitude) to just east of Trail, B.C. (about 117° west
longitude). The regions affected most severely, as depicted by deep reds, lie near the west coast of the continent,
especially in Northern California and Southern Oregon, as well as areas fairly close to Vancouver and Victoria. Areas
affected moderately, as depicted by yellows, stretch inland somewhat in British Columbia, Washington, and Oregon, and
a long way in California. Areas affected slightly, as depicted by blue-greens, are in the northern and eastern parts of
British Columbia, the eastern parts of Washington and Oregon, and the state of Idaho. There are no areas unaffected, as
depicted by white.
(b) Students have to combine the colours on the map with the colours in the legend, and they have to compare the contours
and locations on the map to a conventional atlas of the same region.
PRACTICE
(Page 35)
Understanding Concepts
5.
32
G
vi = 0
G
a = 9.80 m/s2 [down]
G
(a) ∆d = 5.00 m
G
vf = ?
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Making Connections
18. In the relay, the second, third, and fourth runners have already accelerated to vf before they handoff to the next runner.
Thus, the time is shorter than if you include acceleration time for all runners.
1.3 ACCELERATION DUE TO GRAVITY
PRACTICE
(Pages 32–33)
Understanding Concepts
1.
2.
The skydiver’s velocity is much greater than the diver’s velocity. Air resistance increases with velocity and cannot be
neglected for the skydiver.
The disadvantage is that what might seem logical or reasonable does not agree with what actually happens. One reason for
this is some events occur too rapidly for our senses to be able to observe slight differences. Aristotle’s reasoning that
heavy objects fell faster than lighter ones provides an example of the disadvantage of not using experimentation to
determine the dependency of one variable on another, in this case the dependency of the acceleration of a falling body on
the mass of the body.
Applying Inquiry Skills
3.
The experimental setup would require a vacuum chamber in which the coin and the feather are released simultaneously.
(This device is available commercially from scientific supply companies.)
Making Connections
4.
Since there is no atmosphere on the Moon, falling objects do not experience air resistance.
Case Study: Predicting Earthquake Accelerations
(Page 34)
(a) The map stretches from Northern California (40° north latitude) to a few kilometres north of Vancouver (50° north
latitude), and from the middle of Vancouver Island (125° west longitude) to just east of Trail, B.C. (about 117° west
longitude). The regions affected most severely, as depicted by deep reds, lie near the west coast of the continent,
especially in Northern California and Southern Oregon, as well as areas fairly close to Vancouver and Victoria. Areas
affected moderately, as depicted by yellows, stretch inland somewhat in British Columbia, Washington, and Oregon, and
a long way in California. Areas affected slightly, as depicted by blue-greens, are in the northern and eastern parts of
British Columbia, the eastern parts of Washington and Oregon, and the state of Idaho. There are no areas unaffected, as
depicted by white.
(b) Students have to combine the colours on the map with the colours in the legend, and they have to compare the contours
and locations on the map to a conventional atlas of the same region.
PRACTICE
(Page 35)
Understanding Concepts
5.
32
G
vi = 0
G
a = 9.80 m/s2 [down]
G
(a) ∆d = 5.00 m
G
vf = ?
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
G
G
G G
vf 2 = vi 2 + 2a ∆d
vf 2 = 0 + 2(9.80 m/s 2 [down])(5.00 m[down])
vf = 2(9.80 m/s 2 )(5.00 m)
G
vf = 9.90 m/s[down]
= 9.90 m/s ×
1km
× 3600 s/h
1000 m
G
vf = 35.6 km/h [down]
The diver’s velocity is 9.90 m/s [down] or 35.6 km/h [down].
G
(b) ∆d = 10.00 m
G
vf = ?
G
G
G G
vf 2 = vi 2 + 2a ∆d
vf 2 = 0 + 2(9.80 m/s 2 [down])(10.0 m[down])
vf = 2(9.80 m/s 2 )(10.0 m)
G
vf = 14.0 m/s [down]
= 14.0 m/s ×
1km
× 3600 s/h
1000 m
G
vf = 50.4 km/h [down]
The diver’s velocity is 14.0 m/s [down] or 50.4 km/h [down].
Applying Inquiry Skills
6. (a) A good guess would be between 150 ms and 250 ms. The actual answer depends on the student’s hand span.
(b) v yi = 0
a y = 9.8 m/s
2
∆y = 0.20 m (a typical hand span)
∆t = ?
1
∆y = v yi ∆t +
∆y =
∆t =
=
1
2
a y ∆t
2
a y ∆t
2
2
2 ∆y
ay
2(0.20 m)
9.8 m/s
2
∆t = 0.20 s
For a hand span of 0.20 m, the time interval is 0.20 s, or 200 ms.
(c) Comparisons will vary. Students can improve their skills by trying to estimate answers to problems before performing
calculations, and by practising estimating quantities in everyday transactions and activities.
G
7. (a) vi = 0
G
∆d = 1.55 m [down]
∆t = 0.600 s
G
a =?
G G
1G
2
Rearranging the equation ∆d = vi ∆t + a ( ∆t ) :
2
Copyright © 2003 Nelson
Chapter 1 Kinematics
33
G
G 2∆d
a=
2
( ∆t )
=
(
)
2 1.55 m [down ]
(0.600 s )
G
a = 8.61m/s 2 [down ]
Thus, the acceleration of the ball is 8.61 m/s2 [down].
(b) Plot the position-time graph, find the slopes of the tangents at three or more times, plot the corresponding velocity-time
graph, and calculate the slope of the line on that graph to find the acceleration.
G
(c) a = 9.81 m/s2 [down]
measured value − accepted value
% error =
× 100%
accepted value
2
=
8.61 m/s 2 − 9.81 m/s 2
9.81 m/s 2
×100%
% error = −12.2%
The percent error is 12.2 %.
(d) The high percent error results from a variety of possible errors, with air resistance likely the most crucial influence.
(Since the ball is very light, air resistance has a greater effect than if the ball were more massive but the same size.)
Another important source of error is measuring the time interval of the fall.
Making Connections
8.
Answers may vary. The change in the gravitational field strength at various altitudes is quite small, and deciding how
much adjustment would be required might be controversial. Furthermore, if records were adjusted downward for events
that are easier at higher altitudes, then calls would be made for records to be adjusted upward for events that are more
difficult in the rarified atmosphere at higher altitudes. Adjustments are not made for the effects of low winds, so they
would not likely be made for changes in altitude.
PRACTICE
(Pages 37–38)
Understanding Concepts
9. (a)
(b)
(c)
(d)
The time the ball takes to rise equals the time the ball takes to fall since air resistance is negligible.
The initial and final velocities would be equal in magnitude but opposite in direction.
The ball’s velocity at the top of the flight is zero.
The ball’s acceleration during the entire motion is the acceleration due to gravity (9.8 m/s2 [down]), even when it is at
the top of the flight.
(e)
34
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
10. We could solve Sample Problem 1(b) using the following equations:
(viy + vfy )
1
1
∆y = viy ∆t + a y ( ∆t )2
∆y = vfy ∆t − a y ( ∆t )2
∆y =
∆t
2
2
2
11. (a) viy = 0
a y = 9.80 m/s2
∆y = 12.5 m
vfy = ?
vfy 2 = viy 2 + 2 a y ∆y
2
vfy = 0 + 2(9.80 m/s 2 )(12.5 m)
vfy = 245 m 2 /s2
vfy = 15.6 m/s
The shellfish has a speed of 15.6 m/s at impact.
(b) viy = 0
a y = 9.80 m/s2
∆t = 3.37 s
vfy = ?
ay =
vfy − viy
∆t
vfy = viy + a y ∆t
= 0 + (9.80 m/s 2 )(3.37 s)
vfy = 33.0 m/s
The steel ball has a speed of 33.0 m/s at impact.
12. viy = −15.0 m/s
a y = 9.80 m/s2
∆y = 15.0 m
Using the equation ∆y = viy ∆t +
a y ( ∆t ) 2
1 
, rearrange as  a y  ( ∆t )2 + viy ∆t − ∆y = 0 , which is a form of the quadratic
2 
2
equation. Thus,
∆t =
−b ± b 2 − 4ac
2a
 ay 
−viy ± viy 2 − 4   ( −∆y )
 2 
=
a
 y
2 
 2 
∆t =
−viy ± viy 2 + 2a y ∆y
ay
(a) The initial velocity is up. Define +y as down.
∆t = ?
vfy = ?
∆t =
15.0 m/s ±
( −15.0 m/s )2 + 2(9.80 m/s2 )(15.0 m)
= 1.53s ± 2.33s
9.80 m/s 2
(only the positive root is applicable)
∆t = 3.86 s
Copyright © 2003 Nelson
Chapter 1 Kinematics
35
vfy = viy + a y ∆t
(
)
= −15.0 m/s + 9.80 m/s2 (3.86s )
vfy = 22.8 m/s
Thus, the total flight time is 3.86 s and the speed of impact is 22.8 m/s.
(b) The initial velocity is down. Define +y as down.
∆t = ?
vfy = ?
∆t =
−15.0 m/s ±
(15.0 m/s )
= −1.53s ± 2.33s
2
+ 2(9.80 m/s 2 )(15.0 m)
9.80 m/s 2
(only the positive root is applicable)
∆t = 0.794s
vfy = viy + a y ∆t
(
= 15.0 m/s + 9.80 m/s2
) (0.794 s )
vfy = 22.8 m/s
Thus, the total flight time is 0.794 s and the speed of impact is 22.8 m/s if the initial velocity is down. (Notice that the
answers are written to three significant digits in order to compare them to the answers in (a). Rules for rounding off
have not been followed exactly here.)
(c) The final velocity is independent of whether the ball is thrown up or down (at the same speed).
13. Let ∆y1 represent the distance the ball travels from t = 0.0 s to t = 1.0 s and let ∆y2 represent the distance the ball travels
from t = 1.0 s to t = 2.0 s. For ∆y1, the initial velocity viy,1 = 0
1
Using the equation ∆y = viy ∆t + a y ( ∆t )2 :
2
1
∆y1 = viy ,1∆t1 + a y ( ∆t1 )2
2
1
= 0 m/s (1.0 s − 0.0 s) + 9.8 m/s 2 (1.0 s − 0.0 s) 2
2
∆y1 = 4.9 m
For ∆y2, the initial velocity viy,2 is the velocity at t = 1.0 s. Thus,
viy ,2 = viy ,1 + a y ( ∆t1 )
(
)
= 0 m/s + 9.8 m/s2 (1.0 s − 0.0 s)
viy ,2 = 9.8 m/s
1
∆y2 = viy ,2 ∆t2 + a y ( ∆t2 )2
2
1
= 9.8 m/s (2.0s − 1.0s) + 9.8 m/s 2 (2.0s − 1.0s) 2
2
∆y2 = 14.7 m
∆y2 (14.7 m )
=
= 3.0
∆y1 (4.9 m )
Therefore, a ball travels three times as far from 1.0 s to 2.0 s as from 0.0 s to 1.0 s.
14. Define +y as up.
vfy = 0 m/s
4.2s
∆t =
= 2.1 s
2
ay = –9.80 m/s2
36
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(a) viy = ?
ay =
vfy − viy
∆t
viy = vfy − a y ∆t
= 0 m/s − ( −9.80 m/s2 )(2.1s)
viy = 21m/s
Thus, the pitcher threw the ball with a velocity of 21 m/s [up].
(b) ∆y = ?
1
∆y = viy ∆t + a y (∆t ) 2
2
1
= 21 m/s (2.1 s) + (−9.80 m/s 2 )(2.1 s) 2
2
∆y = 22 m
The ball rises 22 m.
15. Define +y as down.
viy = −2.1 m/s
∆t = 3.8 s
ay = 9.80 m/s2
(a) ∆y = ?
a y (∆t ) 2
∆y = viy ∆t +
2
(9.80 m/s 2 )(3.8s)2
= (−2.1m/s)(3.8s) +
2
∆y = 63m
The balloon was 63 m when the ballast was released.
(b) vfy = ?
vfy − viy
ay =
∆t
vfy = viy + a y ∆t
= −2.1 m/s + (9.80 m/s 2 )(3.8 s)
vfy = 35 m/s
The velocity of the ballast at impact was 35 m/s [down].
16. Define +y as down.
∆d = 2.3 m
∆t = 1.7 s
viy = 0
G
(a) g = ?
∆y = viy ∆t +
∆y =
G
g=
=
1
g ( ∆t ) 2
2
1
g ( ∆t ) 2
2
2∆y
( ∆t ) 2
2(2.3 m)
(1.7 s)2
G
g = 1.6 m/s 2 [down]
The acceleration of gravity on the Moon is 1.6 m/s2 [down].
Copyright © 2003 Nelson
Chapter 1 Kinematics
37
(b)
G
g Earth
9.8 m/s 2
=
= 6.1
G
g Moon
1.6 m/s 2
G
G
Thus, the ratio of g Earth to g Moon is 6.1:1.
Applying Inquiry Skills
17. (a) Students can use graphing techniques or uniform acceleration equations to determine the acceleration. Sample
calculations using the final data points are shown below.
Let +y be down.
For mass 1:
viy = 0
∆y = 0.736 m
∆t = 0.40 s
ay1 = ?
∆y = viy ∆t +
∆y =
a y1 =
=
1
2
a y 1 ∆t
1
2
a y 1 ∆t
2
2
2 ∆y
∆t
2
2(0.736 m)
( 0.40 s )2
a y1 = 9.2 m/s
2
For mass 2:
viy = 0
∆y = 0.776 m
∆t = 0.40 s
ay2 = ?
1
∆y = viy ∆t +
∆y =
ay 2 =
=
1
2
2
a y 2 ∆t
a y 2 ∆t
2
2
2 ∆y
∆t
2
2(0.776 m)
( 0.40 s )2
a y 2 = 9.7 m/s
2
The accelerations of the two masses are 9.2 m/s2 and 9.7 m/s2, respectively.
(b) To find the percent difference:
difference in values
× 100%
% difference =
average of values
2
=
9.2 m/s − 9.7 m/s
1
2
(9.2 m/s
2
2
+ 9.7 m/s
2
)
× 100%
% difference = 5.3%
38
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(c) Likely, the lower experimental acceleration, 9.2 m/s2, is attributable to the ticker-tape timer because friction between
the paper strip and the timer would reduce the acceleration.
Making Connections
18. Answers will vary. Walking, running, and going up stairways are just a few examples of the activities that would be more
difficult with a higher acceleration due to gravity. One advantage is that friction would be increased when it might be
helpful, for example for a construction worker on a slanted roof. Another advantage is that objects subject to toppling over
might be more stable.
PRACTICE
(Page 39)
Understanding Concepts
19. The two main factors that affect terminal speed are mass and surface area. (For solid spherical objects, this is equivalent
to saying that the terminal speed depends on the object’s density.) For objects of the same mass, an increased surface area
contacting the air causes the terminal speed to become lower. For objects with the same surface area, the greater the mass
the greater is the terminal speed.
20. There is no “terminal speed” related to air resistance on the Moon because the Moon does not have an atmosphere.
21. There are two terminal speeds in this case, a high
speed followed by a much lower terminal speed when
the parachute is opened. The graph is shown below.
The slope of the line near the beginning of the motion
is equal to the magnitude of the acceleration due to
gravity at that location. The terminal speed values are
found on page 39 of the text (Table 5).
Applying Inquiry Skills
22. (a) Examples of factors to consider in designing the package are the mass, dimensions, and contents of the package, the
materials of the packets within the package, the material of the package itself, the method of securing the outer layer of
the package, the height from which the package will be dropped, the forward speed of the aircraft that drops the
package, the type of terrain on which the package will land, and the ease with which the package can be opened when it
is found.
(b) Since there are so many factors that could be tested, it would be wise to predict what choices would be best for the
design, create a prototype, and then perform a controlled experiment by dropping the sample of the design from various
heights onto various surfaces to determine how well the package can survive the fall. Modifications and further testing
are part of the process.
Making Connections
23. There are two main factors to consider, terminal speed and stopping distance. Once the terminal speed of a body is
reached, the speed remains constant, or it may even drop as the air becomes more dense nearer the ground. Furthermore, a
person falling from a greater height can reduce the terminal speed by using the “spread-eagle” orientation. Increasing the
stopping or deceleration distance reduces the chances of death or serious injury. The distance can be increased if the
landing occurs on the down slope of a hill or if the fall is broken by trees, bushes, a layer of snow, or water.
Section 1.3 Questions
(Page 40)
Understanding Concepts
1.
2.
Air resistance increases dramatically with increased speed, so air resistance is negligible for most common objects that
fall for a short distance. For example, air resistance is negligible for all but the lowest density objects that can fall in a
classroom setting.
Aristotle believed that the central region of Earth was made up of four elements: earth, air, fire, and water. Each element
had its proper place, which was determined by its relative heaviness. Each element moved in a straight line to its proper
place where it could be at rest. For example, similar objects were attracted to each other. Thus, objects made from the
Copyright © 2003 Nelson
Chapter 1 Kinematics
39
earth were attracted down to Earth. Fire tended to rise from Earth. Aristotle also believed that heavier objects fell faster
than less massive objects of the same shape.
Galileo believed that all objects fall toward Earth at the same acceleration, regardless of their mass, size, or shape,
when gravity is the only force acting on them. Evidently, he proved his theory by dropping two objects of different mass
from the top floor of the Leaning Tower of Pisa.
3. In each case, we define +y as down.
(a) viy = 0
ay = 9.8 m/s2
∆y = 36 m
vfy = ?
vfy 2 = viy 2 + 2a y ∆y
vfy 2 = 0 + 2(9.8 m/s 2 )(36 m)
 27 m   1 km   3600 s 
vfy = 



 s   1000 m   1 h 
vfy = 97 km/h
Thus, the landing speed of the diver is 27 m/s or 97 km/h.
(b) viy = 0
ay = 9.8 m/s2
∆t = 3.2 s
vfy = ?
vfy − viy
ay =
∆t
vfy = viy + a y ∆t
= 0 + (9.8 m/s 2 )(3.2 s)
 1 km   3600 s 
= (31m/s) 


 1000 m   1 h 
vfy = 1.1× 102 km/h
4.
Thus, the landing speed of the stone is 31 m/s or 1.1 × 102 km/h.
Define +y as up.
viy = 5.112 m/s
vfy = 0
∆y = ?
In Java, g = −9.782 m/s2:
vfy 2 = viy 2 + 2a y ∆y
2
2a y ∆y = vfy − viy
∆y =
=
−viy
2
2
2a y
−(5.112 m/s)2
2( −9.782 m/s 2 )
∆y = 1.336 m
In London, g = −9.823 m/s2:
∆y =
=
−viy 2
2a y
−(5.112 m/s)2
2( −9.823 m/s 2 )
∆y = 1.330 m
In Java, the jumper achieved a height of 1.336 m; in London, the jumper achieved a height of 1.330 m.
40
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
5.
Let +y be the forward direction of the shuttle, assumed to be a straight line.
viy = 0
ay = 5(9.80 m/s2)
∆t = 1.0 min = 60 s (two significant digits)
vfy = ?
vfy − viy
ay =
∆t
vfy = a y ∆t
= 5(9.80 m/s 2 )(60 s)
 1 km   3600 s 
= (2.9 ×103 m/s) 


 1000 m   1 h 
vfy = 1.1× 10 4 km/h
The shuttle’s speed is 2.9 m/s or 1.1 × 104 km/h.
6. Let +y be up.
∆t = 2.6 s
ay = g = −9.80 m/s2
(a) ∆t = ?
The time for the ball to rise will be half of the total time.
2.6 s
∆t =
2
∆t = 1.3 s
Therefore, the time for the ball to rise is 1.3 s.
(b) vfy = 0
∆t = 1.3 s
viy = ?
vfy − viy
ay =
∆t
viy = vfy − a y ∆t
= 0 − ( −9.8 m/s2 )(1.3s)
viy = 13m/s
The velocity is 13 m/s [up] when the golf ball leaves the person’s hand.
(c) Again, +y is up.
viy = 13 m/s
ay = g = −3.8 m/s2
∆t = ?
vfy − viy
ay =
∆t
vfy − viy
∆t =
ay
=
7.
0 − 13m/s
−3.8 m/s 2
∆t = 3.4 s
The total time in flight would be 2(3.4 s) = 6.8 s.
Let +y be down.
∆t = 0.087 s
∆y = 0.80 m
ay = 9.8 m/s2
Copyright © 2003 Nelson
Chapter 1 Kinematics
41
1
∆y = vfy ∆t − a y (∆t ) 2
2
∆y 1
vfy =
+ a y ( ∆t )
∆t 2
0.80 m 1
=
+ 9.8 m/s 2 (0.087 s)
0.087 s 2
(
)
vfy = 9.6 m/s
8.
The velocity of the ball as it hits the floor is 9.6 m/s [down].
Let +y be down.
viy = 14 m/s
ay = 9.8 m/s2
(a) ∆y = 21 m
∆t = ?
ay
1
Using the equation ∆y = viy ∆t + a y ( ∆t )2 , rearrange as
( ∆t )2 + viy ∆t − ∆y = 0 , which is a form of the quadratic
2
2
equation. Therefore,
∆t =
−b ± b 2 − 4ac
2a
 ay 
−viy ± viy 2 − 4   ( −∆y )
 2 
=
 ay 
2 
 2 
=
=
−viy ± viy 2 + 2a y ∆y
ay
−14 m/s ± (14 m/s) 2 + 2(9.8 m/s 2 )(21m)
9.8 m/s 2
∆t = 1.1s or − 3.9s
The stone will take 1.1 s to reach the water below.
(b) The positive root is the actual answer when the stone is thrown vertically downward. The negative root is the time that
the stone would have travelled had the initial velocity been upward rather than downward (i.e., if vi = −14 m/s in this
case).
9.
10. Note: To make the solution more compact, units are omitted until the final answer is written.
Let +y be down, let F represent the flowerpot, and let B represent the ball.
Since the ball is thrown 1.00 s after the flowerpot is released, after some time ∆tF = ∆tB − 1.00, let us assume the ball will
pass the flowerpot. At this instant, the flowerpot will have travelled 28.5 m − 26.0 m = 2.5 m farther than the ball. Thus,
∆yF = ∆yB + 2.5.
But with uniform acceleration, ∆y = viy ∆t +
1
2
ay∆t 2 .
Combining these relationships, we have:
42
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
∆yF = ∆yB + 2.5
viyF ∆tF +
1
2
a y ∆tF 2 = viyB ∆tB +
1
2
a y ∆tB 2 + 2.5
0 + 4.9∆t F 2 = 12 ( ∆t F − 1) + 4.9 ( ∆t F − 1) + 2.5
2
(
)
4.9 ∆t F 2 = 12 ( ∆t F − 1) + 4.9 ∆t F 2 − 2 ∆t F + 1 + 2.5
4.9 ∆t F 2 = 12 ∆t F − 12 + 4.9 ∆t F 2 − 9.8∆t F + 4.9 + 2.5
0 = 12 ∆t F − 12 − 9.8∆t F + 4.9 + 2.5
0 = 2.2∆t F − 4.6
∆t F =
4.6
2.2
∆tF = 2.09 s
After this time interval, the flowerpot has fallen:
1
2
∆y F = a y ∆t F
2
1
2
2
= 9.80 m/s ( 2.09 s )
2
∆yF = 21.4 m
(
)
The flowerpot is 28.5 m − 21.4 m = 7.10 m above the ground when the ball passes it.
11. The ranking from highest terminal speed to lowest is: pollen; a ping-pong ball; a basketball; a skydiver diving headfirst;
and a skydiver in spread-eagle orientation.
Applying Inquiry Skills
12. (a) 9.809 060 m/s2 has 7 significant digits, a possible error of ±0.000 000 5, and percent possible error of
±0.000 000 5 m/s 2
× 100% = ±0.000 005%.
9.809 060 m/s 2
(b) 9.8 m/s2 has 2 significant digits, a possible error of ±0.05 m/s2, and percent possible error of
±0.05 m/s 2
× 100% = ±0.5%.
9.8 m/s 2
(c) 9.80 m/s2 has 3 significant digits, a possible error of ±0.005 m/s2, and percent possible error of
±0.005 m/s 2
× 100% = ±0.05%.
9.80 m/s 2
(d) 9.801 m/s2 has 4 significant digits, a possible error of ±0.0005 m/s2, and percent possible error of
±0.0005 m/s 2
× 100% = ±0.005%.
9.801 m/s 2
(e) 9.8 × 10–6 m/s2 has 2 significant digits, a possible error of ±0.05× 10–6 m/s2, and percent possible error of
±0.05 ×10 −6 m/s 2
×100% = ±0.5%.
9.8 ×10−6 m/s 2
13. (a) Students may recall performing this activity in a previous grade. Hold a metre stick vertically while your partner holds
his or her separated index finger and thumb ready to catch the stick at a specific mark, such as the 50-cm mark.
Without warning, drop the stick and determine how far the stick falls before the partner catches it. Repeat several times
and take an average of the distances, ∆y. Use this value in the appropriate equation, as shown below.
Let +y be down.
Assume ∆y = 18 cm = 0.18 m
ay = 9.8 m/s2
viy = 0
∆t = ?
Copyright © 2003 Nelson
Chapter 1 Kinematics
43
∆y = viy ∆t +
∆y =
∆t =
=
1
2
a y ∆t
1
2
a y ∆t
2
2
2 ∆y
ay
2(0.18 m)
9.8 m/s 2
∆t = 0.19 s
(b) Talking on a cell phone would likely increase reaction time. Students can simulate this situation by engaging in
distracting conversation with the lab partner whose reaction time is being tested.
Making Connections
14. Aristotle and Galileo influenced the philosophy and scientific thought of their respective eras, and in both cases their
influence lasted long after they died. Students can find information about these science “giants” in books and
encyclopedias, or on the Internet. For example, an advanced word search on the Internet, entering only the words Aristotle
and Galileo, found more than 20 thousand hits, many of which featured discussions of the same topics featured in the text.
15. Deductive reasoning involves using theories to account for specific experimental results. Thus, deductive reasoning uses
ideas to explain observed phenomena. Inductive reasoning involves making and collecting observations, and then
developing general theories or hypotheses to account for the observations.
(a) Aristotle and other ancient scientists used deductive reasoning.
(b) Galileo used the process of inductive reasoning. When he observed that heavy objects fall with increasing speed, he
formed the hypothesis that the speed of the object was directly proportional to the distance the object fell. When this
hypothesis proved false, he hypothesized that the speed of a falling object is directly proportional to the time, not the
distance. Through experiments, he was able to verify his hypothesis.
(c) Various ways are used to contrast these types of reasoning. For example, in deductive reasoning, particular results are
inferred from a general law, whereas in inductive reasoning, a general law is inferred from particular results. Stated
another way for deduction, conclusions follow from premises, that is the reasoning goes from the general to the
specific. In induction, premises lead to the conclusions, or the reasoning goes from the specific to the general. In
mathematics, induction involves proving a theorem using a process in which the theorem is verified for a small value of
an integer, and then extending the verification to greater values of the integer.
16. Many sites can be found by doing an advanced word search on the Internet. For example, in entering the words Luis
Alvarez, Yucatan, and dinosaurs, more than 400 sites were found, many of them highly credible. Two examples are:
www.ceemast.csupomona.edu/nova/alvarez2.html
www.space.com/scienceastronomy/planetearth/deep_impact_991228.html
1.4 PROJECTILE MOTION
PRACTICE
(Page 46)
Understanding Concepts
1.
A projectile is an object that moves through the air without a propulsion system and follows a curved path. An airplane
has a propulsion system and does not follow a trajectory. Thus, an airplane is not a projectile.
2. The projectile experiences constant downward acceleration due to gravity (vertical acceleration) and the horizontal
component of acceleration is zero.
3. Let +x be to the right and +y be downward. The initial position is the position where the marble leaves the table.
(a) Horizontally (constant vix ):
vix = 1.93 m/s
∆x = ?
∆t = ?
44
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
∆y = viy ∆t +
∆y =
∆t =
=
1
2
a y ∆t
1
2
a y ∆t
2
2
2 ∆y
ay
2(0.18 m)
9.8 m/s 2
∆t = 0.19 s
(b) Talking on a cell phone would likely increase reaction time. Students can simulate this situation by engaging in
distracting conversation with the lab partner whose reaction time is being tested.
Making Connections
14. Aristotle and Galileo influenced the philosophy and scientific thought of their respective eras, and in both cases their
influence lasted long after they died. Students can find information about these science “giants” in books and
encyclopedias, or on the Internet. For example, an advanced word search on the Internet, entering only the words Aristotle
and Galileo, found more than 20 thousand hits, many of which featured discussions of the same topics featured in the text.
15. Deductive reasoning involves using theories to account for specific experimental results. Thus, deductive reasoning uses
ideas to explain observed phenomena. Inductive reasoning involves making and collecting observations, and then
developing general theories or hypotheses to account for the observations.
(a) Aristotle and other ancient scientists used deductive reasoning.
(b) Galileo used the process of inductive reasoning. When he observed that heavy objects fall with increasing speed, he
formed the hypothesis that the speed of the object was directly proportional to the distance the object fell. When this
hypothesis proved false, he hypothesized that the speed of a falling object is directly proportional to the time, not the
distance. Through experiments, he was able to verify his hypothesis.
(c) Various ways are used to contrast these types of reasoning. For example, in deductive reasoning, particular results are
inferred from a general law, whereas in inductive reasoning, a general law is inferred from particular results. Stated
another way for deduction, conclusions follow from premises, that is the reasoning goes from the general to the
specific. In induction, premises lead to the conclusions, or the reasoning goes from the specific to the general. In
mathematics, induction involves proving a theorem using a process in which the theorem is verified for a small value of
an integer, and then extending the verification to greater values of the integer.
16. Many sites can be found by doing an advanced word search on the Internet. For example, in entering the words Luis
Alvarez, Yucatan, and dinosaurs, more than 400 sites were found, many of them highly credible. Two examples are:
www.ceemast.csupomona.edu/nova/alvarez2.html
www.space.com/scienceastronomy/planetearth/deep_impact_991228.html
1.4 PROJECTILE MOTION
PRACTICE
(Page 46)
Understanding Concepts
1.
A projectile is an object that moves through the air without a propulsion system and follows a curved path. An airplane
has a propulsion system and does not follow a trajectory. Thus, an airplane is not a projectile.
2. The projectile experiences constant downward acceleration due to gravity (vertical acceleration) and the horizontal
component of acceleration is zero.
3. Let +x be to the right and +y be downward. The initial position is the position where the marble leaves the table.
(a) Horizontally (constant vix ):
vix = 1.93 m/s
∆x = ?
∆t = ?
44
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Vertically (constant ay):
viy = 0
a y = + g = 9.80 m/s 2
∆y = 76.5 m
1
The vertical motion, ∆y = viy ∆t + a y (∆t ) 2 can be simplified to:
2
2 ∆y
∆t = ±
ay
2(0.765 m)
9.80 m/s2
∆t = ±0.395 s
Since only the positive root applies, the marble takes 0.395 s to hit the floor.
(b) ∆x = ?
=±
Using the equation for horizontal motion:
∆x = vix ∆t
= (1.93 m/s)(0.395 s)
∆x = 0.763m or 76.3 cm
The horizontal range is 76.3 cm.
(c) To determine the marble’s final velocity, determine its horizontal and vertical components.
The x-component is constant at 1.93 m/s.
The y-component is:
vfy = viy + a y ∆t
= 0 m/s + (9.8 m/s 2 )(0.395 s)
vfy = 3.87 m/s
Using the law of Pythagoras and trigonometry:
2
2
vf = vfx + vfy
2
vf = (1.93 m/s)2 + (3.87 m/s)2
vf = 4.33 m/s
θ = tan −1
= tan −1
vfy
vfx
3.87 m/s
1.93 m/s
θ = 63.5°
Thus, the final velocity of the marble just prior to landing is 4.33 m/s [63.5° below the horizontal].
4. (a) Let the +x direction be to the right and the +y direction be downward.
Horizontally (constant
vix = 8.0 m/s
∆t = 1.0 s
∆x = ?
vix ):
∆x = vix ∆t
= (8.0 m/s)(1.0 s)
∆x = 8.0 m
The ∆x values at t = 0.0 s, 1.0 s, 2.0 s, and 3.0 s are shown in Table 1.
Copyright © 2003 Nelson
Chapter 1 Kinematics
45
Vertically (constant
a y ):
viy = 0 m/s
ay = +g = 9.8 m/s2
∆t = 1.0 s
∆y = ?
1
∆y = viy ∆t + a y (∆t ) 2
2
1
= a y ( ∆t ) 2
2
(9.8 m/s 2 )(1.0 s) 2
=
2
∆y = +4.9 m
The ∆y values at t = 0.0 s, 1.0 s, 2.0 s, and 3.0 s are shown in Table 1.
Determine the final velocity using its horizontal and vertical components.
The x-component is constant at 8.0 m/s.
The y-component at 1.0 s is:
vfy = viy + a y ∆t
= 0 m/s + (9.8 m/s 2 )(1.0 s)
vfy = 9.8 m/s
Using the law of Pythagoras and trigonometry:
2
2
2
vf = vfx + vfy
vf = (8.0 m/s) 2 + (9.8 m/s) 2
vf = 12.65 m/s
θ = tan −1
= tan −1
vfy
vfx
9.8 m/s
8.0 m/s
θ = 50.8°
G
Therefore, vf = 13 m/s [51° below the horizontal].
The vf values for ∆t = 0.0 s, 1.0 s, 2.0 s, and 3.0 s are shown in Table 1.
Table 1 Calculated Horizontal and Vertical Displacements and Instantaneous Velocity at Select Times
t (s)
∆x (m)
∆y (m)
G
vf (m/s)
46
0.0
1.0
2.0
3.0
0.0
8.0
16
24
0.0
4.9
20
44
0.0
13 [51° below horizontal]
21 [68° below horizontal]
30 [75° below horizontal]
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b), (c)
(d) We can use components to find the average acceleration. The horizontal component of the acceleration is zero because
vx = constant = 8.0 m/s.
v1y = 9.8 m/s
v2y = 19.6 m/s
v3y = 29.4 m/s
∆t = 1.0 s
ay = ?
For the vertical acceleration between 1.0 s and 2.0 s:
ay =
=
v2 y − v1 y
∆t
19.6 m/s − 9.8 m/s
1.0 s
a y = 9.8 m/s2
For the vertical acceleration between 2.0 s and 3.0 s:
ay =
=
v3 y − v2 y
∆t
29.4 m/s − 19.6 m/s
a y = 9.8 m/s
1.0 s
2
Thus, we can conclude that the vertical acceleration is constant, and it equals the acceleration due to gravity.
Copyright © 2003 Nelson
Chapter 1 Kinematics
47
5.
Let +x be forward and +y be downward.
∆y = 83 cm = 0.83 m
∆x = 18.4 m
ay = 9.8 m/s2
vix = ?
Determine ∆t:
2 ∆y
ay
∆t = ±
2(0.83m)
9.8 m/s 2
∆t = ±0.41 s
=±
Now consider the horizontal component of the motion:
∆x
vix =
∆t
18.4 m
=
0.41 s
vix = 45 m/s
The ball’s initial horizontal speed is 45 m/s.
Applying Inquiry Skills
6.
The apparatus shown in the text, page 46, is available commercially and is recommended as a relatively inexpensive way
of performing projectile motion experiments.
(a) Start with the vertical target plate close to the bottom of the launching ramp, say at a separation of 2.0 cm. Allow the
steel ball, initially at rest, to roll down from the top of the ramp. Mark the exact location where the ball strikes the
target plate. Repeat this procedure with the target plate at increasing distances from the ramp (e.g., 4.0 cm, 6.0 cm,
etc.), while always releasing the ball from the same height. After the trials are complete, join the dots with a smooth
curve to observe the path of the projectile. Show that the horizontal component of the motion is a constant speed
(i.e., compare all the horizontal displacements). Show that the vertical component of the motion is a vertical
acceleration of magnitude 9.8 m/s2.
(b) The diagram looks like the projectile part shown in the text, page 41, Figure 3, although the initial velocity is to the left
rather than to the right.
Making Connections
7.
If cartoon characters obeyed the laws of physics, they would follow a projectile path after running off the edge of a cliff.
Of course, it is more fun to have the characters defy physics by appearing to be suspended in midair before plummeting
downward at a great velocity.
Try This Activity: Comparing Horizontal Range
(Page 49)
A spreadsheet program using the correct equations can be used to create the table. Solve for maximum height using
vfy = 0 m/s and +y downward:
vfy 2 = viy 2 + 2a y ∆y
∆y =
viy 2
2a y
− ( 25.00 m/s (sin θ ))
2
∆y =
2(−9.800 m/s 2 )
The values for the maximum height are given in Table 2.
48
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Solve for the total time of flight (twice the time taken to get to maximum height):
1
∆y = vfy ∆t − a y (∆t ) 2
2
−2∆y
∆t =
ay
total time = 2∆t
The values for the time of flight are given in Table 2. Solve for the horizontal range using the total time:
∆x = vix ∆t
∆x = 25 m/s (cos θ )∆t
The values for the horizontal range are given in Table 2.
Table 2
Launch
Angle
3°
6°
9°
12°
15°
18°
21°
24°
27°
30°
33°
36°
39°
42°
45°
48°
51°
54°
57°
60°
63°
66°
69°
72°
75°
78°
81°
84°
87°
Time of
Flight (s)
0.2670
0.5333
0.7981
1.061
1.321
1.577
1.828
2.075
2.316
2.551
2.779
3.000
3.211
3.414
3.608
3.792
3.965
4.128
4.279
4.418
4.546
4.661
4.763
4.852
4.928
4.991
5.039
5.074
5.095
Maximum
Height (m)
0.08734
0.3484
0.7803
1.378
2.136
3.045
4.095
5.275
6.572
7.972
9.459
11.02
12.63
14.23
15.94
17.61
19.26
20.87
22.43
23.92
25.32
26.61
27.79
28.84
29.75
30.51
31.11
31.54
31.80
Horizontal
Range (m)
6.666
13.26
19.71
25.94
31.89
37.49
42.67
47.39
51.60
55.23
58.26
60.65
62.38
63.43
63.78
63.43
62.38
60.65
58.26
55.23
51.60
47.39
42.67
37.49
31.89
25.94
19.71
13.26
6.666
The following conclusions can be made for a projectile with an initial velocity at some angle above the horizontal, assuming
that air resistance can be neglected.
• The greater the angle above the horizontal becomes, the greater the maximum height of the projectile and the greater the
time of flight.
• The maximum horizontal range occurs when the angle of the initial velocity is 45° above the horizontal. Values of the
horizontal range are identical for angles equidistant on either side of 45°.
Copyright © 2003 Nelson
Chapter 1 Kinematics
49
PRACTICE
(Page 50)
Understanding Concepts
8. (a) The vertical component of the ball’s velocity at the top of the flight is zero (vy = 0 m/s).
(b) The ball’s acceleration at the top of the flight is 9.8 m/s2 [down].
(c) The rise time is equal to the fall time when the ball lands at the same level from which it was struck.
9. Let +y be up.
G
(a) vi = 2.2 × 102 m/s [45°above the horizontal]
ay = –9.8 m/s2
∆y = ?
Find the horizontal and vertical components of the initial velocity:
Horizontally (constant vix ):
G
vix = vi cos θ
= (2.2 × 102 m/s)(cos 45°)
vix = 1.6 × 102 m/s
Vertically (constant a y ):
G
viy = vi sin θ
= (2.2 ×102 m/s)(sin 45°)
viy = 1.6 ×102 m/s
At the highest position, the y-component of the instantaneous velocity is zero (vfy = 0 m/s). Thus,
2
2
vfy = viy + 2a y ∆y
2
0 = viy + 2a y ∆y
∆y =
=
viy
2
−2 a y
(1.6 ×102 m/s)2
−2(−9.8 m/s 2 )
∆y = 1.2 × 103 m
The maximum height of the cannonball is 1.2 × 103 m.
(b) ∆t = ?
1
∆y = viy ∆t + a y (∆t ) 2
2
0 =1.6 × 10 2 m/s ∆t − 4.9 m/s 2 ( ∆t )
2
(
0 = ∆t 1.6 × 102 m/s − 4.9 m/s 2 ∆t
∆t = 0
or
)
(1.6 × 10 2 m/s − 4.9 m/s 2 ∆t ) = 0
∆t =
1.6 × 102 m/s
4.9 m/s 2
∆t = 32s
Therefore, the cannonball was fired at ∆t = 0 and the cannonball lands at ∆t = 32 s.
50
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(c) ∆x = ?
∆x = vix ∆t
= (1.6 ×10 2 m/s)(32 s)
∆x = 4.9 × 103 m
The horizontal range is 4.9 × 103 m.
(d) The final velocity has the same magnitude as the initial velocity, but is at an angle of 45° below the horizontal. Thus,
the final velocity is 2.2 × 102 m/s [45° below the horizontal].
10. Let +y be up.
G
(a) vi = 12 m/s [42°above the horizontal]
ay = –9.8 m/s2
∆y = –9.5 m
∆t = ?
Find the horizontal and vertical components of the initial velocity:
Horizontally (constant vix ):
G
vix = vi cos θ
= (12 m/s)(cos 42°)
vix = 8.9 m/s
Vertically (constant a y ):
G
viy = vi sin θ
= (12 m/s)(sin 42°)
viy = 8.0 m/s
1
∆y = viy ∆t + a y ( ∆t )2
2
−9.5 m = 8.0 m/s ∆t − 4.9 m/s2 ( ∆t )2
0 = 4.9 m/s 2 ( ∆t )2 − 8.0 m/s ∆t − 9.5 m
Solve using the quadratic formula,
∆t =
=
−b ± b 2 − 4ac
2a
(where a = 4.9 m/s 2 , b = − 8.0 m/s, and c = − 9.5 m)
−(−8.0 m/s) ± (−8.0 m/s)2 − 4(4.9 m/s 2 )(−9.5 m)
2(4.9 m/s 2 )
∆t = 2.4 s
Thus, the time of flight is 2.4 s.
(b) ∆x = ?
∆x = vix ∆t
= (8.9 m/s)(2.4 s)
∆x = 22 m
The width of the moat is 22 m.
(c) vfx = vfy = 8.9 m/s
vf = ?
vfy 2 = viy 2 + 2 a y ∆y
vfy = (8.0 m/s)2 + 2( −9.8 m/s 2 )( − 9.5 m)
vfy = 16 m/s
Copyright © 2003 Nelson
Chapter 1 Kinematics
51
vf = vfx 2 + vfy 2
= (8.9 m/s) 2 + (16 m/s)2
vf = 18 m/s
tan θ =
vfy
vfx
 16 m/s 
θ = tan −1 

 8.9 m/s 
θ = 60°
The final velocity just before landing is 18 m/s [60° below the horizontal].
Section 1.4 Questions
(Pages 50–51)
Understanding Concepts
1. The vertical acceleration of a projectile is the same throughout its trajectory and is equal to the acceleration due to gravity.
2. (a) For a projectile with the launch point lower than the landing point, the magnitude of the velocity is at a maximum at the
initial part of the flight (initial velocity) and is at a minimum at the top of its trajectory.
(b) For a projectile with the launch point higher than the landing point, the magnitude of the velocity is at a maximum just
before landing (final velocity) and is at a minimum at the top of its trajectory.
3. Let +y be up.
∆x = 16 m
∆y = –1.5 m
ay = –9.8 m/s2
viy = 0 m/s
vi = ?
First we must solve for the change in time:
a y ( ∆t ) 2
∆y = viy ∆t +
2
2∆y
∆t =
ay
=
2(−1.5 m)
−9.8 m/s 2
∆t = 0.55 s
To calculate the initial velocity:
vi = vix
∆x
∆t
16 m
=
0.55s
=
vi = 29 m/s
4.
52
The initial velocity of the projectile is 29 m/s [horizontally].
Let +y be up.
(a) ∆y = –2.5 m
ay = –9.8 m/s2
viy = 0 m/s
∆t = ?
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
∆y = viy ∆t +
1
a y ( ∆t ) 2
2
2 ∆y
ay
∆t =
2( −2.5 m)
−9.8 m/s2
=
∆t = 0.71s
The tennis ball is in the air for 0.71 s.
(b) vix = 24 m/s
∆x = ?
∆x = vix ∆t
= (24 m/s)(0.71s)
∆x = 17 m
The horizontal displacement of the ball is 17 m [fwd].
(c) vf = ?
vfy 2 = viy 2 + 2a y ∆y
vfy = (0 m/s) 2 + 2(−9.8 m/s 2 )( − 2.5 m)
vfx = vix = 24 m/s
vfy = 7.0 m/s
vf = vfx 2 + vfy 2
= (24 m/s) 2 + (7.0 m/s)2
vf = 25 m/s
tan θ =
vfy
vfx
 7.0 m/s 
θ = tan −1 

 24 m/s 
θ = 16°
The ball’s maximum velocity just prior to landing on the court surface is 25 m/s [16° below the horizontal].
(d) To calculate the distance d that the ball clears the net, first calculate the height of the ball above the ground at the net h,
and then subtract the height of the net.
∆x = 12 m
h = 2.5 m + ∆y = ?
d = h – 0.9 m = ?
First we must determine the change in time:
∆x
∆t =
vix
=
12 m
24 m/s
∆t = 0.50s
Copyright © 2003 Nelson
Chapter 1 Kinematics
53
Next, we determine the change in the vertical component:
a y ( ∆t ) 2
∆y = viy ∆t +
2
2
( −9.8 m/s )(0.50s)2
=
2
∆y = −1.22 m
Now, we can determine h and d:
h = 2.5 m − 1.22 m = 1.28 m
5.
d = 1.28 m − 0.90 m = 0.38 m
Thus, the ball clears the net by 0.38 m.
Let +y be down. Determine the horizontal and vertical components of the initial velocity.
Horizontally (constant vix ):
G
vix = vi cos θ
= (3.2 m/s)(cos 33°)
vix = 2.7 m/s
Vertically (constant ay):
G
viy = vi sin θ
= (3.2 m/s)(sin 33°)
viy = 1.7 m/s
(a) ay = 9.8 m/s2
∆t = ?
∆y = 6.2 m − 1.0 m = 5.2 m
1
a y ( ∆t ) 2
2
(9.8 m/s 2 )(∆t )2
0=
+ 1.7 m/s ∆t − 5.2 m
2
∆y = viy ∆t +
Solve using the solution to the quadratic equation:
∆t =
=
−b ± b 2 − 4ac
2a
(where a = 4.9 m/s 2 , b = 1.7 m/s, and c = −5.2 m)
−(1.7 m/s) ± (1.7 m/s)2 − 4(4.9 m/s 2 )(−5.2 m)
2(4.9 m/s 2 )
∆t = 0.87 s
The ball is airborne for 0.87 s.
(b) ∆x = ?
∆x = vix ∆t
= (2.7 m/s)(0.87 s)
∆x = 2.3 m
The child should hold the glove 2.3 m from the edge of the roof.
(c) vf = ?
vfy 2 = viy 2 + 2a y ∆y
vfy = (1.7 m/s) 2 + 2(9.8 m/s 2 )(5.2 m)
vfy = 10.2 m/s
vfx = vix = 2.7 m/s
54
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
2
vf = vfx + vfy
2
= (2.7 m/s)2 + (10.2 m/s)2
vf = 11m/s
tan θ =
6.
7.
vfy
vfx
 10.2 m/s 
θ = tan −1 

 2.7 m/s 
θ = 75°
The ball’s final velocity just before it lands in the glove is 11 m/s [75° below the horizontal].
As determined in the Try This Activity on page 49, the horizontal range in this type of situation is identical for angles
“equidistant” on either side of 45°. Thus, the launch angles are 54° (same range as 36°), 74° (same range as 16°), and
44.4° (same range as 45.6°).
Let +y be up. Find the horizontal and vertical components of the initial velocity:
Horizontally (constant
vix ):
G
vix = vi cos θ
= (1.1 × 103 m/s)(cos 45°)
vix = 7.8 × 102 m/s
Vertically (constant
a y ):
G
viy = vi sin θ
= (1.1 × 103 m/s)(sin 45°)
viy = 7.8 × 102 m/s
(a) Solve for the time of flight using the equation:
1
∆y = viy ∆t + a y ( ∆t ) 2
2
1
0 = viy ∆t + a y (∆t )2
2
1
−viy ∆t = a y ( ∆t )2
2
−2viy ∆t = a y ( ∆t ) 2
∆t =
=
−2viy
ay
(
−2 7.8 × 102 m/s
2
)
−9.8 m/s
∆t = 1.6 × 102 s
Therefore, each shell was airborne for 1.6 × 102 s.
(b) To determine the horizontal range:
∆x = vix ∆t
= (7.8 × 102 m/s)(1.6 × 102 s)
∆x =1.2 × 105 m
The horizontal range is 1.2 × 105 m.
Copyright © 2003 Nelson
Chapter 1 Kinematics
55
(c) At the highest position, the y-component of the instantaneous velocity is zero (vfy = 0 m/s). Thus,
vfy 2 = viy 2 + 2 a y ∆y
2
0 = viy + 2a y ∆y
∆y =
=
8.
viy
2
−2 a y
(7.8 × 102 m/s)2
−2( −9.8 m/s2 )
∆y = 3.1 × 104 m
The maximum height of each shell is 3.1 × 104 m or 31 km.
G
g = 1.6 m/s2
vi = 32 m/s [35° above the Moon’s horizontal]
∆yf = –15 m
(a) Determine the horizontal and vertical components of the initial velocity.
Horizontally (constant vix ):
G
vix = vi cos θ
= (32 m/s)(cos 35°)
vix = 26 m/s
Vertically (constant g):
G
viy = vi sin θ
= (32 m/s)(sin 35°)
viy = 18 m/s
Let +y be down.
G
2
a y = − g = −1.6 m/s
At the highest position, the y-component of the instantaneous velocity is zero (vfy = 0 m/s). Thus,
vfy 2 = viy 2 + 2a y ∆y
2
0 = viy + 2a y ∆y
∆y =
=
−viy
2
2a y
−(18 m/s)2
2( −1.6 m/s2 )
∆y = 1.1 × 102 m
The maximum height of the golf ball is 1.1 × 102 m.
(b) ∆t = ?
Solve for the time of flight using the equation:
1
∆y = viy ∆t + a y ( ∆t )2
2
−15 m = 18 m/s ∆t − 0.80 m/s 2 ( ∆t )
2
0 = 0.80 m/s2 ( ∆t ) − 18 m/s ∆t − 15 m
2
56
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Solve using the solution to the quadratic equation:
∆t =
=
−b ± b2 − 4 ac
where a = 0.80 m/s2 , b = −18 m/s, and c = −15 m
2a
−( −18 m/s) ± ( −18 m/s)2 − 4(0.80 m/s2 )( −15 m)
2(0.8 m/s 2 )
∆t = 24 s
The time of flight is 24 s.
(c) ∆x = ?
∆x = vix ∆t
= (26 m/s)(24 s)
∆x = 6.2 ×10 2 m
The horizontal range is 6.2 × 102 m.
Applying Inquiry Skills
9.
Let +x be the horizontal direction of the initial velocity and +y be down.
ay = 9.8 m/s2
viy = 0
∆x and ∆y can both be measured with the metre stick.
vix = ?
Using the vertical motion:
∆y = viy ∆t +
∆y =
1
2
1
2
a y ∆t 2
a y ∆t 2
2 ∆y
∆t =
ay
Using the horizontal motion:
vix =
=
∆x
∆t
∆x
2 ∆y
ay
vix = ∆x
ay
2∆y
The initial speed can be found by substituting the appropriate values into this equation.
10. If students used the Nelson Science 9 text, they may recall designing and testing the type of device needed for this
question. (In that text, refer to page 505, Part 3 of Investigation 16.8. Figure 1 on that page shows that the device is
simple to make and very inexpensive.) Fold a piece of thin cardboard lengthwise so a vertical component facing upward
separates two horizontal components. Place a coin on either side of the vertical component and rotate the device quickly
on the horizontal plane. One coin will project outward, while the other coin will drop vertically downward.
Making Connections
11. (a) ∆y = 2.2 m
vi = 14 m/s [42° above the horizontal]
g = 9.8 m/s2
∆x = ?
Copyright © 2003 Nelson
Chapter 1 Kinematics
57
 −v sin θ + v 2 sin 2 θ + 2 g ∆y
2
2vi sin θ cos θ
i
i
∆x = 0.30 m +
+ vi cos θ 

g
g





 −(14 m/s) sin 42° + (14 m/s) 2 sin 2 42° + 2(9.8 m/s2 )(2.2 m) 
2(14 m/s)2 sin 42° cos 42°


+
°
14
m/s(cos42
)


9.8 m/s2
9.8 m/s2


= 0.30 m + 19.6 m + 2.3 m
= 0.30 m +
(b)
(c)
∆x = 22 m
The range of the shot is 22 m.
The calculated value using the equation is slightly less than the world record. Perhaps the world record holder has
longer arms than the equation considers and was able to throw the shot with a slightly higher initial velocity than the
value stated in the question.
The equation is set up using g, which is positive, rather than using ay, which can be positive or negative, depending on
which direction is defined as +y. The first term in the equation takes into consideration the fact that the thrower’s hand
goes about 30 cm beyond the line from which the horizontal range is measured. The second term is the same as the
equation derived on page 49 for the situation in which the projectile lands at the same level from which it began. The
third term takes into consideration the fact than the horizontal range increases when the projectile lands at a level lower
than its original level.
1.5 FRAMES OF REFERENCE AND RELATIVE VELOCITY
PRACTICE
(Page 56)
Understanding Concepts
G
G
G
1. (a) vLE = vLD + vDE
G
G
G
(b) vAC = vAB + vBC
G
G
G
(c) vMN = vMT + vTN
G
G
G
vNM = vNT + vTM
G
G
(d) Replace vML with vLM .
G
G
G
G
G
vLP = vLM + vMN + vNO + vOP
2. Use the subscripts S for the ship, W for the water, and T for the tourist group.
G
(a) vSW = 2.8 m/s [fwd]
G
vTS = 1.1 m/s [fwd]
G
vTW = ?
G
G
G
vTW = vTS + vSW
= 1.1m/s [fwd] + 2.8 m/s [fwd]
G
vTW = 3.9 m/s [fwd]
When walking toward the bow, the group’s velocity relative to the water is 3.9 m/s [fwd].
G
(b) vTS = 1.1 m/s [backward] = –1.1 m/s [fwd]
G
vTW = ?
G
G
G
vTW = vTS + vSW
= −1.1m/s [fwd] + 2.8 m/s [fwd]
G
vTW = 1.7 m/s [fwd]
When walking toward the stern, the group’s velocity relative to the water is 1.7 m/s [fwd].
(c) Let the +x direction be forward and the +y direction be to the right.
G
vTS = 1.1 m/s [right]
G
vTW = ?
58
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
G
∆v
G
aav =
∆t
10.2 m/s [45° W of S]
=
2.0s
G
2
aav = 5.1 m/s [45° W of S]
19. (c) constant
vix = 5.5 m/s
CHAPTER 1 REVIEW
(Pages 64–67)
Understanding Concepts
 1× 103 m   1h

= 27.8 m/s
1. (a) 100 km/h 
 
3 
1km
×
3.6
10
s




Thus, 100 km/h is 27.8 m/s.
3
 1km   3.6 × 10 s 
2
(b) 97 m/s 
 = 3.5 × 10 km/h
  1h
3

×
1
10
m



The peregrine falcon can dive at speeds of 3.5 × 102 km/h.
(c) To convert km/h to m/s, divide by 3.6. To convert m/s to km/h, multiply by 3.6.
2. (a) L × T–1 results in speed.
L
 L 
(b)  3  ¯ T = 2 results in acceleration.
T
T


 L 
(c)  2  ¯ T ¯ T = L results in length.
T 
G G
3. (a) ∆d = a (∆t ) 2
Therefore:
?
 L 
L =  2  × T2
T 
?
4.
5.
6.
7.
L=L
The equation is dimensionally correct because both sides of the equation have the same dimensions.
(b) An equation may be dimensionally correct yet still be wrong. In (a), for example, the equation could be one of the
following:
G G
1G
∆d = v ∆t + a ∆t 2
i
2
G 1G 2
or ∆d = a ∆t
2
(a) The instantaneous speed is equal to the average speed throughout the motion.
(b) The instantaneous velocity is equal to the average velocity throughout the motion.
(c) The instantaneous speed is equal to the magnitude of the average velocity throughout the motion.
(a) The displacement is equal to the area under a velocity-time graph.
(b) The instantaneous acceleration is the slope of the line (for constant acceleration) or the slope of the tangent to the curve
(for nonconstant acceleration) on a velocity-time graph.
No, a component of a vector cannot have a magnitude greater than the vector’s magnitude. A vector can be broken into
components using sine and cosine, which have a maximum of one.
(a) Yes, two vectors having the same magnitude can be combined to give a zero resultant vector by adding two vectors that
have opposite directions; for example, 5 m [S] + 5 m [N] = 0.
(b) No, two vectors having different magnitudes cannot be combined to give a zero resultant vector.
(c) Yes, three vectors having different magnitudes can be combined to give zero. Adding three vectors in two dimensions
head-to-tail, can total zero. Even in one dimension three vectors can add to zero; for example,
2 m [S] + 5 m [N] + 3 m [S] = 0.
Copyright © 2003 Nelson
Chapter 1 Kinematics
71
8.
G
d1 = 214 m [E]
G
d 2 = 96 m [28° N of E]
G
d3 = 12 m [25° S of E]
(a)
G
(b) ∆d = ?
Let the +x direction be to the east and the +y direction be to the north.
G
G
G
G
∆d = ∆d1 + ∆d 2 + ∆d 3
∆d y = ∆d1, y + ∆d 2, y + ∆d 3, y
∆d x = ∆d1, x + ∆d 2, x + ∆d 3, x
= (96 m) sin 28° − (12 m) sin 25°
= 214 m + (96 m) cos 28° + (12 m) cos 25°
∆d y = 40 m
∆d x = 310 m
G
∆d = ∆d x 2 + ∆d y 2
= (310 m/s)2 + (40 m/s) 2
G
∆d = 312 m/s
 ∆d y 
tan θ = 

 ∆d x 
9.
 40 m/s 
θ = tan −1 

 310 m/s 
θ = 7.4°
The displacement from the tee needed to get a hole-in-one using components is 312 m/s, or 3.1 × 102 m [7.4° N of E].
Let the +x direction be to the east and the +y direction be north.
G
Let C be the addition of the two vectors.
G G G
C = A+ B
C x = Ax + Bx
C y = Ay + By
= (5.1 km) cos 38° + (6.8 km) sin 19°
C x = 6.2 km
C y = −3.3 km
= (5.1 km) sin 38° − (6.8 km) cos19°
G
(a) The vector that would add to the vector C to give a resultant displacement of zero would be equal in magnitude, but
G
G
opposite in direction to the vector C . Thus, vector C is
72
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
G
C = Cx 2 + C y 2
= (6.2 km)2 + (−3.3km) 2
G
C = 7.0 km
 Cy 

tan θ = 
 Cx 


 3.3 km 
θ = tan −1 

 6.2 km 
θ = 28°
Thus, the vector that must be added is 7.0 km [28° N of W].
G
(b) Let the vector D represent the desired vector.
G G G
4.0 km [W] = A + B + D
G G
4.0 km [W] = C + D
G
G
D = 4.0 km [W] − C
Dx = −4.0 km − 6.2 km
Dy = −C y
Dx = −10.2 km
Dy = −3.3 km
G
D = Dx 2 + Dy 2
= (−10.2 km) 2 + (−3.3 km)2
G
D = 11 km
 Dy 

tan θ = 
 Dx 


 3.3km 
θ = tan −1 

 10.2 km 
θ = 18°
The vector is 11 km [18° N of W].
G
10. (a) At 5 P.M., assume that people are sitting down to dinner. If we assume an average nose-to-toes ∆d value of 1 m
[down] (when person is sitting), for a town of 2000, the resultant displacement vector of the sum of all the nose-to-toes
vectors is about 2000 m or about 2 × 103 m [down].
(b) At 5 A.M., assume that everyone is sleeping lying down. Because the beds face different directions, we can assume that
the vectors point in random directions and average to 0 m. Thus the resultant displacement averages to 0 m.
304.29 km
11. d =
= 4.4100 km
69 laps
∆t = 84.118 s
vav = ?
d
vav =
∆t
4.4100 × 103 m
=
84.118 s
vav = 52.426 m/s
The average speed for the lap is 52.426 m/s.
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Chapter 1 Kinematics
73
12. (a) ∆t = 2.0 s
vav = 115 km/h = 31.9 m/s
d=?
d
∆t
d = vav ∆t
vav =
= (31.9 m/s)(2.0 s)
d = 64 m
The distance in metres if the speed of your car is 115 km/h is 64 m.
(b) Assuming that the average length of a car is 5.0 m, the number of car lengths is
13. (a) v1 = 24 m/s
∆d1 = 1.2 × 103 m
v2 = 18 m/s
∆d2 = 1.2 × 103 m
∆t = ?
64 m
= 13 car lengths.
5.0 m
∆t = ∆t1 + ∆t2
=
∆d1 ∆d 2
+
v1
v2
=
1.2 ×103 m 1.2 × 103 m
+
24 m/s
18 m/s
∆t = 1.2 × 102 s
The time interval is 1.2 × 102 s.
(b) vav = ?
d
vav =
∆t
1.2 × 103 m + 1.2 × 103 m
=
1.2 × 102 s
vav = 21 m/s
The eagle’s average speed during this motion is 21 m/s.
14. (a) Starting from a positive position and a low velocity toward the zero position, the speed increases gradually at first and
then dramatically. Upon reaching the zero position, the velocity changes direction. The motion now is a high velocity
followed by decreasing velocity back to the initial position. The last part of the motion takes about half as long as the
first part.
(b) Starting from rest, the object undergoes uniform acceleration in one direction, followed by a higher uniform
acceleration in the opposite direction for a shorter period of time until coming to a stop.
15. d1 = 4.5 m
d2 = 6.8 m
∆t = 5.0 s
(a) vav = ?
d + d2
vav = 1
∆t
4.5 m + 6.8 m
=
5.0 s
vav = 2.3 m/s
The firefighter’s average speed is 2.3 m/s.
(b) Let the +x direction be horizontal and the +y direction be down. Thus,
∆dx = 6.8 m
∆dy = 4.5 m
G
∆d = ?
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Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
G
∆d = ∆d x 2 + ∆d y 2
= (6.8 m)2 + (4.5 m)2
G
∆d = 1.6 m
 ∆d y 
tan θ = 

 ∆d x 
 4.5 m 
θ = tan −1 

 6.8 m 
θ = 33°
The firefighter’s average velocity is 1.6 m/s [33° below the horizontal].
G
16. ∆d1 = 16 m [35° S of W]
G
∆d 2 = 22 m [15° S of E]
∆t = 6.4 s
G
(a) ∆d = ?
Solve using components. Let the +x direction be to the east and the +y direction be south.
G
G
G
∆d = ∆d1 + ∆d 2
∆d y = ∆d1 y + ∆d 2 y
∆d x = ∆d1 x + ∆d 2 x
= ( −16 m) cos 35° + (22 m) cos 15°
∆d x = 8.1 m
= (16 m) sin 35° + (22 m) sin 15°
∆d y = 15 m
G
∆d = ∆d x 2 + ∆d y 2
= (8.1m)2 + (15 m)2
G
∆d = 17 m
 ∆d y 
tan θ = 

 ∆d x 
 15 m 
θ = tan −1 

 8.1m 
θ = 61°
G
∆ d = 17 m [61° S of E] or 17 m [29° E of S] , thus, the player’s resultant displacement is 17 m [29° E of S].
G
(b) vav = ?
G
∆d
G
vav =
∆t
17 m [29° E of S]
=
6.4 s
G
vav = 2.7 m/s [29° E of S]
Thus, the player’s average velocity is 2.7 m/s [29° E of S].
17. vav = 100 km/h
The directions are found by estimating the directions of the tangents at the positions D, E, and F.
The instantaneous velocity at position D is 100 km/h [10° S of E].
The instantaneous velocity at position E is 100 km/h [10° N of E].
The instantaneous velocity at position F is 100 km/h [75° N of E].
Copyright © 2003 Nelson
Chapter 1 Kinematics
75
18. (a) vi = 42 km/h
vf = 105 km/h
∆t = 26 s = 7.2 × 10–3 h
∆d = ?
∆d = vav ( ∆t )
1
(105 km/h + 42 km/h)(7.2 × 10−3 h)
2
∆d = 0.53km
The car travels 0.53 km in the time interval.
G
(b) aav = ?
G G
vf − vi
G
aav =
∆t
105 km/h [fwd] − 42 km/h [fwd]
=
26 s
G
aav = 2.4 (km/h)/s [fwd]
The magnitude of the car’s average acceleration is 2.4 (km/h)/s.
G
19. vi = 0
G
∆d = 15 m [fwd]
∆t = 1.2 s
G
(a) a = ?
G G
1G
2
∆d = vi ∆t + a ( ∆t )
2
G
1G
2
∆d = 0 + a ( ∆t )
2
G
G 2∆d
a=
2
( ∆t )
=
=
2(15 m [fwd])
(1.2 s) 2
G
a = 21m/s 2 [fwd]
The average acceleration of the cars is 21 m/s2 [fwd].
G
(b) vf = ?
G G G
vf = vi + a ∆t
= 0 + (20.8 m/s 2 [fwd])(1.2 s)
G
vf = 25 m/s [fwd]
The velocity of the cars at 1.2 s is 25 m/s [fwd].
G
(c) The magnitude of the acceleration in terms of g can be determined as:
G
a
20.8 m/s 2
G =
g
9.8 m/s 2
G
G
a = 2.1 g
G
20. vf = 4.0 × 102 m/s [fwd]
G
vi = 0
G
∆d = 0.80 m [fwd]
G
a=?
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Unit 1 Forces and Motion: Dynamics
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G 2 G2
G G
vf = vi + 2a ∆d
G
G G
vf 2 = 0 + 2a ∆d
G
G v2
a= f G
2 ∆d
(4.0 × 102 m/s[fwd])2
=
2(0.8 m[fwd])
G
a = 1.0 × 105 m/s 2 [fwd]
The constant acceleration needed by the bullet is 1.0 × 105 m/s2 [fwd].
G
21. vi = 2.28 × 102 m/s [fwd]
G
a = 6.25 × 101 m/s2 [fwd]
G
∆d = 1.86 × 103 m [fwd]
G
vf = ?
G 2 G2
G G
vf = vi + 2a ∆d
G
vf = (2.28 × 10 2 m/s[fwd]) 2 + 2(6.25 × 101 m/s 2 [fwd])(1.86 × 103 m[fwd])
G
vf = 5.33 × 102 m/s[fwd]
Thus, the rocket’s velocity is 533 m/s [fwd].
G
22. vf = 0
G
a = 9.8 m/s2 [down]
G
∆d =1.9 m [up]
G
vi = ?
G 2 G2
G G
vf = vi + 2a ∆d
G
G G
0 = vi 2 + 2 a ∆ d
G
G G
vi = −2a ∆d
= −2( −9.8 m/s 2 [up])(1.9 m [up] )
G
vi = 6.1 m/s [up]
The minimum vertical velocity needed by the salmon to jump to the top of the waterfall is 6.1 m/s [up].
G
23. ∆d = 2.0 × 102 m [fwd]
G
a =1.6 m/s2 [fwd]
G
(a) vi = 0.0 m/s
∆t = ?
G G
1G
2
∆d = vi ∆ t + a ( ∆ t )
2
G
1G
2
∆d = 0 + a ( ∆t )
2
G
2 ∆d
∆t =
G
a
=
2(2.0 × 102 m[fwd])
1.6 m/s 2 [fwd]
∆t = 16s
The motion takes 16 s.
Copyright © 2003 Nelson
Chapter 1 Kinematics
77
G
(b) vi = 8.0 m/s [fwd]
∆t = ?
G G
1G
2
∆d = vi ∆t + a ( ∆t )
2
G
G
1G
2
0 = vi ∆t + a ( ∆t ) − ∆d
2
G
G
1G
2
0 = a ( ∆ t ) + vi ∆ t − ∆ d
2
This is a quadratic equation with solution:
∆t =
−b ± b 2 − 4ac
2a
G
G
G
G
a
−vi ± vi 2 − 4   ( −∆d )
2
=
G
a
 
2 
2
=
−8.0 m/s[fwd] ± (8.0 m/s [fwd]) 2 + 2(1.6 m/s 2 [fwd])(2.0 × 102 m [fwd])
1.6 m/s 2 [fwd]
∆t = 12 s
The motion takes 12 s.
24. Let the +x direction be east and the +y direction be south.
G
vi = 240 m/s [28° S of E]
G
vf = 220 m/s [28° E of S]
∆t = 35 s
G
aav = ?
∆v y = vfy − viy
∆vx = vfx − vix
= (220 m/s) sin 28° − (240 m/s) cos 28°
∆vx = −109 m/s
= (220 m/s) cos 28° − (240 m/s) sin 28°
∆v y = 82 m/s
G
2
2
∆v = ∆v x + ∆v y
= (−109 m/s)2 + (82 m/s) 2
G
∆v = 136 m/s
G
G ∆v
a=
∆t
136 m/s
=
35 s
G
a = 3.9 m/s 2
tan θ =
vx
vy
 109 m/s 
θ = tan −1 

 82 m/s 
θ = 53°
The airplane’s average acceleration during this time interval is 3.9 m/s2 [53° W of S].
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Unit 1 Forces and Motion: Dynamics
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G
25. vf = 54 m/s [N]
G
aav =19 m/s 2 [45° W of N]
∆t = 4.0 s
G
vi = ?
To find the initial velocity:
G G G
vf = vi + aav ∆t
G G G
vi = vf − aav ∆t
G
G
= vf + ( − aav ∆t )
(
= 54 m/s [N] + −19 m/s 2 [45° W of N](4.0 s
= 54 m/s [N] + ( −76 m/s [45° W of N] )
G
vi = 54 m/s [N] + 76 m/s [45° E of S]
)
Using components with +x east and +y north:
vix = 0 + 76 m/s ( sin 45° )
vix = 54 m/s
viy = 54 m/s − 76 m/s ( cos 45° )
= 54 m/s − 54 m/s
viy = 0 m/s
G
vi = 54 m/s [E]
The car should enter the curve with a velocity of 54 m/s [E].
G
26. vi = 17 m/s [up]
G
∆d = 5.2 m
G
a = 9.8 m/s2 [down]
∆t = ?
G G
1G
2
∆d = vi ∆t + a ( ∆t )
2
G
G
1G
2
0 = vi ∆t + a ( ∆t ) − ∆d
2
G
G
1G
2
0 = a ( ∆ t ) + vi ∆ t − ∆ d
2
This is a quadratic equation with solution:
∆t =
−b ± b 2 − 4ac
2a
G
G
G
G
a
−vi ± vi 2 − 4   ( −∆d )
2
=
G
a
2 
2
=
−17 m/s[up] ± (17 m/s[up]) 2 + 2( −9.8 m/s 2 [up])(5.2 m[up])
−9.8 m/s 2 [up]
= 1.7 s ± 1.4 s
∆t = 0.34 s or 3.1 s
The ball passes the camera at 0.34 s on the way up and at 3.1 s on the way down.
Copyright © 2003 Nelson
Chapter 1 Kinematics
79
27.
(a) The slopes of the line segments on the velocity-time graph indicate the accelerations.
(b) The area between the line and the x-axis up to the specific times indicate the total displacements up to those times.
28. rVenus = 1.08 × 1011 m
TVenus = 1.94 × 107 s
(a) v = ?
2π r
v=
T
2π (1.08 ×1011 m)
=
1.94 × 107 s
v = 3.50 × 10 4 m/s
 1 km   3600 s 
= 3.50 × 10 4 m/s 


 1000 m   1 h 
v = 1.26 ×105 km/h
Venus has an average speed of 3.50 × 104 m/s, or 1.26 × 105 km/h.
G
(b) vav = ?
G
Determine the average velocity after half a revolution using ∆d = 2r. Thus,
G
∆d
G
vav =
∆t
2r
=
T 
2
 
4r
=
T
4(1.08 × 1011 m)
=
1.94 × 107 s
G
vav = 2.23 × 104 m/s
The magnitude of the average velocity of Venus after it has completed half a revolution around the Sun is
2.23 × 104 m/s.
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Unit 1 Forces and Motion: Dynamics
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(c) Use the initial and final velocities to determine the average acceleration after a quarter revolution. Choose directions
such as east and south that are perpendicular to define the directions at these two positions.
G
a =?
G G
G v − vi
a= f
∆t
G G
vf − vi
=
T 
4
 
=
4(3.50 × 10 4 m/s[E] − 3.50 × 104 m/s[S])
1.94 × 107 s
G
a =7.22 × 10 −3 m/s 2 [E] − 7.22 ×10−3 m/s 2 [S]
G
a = ax 2 + a y 2
= 2(7.22 × 10−3 m/s 2 ) 2
G
a = 1.02 × 10−2 m/s 2
The magnitude of its average acceleration during a quarter of a revolution around the Sun is 1.02 × 10–2 m/s2.
29. (a) The horizontal component of the acceleration of a projectile is zero. The vertical component of the acceleration of a
projectile is 9.8 m/s2 toward Earth.
(b) With air resistance affecting both components, the horizontal acceleration would be negative (assuming the direction of
the horizontal component of the velocity is defined as positive), and the vertical component of the acceleration would
be less than 9.8 m/s2.
30. (a) Let +y be down.
G
vi = 18 m/s [horizontal]
ay = 9.8 m/s2
∆x = 9.0 m
∆t = ?
Find the horizontal component of the initial velocity:
Horizontally (constant vix ):
vix = 18.0 m/s
∆t =
∆x
vix
9.0 m
18.0 m/s
∆t = 0.50 s
The snowball will hit the tree after 0.50 s.
(b) Find the vertical component of the initial velocity:
=
Vertically (constant a y ):
viy = 0 m/s
1
∆y = viy ∆t + a y (∆t ) 2
2
1
= a y ( ∆t ) 2
2
(9.8 m/s 2 )(0.50 s) 2
=
2
∆y = 1.2 m
Since the snowball’s original location was 1.5 m above ground, the height that the snowball will hit the tree is
1.5 m – 1.2 m = 0.3 m.
Copyright © 2003 Nelson
Chapter 1 Kinematics
81
G
(c) vf = ?
2
2
vfy = viy + 2a y ∆y
vfy 2 = 0 + 2a y ∆y
vfy = 2a y ∆y
= 2(9.8 m/s 2 )(1.2 m)
vfy = 4.8 m/s
Therefore,
G 2
2
2
vf = vfx + vfy
G
vf = (18 m/s) 2 + (4.8 m/s) 2
G
vf = 19 m/s
θ = tan −1
= tan −1
vfy
vfx
4.8 m/s
18 m/s
θ = 15°
The snowball’s velocity as it strikes the tree is 19 m/s [15° below the horizontal].
31. (a) Let +y be down.
∆y = 1.5 may = 9.8 m/s2
∆x = 16 m
viy = 0 m/s
vix = ?
1
∆y = viy ∆t + a y (∆t ) 2
2
1
∆y = a y ( ∆t ) 2
2
2 ∆y
∆t =
ay
2(1.5 m)
9.8 m/s 2
∆t = 0.55 s
=
∆x
∆t
16 m
=
0.55 s
vix = 29 m/s
Thus, the initial velocity of a ball projected horizontally is 29 m/s [horizontally].
32. (a) In the train’s frame of reference, the path of the ball is straight downward from the release position.
(b) In the other frame of reference (actually Earth’s frame), the path of the ball is parabolic and is shaped just like any
projectile whose initial velocity is horizontal. (The magnitude of the initial velocity relative to Earth is equal in
magnitude to the magnitude of the train’s velocity relative to Earth.)
vix =
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Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
33. Use the subscripts P for the plane, A for the wind (air), and G for the ground. Let the +x direction be east and the +y
direction be south.
G
vPA = 285 km/s [45° S of E]
G
vAG = 75 km/h [22° E of N]
G
vPG = ?
G
G
G
vPG = vPA + vAG
vPG,x = (285 km/h)cos 45° + (75 km/h)sin 22°
vPG,y = (285 km/h)sin 45° − (75 km/h)cos 22°
vPG,x = 230 km/h
vPG,y = 132 km/h
G
2
2
vPG = vPG,x + vPG,y
= (230 km/h)2 + (132 km/h)2
G
vPG = 2.6 ×10 2 km/h
 vPG,x
tan θ = 
 vPG,y





 230 km/h 
θ = tan −1 

 132 km/h 
θ = 60°
The velocity of the plane relative to the ground is 2.6 × 102 km/h [60° E of S].
34. Use the subscripts S for the swimmer, W for the water, and G for the ground. Let the +x direction be downstream from the
initial shore and the +y direction be across the river.
G
vSW = 0.80 m/s [across the river]
∆y = 86 m
∆x = 54 m
G
=?
(a) v
WG
∆y
∆t = G
vSW
=
86 m
0.80 m/s [across the river]
∆t = 108s
G
∆x
vWG =
∆t
54 m
=
108s
G
vWG = 0.50 m/s
The speed of the river current is 0.50 m/s.
G
(b) vSG = ?
G
G
G
vSG = vSW + vWG
vSG,x = vWG = 0.50 m/s
Copyright © 2003 Nelson
vSG,y = vSW = 0.80 m/s
Chapter 1 Kinematics
83
G
vSG = vSG,x 2 + vSG,y 2
= (0.50 m/s)2 + (0.80 m/s)2
G
vSG = 0.94 m/s
 vSG,y
tan θ = 
v
 SG,x



 0.80 m/s 
θ = tan −1 

 0.50 m/s 
θ = 58°
The swimmer’s velocity relative to the shore is 0.94 m/s [downstream, 58° from the initial shore].
(c) θ = ?
G
G
G
vSG = vSW + vWG
G 2 G 2 G 2
vSW = vSG + vWG
G 2 G 2 G 2
vSG = vSW − vWG
G
vSG = (0.80 m/s)2 − (0.50 m/s) 2
G
vSG = 0.62 m/s
G
 vSG
tan θ =  G
v
 WG



 0.62 m/s 
θ = tan −1 

 0.50 m/s 
θ = 51°
The direction in which the swimmer should have aimed to land directly across from the departure position is upstream,
51° from the near shore.
35. Use the subscripts P for the plane, A for the wind (air), and G for the ground. Let the +x direction be south and the +y
direction be east.
G
∆ d = 2.5 × 103 km [18° S of W]
∆t = 5.3 h
G
vAG = 85 km/h [E]
G
vPA = ?
First we find the velocity of the plane relative to the ground that allows it to reach its destination on schedule.
G
∆d PG
G
vPG =
∆t
2.5 × 103 km [18° S of W]
=
5.3h
G
2
vPG = 4.7 × 10 km/h [18° S of W]
G
G
G
vPG = vPA + vAG
G
G
G
vPA = vPG − vAG
Using the relative velocity equation:
G
G
G
vPG = vPA + vAG
G
G
G
vPA = vPG − vAG
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Unit 1 Forces and Motion: Dynamics
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Using components with +x west and +y south:
vPA,x = (4.7 × 102 km/h) cos18° − ( −85 km/h)
vPA,y = (4.7 × 102 km/h) sin18° − 0
vPA,x = 5.34 × 102 km/h
vPA,y = 1.45 × 102 km/h
G
vPA = vPA,x 2 + vPA,y 2
= (5.34 × 102 km/h)2 + (1.45 × 102 km/h)2
G
vPA = 5.5 × 102 km/h
 vPA,y
tan θ = 
v
 PA,x



 1.45 × 10 2 km/h 
θ = tan −1 

2
 5.34 × 10 km/h 
θ = 15°
The velocity of the plane relative to the air is 5.5 × 102 km/h [15° S of W].
36. Let +y be up.
G
vi = 21.0 m/s [47° above the horizontal]
ay = 9.8 m/s2
∆x = 25 m
∆y = ?
Find the horizontal and vertical components of the initial velocity:
Horizontally (constant
G
vix = vi cos θ
vix )
= (21.0 m/s)(cos 47°)
vix = 14.3 m/s
Vertically (constant ay)
G
viy = vi sin θ
= (21.0 m/s)(sin 47°)
viy = 15.4 m/s
First we must determine the change in time:
∆x
∆t =
vix
25 m
14.3 m/s
∆t = 1.7 s
To calculate the vertical height of the football over the bar:
1
∆y = viy ∆t + a y (∆t ) 2
2
1
= (15.4 m/s)(1.7 s) + (9.8 m/s 2 )(1.7 s)2
2
∆y = 12 m
Since the horizontal bar of the goal post is 3.0 above the field, the football will pass above the bar at a height of
12 m – 3.0 m = 8.9 m.
=
Copyright © 2003 Nelson
Chapter 1 Kinematics
85
Applying Inquiry Skills
37. (a) Since the only instrument allowed is a metre stick or a measuring tape, the student could measure the horizontal range
of the ball that is observed to have the maximum range. The angle (θ) of the ball’s initial velocity must be estimated.
The equation derived for the maximum horizontal range for a projectile that lands at the same level as its starting
position can be applied. (See the text, page 49.)
∆x =
vi 2 =
vi 2
g
sin 2θ
g ∆x
sin 2θ
vi =
g ∆x
sin 2θ
(b) The biggest source of random error is in estimating the launch angle of the ball. Another source of random error is in
measuring the horizontal range. A systematic error may occur if the horizontal range is measured to a position below
the level at launch. A source of error that could be either random or systematic, depending on the situation, is the effect
of the wind and/or air resistance.
38. L = 62.0 cm
dU = 9.9 cm
dL = 4.3 cm
(a) θ = ?
 d − dL 
sin θ =  U

L


 9.9 cm − 4.3 cm 
θ = sin −1 

62 cm


θ = 5.2°
The angle of incline of the air table is 5.2° above the horizontal.
(b) a = ?
Refer to Investigation 1.4.1, questions (c) and (j), for background information.
a = g sin θ
= (9.8 m/s 2 )(sin 5.2°)
a = 0.89 m/s 2
The magnitude of the acceleration of the puck parallel to the inclined plane is 0.89 m/s2.
(c) The main random sources of error occur in using a ruler to measure the required distances. A systematic source of error
may occur if the ruler or metre stick has a worn end or if the calibration does not begin at 0.0 cm. (Notice that the
sources of error mentioned here relate to the way in which the data were found and applied in this question. In a
complete experiment, the motion of the puck would be analyzed and there would be many more sources of error.)
39. No matter what values of the launch angle, horizontal range, and initial velocity students choose for their example, their
calculations will show that the distance the target falls equals the difference between its height above the launch level and
∆y of the projectile. A specific example follows.
∆x = 8.50 m
G
vi = 38.0 m/s [33.0° above the horizontal]
To satisfy the condition that the dart launcher is aimed at the stationary target, tan θ =
target above the launch level:
h
∆x
, where h is the level of the
h = ∆x tan θ
= (8.50 m )( tan 33.0° )
h = 5.52 m
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Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
The time of flight can be found by using the horizontal component of the dart’s motion:
∆x
vix =
∆t
∆x
∆t =
vix
=
=
∆x
vi cos θ
8.50 m
(38.0 m )(cos 33.0° )
∆t = 0.267 s
In that time interval, the target falls by an amount ∆yT:
1
∆yT = viy ∆t + a y ∆t 2
2
(
= 0 + 4.9 m/s 2
)(0.267 s )
2
∆yT = 0.349 m
The target is thus 5.52 m − 0.349 m = 5.17 m above the launch level.
Finally, consider the vertical component of the dart’s displacement, ∆yT, using +y up:
1
∆yD = viy ∆t + a y ∆t 2
2
1
= (vi sin θ ) ∆t + a y ∆t 2
2
(
= (38.0 m/s )(sin 33.0° )(0.267 s ) − 4.9 m/s 2
)(0.267 s )
2
∆yD = 5.18 m
Thus, in the time interval that the target falls to a specific height above the launch level, the dart’s motion results in the
dart striking the target. (Rounding error resulted in a slight discrepancy in this example.)
If different students choose different initial conditions, the final result will be the same. This problem can also be
solved for general cases. If you have your students try this, be sure they use a consistent +y direction throughout the entire
solution.
Making Connections
40. v1 = 125 km/h
v2 = 100 km/h
d = 17 km
∆t = ?
(a) ∆t = t2 − t1
=
d d
−
v2 v1
17 km
17 km
−
100 km/h 125 km/h
= 0.17 h − 0.14 h
=
∆t = 0.03 h or 2.0 min
The driver saves 2.0 minutes by breaking the speed limit.
(b) At a higher speed there is more air resistance and more heat produced by the higher amount of friction in the moving
parts of the car and its engines. More fuel per unit distance moved is required to compensate for these effects.
Copyright © 2003 Nelson
Chapter 1 Kinematics
87
41. Let ∆t represent the time it takes for the signal to travel from Earth to the satellite and tD represent the delay time.
d = 4.8 × 107 m
c = 3.0 × 108 m/s
tD = 0.55 s
ttotal = ?
(a) First we must calculate the change in time:
d
∆t =
c
4.8 ×10 7 m
=
3.0 × 108 m/s
∆t = 0.16 s
Now we can determine the total time:
ttotal =∆t + tD + ∆t
= 2(0.16 s) + 0.55 s
ttotal = 0.87 s
The total time interval between sending the signal and receiving the return signal on Earth is 0.87 s.
(b) The time delays mentioned are obvious in live television broadcasts involving large distances between interviewers.
(Time delays are less when transmission is via cable rather than via satellite.)
42. Three-dimensional positions, velocities, and accelerations of the past and present could be analyzed and used to predict
the motion of an asteroid or other body in the future. This applies to both Earth and the asteroid.
43. Typical estimated speed and stopping distance are 5.0 m/s and 0.50 m, respectively, both of which are conservative
values. Using these values, we calculate the acceleration using magnitudes:
vf 2 = vi 2 + 2a∆d
0 = vi 2 + 2 a∆d
a=
=
−vi 2
2 ∆d
−(5.0 m/s) 2
2(0.50 m)
a = −25 m/s 2
G
Since this acceleration is more than 2.5 g , it is recommended that the patient avoid a rigorous racket sport. (Likely the
true acceleration would be even greater than indicated here because most sports enthusiasts can run faster than 5.0 m/s,
and often the stopping distance is less than 50 cm.)
44. At this stage, students are expected to confine their answers to concepts related to Chapter 1.
(a) The main principles are velocity and relative velocity of a flowing liquid (i.e., blood). The equations are:
d
G
G
G
vblood/artery = vblood/sensor + vsensor/artery
v
= blood
and
blood
∆t
G
(b) Answers will vary. One design could be a tiny device that sends signals outside the body so its motion ( vsensor/artery )
could be monitored by a receiver at the same time that a sensor within the device measures the speed of the blood
G
( vblood/sensor ). Both sets of data could be stored in a computer for later analysis.
45. (a) The explosive was fired vertically upward and detonated right at its maximum height.
(b) The explosive was fired vertically upward and detonated while still rising fairly quickly.
(c) The explosive was beginning to fall back downward and was moving slightly to the right when detonation occurred.
(d) The explosive was moving upward and fairly quickly to the left when detonation occurred.
Extension
46. Let d1 represent the distance the helicopter is from the cliff, the subscript S represent the sonar signal, and the subscript H
represent the helicopter.
∆d1 = 7.0 × 102 m
∆tS = 3.4 s
vS = 3.5 × 102 m/s
vH = ?
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Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
First we must calculate the distance the sonar signal travels:
∆dS = vS ∆tS
= (3.5 ×10 2 m/s)(3.4s)
∆dS = 1.2 × 103 m
Next we calculate the distance the helicopter travels:
∆d H = 2∆d1 − ∆dS
= 2(7.0 ×10 2 m) − 1.2 × 103 m
∆d H = 2.1 × 10 2 m
Finally, we can determine the speed of the helicopter:
∆d H
vH =
∆tS
=
2.1 × 10 2 m
3.4 s
vH = 62 m/s
The speed of the helicopter is 62 m/s
47. Let the subscript T represent the truck travelling at constant speed, the subscript P represent the police cruiser.
vT = 18 m/s
aP = 2.2 m/s2
viP = 0 m/s
(a) To calculate how far the cruiser travels before catching the truck, ∆dP = ∆dT.
∆d T = vT ∆t
∆t =
∆d T
vT
∆d P = viP ∆t +
∆d P =
1
a P ( ∆t ) 2
2
1
a P ( ∆t ) 2
2
Substitute for ∆t:
2
1  ∆d 
∆d P = aP  T  which is equal to
2  vT 
∆d P =
1  ∆d P 
aP 

2  vT 
2
2
=
2 vT
aP
=
2(18 m/s) 2
2.2 m/s2
∆d P = 2.9 × 102 m
The cruiser travels 2.9 × 102 m before catching the truck.
(b) ∆t = ?
∆d T
∆t =
vT
=
2.9 ×10 2 m
18 m/s
∆t = 16 s
The pursuit lasts 16 s.
Copyright © 2003 Nelson
Chapter 1 Kinematics
89
48. Let the +x direction be east and the +y direction be south.
vi = 80 km/h [E]
vf = 100 km/h [45° S of E]
a = 5.0 (km/h)/s
(a) vfx =?
vfy = ?
G
vfy = vf sin θ
G
vfx = vf cos θ
= (100 km/h)(sin 45°)
= (100 km/h)(cos 45°)
vfy = 71 km/h
vfx = 71 km/h
The direction of the acceleration is the same as the direction of the change of velocity.
G G G
∆v = vf − vi
∆vx = vfx − vix
1
= 70.7 km/h − 8.0 × 10 km/h
∆vx = −9.3 km/h
tan θ =
∆v x
∆v y
θ = tan −1
(9.3 km/h )
(70.7 km/h )
θ = 7.5°
The direction of the acceleration is [7.5° W of S].
(b) ∆t = ?
From (a) above, we can calculate the change in velocity:
∆v =
=
( ∆vx )2 + ( ∆v y )
2
( −9.3 km/h )2 + (70.7 km/h )2
∆v = 71 km/h
We can now calculate the time interval:
G
G ∆v
a=
∆t
G
∆v
∆t = G
a
71 km/h [7.5° W of S]
=
5.0 (km/h)/s [7.5° W of S]
∆t = 14 s
The time interval of the acceleration is 14 s.
49. Let +y be up, then ay = −g = −9.8 m/s2.
Begin by determining an expression for ∆t using the vertical component of the motion:
1
∆y = viy ∆t + a y ∆t 2
2
(
)
∆y = vi sin θ∆t − 4.9 m/s 2 ∆t 2
(4.9 m/s ) ∆t
2
90
Unit 1 Forces and Motion: Dynamics
2
− vi sin θ∆t + ∆y = 0
Copyright © 2003 Nelson
Using the quadratic formula to solve for ∆t:
∆t =
∆t =
−b ± b 2 − 4 ac
2a
vi sin θ ±
( −vi sin θ )2 − (19.9 m/s 2 ) ∆y
9.8 m/s 2
Substituting this equation for ∆t into the equation involving the horizontal component of the motion:
∆x = vix ∆t
∆x = vi cosθ∆t
 v sin θ ±
i
∆x = vi cosθ 



( −vi sin θ )2 − (19.9 m/s 2 ) ∆y 
9.8 m/s 2



50. Consider the motion in the frame of reference of the flowing river, which moves at a constant velocity downstream
relative to the shore. The sunbather swims directly away from the raft for 15 min at a constant speed, so it will take her
exactly the same length of time to swim directly back to the raft. Thus, the raft drifts for 30 min or 0.50 h while moving
(relative to the shore) 1.0 km downstream. Let R represent the river and S represent the shore.
G
∆d
G
vRS =
∆t
1.0 km [downstream]
=
0.50 h
G
vRS = 2.0 km/h [downstream]
The speed of the current in the river is 2.0 km/h. (A more complex solution from the frame of reference of Earth or the
shore provides the same answer. Refer to the SIN solutions book, question 75-10.)
Copyright © 2003 Nelson
Chapter 1 Kinematics
91
5. (a) The orientations are (i) J, (ii) I, and (iii) K.
(b)
Try This Activity: Predicting Forces
(Page 69)
If spring scales are used for this activity, they should be checked to be sure they are properly zeroed. Springs M and N must be
zeroed while held horizontally. In (b) and (c), the pulleys and strings should be as near to “frictionless” as possible.
(a) Predictions may vary. The readings on all 5 springs are equal, 9.8 N.
(b) If any students had differences, urge them to explain why.
2.1 FORCES AND FREE-BODY DIAGRAMS
PRACTICE
(Page 71)
Understanding Concepts
1.
Table 1 Common Forces
Name of Force
gravity
Type of Force
action-at-a-distance
normal
tension
friction
contact force
contact force
contact force
kinetic friction
static friction
contact force
contact force
air resistance
applied force
contact force
contact force
Example in Daily Life
force exerted by Earth on a ball dropped from your
hand
force exerted by the ground on your feet
force exerted by a dog on a leash
force acting between your shoes and the ground
that allows you to walk without slipping
force of the ice on a moving puck
horizontal force of the floor against a stationary desk
when a horizontal force is applied to the desk but
the desk remains at rest
friction of air molecules against a moving airplane
force applied to an object, such as a push to a
swing
2.
3.
The tension is still 18 N because the tension in a single cord is the same everywhere along the length of the cord.
Answers may vary. One way of rephrasing the statement is: “A rope is useful to apply a force on an object only when it is
under tension.”
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Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
PRACTICE
(Page 73)
Understanding Concepts
4.
5.
The FBD of the ball is the same in all three cases.
6.
7. (a)
Copyright © 2003 Nelson
Chapter 2 Dynamics
95
(b)
(c) The first diagram is more convenient because only one force needs to have its components shown.
Applying Inquiry Skills
8.
(b) The force of air resistance is upward and equal in magnitude to the downward force of gravity. If we estimate the mass
of the skydiver and parachute apparatus to be about 102 kg, we can apply the equation involving the gravitational field
strength (which students learned about in Grade 11 physics):
Fair = mg
= (10 2 kg)(9.8 N/kg)
Fair = 9.8 × 10 2 N
The air resistance is about 103 N [up].
PRACTICE
(Page 75)
Understanding Concepts
G
9. (a) Fapp = 3.74 N [up]
G
Fg =3.27 N [down]
G
Fair = 0.354 N [horizontally]
G G
G
ΣF = Fg + Fapp
Let +x be the horizontal direction of the motion and +y be up.
G
The components of Fapp are:
Fapp,x = 0.0 N
96
Fapp,y = 3.74 N
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
G
The components of Fg are:
Fgx = 0.0 N
Fgy = −3.27 N
G
The components of Fair are:
Fair,x = −0.354 N
Fair,y = 0.0 N
The components of the net force are:
ΣFx = Fapp,x + Fgx + Fair,x
ΣFx = −0.354 N
Thus,
ΣFy = Fapp,y + Fgy + Fair,y
= 3.74 N + ( −3.27 N ) + 0.0 N
ΣFy = 0.47 N
G
ΣF = ( −0.354 N) 2 + (0.47 N)2
G
ΣF = 0.59 N
θ = tan −1
0.47
0.354
θ = 53°
The net force on the bird is 0.59 N [53° above the horizontal].
G
(b) Fapp = 6382 N [28.3° above the horizontal]
G
Fg = 538 N [down]
G G
G
ΣF = Fg + Fapp
Let +x be opposite to the horizontal direction of the jumper’s motion and +y be up.
G
The components of Fapp are:
Fapp,x = (6382 N)(cos 28.3°)
Fapp,y = (6382 N)(sin 28.3°)
Fapp,x = 5619 N
Fapp,y = 3026 N
G
The components of Fg are:
Fgx = 0.0 N
Fgy = −538 N
The components of the net force are:
ΣFx = Fapp,x + Fgx
= 5619 N + 0.0 N
ΣFx = 5619 N
Thus,
ΣFy = Fapp,y + Fgy + Fair,y
= 3026 N + ( −538 N )
ΣFy = 2488 N
G
ΣF = (5619 N) 2 + (2488 N) 2
G
ΣF = 6.15 N
θ = tan −1
2488 N
5619 N
θ = 23.9°
The net force on the jumper is 6.15 × 103 N [23.9° above the horizontal].
G
(c) Fapp,1 = 412 N [27.0° W of N]
G
Fapp,2 = 478 N [36.0° N of E].
G G
G
ΣF = Fg + Fapp
Let +x be east and +y be north.
Copyright © 2003 Nelson
Chapter 2 Dynamics
97
G
The components of Fapp1 are:
Fapp1,x = (−412 N)(sin 27.0°)
Fapp1,y = (412 N)(cos 27.0°)
Fapp1,x = −187 N
Fapp1,y = 367 N
Fapp2,x = (478 N)(cos 36.0°)
Fapp2,y = (478 N)(sin 36.0°)
Fapp2,x = 387 N
Fapp2,y = 281 N
G
The components of Fapp2 are:
The components of the net force are:
ΣFx = Fapp1,x + Fapp2,x
ΣFy = Fapp1,y + Fapp2,y
= −187 N + 387 N
= 367 N + 281 N
ΣFx = 200 N
Thus,
ΣFy = 648 N
G
ΣF = (200 N) 2 + (648 N) 2
G
ΣF = 678 N
θ = tan −1
200 N
648 N
θ = 17.1°
The net force on the quarterback is 678 N [17.1° E of N].
G
10. (a) F1 = 48 N [16° N of E]
G
F2 = 48 N [16° S of E].
Let the +x direction be to the east and the +y direction be to the north.
G G G
ΣF = F1 + F2
G
The components of F1 are:
F1,y = (48 N)(sin 16°)
F1,x = (48 N)(cos 16°)
F1,x = 46 N
F1,y = 13 N
F2,x = (48 N)(cos 16°)
F2,y = (−48 N)(sin 16°)
F2,x = 46 N
F2,y = −13 N
G
The components of F2 are:
The components of the net force are:
ΣFx = F1,x + F2,x
= 46 N + 46 N
ΣFx = 92 N
ΣFy = F1,y + F2,y
= 13 N + (−13 N)
ΣFy = 0 N
Thus, the sum of the tension forces in the two ropes is 92 N [E].
(b) To solve Sample Problem 5(a) using the cosine law, we need to know the angle opposite the resultant force in
G
G
G
Figure 11 on page 75 in the text. In the triangle in that diagram, let side F1 be a, side F2 be b, side ΣF be c, and the
corresponding opposite angles be A, B, and C.
C = 180° − 16° − 16°
C = 148°
c 2 = a 2 + b2 − 2ab cos C
c = a 2 + b 2 − 2ab cos C
= (48 N) 2 + (48 N)2 − 2(48 N)(48 N)(cos148°)
c = 92 N
98
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Applying the sine law to find angle B:
sin B sin C
=
b
c
b sin C
sin B =
c
b sin C
B = sin −1
c
(48
N)(sin 148°)
= sin −1
92 N
B = 16°
Thus, the sum of the tension forces in the two ropes is 92 N [E].
G
11. ΣFapp = 56 N [16° S of E]
G
FT1 = 27 N [E]
G
G
G
ΣFapp = FT1 + FT2
Let +x be east and +y be south.
G
The components of Fapp are:
Fapp, x = (56 N)(cos 16°)
Fapp,y = (56 N)(sin 16°)
Fapp,x = 54 N
Fapp,y = 15 N
G
The components of FT1 are:
FT1,x = 27 N
FT1,y = 0.0 N
Thus,
ΣFapp,x = FT1,x + FT2,x
FT2,x = ΣFapp,x − FT1,x
= 54 N − 27 N
FT2,x = 27 N
ΣFapp,y = FT1,y + FT2,y
FT2,y = ΣFapp,y − FT1,y
= 15 N − 0.0 N
FT2,y = 15 N
G
ΣFT2 = (27 N) 2 + (15 N)2
G
ΣFT2 = 31 N
θ = tan −1
15 N
27 N
θ = 30°
The tension in cord 2 is 31 N [30° S of E].
Section 2.1 Questions
(Page 76)
Understanding Concepts
1. (a) The normal force of the desk on the ruler, the applied force of the hand, and the force of kinetic friction are the contact
forces acting on the ruler. Gravity is the noncontact force.
(b) The electromagnetic force is responsible for the contact forces. (The electric force is also an acceptable answer.)
Copyright © 2003 Nelson
Chapter 2 Dynamics
99
(c)
2.
3. (a) When the book is held stationary in your hand, the net force on the book is zero. The force of gravity is balanced by the
normal force of your hand on the book.
(b) If you suddenly remove your hand, the net force is 18 N [down] (the force of gravity, neglecting air resistance).
G
4. Fg = 1.5 N [down]
G
Fair = 0.50 N [32° above the horizontal]
G G
G
ΣF = Fg + Fair
Let +x be in the horizontal direction of the air resistance and +y be up.
G
The components of Fg are:
Fgx = 0.0 N
Fgy = −1.5 N
Fair,x = (0.50 N)(cos 32°)
Fair,y = (0.50 N)(sin 32°)
Fair,x = 0.42 N
Fair,y = 0.27 N
G
The components of Fair are:
100 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
The components of the net force are:
ΣFx = Fgx + Fair,x
ΣFy = Fgy + Fair,y
= 0.0 N + 0.42 N
= −1.5 N + 0.27 N
ΣFx = 0.42 N
Thus,
ΣFy = −1.24 N
G
ΣF = (0.42N) 2 + ( − 1.24 N) 2
G
ΣF = 1.3 N
θ = tan −1
0.42 N
1.22 N
θ = 19°
The net force on the ball is 1.3 N [71° below the horizontal].
G
5. FA = 3.6 N [28° W of S]
G
FB = 4.3 N [15° N of W]
G
FC = 2.1 N [24° E of S]
G
G
G
(a) FA + FB + FC , using a vector scale diagram
G
G
G
(b) FA + FB + FC , using components
Let +x be east and +y be north.
G
The components of FA are:
FAx = (−3.6 N)(sin 28°)
FAy = (−3.6 N)(cos 28°)
FAx = −1.7 N
FAy = −3.2 N
FBx = (−4.3 N)(cos 15°)
FBy = (4.3 N)(sin 15°)
FBx = −4.2 N
FBy = 1.1N
FCx = (2.1 N)(sin 24°)
FCx = (−2.1 N)(cos 24°)
FCx = 0.9 N
FCx = −1.9 N
G
The components of FB are:
G
The components of FC are:
The components of the net force are:
ΣFx = FAx + FBx + FCx
= −1.7 N − 4.2 N + 0.9 N
ΣFx = −5.0N
Copyright © 2003 Nelson
ΣFy = FAy + FBy + FCy
= −3.2 N + 1.1 N − 1.9 N
ΣFy = −4.0 N
Chapter 2 Dynamics 101
Thus,
G
ΣF = (−5.0N) 2 + ( − 4.0 N)2
G
ΣF = 6.4 N
ΣFy
θ = tan −1
= tan −1
ΣFx
4.0 N
5.0 N
θ = 39°
The net force is 6.4 N [39° S of W].
G
G
G
G
(c) FA − FB = FA + − FB , using a vector scale diagram
(
)
G
G
(d) FA − FB , using trigonometry
As in (b), let +x be east and +y be north.
The components of the net force are:
ΣFx = FAx − FBx
= −1.7 N − ( −4.2 N)
ΣFx = 2.5N
Thus,
ΣFy = FAy − FBy
= −3.2 N − 1.1 N
ΣFy = −4.3 N
G
ΣF = (2.5N)2 + ( − 4.3 N) 2
G
ΣF = 4.9 N
ΣFy
θ = tan −1
= tan −1
ΣFx
4.3 N
2.5 N
θ = 60° (to two significant digits)
6.
The net force is 4.9 N [60° S of E].
G
F1 = 36 N [25° N of E]
G
F2 = 42 N [15° E of S]
G
ΣF = 0 N
102 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Let +x be east and +y be north.
G
The components of F1 are:
F1x = (36 N)(cos 25°)
F1y = (36 N)(sin 25°)
F1x = 33 N
F1y = 15 N
F2x = (42 N)(sin 15°)
F2 y = −(4.3 N)(cos 15°)
F2x = 11N
F2 y = −41N
G
The components of F2 are:
Thus,
ΣFx = F1x + F2 x + F3 x = 0
F3 x = 0 N − F1x − F2 x
ΣFy = F1 y + F2 y + F3 y = 0
F3 y = 0 N − F1 y − F2 y
= 0 N − 33 N − 11N
F3 x = −44 N
= 0 N − 15 N + 41N
F3 y = 26 N
G
ΣF3 = (−44 N)2 + (26 N)2
G
ΣF3 = 50 N
θ = tan −1
= tan −1
ΣFy
ΣFx
26 N
44 N
θ = 30° (to two significant digits)
G
The force F3 that must be added is 5.0 × 101 N [30° N of W].
2.2 NEWTON’S LAWS OF MOTION
PRACTICE
(Page 80)
Understanding Concepts
G
1. (a) The upward lift force on the plane: Flift = 6.6 × 104 N [up]
G
(b) The force due to air resistance on the plane: Fair = 1.3 × 104 N [E]
2. The objects in (c) and (e) are not examples on Newton’s first law of motion. In both cases, the magnitude and direction of
the object’s speed are continually changing. In other words, these two objects are undergoing acceleration.
G
3. Since Fapp = 38 N, the magnitude of the force of friction must also be 38 N since the desk is not moving.
4.
The snowboarder will tend to maintain a constant velocity (according to the first law of motion) while the snowboard will
slow down rapidly. Thus, the snowboarder will likely fall to the snow ahead of the board.
5. The ball is in motion horizontally and that horizontal component will continue (as stated in the first law of motion). Thus,
the ball will not collide with the thrower. In this case, we assume that air resistance on the ball is negligible. (If the
thrower tried the same experiment while on a motorcycle, the result would be much different because of air resistance.)
G
6. In each case, the net force must be zero. Let the required force be FR .
(a) Let +x be east.
G
G G
FR + F1 + F2 = 0
G
G G
FR = − F1 + F2
(
Copyright © 2003 Nelson
)
Chapter 2 Dynamics 103
Let +x be east and +y be north.
G
The components of F1 are:
F1x = (36 N)(cos 25°)
F1y = (36 N)(sin 25°)
F1x = 33 N
F1y = 15 N
F2x = (42 N)(sin 15°)
F2 y = −(4.3 N)(cos 15°)
F2x = 11N
F2 y = −41N
G
The components of F2 are:
Thus,
ΣFx = F1x + F2 x + F3 x = 0
F3 x = 0 N − F1x − F2 x
ΣFy = F1 y + F2 y + F3 y = 0
F3 y = 0 N − F1 y − F2 y
= 0 N − 33 N − 11N
F3 x = −44 N
= 0 N − 15 N + 41N
F3 y = 26 N
G
ΣF3 = (−44 N)2 + (26 N)2
G
ΣF3 = 50 N
θ = tan −1
= tan −1
ΣFy
ΣFx
26 N
44 N
θ = 30° (to two significant digits)
G
The force F3 that must be added is 5.0 × 101 N [30° N of W].
2.2 NEWTON’S LAWS OF MOTION
PRACTICE
(Page 80)
Understanding Concepts
G
1. (a) The upward lift force on the plane: Flift = 6.6 × 104 N [up]
G
(b) The force due to air resistance on the plane: Fair = 1.3 × 104 N [E]
2. The objects in (c) and (e) are not examples on Newton’s first law of motion. In both cases, the magnitude and direction of
the object’s speed are continually changing. In other words, these two objects are undergoing acceleration.
G
3. Since Fapp = 38 N, the magnitude of the force of friction must also be 38 N since the desk is not moving.
4.
The snowboarder will tend to maintain a constant velocity (according to the first law of motion) while the snowboard will
slow down rapidly. Thus, the snowboarder will likely fall to the snow ahead of the board.
5. The ball is in motion horizontally and that horizontal component will continue (as stated in the first law of motion). Thus,
the ball will not collide with the thrower. In this case, we assume that air resistance on the ball is negligible. (If the
thrower tried the same experiment while on a motorcycle, the result would be much different because of air resistance.)
G
6. In each case, the net force must be zero. Let the required force be FR .
(a) Let +x be east.
G
G G
FR + F1 + F2 = 0
G
G G
FR = − F1 + F2
(
Copyright © 2003 Nelson
)
Chapter 2 Dynamics 103
FRx = − ( F1x + F2x )
= − ( 265 N − 122 N )
FRx = −143 N
G
Therefore, FR = 143 N [W] .
(b) Let +x be east and +y be north.
G
G G
FR + F1 + F2 = 0
G
G G
FR = − F1 + F2
(
)
(
FRx = − ( F1x + F2x )
FRy = − F1y + F2y
= − (0 N + 44 N )
)
= − (32 N + 0 N )
FRx = −44 N
FRy = −32 N
G
FR = FRx 2 + FRy 2
=
( −44 N )2 + ( −32 N )2
G
FR = 54 N
θ = tan −1
FRy
FRx
32 N
= tan −1
44 N
θ = 36°
G
Therefore, FR = 54 N [36° S of W] .
(c) Let +x be east and +y be north.
G
G G
G
FR + F1 + F2 + F3 = 0
G
G G
G
FR = − F1 + F2 + F3
(
)
(
FRx = − ( F1x + F2x + F3x )
FRy = − F1y + F2y + F3y
FRx = −2.0 N
FRy = −6.9 N
= − (6.5 N (sin25°) − 4.5 N + 3.9 N (cos15°) )
)
= − (6.5 N (cos25°) + 0 N + 3.9 N (sin15°) )
G
FR = FRx 2 + FRy 2
=
( −2.0 N )2 + ( −6.9 N )2
G
FR = 7.2 N
θ = tan −1
= tan −1
FRx
FRy
2.0 N
6.9 N
θ = 16°
G
Therefore, FR = 7.2 N [16° W of S] .
104 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
7.
m = 30.0 kg
Let +y be up.
10 m
= 5.0 m.
2
The vertical distance is ∆y = 0.40 m. The angle of the line with respect to the horizontal is:
∆y
θ = tan −1
∆x
0.40 m
= tan −1
5.0 m
The horizontal distance from the pole to the middle of the line is ∆x =
Vertically:
θ = 4.6°
ΣFy = 2 FT,y − Fg = 0
G
2 FT sin θ − Fg = 0
G
FT =
Fg
2(sin θ )
294 N
=
2(sin 4.6°)
G
FT = 1.8 ×103 N
The magnitude of the tension in the clothesline is 1.8 × 103 N.
Applying Inquiry Skills
8. (a) Place the coin on the paper on the desk and quickly jerk the paper horizontally. The coin, initially at rest, will remain at
rest relative to the desk because the net force acting on it is essentially zero. (The force of static friction would be
noticed if the paper is pulled slowly, but it is negligible if the paper is pulled quickly.)
(b) Answers will vary. One example is to place an object, such as a toy Teddy bear, on a dynamics cart, and send the cart
at a fairly high, yet safe, speed crashing into a brick barrier. The cart will stop but the Teddy bear, initially in motion,
will continue to move forward, crashing into the barrier.
Making Connections
9.
In any emergency, such as a collision of fast braking, any object in the rear window will continue moving forward, and
could easily injure any person in its path.
PRACTICE
(Page 83)
Understanding Concepts
10. m = 0.16 kg
ax = 32 m/s2
Fx = ma x
= (0.16 kg)(32 m/s 2 )
Fx = 5.1N
The magnitude of the force is 5.1 N.
11. m = 2.95 × 104 kg
G
ΣF = 2.42 × 104 N [fwd]
G
G ΣF
a=
m
2.42 × 104 N [fwd]
=
2.95 × 104 kg
G
a = 0.820 m/s 2 [fwd]
The truck’s acceleration while the force is applied is 0.820 m/s2 [fwd].
Copyright © 2003 Nelson
Chapter 2 Dynamics 105
G
G
12. ΣF = ma
G G
G v − vi
a= f
∆t
G G
G
 vf − vi 
ΣF = m 

 ∆t 
13. m = 7.27 kg
G
vi = 5.78 m/s [W]
G
vf = 4.61 m/s [W]
∆t = 1.2 × 10–3 s
G G
G
 v − vi 
ΣF = m  f

 ∆t 
(4.61 m/s [W] − 5.78 m/s [W])
= (7.27 kg)
1.2 × 10−3 s
= −7.1× 103 N [W]
G
ΣF = 7.1× 103 N [E]
The net force on the ball during the collision is 7.1 × 103 N [E].
Applying Inquiry Skills
14. The experiment involves measuring (or at least observing) the acceleration of an object as two variables, mass and net
force, are controlled to determine their effect on the acceleration. To determine the effect of changing the mass, attach the
elastic band to a single cart and, with the elastic band stretched a convenient amount, observe the cart’s acceleration.
Repeat the experiment using the same stretch of the elastic band but adding a second cart on top of the one being pulled.
Then repeat with the third cart atop the others. To determine the effect of changing the force, use one elastic band
stretched a small amount, then double that amount, and then triple it while pulling on a cart of constant mass. Care must
be taken to ensure that the carts do not fall to the floor or crash into other objects.
Making Connections
15. The method mentioned in the question is proposed in an abstract found at http://www.tsgc.utexas.edu/floatn/1999/99_fall/
teams/charleston.html. To help distinguish useful from non-useful minerals, objects of approximately equal size can be
separated according to their densities by accelerating them on a specially-designed device. General discussion about
mining on asteroids can be found at http://home.earthlink.net/∼nrunner/trav/tirem/belter.htm.
PRACTICE
(Page 84)
Understanding Concepts
G
16. In each case, the magnitude of the weight is Fg = Fg = mg .
(a) g = 9.8 N/kg
m = 2.4 kg
Fg = mg
= (2.4 kg)(9.8 N/kg)
Fg = 24 N
The magnitude of the weight of a horseshoe is 24 N.
(b) g = 9.8 N/kg
m = 1.3 × 106 kg
Fg = mg
= (1.3 × 106 kg)(9.8 N/kg)
Fg = 1.3 ×107 N
The magnitude of the weight of an open-pit coal-mining machine is 1.3 × 107 N.
106 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(c) g = 9.80 N/kg
m = 2.50 g
Fg = mg
= (2.50 × 10−3 kg)(9.80 N/kg)
Fg = 2.45 × 10 −2 N
The magnitude of the weight of a table tennis ball is 2.45 × 10−2 N.
(d) g = 9.80 N/kg
m = 1.81 × 10−9 kg
Fg = mg
= (1.81 ×10−9 kg)(9.80 N/kg)
Fg = 1.77 × 10−8 N
The magnitude of the weight of a speck of dust is 1.77 × 10−8 N.
(e) g = 9.8 N/kg
m = 55 kg
Fg = mg
= (55 kg)(9.8 N/kg)
Fg = 5.4 × 102 N
The magnitude of the weight of a 55-kg student would be 5.4 × 102 N.
G
17. (a) Fg = 1.53 N [down]
G
g = 9.80 N/kg [down]
G
G
Fg = mg
G
Fg
m= G
g
1.53 N [down]
=
9.80 N/kg [down]
m = 1.56 ×10 −1 kg
The mass of a field hockey ball is 1.56 × 10−1 kg.
G
(b) Fg = 1.16 × 106 N [down]
G
g = 9.80 N/kg [down]
G
Fg
m= G
g
=
1.16 × 106 N [down]
9.80 N/kg [down]
m = 1.18 × 105 kg
The mass of the payload capacity of a C-5 Galaxy cargo plane is 1.18 × 105 kg.
18. m = 76 kg
G
g = 3.7 N/kg [down]
G
G
Fg = mg
= (76 kg)(3.7 N/kg[down])
G
Fg = 2.8 ×102 N [down]
The weight of the astronaut is 2.8 × 102 N [down].
Copyright © 2003 Nelson
Chapter 2 Dynamics 107
Applying Inquiry Skills
N
m
=
19. The units N/kg and m/s are equivalent since 1 N = 1 kg  2  . Thus,
kg
s 
2
m
kg  2 
s  = m .
kg
s2
PRACTICE
(Page 86)
Understanding Concepts
20. (a) The action force is the downward force of the rocket engine on the expanding gases as they exit the bottom of the
rocket. The reaction force is the upward force of the expanding gases on the engine (and thus the rocket), causing the
rocket to accelerate forward.
(b) The action force is the downward force of the rotating propeller blades on the air. The reaction force is the upward
force of the air on the rotating propeller blades.
(c) The action force is the force of the interior walls of the east end of the balloon pushing westward on the air as it exits
the west end of the balloon. The reaction force is the force of the air eastward on the wall of the balloon, pushing the
balloon eastward.
21. (a)
(b) There are two forces acting on the pencil, and those forces are equal in magnitude but opposite in direction. Since the
net force is zero, no acceleration occurs.
Applying Inquiry Skills
22. Answers will vary. Any action-reaction toys are acceptable, as long as they are demonstrated safely. Examples include a
balloon, a water rocket (for outdoor use only), a toy projectile launcher that jerks backward when the projectile is
launched forward, and a toy car or truck with wind-up or battery-powered propulsion to show action-reaction forces
between the wheels and the floor. (In the last example, the forces can be shown more readily if the wheels are “wound up”
before placing the vehicle gently down on a piece of cardboard that is resting on several straws or round pencils. The
vehicle moves forward and the cardboard moves backward.)
108 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Section 2.2 Questions
(Page 87)
Understanding Concepts
1.
2.
Since the mallard duck is flying at a constant velocity, the net force acting on the duck must be 0 N.
Let +y be up.
ΣFy = FN − Fg = 0
FN = Fg
G
= mg
= (1.9 kg)(9.8 m/s 2 )
3.
FN = 19 N
G
FN = 19 N [up]
The normal force acting on a stationary 1.9-kg carton of juice is 19 N [up].
m = 67 kg
G
v = 85 cm/s [down]
Let +y be up.
Since the velocity is constant, ΣFy = 0 . Thus,
ΣFy = FN − Fg = 0
FN = Fg
G
= mg
= (67 kg)(9.8 m/s 2 )
4.
FN = 6.6 ×102 N
G
FN = 6.6 × 102 N [up]
The normal force exerted by the cherry picker on the worker is 6.6 × 102 N [up].
Let +x be the direction of the net force and the acceleration.
me = 9.11 × 10−31 kg
ΣFx = 3.20 × 10−15 N
ΣFx
m
3.20 × 10 −15 N
=
9.11 ×10 −31 kg
ax =
a x = 3.51× 1015 m/s 2
The magnitude of the resulting acceleration of the electron is 3.51 × 1015 m/s2.
G
5. vf =0 m/s
G
vi = 13 m/s [down]
∆t = 3.0 ms = 3.0 × 10–3 s
G G
G v − vi
(a) a = f
∆t
0 m/s − 13 m/s[down]
=
3.0 × 10−3 s
G
a = 4.3 ×103 m/s 2 [up]
The acceleration of the hand is 4.3 × 103 m/s2 [up].
Copyright © 2003 Nelson
Chapter 2 Dynamics 109
(b) m = 0.65 kg
G
G
ΣF = ma
= (0.65 kg)(4.3 × 103 m/s 2 [up])
G
ΣF = 2.8 × 103 N [up]
The net force acting on the hand is 2.8 × 103 N [up], which is exerted by the brick. (Notice that the force of gravity on
the hand has been ignored because it is negligible compared to the force of the brick on the hand.)
(c) m = 65 kg.
G
G
ΣF = mg
= (65 kg)(9.8 N/kg[down])
G
ΣF = 6.4 × 102 N [down]
G
ΣFbrick
2.8 × 103 N [up]
G
=
ΣFexpert 6.4 × 102 N [down]
G
ΣFbrick
G
= 4.4
ΣFexpert
6.
Thus, the ratio of the magnitude of the force of the brick to the magnitude of the expert’s weight is 4.4:1.
Let +x be the direction of the net force and the acceleration.
ΣFx = 1.24 ×102 N
ax = 4.43 × 103 m/s2
ΣF
m= x
ax
=
1.24 × 102 N
4.43 ×103 m/s 2
= 0.0280 kg
7.
m = 28.0 g
The mass of the arrow is 28.0 g.
G
g = 8.9 N/kg
m = 55 kg (assumed mass of student)
G
G
(a) Fg = m g
= (55 kg)(8.9 N/kg)
G
Fg = 4.9 ×10 2 N
The magnitude of a 55-kg student’s weight on Venus is 4.9 × 102 N.
(b) On Earth:
G
G
Fg = m g
= (55 kg)(9.8 N/kg)
G
Fg = 5.4 ×102 N
G
Fg,Venus
4.9 ×102 N
=
G
2
Fg,Venus 5.4 × 10 N
G
Fg,Venus
= 0.91
G
Fg,Venus
The percentage change in weight is (1.00 – 0.91) × 100% = 9%. Since the weight on Venus is less than the weight on
Earth, a 55-kg student’s weight on Venus would decrease by 9% compared to its magnitude on Earth.
8. (a) The baking pan exerts a pulling force on the chef toward the oven.
(b) Saturn exerts a gravitational force on the Sun toward Saturn.
(c) The water exerts a forward normal force on the hands.
110 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(d) The watermelon exerts an upward gravitational force on Earth.
(e) The hailstone exerts a downward applied or normal force on the air.
9. m = 0.200 kg
G
g = 9.8 N/kg
G
G
Fg = m g
= (0.200 kg)(9.80 N/kg)
G
Fg = 1.96 N
This downward force of gravity on each bag is balanced by the tension in the string. This tension in turn equals the
reading on the scale, 1.96 N.
Applying Inquiry Skills
10. (a) Answers will vary. The example used here is 35 g or 3.5 × 10−2 kg.
(b) m = 3.5 × 10−2 kg
G
g = 9.8 N/kg [down]
G
G
Fg = mg
(
)
= 3.5 × 10−2 kg (9.8 N/kg )
G
Fg = 34 N [down]
(c) Answers will vary. For this example, assume the value is 45 N [down].
(d) accepted value = 45 N [down]
estimated value = 34 N [down]
estimated value − accepted value
% error =
× 100%
accepted value
45 N − 34 N
=
45 N
% error = 24%
Making Connections
11. (a) The astronaut, like all objects aboard an orbiting vehicle, is undergoing constant free fall, so there can be no normal
force pushing up from a bathroom scale, a floor, a chair, or a wall.
G
(b) ΣF = 87 N [fwd]
G
a = 1.5 m/s2 [fwd]
G
G
ΣF = ma
G
ΣF
m= G
a
87 N [fwd]
=
1.5 m/s 2 [fwd]
m = 58 kg
The mass of the astronaut is 58 kg.
(c) ∆t = 1.2 s
G G
1G
2
∆d = vi ∆t + a ( ∆t )
2
1
2
= 0 + 1.5 m/s 2 [fwd] (1.2 s )
2
G
∆d = 1.1 m [fwd]
The astronaut moved 1.1 m.
(d) To find out more about the inertial device, students can refer to a variety of NASA sites, two of which are listed here:
http:/nasaexplores.com/lessons/o1-044/9-12_1.html
http://www-istp.gsfc.nasa.gov/stargaze/Smass.htm
(
Copyright © 2003 Nelson
)
Chapter 2 Dynamics 111
12. There are numerous resources on the life and times of Isaac Newton, both in resource centres and on the Internet. Students
can work in groups, with each member discovering something different to share with the other members.
2.3 APPLYING NEWTON’S LAWS OF MOTION
PRACTICE
(Page 92)
Understanding Concepts
1.
G
Fapp = 0.35 N [up]
G
a = 0.15 m/s2 [up]
Let +y be up.
G
G
ΣF = ma
ΣFy = ma y
Fapp + (− Fg ) = ma y
Fapp − mg = ma y
m=
=
Fapp
ay + g
0.35 N
0.15 m/s 2 + 9.8 m/s 2
m = 3.5 × 10−2 kg
2.
The mass of the fork is 35 g.
G
a = 1.10 m/s2 [down]
m = 315 kg
Let +y be up.
G
G
(a)
ΣF = ma
ΣFy = ma y
FB + (− Fg ) = −ma y
FB − mg = −ma y
FB = m( g − a y )
= (315 kg)(9.80 m/s 2 − 1.10 m/s 2 )
FB = 2.74 × 103 N
The upward (buoyant) force on the balloon, basket, and its contents is 2.74 × 103 N [up].
ΣFy = ma y
(b)
FB + (− Fg ) = 0
FB = mg
m=
=
FB
g
2.74 × 103 N
9.80 m/s 2
m = 280 kg (3 significant digits)
The mass of the ballast that must be discarded overboard is 315 kg – 280 kg = 35 kg.
112 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
12. There are numerous resources on the life and times of Isaac Newton, both in resource centres and on the Internet. Students
can work in groups, with each member discovering something different to share with the other members.
2.3 APPLYING NEWTON’S LAWS OF MOTION
PRACTICE
(Page 92)
Understanding Concepts
1.
G
Fapp = 0.35 N [up]
G
a = 0.15 m/s2 [up]
Let +y be up.
G
G
ΣF = ma
ΣFy = ma y
Fapp + (− Fg ) = ma y
Fapp − mg = ma y
m=
=
Fapp
ay + g
0.35 N
0.15 m/s 2 + 9.8 m/s 2
m = 3.5 × 10−2 kg
2.
The mass of the fork is 35 g.
G
a = 1.10 m/s2 [down]
m = 315 kg
Let +y be up.
G
G
(a)
ΣF = ma
ΣFy = ma y
FB + (− Fg ) = −ma y
FB − mg = −ma y
FB = m( g − a y )
= (315 kg)(9.80 m/s 2 − 1.10 m/s 2 )
FB = 2.74 × 103 N
The upward (buoyant) force on the balloon, basket, and its contents is 2.74 × 103 N [up].
ΣFy = ma y
(b)
FB + (− Fg ) = 0
FB = mg
m=
=
FB
g
2.74 × 103 N
9.80 m/s 2
m = 280 kg (3 significant digits)
The mass of the ballast that must be discarded overboard is 315 kg – 280 kg = 35 kg.
112 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
3. (a) vi = 0 m/s
∆t = 2.00 s
G
∆d = 3.10 m [down]
m = 35.7 kg
Let +y be down.
∆y = vi ∆t +
1
a y ( ∆t ) 2
2
2 ∆y
( ∆t ) 2
2(3.10 m)
=
(2.00 s)2
ay =
a y = 1.55 m/s 2
The girl’s (constant) downward acceleration is 1.55 m/s2.
(b) Let +y be down.
ΣFy = may
Fg − Ff = ma y
Ff = Fg − ma y
= mg − ma y
= m( g − a y )
= (35.7 kg)(9.80 m/s 2 − 1.55 m/s 2 )
4.
Ff = 295 N
The upward force of friction exerted by the pole on the girl is 295 N.
Let m be the unknown mass and m2 be the second mass.
m2 = m + 2.0 kg
a2 = 0.37a
ΣFx = ma x
ΣFx = m2 a2 x
ΣFx = m2 (0.37 a x )
ma x = m2 (0.37a x )
m = 0.37 m2
m = 0.37(m + 2.0 kg)
m − 0.37 m = 0.74 kg
0.63m = 2.0 kg
m = 1.2 kg
5.
The mass m is 1.2 kg.
FK = 5.7 N
mA = 2.7 kg
mB = 3.7 kg
Let the magnitude of the acceleration of the blocks be ay. For block B let +y be down and for block A let +y be to the
right.
Copyright © 2003 Nelson
Chapter 2 Dynamics 113
(a)
ΣFy = ma y
FgB − FT = mB a y
FT − Ff = mA a y
FgB − Ff = (mA + mB )a y
ay =
=
mB g − Ff
mA + mB
(3.7 kg)(9.8 m/s 2 ) − 5.7 N
(2.7 kg + 3.7 kg)
a y = 4.8 m/s 2
The magnitude of the acceleration of the blocks is 4.8 m/s2.
(b) To calculate the tension in the string, use either equation in part (a):
FT − Ff = mA a y
FT = mA a y + Ff
= (2.7 kg)(4.8 m/s 2 ) + 5.7 N
6.
FT = 19 N
The magnitude of the tension is 19 N.
m = 17.9 kg
Fapp = 32.9 N [35.1° above the horizontal]
a = 1.37 m/s2
∆t = 0.58 s
G
vi = 0 m/s
Let +x be in the horizontal direction of the applied force and +y be down.
(a)
ΣFy = 0
Fapp,y + Fg − FN = 0
FN = Fapp,y + Fg
= Fapp sin θ + mg
= (32.9 N) sin 35.1° + (17.9 kg)(9.80 m/s 2 )
FN = 194 N
The magnitude of the normal force on the mower is 194 N.
(b)
ΣFx = ma x
Fapp,x − Ff = ma x
Ff = Fapp,x − ma x
= Fapp cos θ + ma x
= (32.9 N) cos 35.1° + (17.9 kg)(1.37 m/s 2 )
Ff = 2.4 N
The magnitude of the frictional force on the mower is 2.4 N.
v − vi
(c) a x = f
∆t
vf = vi + ax ∆t
= 0 + (1.37 m/s 2 )(0.58s)
vf = 0.79 m/s
The magnitude of the maximum velocity of the mower is 0.79 m/s.
114 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(d)
ΣFx = 0
Fapp,x − Ff = 0
Fapp,x = Ff
Fapp cos θ = Ff
Ff
cos θ
2.4 N
=
cos 35.1°
= 2.9 N
Fapp =
Fapp
7.
The magnitude of the force applied by the boy to maintain the constant velocity is 2.9 N.
m = 65 kg
θ = 12°
(a) Let the +x direction be downward parallel to the hillside and the +y direction be downward, parallel to the hillside.
ΣFy = 0
mg cos θ − FN = 0
FN = mg cos θ
= (65 kg)(9.8 m/s 2 ) sin12°
FN = 6.2 × 102 N
The magnitude of the normal force on the skier is 6.2 × 102 N.
(b)
ΣFx = ma x
mg sin θ = ma x
a x = g sin θ
= (9.8 m/s 2 ) sin12°
a x = 2.0 m/s 2
The magnitude of the skier’s acceleration is 2.0 m/s2.
Applying Inquiry Skills
8. (a)
(b) With +y up, the equation for the magnitude of the normal force on the block in terms of the given parameters FA, g, m,
and θ is as follows:
ΣFy = 0
FA sin θ + FN − Fg = 0
FN = mg − FA sin θ
(c) With +x in the horizontal direction of the applied force, the equation for the magnitude of friction on the block in terms
of FA and θ is as follows.
ΣFx = 0
FA cos θ − Ff = 0
Ff = FA cos θ
Copyright © 2003 Nelson
Chapter 2 Dynamics 115
Making Connections
9.
m = 55.3 kg
(a) Let +y be up.
G
a = 1.08 m/s2 [up]
ΣFy = ma y
FN − Fg = ma y
FN = m( g + a y )
= (55.3kg)(9.80 m/s 2 + 1.08 m/s 2 )
FN = 602 N
The reading on the scale is 602 N.
(b) The student’s true weight (Fg = mg = 542 N) is less than the apparent weight when the acceleration is upward. The
apparent weight is less than the true weight when the acceleration is downward. When the elevator is undergoing free
fall, the apparent weight is zero since there is no normal force.
(c) Let +y be down.
G
a1 = 1.08 m/s2 [down]
ΣFy = ma y
Fg − FN = ma y
FN = m( g − a y )
FN = (55.3kg)(9.80 m/s 2 − 1.08 m/s 2 )
FN = 482 N
G
2
If a2 = 9.80 m/s [down], then
FN = (55.3kg)(9.80 m/s 2 − 9.80 m/s 2 )
FN = 0 N
Thus, at 1.08 m/s2 [down], the student’s apparent weight is 482 N and at 9.80 m/s2 [down], the student’s apparent
weight is 0 N.
(d) The term “weightless” is used because that is what a person feels during free-fall acceleration. The term is not valid
from the physics point of view because there is still weight (i.e., the force of gravity) acting on the body in free fall.
(e) Let +y be up.
G
v = 1.08 m/s [up]
ΣFy = 0
FN − Fg = 0
FN = mg
= (55.3kg)(9.80 m/s 2 )
FN = 542 N
The reading on the scale is 542 N.
PRACTICE
(Page 94)
Understanding Concepts
G
10. a = 0.33 m/s2 [fwd]
m1 = m2 = 3.1 × 104 kg
Let the +x direction be the direction of the acceleration.
116 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(a) ΣFx = m2 a x
Fapp = m2 a x
= (3.1× 104 kg)(0.33m/s 2 )
Fapp = 1.0 × 104 N
The force exerted by the first car on the second car is 1.0 × 104 N [fwd].
(b) ΣFx = ( m1 + m2 ) ax
Fapp = 2m2 a x
= 2(3.1× 104 kg)(0.33 m/s 2 )
Fapp = 2.0 ×10 4 N
The force exerted by the locomotive on the first car is 2.0 × 104 N [fwd].
11. Fapp = 0.58 N
G
a = 0.21 m/s2 [horizontally]
m1 = 1.0 kg
Let the +x direction be the direction of the acceleration.
(a) ΣFx = ( m1 + m2 ) ax
Fapp = m1a x + m2 a x
m2 =
=
Fapp − m1a x
ax
0.58 N − (1.0 kg)(0.21m/s 2 )
0.21m/s 2
m2 = 1.8 kg
The mass of the second book is 1.8 kg.
(b) ΣFx = m2 a x
Fapp = (1.8 kg)(0.21m/s 2 )
Fapp = 0.37 N
The magnitude of the force exerted by one book on the other is 0.37 N.
Section 2.3 Questions
(Pages 95–96)
Understanding Concepts
1.
All the FBDs are identical since there is only one force acting on the basketball.
2. The net force acting on the shark is zero since the acceleration is zero (constant velocity).
3. (a) The bottom mass
ΣFy = 0
FT − m3 g = 0
FT = m3 g
= (1.00 kg)(9.80 m/s 2 )
FT = 9.80 N
The magnitude of tension in the lowest thread is 9.80 N.
Copyright © 2003 Nelson
Chapter 2 Dynamics 117
(b) The middle mass
ΣFy = 0
FT − (m2 + m3 ) g = 0
FT = ( m2 + m3 ) g
= (2.00 kg +1.00 kg)(9.80 m/s 2 )
FT = 29.4 N
The magnitude of tension in the middle thread is 29.4 N.
(c) The top mass
ΣFy = 0
FT − (m1 + m2 + m3 ) g = 0
FT = (m1 + m2 + m3 ) g
= (5.00 kg +2.00 kg +1.00 kg)(9.80 m/s 2 )
FT = 78.4 N
4.
The magnitude of tension in the highest thread is 78.4 N.
G
a = 0.50 g [up]
m = 2.0 × 106 kg.
Let +y be up.
ΣFy = ma y
(a)
Fapp − mg = ma y
Fapp = m( g + a y )
= m( g + 0.5 g )
= 1.5(2.0 × 106 kg)(9.8 m/s 2 )
Fapp = 2.9 × 107 N
The approximate magnitude of the upward force on the shuttle is 2.9 × 107 N.
(b) The upward force is caused by the gases as they are expelled from the base of the rocket engine. This force is the
reaction force to the action force of the engine pushing downward on the expanding gases.
118 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
5.
m1 = 35 kg
m2 = 45 kg
Let +y be up for m1 and down for m2 (as the rope slides over the metal rod).
ΣFy = ma y
(a)
FT − m1 g = m1a y
m2 g − FT = m2 a y
Adding these two equations gives:
m2 g − m1 g = m1a y + m2 a y
ay =
=
(m2 − m1 ) g
(m1 + m2 )
(45 kg − 35 kg)(9.8 m/s 2 )
(45 kg + 35 kg)
a y = 1.2 m/s 2
The magnitude of the acceleration of the boxes is 1.2 m/s2.
(b) Substituting ay into either of the equations in part (a) gives:
FT − m1 g = m1a y
FT = m1 ( g + a y )
= (35 kg)(9.8 m/s 2 + 1.2 m/s 2 )
FT = 3.9 × 102 N
The magnitude of the tension in the rope is 3.9 × 102 N.
(c) ∆t = 0.50 s
1
∆d = vi ∆t + a y (∆t ) 2
2
1
= (1.2 m/s 2 )(0.5s)2
2
∆d = 0.15 m
The magnitude of each box’s displacement is 0.15 m.
6. FfA = 1.8 N
FgA = 6.7 N
FgB = 2.5 N
(a) Block B
ΣFy = 0
FT1 − FgB = 0
FT1 = FgB
FT1 = 2.5 N
The magnitude of the tension in the vertical rope is 2.5 N.
(b) Block A
ΣFx = 0
FT2 − FfA = 0
FT2 = FfA
FT2 = 1.8 N
The magnitude of the tension in the horizontal rope is 1.8 N.
Copyright © 2003 Nelson
Chapter 2 Dynamics 119
ΣFy = 0
FNA − FgA = 0
FNA = FgA
FNA = 6.7 N
The magnitude of the normal force acting on block A is 6.7 N.
(c) Point P
ΣFy = 0
ΣFx = 0
FT3 sin θ − FT1 = 0
FT3 cos θ − FT2 = 0
FT3 =
FT1
sin θ
FT3 =
FT2
cos θ
FT1
F
= T2
sin θ cos θ
F
sin θ
= T1
cos θ FT2
tan θ =
2.5 N
1.8 N
θ = 54°
Substitute angle into either equation for FT3:
F
FT3 = T1
sin θ
2.5 N
=
sin 54°
FT3 = 3.1N
7.
The tension in the third rope is 3.1 N [54° above the horizontal].
m1 = 15.0 kg
m2 = 13.2 kg
m3 = 16.1 kg
Let +x be to the right.
(a)
ΣFx = ( m1 + m2 + m3 ) ax
FT cos θ = ( m1 + m2 + m3 ) a x
ax =
=
FT cos θ
(m1 + m2 + m3 )
(35.3 N) cos 21.0°
(15.0 kg + 13.2 kg + 16.1 kg)
a x = 0.744 m/s 2
The magnitude of the acceleration of the carts is 0.744 m/s2.
120 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b) ΣFx = m3 a x
FT3 = m3 ax
= (16.1 kg)(0.744 m/s 2 )
FT3 = 12.0 N
The magnitude of the tension in the last cord is 12.0 N.
(c) ΣFx = ( m2 + m3 ) ax
FT2 = (m2 + m3 )a x
= (13.2 kg + 16.1kg)(0.744 m/s 2 )
FT2 = 21.8 N
The magnitude of the tension in the middle cord is 21.8 N.
G
8. Fapp = 91 N [15° above the horizontal]
G
FN = 221 N [up]
G
a = 0.076 m/s2 [fwd]
(a) Let +y be up.
ΣFy = 0
FN + Fapp sin θ − mg = 0
m=
FN + Fapp sin θ
g
221 N + (91 N) sin15°
=
9.8 m/s 2
m = 25 kg
The mass of the chair is 25 kg.
(b) Let +x be the direction of the acceleration.
ΣFx = max
Fapp cos θ − Ff = max
Ff = Fapp cos θ − max
= (91N) cos15° − (25 kg)(0.076 m/s 2 )
9.
Ff = 86 N
The magnitude of the friction force on the chair is 86 N.
ax = 1.5 m/s2
Let +x be downward, parallel to the hillside.
ΣFx = ma x
mg sin θ = ma x
sin θ =
ax
g
 1.5 m/s 2 
θ = sin −1 
2 
 9.8 m/s 
θ = 8.8°
The angle between the hill and the horizontal is 8.8°.
10. mX = 5.12 kg
mY = 3.22 kg
(a) For block X:
Let the +x direction be defined as up the slope parallel to the slope.
ΣFx = ma
FT − mX g sin θ = mX a
FT = mX ( g sin θ + a )
Copyright © 2003 Nelson
Chapter 2 Dynamics 121
For block Y:
Let the +y direction be defined as vertically down.
ΣFy = ma
mY g − FT = mY a
FT = mY ( g − a )
Since the two equations for FT are equal:
mX ( g sin θ + a ) = mY ( g − a )
mX a + mY a = mY g − mX g sin θ
a=
=
(mY − mX sin θ ) g
mX + mY
(3.22 kg − (5.12 kg)sin35.7°)(9.80 m/s 2 )
5.12 kg + 3.22 kg
a = 0.273m/s 2
The magnitude of the acceleration is 0.273 m/s2.
(b) Substitute a into either equation for tension:
FT = mY ( g − a)
= (3.22 kg)(9.80 m/s 2 − 0.273 m/s 2 )
FT = 30.7 N
The magnitude of the tension in the fishing line is 30.7 N.
11. m = 56 kg
∆t = 0.75 s
vfx = 75 cm/s
vix = 0 cm/s
(a) Let +x be the direction of the acceleration away from the boards.
v −v
a x = fx ix
∆t
0.75 m/s − 0 m/s
=
0.75s
a x = 1.0 m/s 2
The magnitude of the skater’s acceleration is 1.0 m/s2.
(b) Let +x be opposite to the direction of the acceleration, that is, toward the boards.
Fx = ma x
= (56 kg)(1.0 m/s 2 )
Fx = 56 N
The magnitude of the force the skater exerted on the boards is 56 N.
(c) The magnitude of the force the boards exerted on the skater is 56 N.
(d) The figure skater accelerates for ∆t1 = 0.75 s then travels at a constant velocity (vfx) for ∆t2 = 0.75 s.
1
∆d1 = vix ∆t1 + a(∆t1 )2
2
∆d 2 = vfx ∆t2
1
2
2
= (1.0 m/s )(0.75s)
= (0.75 m/s)(0.75s)
2
∆d 2 = 0.56 m
∆d1 = 0.28 m
∆d = ∆d1 + ∆d 2
= 0.28 m + 0.56 m
∆d = 0.84 m
The magnitude of the skater’s displacement from the boards after 1.50 s is 0.84 m.
122 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Applying Inquiry Skills
G 1
12. (a) Acceleration versus mass: a ∝
m
G
G
(b) Acceleration versus net force: a ∝ ΣF
Making Connections
13. In a front-end collision, even at a relatively low speed, the airbags deploy very quickly, and they are capable of exerting a
strong force on the driver or passenger in the front seat. Any passenger not wearing a seatbelt will be moving forward and
will continue to do so after the collision, resulting in a greater impact with the deployed airbag. This could result in
serious injury or even death. In the case of a child, it is the child’s head that may impact with the airbag, which would be
even more dangerous than a chest impact.
2.4 EXPLORING FRICTIONAL FORCES
PRACTICE
(Page 97)
Understanding Concepts
1.
Table 1 lists examples of situations when it would be advantageous to have increased friction and decreased friction.
Table 1
Increased Friction
basketball shoes, mountain bike wheels, mountain
climbing equipment, safety boots for roofing installers,
vehicle braking systems
Decreased Friction
cookware (cake pans, cookie sheets), rollerblade
wheels, racing bicycle wheels, truck engine,
airplanes
Try This Activity: Observing Triboluminescence
(Page 101)
Viewing the flashes of light produced when the crystals are crushed is much better in a very dark room. It is important that the
viewer’s eyes be dark-adapted since the light emitted is very dim. Some students may want to try to crush the WintOGreen
Lifesaver in their partially open mouth, but for safety reasons, that is not recommended. The light is caused by the excitation
of molecules due to electric charge differences on the planes of the crystals as the crystals are crushed against each other. For
more information, students can refer to various Web sites, one of which is given here:
http://www.geocities.com/RainForest/9911/tribo.htm
Copyright © 2003 Nelson
Chapter 2 Dynamics 123
Applying Inquiry Skills
G 1
12. (a) Acceleration versus mass: a ∝
m
G
G
(b) Acceleration versus net force: a ∝ ΣF
Making Connections
13. In a front-end collision, even at a relatively low speed, the airbags deploy very quickly, and they are capable of exerting a
strong force on the driver or passenger in the front seat. Any passenger not wearing a seatbelt will be moving forward and
will continue to do so after the collision, resulting in a greater impact with the deployed airbag. This could result in
serious injury or even death. In the case of a child, it is the child’s head that may impact with the airbag, which would be
even more dangerous than a chest impact.
2.4 EXPLORING FRICTIONAL FORCES
PRACTICE
(Page 97)
Understanding Concepts
1.
Table 1 lists examples of situations when it would be advantageous to have increased friction and decreased friction.
Table 1
Increased Friction
basketball shoes, mountain bike wheels, mountain
climbing equipment, safety boots for roofing installers,
vehicle braking systems
Decreased Friction
cookware (cake pans, cookie sheets), rollerblade
wheels, racing bicycle wheels, truck engine,
airplanes
Try This Activity: Observing Triboluminescence
(Page 101)
Viewing the flashes of light produced when the crystals are crushed is much better in a very dark room. It is important that the
viewer’s eyes be dark-adapted since the light emitted is very dim. Some students may want to try to crush the WintOGreen
Lifesaver in their partially open mouth, but for safety reasons, that is not recommended. The light is caused by the excitation
of molecules due to electric charge differences on the planes of the crystals as the crystals are crushed against each other. For
more information, students can refer to various Web sites, one of which is given here:
http://www.geocities.com/RainForest/9911/tribo.htm
Copyright © 2003 Nelson
Chapter 2 Dynamics 123
PRACTICE
(Page 101)
Understanding Concepts
2. (a) The direction of the frictional force is southward on the tires. This force is a reaction force caused as the tires apply a
force of friction (the action force) northward against the road.
(b) The force is static friction, assuming the tires do not spin on the road surface.
3. m = 23 kg
µS = 0.43
µK = 0.36
(a) Let +y be up.
ΣFy = ma y = 0
FN + (−mg ) = 0
FN = mg
= (23 kg)(9.8 N/kg)
FN = 225 N
Now determine the magnitude of the maximum static friction:
FS, max = µS FN
= (0.43)(225 N)
FS, max = 97 N
Thus, the magnitude of the minimum horizontal force needed to set the mat in motion is 97 N.
(b) Let +x be the direction of the applied force, Fapp,x.
ΣFx = max = 0
Fapp, x − FK = 0
Fapp, x = FK
Fapp, x = µ K FN
= (0.36)(225 N)
Fapp, x = 81 N
4.
A horizontal force of magnitude 81 N will keep the mat moving at a constant velocity.
G
Fapp = 17 N [W]
m = 5.1 kg
G
a = 0.39 m/s2 [W]
Let +x be the direction of the acceleration.
ΣFx = ma x
Fapp + (− FK ) = ma x
FK = Fapp − ma x
= 17 N − (5.1 kg)(0.39 m/s 2 )
FK = 15 N
µK =
=
FK
FN
15 N
(5.1kg)(9.8 m/s 2 )
µK = 0.30
The coefficient of kinetic friction between the case and the table is 0.30.
124 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
5. (a)
(b) Static friction between the two boxes causes the small box to accelerate horizontally.
(c) Let +x be the direction of the acceleration of the boxes and +y be up, and let m be the mass of the smaller box.
In the vertical direction:
ΣFy = ma y = 0
FN − Fg = 0
FN = Fg
FN = mg
In the horizontal direction:
ΣFx = ma x
FS,max = ma x
µS FN = max
µS mg = max
ma x
µS =
mg
a
= x
g
2.5 m/s 2
9.8 m/s 2
µS = 0.26
The smallest coefficient of static friction is 0.26.
=
6.
Copyright © 2003 Nelson
Chapter 2 Dynamics 125
7.
m = 47 kg
θ = 23°
µK = 0.11
Let +x be the direction of the velocity and +y be up.
ΣFx = 0
ΣFy = 0
FT cos θ + (− Ff ) = 0
FN + FT sin θ + (−mg ) = 0
FT cos θ = µK FN
FN = mg − FT sin θ
FN =
FT cos θ
µK
Equate two equations for the normal force:
mg − FT sin θ =
FT cos θ
µK
 cos θ

+ sin θ  = mg
FT 
 µK

FT =
mg
 cos θ

+ sin θ 

µ
 K

(47 kg)(9.8 m/s 2 )
 cos 23°

+ sin 23° 

 0.11

FT = 53 N
Thus, the magnitude of the tension in the rope is 53 N.
=
Applying Inquiry Skills
8. (a) Using a straight board that can be raised at one end, students can place the runner at one end of the board and raise that
end gradually until they discover an angle at which the runner moves at a constant speed down the board. They can
then use the metre stick to determine the rise and run of the board and calculate the coefficient of kinetic friction using
rise
the relation µK = tan θ =
. (Refer to Sample Problem 2 on page 100 of the textbook for the derivation of the
run
corresponding equation for the coefficient of static friction.)
(b) A major source of random error is that the runner tends to get “stuck” and “unstuck” in an irregular fashion as it moves
down the inclined board. Another major source of random error is that it is very difficult to judge when an object
moving down the board experiences a constant speed. Another source of random error is parallax error in measuring
the rise and run of the board when it is at an angle to the desk or floor. A minor systematic error occurs if the metre
stick has a worn end.
Making Connections
9. (a) One way to increase friction on the lid is to place a rubber pad over the lid or use a high-friction gloves or mitts.
Another way is to use two hands to apply forces to the lid while someone else holds the jar steady.
(b) Nonstick surfaces of frying pans and other cooking utensils decrease friction. Butter, grease, cooking oils, and cooking
sprays also decrease friction.
Try This Activity: Oil Viscosity
(Page 102)
To perform this activity, each group of 3 or 4 students needs a stopwatch, a test tube rack, 6 test tubes with solid rubber
stoppers, 2 beakers (one with cold water and the other with water from the hot water tap), and 3 grades of motor oil. Two sets
of 3 test tubes containing the 3 different samples of motor oil can be set up, labelled appropriately, and stored for future use.
The observed results depend partly on the temperatures of the water baths and the viscosities of the oils chosen. In
general, the bubbles in lower viscosity oils (e.g., SAE20) travel faster than the bubbles in higher viscosity oils (e.g., SAE40).
126 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Try This Activity: How Will the Cans Move?
(Page 105)
The air blown between the empty cans should travel horizontally and fairly close to the straws. As soon as the air moves
quickly through the gap between the cans, the cans move toward each other. This occurs because the pressure between the
cans, where the speed of the air is high, is reduced, while the pressure on the outside of the cans, where the speed of the air is
low or nil, is greater. The pressure difference causes the forces on the cans.
PRACTICE
(Page 105)
Understanding Concepts
10. Answers will vary. Some liquids, listed in order of lowest to highest viscosity, are vinegar, tomato juice, hair shampoo,
and hair cream rinse.
11. (a) This phrase describes something moving very slowly, just as molasses flows when it is at a low temperature. The
viscosity of molasses increases as the temperature drops.
(b) The implication is that blood has a higher viscosity than water. The phrase refers to the tendency for family members to
defend one another at all costs, with blood relatives more closely knit than non-relatives.
12. The speed of the syrup at the top of the bulge is higher than the speed where the syrup leaves the jar. This provides an
example of laminar flow as the liquid flows smoothly with the layers nearest the jar experiencing more friction than layers
farther from the jar.
13. (a) Most of the components are smooth and curved, and above the cab the air deflector reduces the impact of the air
resistance on the trailer behind the cab.
(b) The rockets are pointed and smooth.
(c) All components are smooth and curved, including the windscreen that prevents the air from striking the rider directly.
(d) The front of locomotives is curved and smooth.
14. (a) The speed of the air above the convertible top is higher than the speed of the air below (relative to the top). Thus, the
air pressure above the top is less than the air pressure in the car, so the top bulges upward.
(b) Moving air across the top of the chimney reduces the pressure there while the pressure in the room at the fireplace
remains higher. The pressure difference causes the air to move more readily up the chimney, resulting in a better draft.
15. The diagram below shows that the ball will curve away from the high pressure toward the low pressure, in this case
downward.
Applying Inquiry Skills
16. According to the hint, the procedure begins by determining the time it takes to fill a container of measurable volume with
water flowing from a nozzle of measurable diameter. A specific example will show how to use the data to calculate the
speed of the water.
diameter of nozzle = 0.96 cm, so radius = 0.48 cm
area of nozzle = πr2 = π(0.48 cm)2 = 0.72 cm2
volume of water collected = 3800 mL = 3.0 × 103 cm3
time interval to collect water = ∆t = 85 s
Copyright © 2003 Nelson
Chapter 2 Dynamics 127
To find the speed:
d
∆t
(volume ÷ area)
=
∆t
v=
(3.8 ×10 cm ) ÷ (0.72 cm2 )
=
3
85 s
v = 62 cm/s
Making Connections
17. The speed of the water from the nozzle in this example is 62 cm/s or 0.62 m/s. As shown in the diagram below, the speed
of the air moving across the top of the mound is likely higher than the speed along the ground surrounding the lower
entrance. With a lower pressure above the front entrance, air will rise there, causing a circulation from the rear entrance to
the front entrance. The burrowing animal is applying Bernoulli’s principle.
Section 2.4 Questions
(Pages 106–107)
Understanding Concepts
1.
2.
The rubbing involves kinetic friction between the hands, causing agitation of the surface molecules, which in turn causes
an increase in thermal energy. (Students have studied the kinetic molecular theory at previous grade levels.)
m = 2.1 × 10 3 kg
µK = 0.18
G
Fapp = 5.3 × 103 N [horizontally]
Let +x be the direction of the applied force and +y be up.
ΣFy = 0
FN + (−mg ) = 0
FN = mg
Ff = µK FN
= µ K mg
= (0.18)(2.1× 103 kg)(9.8 m/s 2 )
Ff = 3.7 ×103 N
The magnitude of the frictional force is 3.7 × 103 N.
128 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
ΣFx = ma x
Fapp + (− Ff ) = max
ax =
Fapp − Ff
m
5.3 × 103 N − 3.7 × 103 N
=
2.1× 103 kg
a x = 0.76 m/s 2
The magnitude of the acceleration of the sled is 0.76 m/s2.
3. Let +x be parallel to and down the hillside and +y be perpendicular to and up from the hillside.
(a) The equation for the coefficient of kinetic friction in terms of the angle φ can be derived as follows:
ΣFx = 0
ΣFy = 0
mg sin φ + (− Ff ) = 0
FN + (−mg cos φ ) = 0
FN = mg cos φ
mg sin φ = µK FN
FN =
mg sin φ
µK
Equate the two equations for the normal force:
mg sin φ
mg cos φ =
µK
sin φ
= µK
cos φ
µ K = tan φ
(b) The equation for the magnitude of the acceleration in terms of g, φ, and µK can be derived as follows:
ΣFy = 0
FN + (−mg cos φ ) = 0
FN = mg cos φ
ΣFx = ma x
mg sin φ + (− Ff ) = ma x
ax =
mg sin φ − µ K FN
m
Substitute for FN:
mg sin φ − µ K FN
m
mg sin φ − µ K mg cos φ
=
m
a x = g (sin φ − µ K cos φ )
(c) The skier’s mass has no affect on the magnitude of the acceleration. As shown in (a) and (b) above, m cancels out.
4. µS = 0.88
(a)
ax =
Copyright © 2003 Nelson
Chapter 2 Dynamics 129
(b) Let +x be the direction of the acceleration and +y be up.
Considering the vertical forces:
ΣFy = 0
FN + (−mg ) = 0
FN = mg
Considering the horizontal forces:
ΣFx = max
Ff = max
µS FN
m
µS mg
=
m
= µS g
ax =
= (0.88)(9.8 m/s 2 )
5.
6.
7.
a x = 8.6 m/s 2
The magnitude of the maximum acceleration is 8.6 m/s2.
The viscosity of a liquid decreases as the temperature increases, so the terminal speed of the ball in glycerine would be
higher at 60°C than at 20°C.
As the gas moves through the pipeline, it experiences both internal friction and friction with the interior walls of the pipe.
The friction causes the speed of the gas to slow down, and pumping stations are needed to get the speed back up to an
acceptable level. (Students may present a more complete explanation by considering the law of conservation of energy,
which they have studied in previous classes.
Where the diameter of the tube is large, the speed of the liquid is low and the pressure is high. Like a river than narrows,
where the diameter of the tube becomes small, the speed of the liquid increases and the pressure decreases, which
coincides with Bernoulli’s principle. The difference in pressures causes the liquid mercury in the flowmeter tube to drop
where the pressure above it is high and rise where the pressure above it is low. The greater the difference in heights, the
greater must be the difference in the slow and fast speeds in the tube. The device can be calibrated to indicate actual
speeds.
Applying Inquiry Skills
8. (a) Estimates will vary, but a good estimate would be between 0.5 and 0.6.
(b) Again, estimates will vary. The important thing here is to estimate a lower value than in part (a). A good estimate
would be between 0.3 and 0.4.
(c) The experimental values can be found by using Sample Problem 2 on page 100 of the textbook as reference. The
difference is that to measure the coefficient of kinetic friction, students must try to get their fingers and then their
fingernails to move at a constant, low speed along the bottom of the text’s cover. This is somewhat difficult because the
mass of the book is fairly high. However, it is easier if the lower end of the book is held still or propped up against
some barrier.
For the fingers, rise = 14 cm and run = 24 cm.
rise
µK =
run
14 cm
=
24 cm
µK = 0.58
130 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
For the fingernails, rise = 10 cm and run = 26 cm.
rise
µK =
run
10 cm
=
26 cm
µK = 0.38
Students can compare their estimated and experimental values by finding the percent error, assuming that the
experimental values are the “accepted” values.
estimated value − experimental value
× 100%
% error =
experimental value
(d) It takes just a few activities, like the one described in this question, to gain proficiency in estimating coefficients of
friction.
9. This is a simple (and somewhat wet!) demonstration of Bernoulli’s principle. With air moving at a high speed across the
top of the straw, the air pressure there is low. The air pressure on the surface of the water in the beaker remains higher,
pushing down on the water in the beaker and thus upward on the water in the straw. As the water exits the top of the
straw, it sprays onto the paper, simulating the action of a paint sprayer.
Making Connections
10. Gauge blocks are end gauges that can be wrung together to provide a variety of specified lengths. The end surfaces of the
blocks are manufactured with extreme precision, allowing the blocks to stick together as they are built up to obtain the
desired length. These blocks are used to check the dimensions of items in a workshop or manufacturing process, as
guidelines for measuring tools, and to calibrate other gauge blocks. More information can be found at the following Web
site: http://www.cej.se/trading_eng/broschrer/701/passbitar.htm
11. In simple terms, a slice is a golf shot that carries the ball to the right of the target line. It is caused by a club swing and
impact that is at an angle to the target line forcing the ball to spin clockwise as it travels away from the tee. A hook carries
the ball to the left of the target line, and is caused by a swing and impact that sends the ball spinning counterclockwise
after it leaves the tee. There are many Web sites devoted to this topic. The ones listed here provide more detail about
different types and causes of slices and hooks, as well as ways of preventing them. The last site listed describes details of
why golf balls are dimpled.
http://www.floridagolfing.com/forefl/9710/wiren9710.html
http://www.izgolf.com/ixaccuracy
http://www.mrgolf.com/slice.html
http://math.ucr.edu/home/baez/physics/General/golf.html
12. Technically, running shoes are much more advanced than in the past, and improvements continue to be made. One
example, taken from the first Web site listed below, is that early track event running shoes were simple leather shoes with
nails driven through them, whereas today’s outsoles for track shoes have moulded plastic plates with interchangeable sets
of spikes chosen for different events. The following Web sites provide details of the importance of considering friction in
the design of running shoes:
http://scire.com/sds/Pages/partout.html
http://stravinsky.ucsc.edu/~josh/5A/book/forces/node21.html
13. Near-frictionless carbon (NFC) is an extremely hard carbon film that resembles smooth diamond films but has a much
lower coefficient of friction, as low as 0.001 or even less in dry inert gases. The hard film is durable and relatively easy to
deposit on surfaces, and provides high efficiency and quiet performance in airline and other transportation applications.
The following Web site provides more details about NFC as well as links to other related sites:
http://www.techtransfer.anl.gov/techtour/nfc-faqs.html
Copyright © 2003 Nelson
Chapter 2 Dynamics 131
2.5 INERTIAL AND NONINERTIAL FRAMES OF REFERENCE
PRACTICE
(Page 110)
Understanding Concepts
1.
Once your hand no longer touches the puck, you will observe the puck moving at a constant velocity relative to the truck.
The truck is an inertial frame of reference, and once the puck is in motion it remains in motion because the net force
acting on it is zero.
2. (a) The ball remains stationary relative to the bus because the net force acting on it is zero. However, the ball moves at a
constant velocity of 12 m/s [E] relative to the road, the same as the velocity of the bus relative to the road.
(b) The FBD is the same in each frame of reference.
(c) As the bus accelerates forward relative to the road, the ball begins to roll backward relative to the bus, experiencing
acceleration.
G
(d) The FBD in the noninertial frame of reference has a fictitious force, Ffict .
3.
The bus is accelerating but it is the noninertial frame of reference.
m = 25 g
θ = 13°
(a) Consider the vertical components of the forces, with +y up.
∑ Fy = ma y = 0
FT cos θ − Fg = 0
FT cos θ = Fg where Fg = mg
FT =
132 Unit 1 Forces and Motion: Dynamics
mg
cos θ
Copyright © 2003 Nelson
Substitute into the equation for the horizontal components, with +x east:
∑ Fx = ma x
FT sin θ = ma x
 sin θ 
a x = ( FT ) 

 m 
 mg   sin θ 
=


 cos θ   m 
 sin θ 
= g

 cos θ 
= g tan θ
(
= 9.8 m/s 2
)(tan 13 )
D
a x = 2.3 m/s 2
Thus, the acceleration of the train is 2.3 m/s2 [E]. The mass of the rubber stopper is not needed since the force of
tension depends on the mass and, thus, cancels during the calculation.
(b) Consider the vertical components of the forces, with +y up.
ΣFy = 0
FT cos θ + (−mg ) = 0
mg
cos θ
(0.025 kg)(9.8 m/s 2 )
=
cos13°
FT = 0.25 N
The magnitude of the tension in the string is 0.25 N. The mass of the rubber stopper is needed for this calculation.
FT =
Section 2.5 Questions
(Page 111)
Understanding Concepts
1.
A noninertial frame of reference is also called an accelerating frame of reference, and can be described as a frame of
reference in which Newton’s law of inertia does not hold.
2. (a) The plane of the accelerometer must align in the north-south plane.
(b) (i) The beads remain at rest in their lowest position.
(ii) The beads move backward, away from the 0° mark to some angle that depends on the magnitude of the
acceleration.
(iii) The beads drop down to the 0° mark and stay there as long as the velocity is constant.
(iv) The move forward, away from the 0° mark to some angle that depends on the magnitude of the acceleration.
(c) In the frame of reference of the road:
Copyright © 2003 Nelson
Chapter 2 Dynamics 133
(d) In the frame of reference of the vehicle:
(e) Let +y be up and +x be the direction of the acceleration.
Consider the vertical components of the forces:
ΣFy = ma y = 0
FN cos θ − Fg = 0
FN cos θ = Fg
FN =
mg
cos θ
Next consider the horizontal components of the forces:
∑ Fx = ma x
FN sin θ = ma x
 sin θ 
a x = ( FN ) 

 m 
 mg   sin θ 
=


 cos θ   m 
 sin θ 
= g

 cos θ 
= g tan θ
(
= 9.8 m/s 2
)(tan 11 )
D
a x =1.9 m/s 2
The magnitude of the acceleration is 1.9 m/s2.
(f) m = 2.2
g = 2.2 × 10−3 kg
From the vertical components considered in (e):
mg
FN =
cos θ
(2.2 × 10−3 kg)(9.8 N/kg)
=
cos11°
−2
FN = 2.2 × 10 N
The magnitude of the normal force acting on the middle bead is 2.2 × 10−2 N.
134 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Applying Inquiry Skills
3. (a)
(b) The FBD of the ornament in the frame of reference of the road can be used to determine the acceleration. As shown in
the second diagram above, the forces acting on the object are gravity and tension. Since the only measurement required
is tan of the angle to the vertical, a ruler can be used to perform the measurement. (A protractor could be used to
measure that angle, but it is not necessary.) Using the equation derived in Sample Problem 2, pages 109 and 110, we
have a x = g tan θ , where ax is the magnitude of the acceleration, g is the magnitude of the acceleration due to gravity,
and θ is the angle from the vertical that the pendulum ornament hangs from the vertical during acceleration.
Making Connections
4. (a) The observation that your car is moving backward results from the adjacent car move forward slightly. Relative to that
car’s frame of reference, you are moving backward, but if you shift your frame of reference to the road (i.e., Earth’s
frame of reference), you soon realize that you are stationary and the adjacent car is moving.
(b) If you are seated in a theatre seat with very dim lighting surrounded by images of moving objects, you will get the
sensation that you are moving. You place yourself in the frame of reference of the moving images rather than in the
frame of reference of the theatre and the seat.
Copyright © 2003 Nelson
Chapter 2 Dynamics 135
19. (b) During the interaction, the acceleration of the double cart is 1.0 m/s2 [E]:
G
G
ΣF = ma
G
G ΣF
a=
m
2.0 N [E]
=
2.0 kg
G
a = 1.0 m/s 2 [E]
20. (e) The velocity of the cart at 0.20 s is less than 0.25 cm/s [E], but greater than zero.
G G G
vf = vi + a ∆t
= 0 m/s + (1.0 m/s 2 [E])(0.20 s)
G
vf = 0.20 m/s [E]
21. (e) After the spring interaction is complete and the carts are separated, the net force acting on the single cart is zero.
CHAPTER 2 REVIEW
(Pages 117–118)
Understanding Concepts
1.
2.
3.
4.
5.
6.
In order for an object to be at rest, it must have a net force of zero acting on it. This can only happen if there are no forces
on the object or if the sum of all the forces acting on it is zero. Thus, there cannot be just one force since the sum would
not be zero.
Whiplash occurs when a vehicle stops suddenly and the head of a person in the vehicle continues to move forward
(according to Newton’s first law of motion), and then jerks backward as the overstretched neck muscles pull backward
(according to Newton’s second law of motion). Shoulder restraints and airbags help to limit the person’s forward motion
after a collision. A properly located headrest helps to limit how far back a person’s head goes after jerking backward.
In order to move toward the front of the canoe, your feet must exert a force on the inside base of the canoe that is
downward and backward (the reaction force). Simultaneously, the canoe exerts a force on you that is upward and forward
(the reaction force). Since the canoe is relatively free to move in the water, it moves in the direction of the force applied to
it, namely in the opposite direction to your motion.
Newton’s second law of motion can be applied to determine the mass of an object in interstellar space. A known net force
is applied to the object and the object’s acceleration is measured. Then the mass is calculated:
G
G
ΣF = ma
G
ΣF
m= G
a
The physics is not good but the drama is greater than it would be with proper physics. In order for the people’s heads to be
pressed against the ceiling, the elevator would have to be accelerating downward faster than the people in it. Since the
elevator and people in it undergo free fall at the same time, they should be observed falling together.
θ = 36°
Let +x be the direction of the acceleration down the hillside.
a x = g sin θ
= (9.8 m/s 2 ) sin 36°
7.
a x = 5.8 m/s 2
The magnitude of the acceleration of the cars down the incline is 5.8 m/s2.
m1 = 113 g
m2 = 139 g
G
Fapp = 5.38 × 10−2 N [horizontal]
Copyright © 2003 Nelson
Chapter 2 Dynamics 141
(a) Let +x be the direction of the applied force and the acceleration.
ΣFx = (m1 + m2 )a x
Fapp = (m1 + m2 )a x
ax =
=
Fapp
(m1 + m2 )
5.38 × 10 −2 N
(0.113kg + 0.139 kg)
a x = 0.213m/s 2
The magnitude of the acceleration of the two-burger system is 0.213 m/s2.
(b) The two forces named are equal in magnitude but opposite in direction (i.e., they are an action-reaction pair). To find
the magnitude of the forces, we can determine the force applied by m2 on m1. Thus, for m1,
ΣFx = m1a x
Fapp = m1a x
= (0.113 kg)(0.213 m/s 2 )
Fapp = 2.41× 10−2 N
The magnitude of the force exerted by each burger on the other burger is 2.41 × 10−2 N.
8. m1 = 26 kg
m2 = 38 kg
m3 = 41 kg
(a) Let the subscripts TL and TR represent the tension in the left and right strings, and let a be the magnitude of the
acceleration of all three masses. Also, let the positive direction for the system be clockwise; thus, for m1 y is up, for m2
y is to the right, and for m3 y is down.
Consider m1:
ΣFy = m1a
FTL − m1 g = m1a
FTL = m1 ( g + a )
Consider m3:
ΣFy = m3 a
m3 g − FTR = m3 a
FTR = m3 ( g − a)
Consider m2:
ΣFx = m2 a
FTR − FTL = m2 a
Substitute for the tensions in the two strings FTR and FTL:
m3 ( g − a ) − m1 ( g + a ) = m2 a
(m3 − m1 ) g = (m1 + m2 + m3 )a
a=
=
(m3 − m1 ) g
(m1 + m2 + m3 )
(41kg − 26 kg)(9.8 m/s 2 )
(26 kg + 38 kg + 41kg)
a = 1.4 m/s 2
The magnitude of the acceleration of the blocks is 1.4 m/s2.
142 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b) Substitute the magnitude of the acceleration into the equations for the tensions developed in (a):
FTL = m1 ( g + a )
= (26 kg)(9.8 m/s 2 + 1.4 m/s 2 )
FTL = 2.9 ×10 2 N
The magnitude of the tension in the left string is 2.9 × 102 N.
FTR = m3 ( g − a)
= (41kg)(9.8 m/s 2 − 1.4 m/s 2 )
FTR = 3.4 ×102 N
9.
The magnitude of the tension in the right string is 3.4 × 102 N.
Let +x be to the left, +y be up, and FA be the force applied by the cliff on the climber. The force of the cliff on the
climber’s feet has both a vertical component and a horizontal component.
Considering the vertical components of the forces:
ΣFy = ma y = 0
FT sin θ + FA sin φ − Fg = 0
FA sin φ = mg − FT sin θ
FA =
mg − FT sin θ
sin φ
Considering the horizontal components of the forces:
ΣFx = ma x = 0
FA cos φ − FT cos θ = 0
FA cos φ = FT cos θ
FA =
FT cos θ
cos φ
Equating these equations for FA:
FT cos θ mg − FT sin θ
=
cos φ
sin φ
sin φ mg − FT sin θ
=
FT cos θ
cos φ
tan φ =
mg − FT sin θ
FT cos θ
φ = tan −1
= tan −1
mg − FT sin θ
FT cos θ
(67.5 kg)(9.80 N/kg) − (729 N)(sin 27.0°)
(729 N)(cos 27.0°)
φ = 27.0°
Substituting this angle into either of the first equations:
F cos θ
FA = T
cos φ
(729 N)(cos 27.0°)
=
(cos 27.0°)
FA = 729 N
The force of the cliff on the climber’s feet is 729 N [27.0° above the horizontal].
Note: This question can also be solved by applying the fact that the horizontal component of the applied force is equal in
magnitude to the horizontal component of the tension, and the vertical component of the applied force is equal in
magnitude to the vertical component of the tension.
Copyright © 2003 Nelson
Chapter 2 Dynamics 143
10. m = 7.38 kg
θ = 14.3°
G
a = 6.45 cm/s2 [up the hill]
(a) Let +x be the direction of the acceleration.
ΣFx = max
Fapp + (− mg sin θ ) = max
Fapp = m(a x + g sin θ )
Fapp
= 7.38 kg(6.45 × 10−2 m/s 2 + (9.8 m/s 2 ) sin14.3°)
= 18.3 N
The magnitude of the force applied by the child is 18.3 N.
(b) Let +y be perpendicular to and up from the hillside.
ΣFy = 0
FN + (−mg cos θ ) = 0
FN = mg cos θ
= (7.38 kg)(9.8 m/s 2 ) cos14.3°
FN = 70.1N
The magnitude of the normal force on the wagon is 70.1 N.
11. The diagrams below show the sagging and tightly stretched clotheslines.
Let +y be up.
For the sagging clothesline:
ΣFy = ma y = 0
2 FT1 sin θ − mg = 0
FT1 sin θ =
144 Unit 1 Forces and Motion: Dynamics
mg
2
Copyright © 2003 Nelson
For the tightly stretched clothesline:
ΣFy = ma y = 0
2 FT2 sin φ − mg = 0
FT2 sin φ =
mg
2
Equating these quantities:
FT1 sin θ = FT2 sin φ
FT1 sin φ
=
FT2 sin θ
F
sin φ
< 1 , so T1 < 1 and FT1 < FT2 .
FT2
sin θ
Thus, the tension is greater in the clothesline that is tightly stretched, and it is more likely to break.
12. Some sports would be difficult or perhaps unsafe if the athletes wore high-friction shies. Examples are the trampoline,
sports acrobatics (especially with group work), curling, beach volleyball, roller hockey, ice hockey, bandy, speed skating,
and luge tobogganing. In most track and field events, spiked shoes provide an advantage. However, in events in which the
athlete rotates within the throwing circle, spikes are a disadvantage. Field events in this category are the shot put, the
discuss throw, and the hammer throw. In some cross-country running events, spikes are a disadvantage, especially if some
of the terrain is rocky. In wrestling, shoes must not have heels, buckles, or nailed soles, and no shoes are worn in judo,
karate, kendo, aikido, and jiu jitsu.
13. Newton’s third law of motion is applied in order to walk. The action force is the foot pushing downward and backward on
the walking surface, and the reaction force is the surface pushing upward and forward on the foot. On a slippery surface,
the backward component of the action force must be small to prevent slipping and falling, so small steps must be taken.
14. m = 16 kg
µK = 0.61
Since φ < θ,
Let +y be up and +x be the direction of the applied force.
(a) Considering the vertical components of the forces:
ΣFy = ma y = 0
FN − mg = 0
FN = mg
Considering the horizontal components of the forces:
ΣFx = 0
Fapp + (− Ff ) = 0
Fapp = µ K FN
= µ K mg
= (0.61)(16 kg)(9.8 m/s 2 )
Fapp = 96 N
The magnitude of the applied force is 96 N.
(b) Fapp = 109 N
∆d = 75 cm
ΣFx = ma x
Fapp + (− Ff ) = max
ax =
Fapp − µK mg
m
109 N − 96 N
=
16 kg
a x = 0.84 m/s 2
Copyright © 2003 Nelson
Chapter 2 Dynamics 145
∆d = vi ∆t +
1
a x ( ∆t ) 2
2
1
a x ( ∆t ) 2
2
2 ∆d
∆t =
ax
∆d =
=
2(0.75 m)
0.84 m/s 2
∆t = 1.3s
Thus, the table would take 1.3 s to travel 75 cm.
G
15. FT = 21 N [31° above the horizontal]
µS = 0.55
µK = 0.50
Let +y be up and +x be in the direction of the horizontal component of the tension force.
Considering the vertical components of the forces:
ΣFy = 0
FT sin θ + FN + (−mg ) = 0
FN = mg − FT sin θ
Considering the horizontal components of the forces:
ΣFx = 0
FT cos θ + (− Ff ) = 0
FT cos θ = µS FN
FT cos θ = µS (mg − FT sin θ )
 cos θ

+ sin θ 
FT 
 µS

m=
g
 cos 31°

+ sin 31° 
(21N) 
0.55


=
2
9.8 m/s
m = 4.4 kg
The smallest possible mass of the box if it remains at rest is 4.4 kg.
16. θ = 4.7°
G
vi = 2.7 m/s [down]
µK = 0.11
G
vf = 0 m/s
Let +x be down the hillside and +y be perpendicular to and up from the hillside.
Considering the vertical components of the forces:
ΣFy = 0
FN + (−mg cos θ ) = 0
FN = mg cos θ
146 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Considering the horizontal components of the forces:
ΣFx = ma x
mg sin θ + (− Ff ) = ma x
ma x = mg sin θ − µ K FN
ma x = mg sin θ − µ K mg cos θ
a x = g (sin θ − µ K cos θ )
= 9.8 m/s 2 (sin 4.7° − 0.11cos 4.7°)
a x = −0.27 m/s 2
vf 2 = vi 2 + 2ax ∆d
vf 2 − vi 2
2a
(0 m/s) 2 − (2.7 m/s)2
=
2(−0.27m/s 2 )
∆d = 13 m
The skier will slide 13 m down the slope before coming to rest.
17. (a) In Earth’s frame of reference:
∆d =
(b) In the train’s frame of reference:
(c) Let +y be up and +x be the direction of the passenger’s acceleration.
Considering the vertical components of the forces:
ΣFy = ma y
FN − mg = 0
FN = mg
Copyright © 2003 Nelson
Chapter 2 Dynamics 147
Considering the horizontal components of the forces:
ΣFx = ma x
Ff = ma x ,max
µS FN = max ,max
µS mg = max ,max
a x ,max = µS g
= (0.47)(9.8 m/s 2 )
a x ,max = 4.6 m/s 2
The magnitude of the maximum acceleration of the train is 4.6 m/s2.
18. As the ball travels eastward through the air, the air moves westward relative to the ball. With the ball spinning clockwise
when viewed from above, the ball’s surface drags air near it eastward on the left and westward on the right. Thus, the
speed of the air near the ball’s surface is slower on the left side, making the pressure there greater. Simultaneously, the
speed of the air is greater on the right side, making the pressure there lower. The pressure difference forces the ball toward
the right, or southward.
19. ∆t = 30.0 s
V = 2.00 L = 2.00 × 103 mL = 2.00 × 103 cm3
r = 1.00 cm
Let R be the rate at which the water ejects from the hose:
2.00 L
R=
30.0 s
= 6.67 × 10−2 L / s
R = 66.7 cm3 /s
To determine the speed of the water, divide the rate by the cross-sectional area of the hose:
A = π r2
= π (1 cm)2
A = 3.14 cm 2
R
A
66.7 cm3 /s
=
3.14 cm 2
v=
v = 21.2 cm/s
Thus, the speed of the water being ejected from the hose is 21.2 cm/s.
Applying Inquiry Skills
20.
148 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
The magnitude of the slope of the line is:
G
∆a
slope =
G
∆ΣF
=
1.85 m/s 2
(
50.0 kg m/s 2
slope = 0.0370 kg
)
−1
Using unit analysis, it is evident that the mass must be the reciprocal of the slope.
1
m=
slope
1
=
0.0370 kg −1
m = 27.0 kg
The mass of the loaded wagon is 27.0 kg.
21. (a)
Let +x be the direction of the applied force and +y be up.
Considering the horizontal components of the forces:
ΣFx = ma x = 0
Fapp − FN = 0
Fapp = FN
Considering the vertical components of the forces:
ΣFy = ma y = 0
FS − Fg = 0
FS = Fg
µS FN = mg
µS Fapp = mg
µS =
mg
Fapp
The coefficient of static friction is µS =
mg
.
Fapp
(b) At this stage, students should be able to determine the coefficient of static friction using these three methods:
(i) Determine the minimum applied horizontal force needed to prevent an object from sliding down a vertical surface.
mg
µS =
Fapp
(ii) Determine the minimum horizontal force needed to start an object moving on a horizontal surface. µS =
(iii) Determine the angle of incline of a ramp such that an object just begins to slide down the ramp. µS =
Copyright © 2003 Nelson
Fapp
mg
rise
run
Chapter 2 Dynamics 149
The advantage of using a ramp is that the only measurements needed are lengths. The other two methods require
measuring both force and mass. A disadvantage of using a vertical surface is that the object may twist around slightly
rather than begin moving downward as the applied force is gradually reduced during experimentation.
22. The demonstration described is a simple but very effective way of demonstrating Bernoulli’s principle. The design in
Figure 5(b) does not produce lift on the paper wing. However, the design in Figure 5(c) produces a noticeable lift when
air blows across it. The air moves across the top or curved part of the wing more quickly than beneath the wing because
the air has a greater distance to cover to reach the far end of the wing. Where the speed of the air is high, the air pressure
is low, and the pressure difference between the bottom and top of the wing causes an upward force on the wing.
Making Connections
23. ∆t = 1.0 ms
vi = 0.0 m/s
vf = 65 m/s
m = 45 g = 4.5 × 10−2 kg
v − vi
(a) aav = f
∆t
65 m/s − 0.0 m/s
=
1 ×10 −3 s
aav = 6.5 × 10 4 m/s 2
Fav = maav
= (4.5 × 10−2 kg)(6.5 × 10 4 m/s 2 )
Fav = 2.9 × 103 N
The magnitude of the average force exerted by the club on the ball is 2.9 × 103 N.
(b) The force of gravity is so much smaller than the average force applied by the club. The ratio of the magnitudes of the
forces is:
Faverage Faverage
=
Fg
mg
=
Faverage
Fg
2.9 ×103 N
(4.5 ×10
−2
)
kg (9.8 N/kg )
= 6.6 × 103 :1.0
(c) High-speed movies are taken with many frames of the film per second. When these movies are viewed at lower speeds,
frame-by-frame analysis can be done with very short time intervals between frames. In the case of the golf ball, a very
short time interval between two frames would be needed to observe a contact time of about 1.0 ms and to determine a
speed of 65 m/s.
Extension
24. (a) m = 72 kg
vi = 0.0 m/s
∆d = 92 cm = 0.92 m
vf 2 = vi 2 + 2a∆d
vf = 2a∆d
= 2(9.8 m/s 2 )(0.92 m)
vf = 4.2 m/s
The gymnast’s speed at impact is 4.2 m/s.
(b) Let +y be down.
vi = 4.2 m/s
vf = 0 m/s
∆d = 35 cm = 0.35 m
150 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
vf 2 = vi 2 + 2a y ∆d
−vi 2
2∆d
−(4.2 m/s) 2
=
2(0.35 m)
ay =
a y = −26 m/s 2
ΣF = ma y
= (72 kg)(−26 m/s 2 )
ΣF = −1.9 × 103 N
The magnitude of the force exerted by the floor is 1.9 × 103 N.
25. m = 22 kg
θ = 45°
µS = 0.78
µK = 0.65
(a) The FBD for this situation is shown below.
ΣFy = 0
FN + (−mg cos θ ) = 0
FN = mg cos θ
ΣFx = 0
Fapp + (− Ff ) + (− mg sin θ ) = 0
Fapp = µS FN + mg sin θ
= µS mg cos θ + mg sin θ
= mg ( µS cos θ + sin θ )
= (22 kg)(9.8 m/s 2 )(0.78 cos 45° + sin 45°)
Fapp = 2.7 ×10 2 N
The magnitude of the largest force that can be applied upward is 2.7 × 102 N.
(b) The FBD for this situation is shown below.
ΣFx = 0
Ff + (− mg sin θ ) = 0
µS FN = mg sin θ
mg sin θ
FN =
µS
Copyright © 2003 Nelson
Chapter 2 Dynamics 151
ΣFy = 0
FN + (− Fapp ) + (−mg cos θ ) = 0
Fapp = FN − mg cos θ
=
mg sin θ
− mg cos θ
µS
 sin θ

= mg 
− cos θ 
 µS

Fapp
 sin 45°

= (22 kg)(9.8 m/s 2 ) 
− cos 45° 
 0.78

= 43 N
The magnitude of the smallest force that can be applied onto the top of the box is 43 N.
G
26. Fapp 3.20 × 10−15 N [down]
m = 9.11 × 10−31 kg
vix = 2.25 × 107 m/s
viy = 0.0 m/s
This problem must be separated into two parts: determine the projectile motion in the region ABCD; and determine the
constant velocity for the remaining 13.0 cm to the screen.
For the projectile motion, with +y down and +x to the right:
ΣFy
ay =
m
Fapp
=
m
3.20 × 10−15 N
=
9.11× 10−31 kg
a y = 3.51× 1015 m/s 2
Determine the distance below the initial axis (∆y1) and the final velocity as the electron exits the region ABCD:
Horizontally (constant velocity):
∆x
∆t1 =
vx
=
3.0 × 10−2 m
2.25 × 107 m/s
∆t1 = 1.3 × 10 −9 s
Vertically (constant acceleration):
∆y1 = viy ∆t1 +
1
a y (∆t1 ) 2
2
1
(3.51× 1015 m/s 2 )(1.3 × 10−9 s)2
2
∆y1 = 3.1× 10 −3 m
=
vfy = viy + a y ∆t1
= (3.51× 1015 m/s 2 )(1.3 × 10−9 s)
vfy = 4.67 × 106 m/s
152 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
When the electron exits the region ABCD, it travels with constant velocity. Determine the vertical distance (∆y2) as the
electron travels the remaining horizontal distance of 13.0 cm (1.30 × 10−3 m) to the screen:
∆x
∆t 2 =
vx
=
1.30 × 10 −3 m
2.25 ×107 m/s
∆t2 = 5.78 ×10 −9 s
∆y2 = vfy ∆t2
= (4.67 × 106 m/s)(5.78 × 10−9 s)
∆y2 = 2.70 × 10 −2 m
Determine the total vertical distance below the axis of entry:
∆y = ∆y1 + ∆y2
= 3.1× 10−3 m + 2.70 × 10−2 m
∆y = 3.02 × 10−2 m
Thus, the electron is 3.02 × 10−2 m below the axis of entry when it hits the screen.
27. m1 = 100 kg
m2 = 50 kg
Let the positive direction of the acceleration of the system of masses be clockwise (i.e., for Tarzan +x is to the right and
for Jane +y is down).
Considering the horizontal components of the forces on Tarzan:
ΣFx = m1a
FT = m1a
(Equation 1)
Considering the vertical components of the forces on Jane:
ΣFy = m2 a
mg − FT = m2 a
(Equation 2)
Adding Equations 1 and 2:
m2 g = m1a + m2 a
m2 g = a (m1 + m2 )
a=
m2 g
m1 + m2


50 kg
=
g
 100 kg + 50 kg 
1
a= g
3
1
Tarzan’s acceleration is a = g .
3
28. m1 = 2.0 kg
m2 = 1.0 kg
G
F = 3.0 N
ax =
=
ΣFx
m1 + m2
3.0 N
3.0 kg
a x = 1.0 m/s 2
Copyright © 2003 Nelson
Chapter 2 Dynamics 153
For m2:
ΣFx = m2 ax
= (1.0 kg)(1.0 m/s 2 )
ΣFx = 1.0 N
The force of m1 on m2 = the force of m2on m1 = 1.0 N. Thus, the force of contact between the blocks is 1.0 N.
29. The diagram below shows the FBD of the situation. Although there is only one clothesline, there are actually two tensions
in the line, with magnitudes of FT1 and FT2. This differs from the situation in which there is no friction between the object
and the clothesline. Here, the chicken’s claws grab onto the line.
Considering the vertical components of the forces:
ΣFy = ma y = 0
FT1 sin θ + FT2 sin φ − mg = 0
(Equation 1)
Considering the horizontal components of the forces:
ΣFx = ma x = 0
FT2 cos φ − FT1 cos θ = 0
FT1 =
FT2 cos φ
cosθ
(Equation 2)
Substitute Equation 2 into Equation 1:
 FT2 cos φ 
 cosθ  (sin θ ) + FT2 sin φ = mg


 cos φ sin θ

+ sin φ  = mg
FT2 
 cos θ

FT2 (cos φ tan θ + sin φ ) = mg
FT2 =
mg
(cos φ tan θ + sin φ )
(2.0 kg)(9.8 N/kg)
cos 45° tan 30° + sin 45°
= 17.57 N
=
FT2 = 18 N
To show that this tension is the limiting factor, we can determine FT1 by substituting FT2 into Equation 1:
FT1 sin θ + FT2 sin φ = mg
mg − FT2 sin φ
sin θ
(2.0 kg)(9.8 N/kg) − (17.57 N) sin 45°
=
sin 30°
FT1 = 14 N
Thus, the minimum breaking strength of the line is 18 N.
FT1 =
154 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b) The system diagram at a faster rotation
The system diagram at a smaller distance
3. (a) Answers may vary. A common way is to set the spacecraft rotating at an appropriate speed so that the components
inside tend to move toward the inside of the outer wall. The normal force of the wall on the occupants causes a
sensation similar to standing on Earth.
(b) One example is a set of two huge cylinders tied together. The cylinders rotate around each other to create artificial
gravity. The neat feature of this design is that in moving from one cylinder to another, astronauts pass through a zone
of zero gravitational effects. Another design featured in older science fiction movies is the wheel-shaped design,
referred to as the Stanford torus (or doughnut), named in honour of a Stanford University research group. Like other
designs, it rotates to create artificial gravity.
Try This Activity: A Challenge
(Page 121)
The device shown in Figure 4 of the text must be made so the marbles experience negligible friction. Otherwise they might
lodge in the two upper holes for the “wrong reason.”
(a) Assuming the device is well made, the only way to get the marbles to stay in the upper two holes is to rotate the device in
the horizontal plane.
(b) After realizing that the device is based on rotation, most students will realize the connection to other rotating devices,
such as a centrifuge and a rotating lettuce drier.
3.1 UNIFORM CIRCULAR MOTION
PRACTICE
(Page 122)
Understanding Concepts
1. (a) In uniform circular motion, the word “uniform” means that the radius of the circle remains constant and the motion is
at a constant speed.
(b) Answers will vary. Examples are the wheel of an exercise bike, a coffee grinder, a rotating lawn sprinkler, a vinyl
record on a record player, a microwave turntable, a rotating spherical mirror on the ceiling of a ballroom, and a chicken
on a rotating spit on a barbeque.
2. A car moving at a constant speed can be accelerating at the same time if the direction of the velocity is changing.
156 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
PRACTICE
(Page 123)
Understanding Concepts
3. (a) The direction of the instantaneous acceleration at every point in the arc is toward the centre of the circle.
(b) The velocity and acceleration directions are shown on the sketch below.
4.
The direction of centripetal acceleration is toward the centre of the circle. Thus, if an object moving with uniform circular
motion reverses direction, the direction of the centripetal acceleration stays the same.
PRACTICE
(Page 126)
Understanding Concepts
5. (a) The direction of the velocity vector is east, the acceleration vector is north, and the direction of the radius vector is
south.
(b) r = 0.80 m
v = 4.0 m/s
v2
ac =
r
(4.0 m/s) 2
=
0.80 m
ac = 2.0 × 101 m/s 2
The magnitude of the centripetal acceleration is 2.0 × 101 m/s2.
v2
1
, ac ∝ , so if the speed of the ball remains constant but the radius of the circle doubles, the magnitude
6. (a) From ac =
r
r
of the centripetal acceleration of the ball will be half its original value.
v2
(b) From ac =
, ac ∝ v 2 , so if the radius of the circle remains constant but the speed doubles, the magnitude of the
r
centripetal acceleration of the ball will be four times its original value.
7. r = 25 km = 2.5 × 104 m
v = 180 km/h = 50 m/s (to two significant digits)
v2
ac =
r
(50 m/s) 2
=
2.5 × 10 4 m
ac = 0.10 m/s 2
The magnitude of the centripetal acceleration of the particles that make up the wind is 0.10 m/s2.
8. (a) v = 2.18 × 106 m/s
d = 1.06 × 10−10 m
1.06 × 10−8 m
= 5.30 × 10−11 m
r=
2
Copyright © 2003 Nelson
Chapter 3 Circular Motion 157
v2
r
(2.18 × 106 m/s)2
=
5.30 × 10−11 m
ac = 8.97 × 1022 m/s 2
The magnitude of the centripetal acceleration is 8.97 × 1022 m/s2.
(b) T = 1.2 s
r = 4.3 m
4π 2 r
ac = 2
T
4π 2 (4.3 m)
=
(1.2 s)2
ac =
ac = 1.2 ×10 2 m/s 2
The magnitude of the centripetal acceleration is 1.2 × 102 m/s2.
(c) r = 13 cm = 0.13 m
f = 33 13 rpm = 0.56 s–1
ac = 4π 2 rf 2
= 4π 2 (0.13 m)(0.56 s −1 ) 2
9.
ac = 1.6 m/s 2
The magnitude of the centripetal acceleration is 1.6 m/s2.
r = 2.0 m
ac = 15 m/s2
v2
ac =
r
v = rac
= (2.0 m)(15 m/s 2 )
v = 5.5 m/s
The speed of the ball is 5.5 m/s.
10. r = 5.79 × 1010 m
ac = 4.0 × 10−2 m/s2
4π 2 r
ac = 2
T
T=
=
4π 2 r
ac
4π 2 (5.79 × 1010 m)
4.0 × 10−2 m/s 2
T = 7.6 ×106 s
Converting:
 1.0 h   1.0 d 
T = 7.6 ×106 s 


 3600 s   24 h 
T = 88 d
Mercury’s period of revolution around the Sun is 7.6 × 106 s or 88 d.
(
158 Unit 1 Forces and Motion: Dynamics
)
Copyright © 2003 Nelson
Applying Inquiry Skills
11.
Section 3.1 Questions
(Page 127)
Understanding Concepts
1. (a) Some examples are an egg beater, a microwave or oven turntable, a rotisserie, an electric meat slicer, a meat grinder, a
juicer, a food processor, a lettuce drier, a coffee grinder, and a dial clock.
(b) Some examples are a drill, an electric screwdriver, a circular saw, a lathe, a router, a paint stirrer, a rotating sander, and
all electric motors.
(c) Some examples are a car travelling at a constant speed around a smooth curve in a highway, a racket or bat in a small
part of its swing during which the speed is constant, a looping roller coaster in a portion of the loop or going around a
corner where the speed is constant for at least a short time interval, and an aircraft in an air show at the bottom of an arc
where the speed is constant.
1
2. Since ac ∝ , the magnitude of the centripetal acceleration of the shorter string is three times the magnitude of the
r
centripetal acceleration of the longer string.
3. (a) v = 7.77 × 103 m/s
r = 6.57 × 106 m
v2
ac =
r
(7.77 × 103 m/s) 2
=
6.57 × 106 m
ac = 9.19 m/s 2
The magnitude of the centripetal acceleration is 9.19 m/s2.
(b) v = 25 m/s
r = 1.2 × 102 m
v2
ac =
r
(25 m/s)2
=
1.2 ×102 m
ac = 5.2 m/s 2
The magnitude of the centripetal acceleration is 5.2 m/s2.
4. (a) r = 6.38 × 106 m
T = 24 h = 8.6 × 104 s
4π 2 r
ac = 2
T
4π 2 (6.38 × 106 m)
=
(8.6 × 104 s)2
ac = 3.4 × 10 −2 m/s 2
The magnitude of centripetal acceleration due to the daily rotation of an object at the Earth’s equator is 3.4 × 10−2 m/s2.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 159
(b) The centripetal acceleration of magnitude 0.034 m/s2 is much smaller than the magnitude of the acceleration due to
gravity, which is approximately 9.8 m/s2. However, it does cause a very small weight reduction as the rotation causes
a person to tend to move away from Earth’s centre. (The reason for the reduction is described in more detail in
Section 3.2 of the textbook, pages 135–136.)
5. ac = 25 m/s2
5.0 m
r=
= 2.5 m
2
ac = 4π 2 rf 2
f =
=
6.
4π 2 r
25 m/s 2
4π 2 (2.5 m)
f = 0.50 Hz
The Rotor revolves with a minimum frequency of 0.50 Hz.
v = 25 m/s
ac = 8.3 m/s2
v2
ac =
r
v2
r=
ac
=
7.
ac
(25 m/s) 2
8.3m
r = 75 m
The radius of the curve is 75 m.
T = 27.3 d = 2.36 × 106 s
ac = 2.7 × 10−3 m/s2
4π 2 r
ac = 2
T
T 2 ac
r=
4π 2
(2.36 × 106 s)2 (2.7 × 10 −3 m/s 2 )
=
4π 2
8
r = 3.8 × 10 m
The average distance from Earth to the Moon is 3.8 × 108 m.
Applying Inquiry Skills
8. (a) Students are asked to describe an experiment to verify the relationships between the variables. (This differs from
Investigation 3.1.1 in which students discover or derive the relationships between the variables.) Thus, we begin by
considering the variables involved. They are centripetal acceleration, the speed of motion, radius, frequency, and
period. Students can twirl a rubber stopper in a horizontal circle of known radius and determine the time for 20
complete revolutions. They can calculate the speed, the frequency, and the period, and verify that the following
equations yield the same result:
v2
4π 2 r
ac = 4π 2 rf 2
ac =
ac = 2
r
T
160 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b) The biggest source of random error occurs in trying to keep the stopper moving at a constant speed. This takes practice
and coordination. Other sources of random error include measuring the radius of the circle and the time for 20
revolutions of the stopper. Systematic error occurs because it is almost impossible to twirl the stopper in the horizontal
plane. Gravity pulls the stopper downward slightly, affecting the direction of the tension in the string attached to the
stopper. Friction may also affect the results in a systematic way. To keep the sources of error within reasonable bounds,
students must perform several trials and then average the results. Another idea is to use the “count-down” method to
know when to start timing the motion (i.e., as the stopper revolving in a circle moves past a specific location, count
3−2−1−0 and simultaneously start the stopwatch at 0).
Making Connections
9. (a) r = 8.4 cm = 8.4 × 10–2 m
f = 6.0 × 104 rev/min = 1.0 × 103 s–1.
ac = 4π 2 rf 2
= 4π 2 (8.4 × 10−2 m)(1.0 × 103s −1 )2
ac = 3.3 × 106 m/s 2
ac
3.3 × 106 m/s 2
=
g
9.8 m/s 2
ac = 3.4 × 105 g
The magnitude of the centripetal acceleration is 3.4 × 105 g.
(b) Some examples are to prepare serums and blood sample, to separate milk from cream, to separate U-238 from U-235
for nuclear material enrichment, to study organic molecules, and to determine molecular weights. (Centrifugal pumps,
like those used in car cooling systems, are rotating pumps that operate in a way similar to a centrifuge.)
3.2 ANALYZING FORCES IN CIRCULAR MOTION
PRACTICE
(Page 132)
Understanding Concepts
1. (a) FN = 2mg
r = 12 m
ΣF = mac
mv 2
r
mv 2
2mg + mg =
r
v = 3 gr
FN + mg =
= 3(9.8 m/s 2 )(12 m)
= 19 m/s
v = 68 km/h
The speed required by a coaster is 19 m/s or 68 km/h.
(b) Using the equation on page 132,
vcircular = 1.4vclothoid
= (1.4)(68 km/h)
vcircular = 95 km/h
Thus, a coaster on a circular loop would have to travel at a speed of 95 km/h.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 161
(b) The biggest source of random error occurs in trying to keep the stopper moving at a constant speed. This takes practice
and coordination. Other sources of random error include measuring the radius of the circle and the time for 20
revolutions of the stopper. Systematic error occurs because it is almost impossible to twirl the stopper in the horizontal
plane. Gravity pulls the stopper downward slightly, affecting the direction of the tension in the string attached to the
stopper. Friction may also affect the results in a systematic way. To keep the sources of error within reasonable bounds,
students must perform several trials and then average the results. Another idea is to use the “count-down” method to
know when to start timing the motion (i.e., as the stopper revolving in a circle moves past a specific location, count
3−2−1−0 and simultaneously start the stopwatch at 0).
Making Connections
9. (a) r = 8.4 cm = 8.4 × 10–2 m
f = 6.0 × 104 rev/min = 1.0 × 103 s–1.
ac = 4π 2 rf 2
= 4π 2 (8.4 × 10−2 m)(1.0 × 103s −1 )2
ac = 3.3 × 106 m/s 2
ac
3.3 × 106 m/s 2
=
g
9.8 m/s 2
ac = 3.4 × 105 g
The magnitude of the centripetal acceleration is 3.4 × 105 g.
(b) Some examples are to prepare serums and blood sample, to separate milk from cream, to separate U-238 from U-235
for nuclear material enrichment, to study organic molecules, and to determine molecular weights. (Centrifugal pumps,
like those used in car cooling systems, are rotating pumps that operate in a way similar to a centrifuge.)
3.2 ANALYZING FORCES IN CIRCULAR MOTION
PRACTICE
(Page 132)
Understanding Concepts
1. (a) FN = 2mg
r = 12 m
ΣF = mac
mv 2
r
mv 2
2mg + mg =
r
v = 3 gr
FN + mg =
= 3(9.8 m/s 2 )(12 m)
= 19 m/s
v = 68 km/h
The speed required by a coaster is 19 m/s or 68 km/h.
(b) Using the equation on page 132,
vcircular = 1.4vclothoid
= (1.4)(68 km/h)
vcircular = 95 km/h
Thus, a coaster on a circular loop would have to travel at a speed of 95 km/h.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 161
PRACTICE
(Page 133)
Understanding Concepts
2.
The cause of the centripetal acceleration in each case is:
(a) the force of gravity of Earth on the Moon
(b) the electric force (or electromagnetic force) of the proton in the nucleus on the electron
(c) the difference between the normal force and Earth’s force of gravity on the snowboarder
3. (a) The force that causes the centripetal acceleration is gravity.
(b) r = 2.87 × 1012 m
v = 6.80 × 103 m/s
m = 8.80 × 1025 kg
mv 2
Fg =
r
(8.80 × 1025 kg)(6.80 ×103 m/s) 2
=
2.87 × 1012 m
Fg = 1.42 ×10 21 N
The magnitude of the force of gravity is 1.42 × 1021 N.
4π 2 mr
(c) Fg =
T2
T=
=
4π 2 mr
Fg
4π 2 (8.80 × 1025 kg)(2.87 × 1012 m)
1.42 × 1021 N
= 2.65 × 109 s
T = 84.1a
The orbital period of Uranus is 2.65 × 109 s, or 84.1 Earth years.
4. m = 0.211 kg
r = 25.6 m
v = 21.7 m/s
ΣFy = mac
(a)
mv 2
r

v2 
F = m  g + 
r 

F + (− mg ) =

(21.7 m/s) 2
= (0.211kg)  9.8 m/s 2 +
25.6 m

F = 5.95 N



The magnitude of the upward lift on the bird’s wings at the bottom of the arc is 5.95 N.
162 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b) The centripetal acceleration is caused by the difference between the upward normal force of the air on the bird and the
downward for of gravity.
5. r = 450 m
v = 97 km/h = 27 m/s
Let +x be the direction of the acceleration. As shown in Sample Problem 2, page 130:
ΣFx = mac
mg tan θ =
6.
mv 2
r
 v2 
θ = tan −1  
 gr 


(27 m/s) 2
= tan −1 

2
 (9.8 m/s )(450 m) 
θ = 9.3°
The proper banking angle for a car travelling at 97 km/h is 9.3°.
m = 2.00 kg
r = 4.00 m
2.00 s
T=
= 0.400s
5.00 revolutions
4π 2 mr
T2
4π 2 (2.00 kg)(4.00 m)
=
(0.400s)2
FT =
7.
FT = 1.97 × 103 N
The magnitude of tension in the rope is 1.97 × 103 N.
r = 1.50 km = 1.50 × 103 m
ΣFy = mac
mv 2
r
v = gr
mg =
= (9.8 m/s 2 )(1.50 ×103 m)
= 121 m/s
v = 436 km/h
The speed is the plane at the top of the loop is 121 m/s or 436 km/h.
8. v = 540 km/h = 150 m/s
(a) a = 7.0 g
mv 2
mac =
r
v2
r=
ac
=
v2
7g
=
(150 m/s) 2
7(9.8 m/s 2 )
r = 3.3 × 10 2 m
The minimum radius of the plane’s circular path is 3.3 × 102 m.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 163
(b) m = 82 kg
Let +y be up.
ΣFy = mac
FN + (−mg ) = mac
FN = m( g + ac )
= m(8 g )
= 8(82 kg)(9.8 m/s 2 )
FN = 6.4 ×103 N
A force of 6.4 × 103 N is applied at the lowest point of the pullout.
Applying Inquiry Skills
9. (a) At the top of the vertical circle, gravity and tension are both downward. At the bottom of the circle, gravity is still
downward but the tension is upward, acting against gravity. Thus, the tension must be greater at the bottom of the
circle. (This is verified in Sample Problem 3 for an object travelling at a constant speed in a vertical circle.)
(b) If the stopper is twirled in a vertical circle starting with a fairly high (but safe) speed and gradually sowing down, the
motion will eventually get to the stage where the tension in the string approaches zero at the top of the circle. When the
speed is slow enough that the string becomes slack at the top of the circle, it is obvious in the demonstration that the
tension is much greater at the bottom. The demonstrator can feel the difference in the tensions.
Making Connections
10. (a) The banking angles for on- and off-ramps are larger than banking angles for gradual highway turns. In Sample
Problem 2, the relationship shown is v = gr tan θ . Since the on- and off-ramps must be designed with a smaller
radius of curvature than a gradual highway turn, it is evident from this relationship that θ must increase (tan θ increases
as θ increases), especially at the part of the curve where vehicles may be travelling almost as fast as highway speeds.
(b) It is dangerous to travel around a curve with a relatively small radius at a high speed, especially for trucks and other
vehicles with a high centre of mass. This is even more crucial when the road conditions are slippery.
11. (a) θ = 5.7°
r = 5.5 × 102 m
ΣFx = mac
mv 2
r
v = rg tan θ
mg tan θ =
= (5.5 × 10 2 m)(9.8 m/s 2 ) tan 5.7°
v = 23 m/s
The ideal speed for a train rounding a curve is 23 m/s.
(b) In order for the train to travel in an arc, it must experience a sideways centripetal acceleration caused by a sideways
force. If the tracks are not banked, that sideways force must come from the horizontal force of the tracks on the wheels
of the train. If, however, the tracks are banked, the sideways force comes from the horizontal component of the normal
force on the train. Tracks are much more stable in the plane perpendicular to the surface than parallel to it, so wear and
stress are much less on tracks with banked curves.
Try This Activity: The Foucault Pendulum
(Page 136)
(a) As shown in the text, the globe must be rotated counterclockwise when viewed from above the North Pole. In Earth’s
frame of reference, a person at the North Pole looking down at a Foucault pendulum would observe that its path of swing
appears to move clockwise, taking close to 24 h (specifically, 23 h and 56 min relative to the Sun’s frame of reference) to
complete one rotation. (If the pendulum were attached to the ceiling such that the observer would have to look straight
upward to see it, the motion would appear to be counterclockwise.)
(b) At the equator, the Foucault pendulum’s motion would remain fixed relative to Earth’s frame of reference. At latitudes
between the equator and the North Pole, the motion observed is between 23 h, 56 min at the North Pole and infinity at 0°
164 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
latitude. For example, at 45° North latitude, which is southern Ontario’s latitude, the pendulum will appear to rotate
clockwise when viewed from above and would take more than 23 h, 56 min to complete one rotation. (The time interval
23.93 h
would be about 34 h, which can be found using the relationship ∆t =
.)
sinθ
(c) The easiest way to set up this demonstration is to place a globe with its axis oriented vertically beneath a pendulum that
vibrates in the fixed frame of the classroom as the globe rotates.
PRACTICE
(Pages 136–137)
Understanding Concepts
12. (a) In Earth’s frame of reference, you are experiencing centripetal acceleration toward the centre of the merry-go-round.
The force that causes this acceleration is static friction exerted by the floor of the ride on your feet. The tendency you
feel to move away from the centre of the rotating ride is simply an example of Newton’s first law of motion: you are in
motion and would continue in a straight line at a constant speed if the static friction did not force you toward the centre
of the circle.
(b) In the accelerating frame of reference of the merry-go-round, you feel a force we can call the fictitious force or
centrifugal force pushing you away from the centre of rotation. This force is equal in magnitude but opposite in
direction to the force of static friction, so the net horizontal force is zero.
13. (a) The system diagram (as viewed from above)
Copyright © 2003 Nelson
Chapter 3 Circular Motion 165
(b) The FBD in Earth’s frame of reference
(c) The FBD in the frame of reference of the ride
(d) r = 2.9 m
T= 4.1 s
Considering the vertical components of the forces:
ΣFy = ma y = 0
FT cos θ − mg = 0
FT cos θ = mg
FT =
mg
cos θ
(Equation 1)
Considering the horizontal components of the forces:
4π 2 mr
ΣFx = ma x =
T2
2
4π mr
FT sin θ =
T2
mg
4π 2 mr
sin θ =
cos θ
T2
4π 2 mr
mg tan θ =
T2
4π 2 r
θ = tan −1
gT 2
= tan −1
4π 2 (2.9 m)
(9.8 m/s 2 )(4.1 s)2
θ = 35°
The angle the string makes with the vertical is 35°.
166 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(e) m = 45 g = 4.5 × 10−3 kg
From Equation 1 in part (d):
mg
FT =
cos θ
(4.5 × 10−3 kg)(9.8 N/kg)
=
cos 35°
FT = 5.4 ×10−2 N
The magnitude of the tension in the string is 5.4 × 10−2 N.
14. (a) The reduction is the acceleration due to gravity is caused by Earth’s rotation. From Appendix C, page 776:
rE = 6.38 × 106 m
T = 8.64 × 104 s
4π 2 r
ac = 2
T
=
(
4π 2 3.38 × 106 m
(8.64 ×10 s )
4
)
2
ac = 3.37 ×10−2 m/s 2
The magnitude of the centripetal acceleration is 3.37 × 10−2 m/s2. (In Earth’s accelerating frame of reference, this is the
magnitude of the centrifugal acceleration.) Relating this value to the acceleration due to gravity at the equator (as stated
in Table 1, text page 33):
a
% of g = c ×100%
g
3.37 ×10 −2 m/s 2
×100%
9.78 m/s 2
% of g = 0.34%
The magnitude of the acceleration of the object is 0.34% less than g.
(b) Answers will vary, depending on the mass of the student. If m = 65 kg:
weight difference = mac
=
(
= (65 kg ) 3.37 × 10−2 m/s 2
)
weight difference = 2.2 N
The difference in weight is only 2.2 N for a person of mass 65 kg.
Applying Inquiry Skills
15. (a) You would hold the accelerometer just as it appears in that figure (i.e., perpendicular to the direction of your
instantaneous velocity).
(b) r = 4.5 m
f = 0.45 Hz
ac = 4π 2 rf 2
= 4π 2 ( 4.5 m )(0.45 Hz )
2
ac = 36 m/s 2
The magnitude of the centripetal acceleration is 36 m/s2.
(c) The FBD of the bead is shown below.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 167
Considering the vertical components of the forces:
ΣFy = ma y = 0
FN cos θ − mg = 0
FN cos θ = mg
FN =
mg
cos θ
Considering the horizontal components of the forces:
ΣFx = max
FN sin θ = mac
mg
sin θ = mac
cos θ
g tan θ = ac
θ = tan −1
= tan −1
ac
g
36 m/s 2
9.8 m/s 2
θ = 75°
The angle from the vertical is 75°.
(d) m = 1.1 g = 1.1 × 10−3 kg
mg
FN =
cos θ
(1.1× 10−3 kg)(9.8 N/kg)
=
cos 75°
FT = 4.1× 10 −2 N
The magnitude of the normal force is 4.1 × 10−2 N.
Making Connections
16. Information about Foucault pendulums can be found in encyclopedias and astronomy reference books and on the Internet.
In 1851 Jean Foucault, a French physicist, suspended a 27-kg pendulum from the domed ceiling of the Pantheon in Paris.
The length of the pendulum was more than 60 m. He pulled the bob to one side and secured it with a cord. When he
burned the cord, the pendulum began swinging without any sideways influences. As the swinging continued, a record of
the path was recorded on a ring of sand beneath the pendulum. From the beginning it was evident that the motion of the
pendulum was changing relative to Earth’s frame of reference.
Foucault pendulums on display at universities and science centres are not nearly as long as the one built by Foucault.
The one in the physics department at the University of Guelph, for example, is 83 cm long with a bob of mass 4.5 kg.
With each swing, the bob touches a ring that surrounds the setup; permanent magnets, one in the bob and the other
beneath it in line with the point of suspension, help prevent unwanted deflections of the bob. A key word search on the
Internet (using “Foucault Pendulum Guelph”) will yield much more information about the University of Guelph
pendulum and links to other related sites.
Section 3.2 Questions
(Page 138)
Understanding Concepts
1.
The banked curve is a better design. In order for the car to experience acceleration toward the centre of curvature, there
must be a net force in that direction. With the flat curve, the net force is created by static friction only. With the banked
curve, the net force is created by a component of the normal force of the road on the car. This is a great advantage,
especially in slippery conditions when static friction is greatly reduced.
168 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
2.
m = 1.00 kg
r = 1.00 m string
FT = 5.00 × 102 N
ΣFx = max
mv 2
r
FT r
v=
m
FT =
=
3.
(5.00 ×10 2 N)(1.00 m)
1.00 kg
v = 22.4 m/s
The maximum speed the stone can attain without breaking the string is 22.4 m/s.
m = 0.20 kg
r = 10.0 m
5.0 s
T=
= 0.50 s
10 rotations
ΣFx = max
FT = mac
4π 2 mr
T2
4π 2 (0.20 kg)(10.0 m)
=
(0.50s) 2
=
FT = 3.2 ×102 N
The magnitude of the tension in the string is 3.2 × 102 N.
4. r = 5.3 × 10−11 m
m = 9.1 × 10−31 kg
T = 1.5 × 10−16 s
4π 2 r
(a) ac = 2
T
4π 2 (5.3 ×10 −11 m)
=
(1.5 × 10−16 s)2
ac = 9.3 × 10 22 m/s 2
The magnitude of the acceleration of the electron is 9.3 × 1022 m/s2.
(b) F = mac
= (9.1× 10−31 kg)(9.3 × 10 22 m/s 2 )
F = 8.4 ×10 −8 N
The magnitude of the electric force acting on the electron is 8.4 × 10−8 N.
5. r = 1.12 m
m = 0.200 kg
(a) Let +y be up.
ΣFy = 0
FT + (− mg ) = 0
FT = (0.200 kg)(9.80 m/s 2 )
FT = 1.96 N
The magnitude of the tension in the string when the pendulum is at rest is 1.96 N.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 169
(b) v = 1.20 m/s
ΣFy = mac
mv 2
r

v2
FT = m  g +
r

FT + (− mg ) =




(1.20 m/s)2 
= (0.200 kg)  9.80 m/s 2 +

1.12 m 

FT = 2.22 N
The magnitude of the tension at the bottom of the swing is 2.22 N.
6. (a) m = 15 g
r = 1.5 m
FT = 0 N
Let +y be down.
ΣFy = mac
mv 2
r
mv 2
mg =
r
v = rg
mg − FT =
= (1.5 m)(9.8 m/s 2 )
v = 3.8 m/s
The speed of the stopper is 3.8 m/s.
(b) If the mass of the stopper doubles, the speed will not change since speed is independent of the mass.
7. m = 0.030 kg
r = 1.3 m
v = 6.0 m/s
ΣFy = mac
mv 2
r
 v2

FT = m  ± g 
 r

FT + (± mg ) =
8.
 (6.0 m/s)2

= (0.030 kg) 
± 9.80 m/s 2 
 1.3m

FT = 1.1N or 0.54 N
The maximum tension in the string is 1.1 N and the minimum tension in the string is 0.54 N.
The statement is wrong. In Earth’s frame of reference, the child is moving in a circle and must be experiencing an
acceleration toward the centre of that circle. This centripetal acceleration caused by the net, nonzero force toward the
centre of the circle.
Applying Inquiry Skills
9. (a) One force is the force of gravity and the other force is the normal force of the seat on the rider.
(b) Let +y be down.
170 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
ΣFy = mac
mv 2
r
mv 2
2mg + mg =
r
v = 3 gr
FN + mg =
= 3(9.8 m/s 2 )(15 m)
v = 21m/s
The speed of the coaster at the top of the loop is 21 m/s.
(c) One force is gravity and the other is the tension in the accelerometer’s spring.
(d) At the top of the loop, the accelerometer is inverted (in Earth’s frame of reference), and the spring pulls downward on
the bob. If students draw an FBD of the bob, both the tension and gravity are downward. If the coaster is moving such
that the tension is zero, then the force on the bob would be 1mg, and the accelerometer would read 0g. However, the
coaster is moving at a speed that causes the net force on the bob to be 3mg, as discussed in (b) above. Thus, the tension
in the spring is 2mg and the reading on the accelerometer is 2g. (A shortcut answer can be stated by realizing that the
accelerometer is calibrated to indicate the same value felt by the rider due to the normal force, which in this case is 2mg
for force and 2g for acceleration.)
(e) There are several random sources of error in using a vertical accelerometer on a roller coaster:
• Coaster rides are fast and jerky, and the bob gets tossed around a lot.
• It is very difficult to judge when you, as an excited and sometimes frightened rider, reach a particular location in the
ride.
• It is almost impossible to remember readings for more than a small number of locations on the ride.
• It is difficult to hold the accelerometer so it is vertical relative to the ground.
• Values between the calibration marks must be estimated.
There are some sources of systematic error:
• The calibration markings may be inaccurate.
• The spring may be overstretched or tangled, causing the bob to be at the wrong position when at equilibrium.
Making Connections
10. Answers will vary. Examples in the home are a clothes washer during the spin cycle, a food processor, a weed whacker,
and an electric drill. In a car, the pulleys linked by fan belts rotate rapidly when the engine is running. Examples in the
workplace are a centrifuge, a meat slicer, electric fans, and all electric motors. The precautions are specific to the
examples. For instance, when operating a weed whacker, a person should wear goggles and protective shoes and gloves.
11. Numerous reference sites can be found on the Internet. Because there is so much information, students can share the
research by choosing one of the categories of uses suggested. For example, each small group can begin their research by
conducting a key word search using centrifuge and one of the following choices: blood analysis; DNA; proteins; dairy
products; geology. Some examples of sites are:
http://www.iptq.com/bloodanalysis.htm
http://www.azduiatty.com/DUIBloodIssues.htm
http://hdklab.wustl.edu/lab_manual/yeast/yeast3.html
3.3 UNIVERSAL GRAVITATION
PRACTICE
(Page 141)
Understanding Concepts
1.
2.
Both the third law of motion and the law of universal gravitation relate to two objects. If the action force is the force of
gravity of object 1 on object 2, then the reaction force is the force of gravity of object 2 on object 1. The forces are equal
in magnitude but opposite in direction, which is what the third law indicates.
The direction of the force of A on B is toward A.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 171
ΣFy = mac
mv 2
r
mv 2
2mg + mg =
r
v = 3 gr
FN + mg =
= 3(9.8 m/s 2 )(15 m)
v = 21m/s
The speed of the coaster at the top of the loop is 21 m/s.
(c) One force is gravity and the other is the tension in the accelerometer’s spring.
(d) At the top of the loop, the accelerometer is inverted (in Earth’s frame of reference), and the spring pulls downward on
the bob. If students draw an FBD of the bob, both the tension and gravity are downward. If the coaster is moving such
that the tension is zero, then the force on the bob would be 1mg, and the accelerometer would read 0g. However, the
coaster is moving at a speed that causes the net force on the bob to be 3mg, as discussed in (b) above. Thus, the tension
in the spring is 2mg and the reading on the accelerometer is 2g. (A shortcut answer can be stated by realizing that the
accelerometer is calibrated to indicate the same value felt by the rider due to the normal force, which in this case is 2mg
for force and 2g for acceleration.)
(e) There are several random sources of error in using a vertical accelerometer on a roller coaster:
• Coaster rides are fast and jerky, and the bob gets tossed around a lot.
• It is very difficult to judge when you, as an excited and sometimes frightened rider, reach a particular location in the
ride.
• It is almost impossible to remember readings for more than a small number of locations on the ride.
• It is difficult to hold the accelerometer so it is vertical relative to the ground.
• Values between the calibration marks must be estimated.
There are some sources of systematic error:
• The calibration markings may be inaccurate.
• The spring may be overstretched or tangled, causing the bob to be at the wrong position when at equilibrium.
Making Connections
10. Answers will vary. Examples in the home are a clothes washer during the spin cycle, a food processor, a weed whacker,
and an electric drill. In a car, the pulleys linked by fan belts rotate rapidly when the engine is running. Examples in the
workplace are a centrifuge, a meat slicer, electric fans, and all electric motors. The precautions are specific to the
examples. For instance, when operating a weed whacker, a person should wear goggles and protective shoes and gloves.
11. Numerous reference sites can be found on the Internet. Because there is so much information, students can share the
research by choosing one of the categories of uses suggested. For example, each small group can begin their research by
conducting a key word search using centrifuge and one of the following choices: blood analysis; DNA; proteins; dairy
products; geology. Some examples of sites are:
http://www.iptq.com/bloodanalysis.htm
http://www.azduiatty.com/DUIBloodIssues.htm
http://hdklab.wustl.edu/lab_manual/yeast/yeast3.html
3.3 UNIVERSAL GRAVITATION
PRACTICE
(Page 141)
Understanding Concepts
1.
2.
Both the third law of motion and the law of universal gravitation relate to two objects. If the action force is the force of
gravity of object 1 on object 2, then the reaction force is the force of gravity of object 2 on object 1. The forces are equal
in magnitude but opposite in direction, which is what the third law indicates.
The direction of the force of A on B is toward A.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 171
3.
F1 = 36 N
m1′ = 2m1
r′ = 3r
Gm1′ m2
2
F2
= r′
Gm
F1
1 m2
r2
F2  m1′   r 2 
 
=
F1  r ′2   m1 


4.
 2m   r 2 
1 
F2 = F1 
 
 (3r )2   m1 


2
= 36 N  
9
F2 = 8.0 N
The magnitude of the force would be 8.0 N.
rM = 0.54rE
mM = 0.11mE
FE = 6.0 × 102 N
GmmM
FM
r 2
= M
GmmE
FE
rE 2
FM  mE
=
FE  rM 2
 rE 2

 m
 E



 0.11m
E
FM = FE 
 (0.54r )2
E

5.
 r 2
 E
 m
 E



 0.11
= 6.0 × 102 N 
 (0.54 )2





FM = 2.3 × 10 2 N
Thus, the magnitude of the force of gravity on a body on Mars is 2.3 × 102 N.
F1 = 14 N
r1 = 8.5 m
F2 = 58 N
Gm1m2
Gm1m2
F1 =
and
F2 =
2
r1
r2 2
F1 r2 2
=
F2 r12
r2 =
=
r12 F1
F2
(8.5 m)2 (14 N)
58 N
r2 = 4.2 m
The centres of the masses are 4.2 m apart.
172 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Applying Inquiry Skills
6.
Making Connections
7.
Pluto is very small and extremely far away from Earth, making its discovery more than 70 years ago all that more
amazing. At the beginning of the twentieth century, some astronomers who had observed very slight perturbations in the
orbits of Neptune and Uranus, began searching for the cause of the perturbations. Percival Lowell, an American
astronomer, searched for the unknown planet from 1906 until his death in 1916. By 1929, his brother donated newer
instruments to continue the search. Each photograph of the search portion of the sky had about 300 000 stars, so a device
called a blink microscope was used to compare photos taken several days or even weeks apart. Finally in February 1930,
Clyde Tombaugh, comparing photos made on January 23 and 29 of that year, discovered the planet. It was named after
Pluto, the god of the underworld.
Toward, the end of the twentieth century, astronomers began discovering other bodies orbiting the Sun beyond Pluto,
causing some astronomers to question whether these newly observed bodies should be called planets, or perhaps Pluto
should be downgraded to something less than a planet. Some of the reasons for the controversy are:
• Pluto is small than several solar system moons.
• Pluto’s own moon is larger in proportion to the size of the planet than any other moon–planet example in the solar
system.
• Pluto’s orbit is unusual when compared to the orbits of the other planets.
• All the other planets far from the Sun are gas giants.
• There are more than 100 objects discovered beyond Pluto that have properties that resemble Pluto’s, yet they are not
classified as planets.
Despite the controversy, Pluto remains a sentimental favourite of many astronomers and others, so its classification is
unlikely to change, at least for now. For more information, refer to the Internet. One suitable Web site is
http://science.nasa.gov/newhome/headlines/ast17feb99_1.htm.
PRACTICE
(Page 143)
Understanding Concepts
8.
m = 1.8 × 108 kg
r = 94 m
Gm 2
r2
(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(1.8 × 108 kg) 2
=
(94 m)2
FG =
9.
FG = 2.4 × 102 N
The magnitude of the force of gravitational attraction is 2.4 × 102 N.
r = 6.38 × 106 m
m = 50.0 kg
mE = 5.98 × 1024 kg
Copyright © 2003 Nelson
Chapter 3 Circular Motion 173
FG =
GmmE
r2
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(50.0 kg)(5.98 × 1024 kg)
=
(6.38 × 106 m)2
FG = 4.90 × 102 N
The magnitude of the force of gravity on the student is 4.90 × 102 N.
10. mJ = 1.90 × 1027 kg
r = 7.15 × 107 m
GmmJ
mg =
r2
Gm
g = 2J
r
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.90 × 1027 kg)
=
(7.15 ×107 m) 2
g = 24.8 m/s 2
The magnitude of the acceleration due to gravity on Jupiter is 24.8 m/s2.
11. (a) FG = 255 N
m = 555 kg
mE = 5.98 × 1024 kg
GmmE
FG =
r2
GmmE
r=
FG
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(555 kg)(5.98 × 1024 kg)
255 N
r = 2.95 × 107 m
The vehicle is 2.95 × 107 m from the centre of Earth.
(b) h = r − rE
= 2.95 ×107 m − 6.38 × 106 m
h = 2.31× 107 m
The vehicle is 2.31 × 107 m above the surface of Earth.
12. m1 = 3.0 kg
m2 = m4 = 1.0 kg
m3 = 4.0 kg
Gm1m2
FG2 =
r12 2
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(3.0 kg)(1.0 kg)
(1.0 m)2
FG2 = 2.0 ×10 −10 N
FG3 =
=
Gm1m3
r132
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(3.0 kg)(4.0 kg)
( 2 m)2
FG3 = 4.0 × 10−10 N
174 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Considering the horizontal components of the force on m1, with +x to the right:
ΣFGx ,1 = ΣFGx,2 + ΣFGx,3 + ΣFGx ,4
(
)
= 2.0 ×10−10 N + 4.0 × 10−10 N (cos 45° ) + 0.0 N
ΣFGx ,1 = 4.83 N
ΣFGy ,1 = ΣFGy ,2 + ΣFGy ,3 + ΣFGy ,4
(
)
= 0.0 N + 4.0 × 10−10 N (sin 45° ) + 2.0 × 10 −10 N
ΣFGy ,1 = 4.83 N
FG1 =
=
( FGx,1 ) + ( FGy,1 )
2
2
( 4.83 N )
2
+ ( 4.83 N )
2
FG1 = 6.8 N
The magnitude of the net gravitational force on m1 is 6.8 N.
Making Connections
13. (a) r = 6.38 × 106 m
mg =
GmmE
r2
gr 2
mE =
G
(9.80 m/s 2 )(6.38 × 106 m)2
=
(6.67 × 10 −11 N ⋅ m 2 /kg 2 )
mE = 5.98 × 10 24 kg
(b) The first relatively accurate calculation of Earth’s mass could not have been made until after 1798 when Henry
Cavendish determined a fairly accurate value of the universal gravitation constant.
(c) Knowing Earth’s mass accurately helps scientists determine the density and thus the composition of Earth’s interior. It
helps in determining factors that influence the launching and orbits of space vehicles. It also helps scientists understand
more about our place in the solar system.
Section 3.3 Questions
(Page 144)
Understanding Concepts
1.
2.
1
where r is the distance between the centres of any two distant objects in the universe, it
r2
is evident that although FG might approach zero as r approaches infinity, it never reaches zero. Thus, the statement is true.
F1 = 26 N
1
r′ = r
2
m2′ = 3m 2
Considering the fact that FG ∝
Copyright © 2003 Nelson
Chapter 3 Circular Motion 175
Let F2 represent the force of attraction when m2 is tripled and the distance between m2 and m1 is halved.
Gm1m2′
2
F2
= r′
Gm1m2
F1
r2
F2  m2′   r 2 

=

F1  r ′2   m2 






3m2   r 2

F2 = F1
 1 2   m
  r    2
 2

 

= (26 N)(12)



F2 = 3.1 × 10 2 N
The magnitude of the force of attraction is 3.1 × 102 N.
3.
Let F1 represent your weight on Earth’s surface (i.e., at rE) and F2 represent your weight at an altitude r where
F2 1
= .
F1 2
 GmmE 

2 
F2  (r + rE )  1
=
=
F1
2
 GmmE 
 r 2 
 E 
rE 2
(r + rE )
2
=
1
2
(r + rE )2 = 2rE 2
r 2 + 2rE r − rE 2 = 0
This is a quadratic equation with solution r =
r=
−b ± b 2 − 4ac
where only the positive root applies.
2a
−2rE ± 4rE 2 + 4rE 2
2
8
rE
= −rE ±
2
2 2

=
− 1  rE
 2



r=
4.
(
)
2 − 1 rE
At an altitude of ( 2 − 1)rE , your weight is half of your weight on the surface.
mp = 1.67 × 10−27 kg
me = 9.11 × 10−31 kg
r = 5.0 × 10−11 m
Gme mp
FG =
r2
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(9.11× 10−31 kg)(1.67 × 10−27 kg)
=
(5.0 × 10 −11 m)2
FG = 4.06 × 10 −47 N
The magnitude of the force of gravitational attraction is 4.1 × 10−47 N.
176 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
5.
mA = 55 kg
mB = 75 kg
mC = 95 kg
rAB = 0.68 m
rBC = 0.95 m.
(a) Let +x be west.
FABx =
=
GmA mB
rAB2
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(55 kg)(75 kg)
(0.68 m)2
FABx = 5.95 × 10−7 N
FCBx =
=
−GmC mB
rCB 2
−(6.67 × 10−11 N ⋅ m 2 /kg 2 )(95 kg)(75 kg)
(0.95 m)2
FCBx = −5.27 ×10−7 N
ΣFx = FABx + FCBx
= 5.95 × 10 −7 N − 5.27 × 10−7 N
ΣFx = 6.8 × 10−8 N
Thus, the net force acting on B is 6.8 × 10−8 N [W].
(b) Let +x be west and +y be south.
G
FAB = 5.95 × 10−7 N [W]
G
FCB = 5.27 × 10−7 N [W]
ΣFx = FABx + FCBx
ΣFy = FABy + FCBy
= 5.95 × 10 −7 N − 0.0 N
ΣFx = 5.95 × 10 −7 N
= 0.0 N + 5.27 × 10−7 N
ΣFy = 5.27 × 10−7 N
ΣF = ΣFx 2 + ΣFy 2
= (5.95 × 10−7 N) 2 + (5.27 × 10−7 N) 2
ΣF = 7.9 × 10−7 N
tanθ =
ΣFy
ΣFx
 5.27 × 10−7 N 
θ = tan −1 

−7
 5.95 × 10 N 
θ = 42°
Thus, the net force acting on B is 7.9 × 10−7 N [42° S of W].
6. (a) rEM = 3.84 × 105 km
mM = 0.012mE
Let r be the distance between the object and Earth’s centre. Then the distance of the object from the Moon’s centre is
( rEM − r ) .
Copyright © 2003 Nelson
Chapter 3 Circular Motion 177
FE =
FM =
GmmE
r2
GmmM
(rEM − r ) 2
FE = FM
GmmE
r2
mE
r
2
=
=
GmmM
( rEM − r )2
0.012mE
(rEM − r ) 2
0.012r 2 = (rEM − r )2
0.012r 2 = rEM 2 − 2rEM r + r 2
0.988r 2 − 2rEM r + rEM 2 = 0
This is a quadratic equation with solution r =
r=
−b ± b 2 − 4ac
.
2a
2rEM ± 4rEM 2 − 4(0.988)rEM 2
2(0.988)
2r ± 0.219rEM
= EM
1.976
r = 1.12rEM or 0.90rEM
Since the distance must be less than the Earth–Moon distance (rEM = 3.84 × 105 km):
r = 0.90rEM
= 0.90(3.84 × 105 m)
r = 3.5 × 105 m
At a distance of 3.5 × 105 m from Earth’s centre the net gravitational force exerted on an object by Earth and the Moon
is zero. There is no other such point located between Earth and the Moon. (The location represented by the other root
of the equation lies beyond the Moon.)
(b) There is a location beyond the Moon, at r = 1.12rEM , where the magnitudes of the two forces are equal. However, these
forces are in the same direction, which can be shown on an FBD of the object.
Applying Inquiry Skills
7.
The values on the graph depend on the student’s mass.
178 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
For the sample graph, the mass used is 82 kg. At rE, the force is:
GmS mE
FG =
rE 2
(6.67 ×10
=
−11
)
(
N ⋅ m 2 /(kg) 2 (82 kg ) 5.98 × 1024 kg
(6.38 × 10 m )
6
)
2
FG = 8.0 × 102 N
The data for the remaining points on the graph are:
rE
r
2
8.0 × 10
FG (N)
3rE
5rE
7rE
89
32
16
Making Connections
8. (a) A geosynchronous satellite must have the same period as Earth to remain at the same location above Earth’s equator:
T = 1 day = 8.64 × 104 s.
(b) FG = Fc
GmmE
r
2
=
r3 =
r=
4π 2 mr
T2
GmET 2
4π 2
3
GmET 2
24
4π 2
(c) mE = 5.98 × 10 kg
T = 8.64 × 104 s
r=
3
=
3
GmET 2
4π 2
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(8.64 × 104 s) 2
4π 2
r = 4.23 × 107 m
The satellite is 4.23 × 107 m above Earth’s centre.
(d) The satellite dishes on Earth’s surface that receive the electromagnetic signals from a satellite must remain aimed at
that satellite. If the satellite moved across the sky (relative to the dishes), then the dishes would have to continually
track it. That would be impractical, and would be impossible once the satellite “set” below the horizon.
(e) The spacing of satellites in a geosynchronous orbit directly above the equator is limited to 0.5° per satellite. This means
that in a full circle of 360°, the maximum number of such satellites is 720. If the spacing decreases too much, the
signals from the adjacent satellites can interfere with each other. If students research these types of satellites on the
Internet, inform them that some sites distinguish geostationary and geosynchronous satellites. Both types take 24 h to
travel once around Earth, but a geostationary satellite is in an orbit directly above the equator. Some of the sites are:
http://www.msu.edu/course/tc/850/scripts/script8.htm
http://www.spaceconnection.org/satellites
http://www.geo-orbit.org/sizepgs/geodef.htm
http://www2.crl.go.jp/ka/control/Kansyo01/konzatu-e.html
http://www.cs.wpi.edu/∼cs4514/b98/week2-physical.html
Copyright © 2003 Nelson
Chapter 3 Circular Motion 179
10. T
11. T
Multiple Choice
12. (c) At the top of the circle, the net force that causes acceleration is vertically downward and greater in magnitude than mg.
If FT is the tension in the string and +y is down:
ΣFy = ma y
FT + mg = ma y
The net force is FT + mg.
13. (c) At the bottom of the circle, the net force that causes acceleration is vertically upward and greater in magnitude than mg.
Let +y be up.
ΣFy = ma y
FT − mg = ma y
Since the acceleration is upward (i.e., toward the centre of the circle), FT −mg must be upward, so FT must be greater
than mg.
14. (e) At the top of the circle, the net force toward the centre of the circle is vertical and equal in magnitude to mg. At the
instant describe, the tension in the string is zero. Let +y be down.
ΣFy = ma y
FT + mg = ma y
mg = ma y
15. (a) The direction of the centrifugal force you feel is west. Since the acceleration is eastward, the centripetal force must be
eastward, and the centripetal force must be opposite in direction, or westward.
16. (e) The direction of the acceleration of the tip is the same as vector 10, which is the direction of the instantaneous
acceleration toward the centre of the circle.
17. (c) The direction of the centripetal force is the same as vector 12. At the lowest position of the swing, the direction of the
acceleration is toward the centre of the circle, which is vertically upward.
18. (d) The direction of the normal force acting on the car is vector 11, which is perpendicular to the banked curve. The
direction of the centripetal acceleration of the car is vector 9, which is horizontal and directed toward the centre of
curvature of the curve.
19. (b) The direction of the skier’s instantaneous velocity is to the left, or vector 9, and the direction of the net force acting on
the skier is the same as the direction of the centripetal acceleration, in other words downward, or vector 6.
20. (b) From ΣFc = max , ΣFc ∝ m .
21. (c) From FG =
Gm1m2
r2
, FG ∝
1
.
r2
v2
, ac ∝ v 2 .
r
Gmplanet
23. (d) From v =
, v ∝ mplanet .
r
22. (e) From ac =
CHAPTER 3 REVIEW
(Pages 159–160)
Understanding Concepts
1.
2.
Centripetal acceleration is an instantaneous acceleration whose direction toward the centre of the circle is constantly
changing as the object travels in a circle (or arc) of constant radius. The equation defining the instantaneous acceleration
G
∆v
G
is a = lim
.
∆t → 0 ∆t
No, the parts farther from the centre of the circle experience a greater centripetal acceleration because ac ∝ r for a
constant period of revolution of the circular motion, as seen in the equation ac =
194 Unit 1 Forces and Motion: Dynamics
4π 2 r
.
T2
Copyright © 2003 Nelson
3.
If the speed of a particle in circular motion is increasing, the particle’s acceleration will be at an angle away from the line
to the centre of the circle. The net acceleration is the vector addition of the acceleration toward the centre of the circle and
the tangential acceleration perpendicular to that acceleration.
4.
ac = 4.4 m/s2
v = 25 m/s
v2
r
v2
r=
ac
ac =
=
(25 m/s)2
4.4 m/s 2
r = 1.4 ×10 2 m
The minimum radius of curvature of this curve is 1.4 × 102 m.
5. (a) For the second hand:
r = 9.8 cm
T = 60 s
4π 2 r
ac = 2
T
4π 2 (9.8 cm )
=
(60 s)2
ac = 0.11cm/s 2
The magnitude of the centripetal acceleration of the tip of the second hand is 0.11 cm/s2.
(b) For the minute hand:
r = 8.0 cm
T = 60 min = 3.6 × 103 s
4π 2 r
ac = 2
T
4π 2 (8.0 cm )
=
(3.6 × 103 s)2
ac = 2.4 × 10−5 cm/s 2
The magnitude of the centripetal acceleration of the tip of the minute hand is 2.4 × 10−5 cm/s2.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 195
(c) For the hour hand:
r = 6.0 cm
T = 12 h = 4.3 × 104 s
4π 2 r
T2
4π 2 (6.0 cm )
=
(4.3 × 10 4 s) 2
ac =
6.
ac = 1.3 × 10 −7 cm/s 2
The magnitude of the centripetal acceleration of the tip of the hour hand is 1.3 × 10−7 cm/s2.
r = 16 cm = 1.6 × 10–1 m
ac = 0.22 m/s2
4π 2 r
ac = 2
T
T=
=
4π 2 r
ac
4π 2 (0.16 m )
0.22 m/s 2
T = 5.4s
The period of rotation of the plate is 5.4 s.
7. The force(s) causing the centripetal acceleration is (are):
(a) static friction of the road on the tires
The truck: +x is toward the centre of curvature of the curve.
(b) horizontal component of the normal force of the road on the bus
The bus: +x is toward the centre of curvature of the banked curve.
(c) force of gravity of the Sun on the planet
The planet: +x is toward the Sun.
196 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(d) force of gravity of Earth on the satellite
The satellite: +x is toward Earth
8.
m = 0.65 kg
r = 26 cm = 0.26 m
f = 4.6 Hz
(a) The force that causes the centripetal acceleration is the normal force exerted by the wall of the rotating tub on the wet
towel.
(b) v = 2π rf
= 2π (0.26 m)(4.6 Hz)
v = 7.5 m/s
The speed of the towel is 7.5 m/s.
mv 2
(c) ∑ F =
r
(0.65 kg)(7.5 m/s)2
=
0.26 m
∑ F = 1.4 × 102 N
The magnitude of the centripetal force on the towel is 1.4 × 102 N.
9.0 ×1012 m
= 4.5 × 1012 m
9. r =
2
mN = 1.0 × 1026 kg
FG = 6.8 × 1020 N
m v2
(a) FG = N
r
FG r
v=
mN
=
(6.8 ×10 20 N)(4.5 × 1012 m)
1.0 ×10 26 kg
v = 5.5 × 103 m/s
The speed of Neptune is 5.5 × 103 m/s.
2π r
(b) v =
T
2π r
T=
v
2π (4.5 × 1012 m)
=
5.5 ×103 m/s
= 5.1 × 109 s
T = 1.6 × 102 a
Thus, Neptune’s period of revolution around the Sun in Earth years is 1.6 × 102 a.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 197
10.
At the specific points, the force(s) causing the centripetal acceleration is (are):
A: the difference between the normal force and the force of gravity
B: the difference between the normal force and the component of the force of gravity parallel to the normal force
C: the normal force
D: the sum of the normal force and the component of the force of gravity parallel to the normal force
E: the sum of the normal force and the component of the force of gravity parallel to the normal force
11. m = 45.7 kg
r = 3.80 m
v = 2.78 m/s
Let the positive direction be upward. The tension in each of the two vertical support chains is equal. Since
mv 2
, then:
ΣF = 2 FT + (− Fg ) and ∑ F =
r
mv 2
2 FT =
− (−mg )
r
(45.7 kg)(2.78 m/s) 2
=
+ (45.7 kg)(9.80 m/s 2 )
3.80 m
2 FT = 541 N
FT = 2.70 ×102 N
The magnitude of the tension in each of the two vertical support chains is 2.70 × 102 N.
12. m = 2.1 × 103 kg
r = 275 m
v = 26 m/s
(a) Considering the vertical components of the forces, with +y up:
ΣFy = ma y = 0
FN − mg = 0
FN = mg
198 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Considering the horizontal components of the forces, with +x the direction of the acceleration:
ΣFx = max
mv 2
r
mv 2
µS FN =
r
mv 2
µS mg =
r
2
v
µS =
gr
FS,max =
=
( 26 m/s )2
(9.8 m/s )(2.75 ×10 m )
2
2
µS = 0.25
The minimum coefficient of static friction between the tires and the road that will allow the SUV to go around the
curve without sliding is 0.25.
(b) As can be seen in (a) above, the mass m cancels out, so increasing the mass has no effect on the minimum coefficient
of static friction.
1
(c) In (a) above, µS ∝ , so decreasing the radius would cause an increase in the minimum coefficient of static friction.
r
13. m = 0.23 kg
r = 75 cm = 0.75 m
(a)
(b) v = 3.6 m/s
Let the +y direction be toward the centre of the circle.
At the top of the circle:
ΣFy = mac
mv 2
r
mv 2
FT =
− mg
r
(0.23 kg)(3.6 m/s)2
=
− (0.23 kg)(9.8 N/kg)
0.75 m
FT = 1.7 N
FT + mg =
Copyright © 2003 Nelson
Chapter 3 Circular Motion 199
At the bottom of the circle:
ΣFy = mac
mv 2
r
mv 2
FT =
+ mg
r
(0.23 kg)(3.6 m/s) 2
=
+ (0.23 kg)(9.8 N/kg)
0.75 m
FT = 6.2 N
The magnitude of the tension in the string at the top of the circle is 1.7 N. The magnitude of the tension in the string at
the bottom of the circle is 6.2 N.
(c) The minimum speed of the ball at the top of the path would be the speed required for Fg to provide the force that causes
the centripetal acceleration (tension would be zero). Thus,
ΣFy = mac
FT − mg =
mv 2
r
v = gr
mg =
= (0.75 m)(9.8 N/kg)
v = 2.7 m/s
The minimum speed of the ball at the top of the path is 2.7 m/s.
GmA mB
may be applied in situations (a) and (c), but not in situations (b) and (d). The equation
14. The equation FG =
r2
applies only to two spherical objects and to situations in which at least one of object is of a smaller size than their
separation distance. The objects in (b) and (d) are not spheres and are large compared to the separation distance.
r
15. 1 = 3.9
r2
 GmE m 


2
F2  r2 
=
F 1  GmE m 


2
 r1 
F2  r1 
= 
F 1  r2 
2
= (3.9) 2
F2
= 15
F1
The force between Earth and the meteor increases by a factor of 15 times.
200 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
16. Let F1 be the gravitational force at Earth’s surface and F2 be the gravitational force at distance r. Thus,
F2
= 0.028.
F1
 GmmE 

2 
F2  (r + rE ) 
=
F1
 GmmE 


2
 rE 
 r 
0.028 =  E 
 r + rE 
2
 rE 

 = 0.17
 r + rE 
0.17 r = (1 − 0.17)rE
r = 5.0rE
The required distance above Earth’s surface would be 5.0 rE.
17. m1 = m2 = 1.62 kg
r = 64.5 cm = 0.645 m
Gm1m2
FG =
r2
(6.67 ×10
=
−11
)
N ⋅ m 2 / kg 2 (1.62 kg)(1.62 kg)
(0.645 m) 2
FG = 4.21× 10 −10 N
The magnitude of the gravitational force between two bowling balls is 4.21 × 10−10 N.
18. mS = 1.99 × 1030 kg
mV = 4.83 × 1024 kg
r = 1.08 × 108 km = 1.08 × 1011 m
Since gravity causes the centripetal acceleration:
GmS mV
mV ac =
r2
Gm
ac = 2 S
r
(6.67 ×10
=
−11
)
N ⋅ m 2 / kg 2 (1.99 × 1030 kg)
11
(1.08 ×10 m) 2
ac = 1.13 ×10−2 m/s 2
The magnitude of the centripetal acceleration of Venus is 1.13 × 10−2 m/s2.
19. mS = 1.99 × 1030 kg
mE = 5.98 × 1024 kg
mM = 7.35 × 1022 kg
r1 = 3.84 × 105 km = 3.84 × 108 m
r2 = 1.49 × 108 km = 3.84 × 1011 m
GmE mM
FG1 =
r2
=
(6.67 ×10
−11
)
N ⋅ m 2 /kg 2 (5.98 ×10 24 kg)(7.35 ×1022 kg)
(3.84 ×108 m)2
FG1 = 1.99 × 1020 N
Copyright © 2003 Nelson
Chapter 3 Circular Motion 201
FG2 =
=
GmS mM
r2
(6.67 ×10
−11
)
N ⋅ m 2 /kg 2 (1.99 ×1030 kg)(7.35 ×1022 kg)
(1.49 ×1011 m)2
FG2 = 4.39 × 1020 N
FG = FG12 + FG2 2
=
(1.99 × 1020 N) 2 + (4.39 × 10 20 N) 2
FG = 4.82 × 1020 N
The direction of the net gravitational force is found by applying trigonometry:
F
θ = tan −1 G1
FG2
= tan −1
1.99 × 10 20 N
4.39 × 1020 N
θ = 24.4°
The net gravitational force on the Moon caused by the gravitational forces of Earth and the Sun is 4.82 × 1020 N at an
angle of 24.4° from the line to the Sun.
20. m = 1.80 × 103 kg
mmax load = 1.16 × 105 kg
mE = 5.98 × 1024 kg
rE = 6.38 × 106 m
r = 4.50 × 105 m + rE = 6.83 × 106 m
GmE mM
(a) FG =
r2
=
(6.67 ×10
−11
)
N ⋅ m 2 / kg 2 (5.98 ×1024 kg)(1.16 ×105 kg)
(6.83 ×106 m)2
FG = 9.92 × 105 N
The magnitude of the force of gravity acting on the maximum load that the arm can move is 9.92 × 105 N.
(b) Like the robotic arm and everything moving along with the ISS, the large mass is experiencing free fall. Since the arm
and the mass are falling toward Earth together, the force of gravity on the mass is not noticed. The only force that
affects the components of the arm is the force needed to cause the mass to accelerate slightly as it either starts to move
or comes to a stop.
Applying Inquiry Skills
21. m1 = 1 ball
m2 = 3 balls
r1 = 0.75 m
r2 = 1.50 m
f1 = 1.5 Hz
f2 = 3.0 Hz
Fc1 = 8.0 units
Fc1 = 4π 2 m1r1 f12
202 Unit 1 Forces and Motion: Dynamics
and
Fc2 = 4π 2 m2 r2 f 2 2
Copyright © 2003 Nelson
Fc2 4π 2 m2 r2 f 2 2
=
Fc1 4π 2 m1r1 f12
=
(3 balls)(3.0 m)(3.0 Hz)2
(1 ball)(1.5 m)(1.5 Hz)2
Fc2
= 24
Fc1
Therefore,
Fc2 = 24 Fc1
= 24(8.0 units)
Fc2 = 1.9 × 102 units
The new value for the centripetal force is 1.9 × 102 units.
22. (a) Students should describe a controlled experiment in which the dependent variable (the frequency of revolution) is
measured as the independent variables (the length of the pendulum, the mass of the bob, and either the radius of the
circle or the angle of the string) are varied in a controlled way, one at a time. (Students would discover that the
frequency is independent of the mass and that the variables length, angle, and radius are interrelated, as shown in the
equations in the remaining solutions.)
(b) The centripetal acceleration is caused by the horizontal component of the tension, FT sin θ.
(c) To determine an equation for the speed of the conical pendulum, we begin by considering the vertical components of
the forces, with +y up.
ΣFy = ma y = 0
FT cos θ − mg = 0
mg
cos θ
Considering the horizontal components of the forces, with +x the direction of the centripetal acceleration:
ΣFx = ma x
FT =
mv 2
r
mv 2
 mg 

 sin θ =
r
 cos θ 
FT sin θ =
v2
r
2
v = gr tan θ
g tan θ =
v = gr tan θ
= g ( L sin θ ) tan θ
=
(9.8 m/s ) (1.15 m )(sin 27.5°)( tan 27.5°)
2
v = 1.65 m/s
The speed of the bob is 1.65 m/s.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 203
(d) Using the data from (c):
v = 2π rf
f =
=
v
2π r
1.65 m/s
2π (1.15 m )(sin 27.5° )
f = 0.567 Hz
The frequency is 0.567 Hz.
23. On Earth’s surface, the device shown in the text must be rotated at a fairly high frequency in order for the marbles to
move up the grooves and lodge in the holes. The force of gravity helps the marbles follow the grooves. However, in space
where the marbles are in free fall with everything else in the ISS, the device would work in any orientation, even if it were
rotated slowly. To operate in space, the sides would have to be covered (for example, with clear plastic barriers) to
prevent the marbles from floating away.
Making Connections
24. The decision to cancel the race mentioned was made by CART’s officials about 75 min before the scheduled start. (CART
stands for Championship Auto Racing Teams.) The race was called the Firestone Firehawk 600, located at the Texas
Motor Speedway. At racing speeds of about 380 km/h, the cars took approximately 22 s to cover the 2.4-km oval track
with a banking angle of approximately 24°. At such a steep angle the drivers experienced accelerations up to 5.5g for
nearly 18 s in each lap. In most races, the accelerations would be around a much safer 3g. Drivers felt the effects of such
high accelerations after several practice laps, explaining they felt ready to pass out. Evidently the track had been designed
for slower-speed NASCAR races, and there had been too little testing of the track prior to this particular CART race.
The following references and Web sites provide more information:
“CART cancels Texas race” The Toronto Star, Monday, Apr. 30, 2001, page C8.
“CART assailed by money woes” The Toronto Star, Tuesday, May 1, 2001, page C12.
http://www.forsthe-racing.com/News/Article.asp?ID=587
http://www.forsthe-racing.com/News/Article.asp?ID=603
http://www.howstuffworks.com/question633.htm
http://dallas.bizjournals.com/dallas/stories/2001/05/07daily17.html
http://dallas.bizjournals.com/dallas/stories/2001/10/15daily18.html
Extension
25. TM = 27 d 8 h = 2.3616 × 106 s
rE = 6.38 × 106 m
rEM = 60.1 rE
Consider first that the force that causes the centripetal acceleration of the Moon around Earth is the force of gravity.
ΣFM = Fg = mac
GmM mE
rEM 2
=
mE =
4π 2 mM rEM
TM 2
4π 2 rEM 3
(Equation 1)
GTM 2
Next consider that the force of universal gravitation exerted by Earth on an object of mass m on Earth’s surface equals the
object’s weight:
Fg = mg
GmmE
rE 2
GmE
rE 2
= mg
=g
mE =
204 Unit 1 Forces and Motion: Dynamics
grE 2
G
(Equation 2)
Copyright © 2003 Nelson
Equating Equations 1 and 2:
grE 2 4π 2 rEM 3
=
G
GTM 2
 4π 2
g =  2
 TM
 4π 2
=  2
 TM
  rEM 3 
  2 
  rE 
  (60.1rE )

  r 2

E
3




 4π 2 
=  2  60.13 rE
 TM 
(
)

4π 2
= 
 2.3616 × 106 s

(
)
2

 60.13 6.38 × 106 m



(
)(
)
g = 9.80 m/s 2
The value of g at Earth’s surface using the motion of the Moon is 9.80 m/s2.
26. v = 180 km/h = 50 m/s (Assume two significant digits.)
Let +y be up.
ΣFy = ma y
FN − Fg =
FN − mg =
4mg − mg =
4g − g =
3g =
r=
=
The radius of the loop is 85 m.
27. r = 24 m
FT = 2mg
vi = 0
mv 2
r
mv 2
r
mv 2
r
v2
r
v2
r
v2
3g
(50 m/s )2
(
3 9.8 m/s 2
)
r = 85 m
Note: If you assign this question at this stage, give your students the hint that they must apply the law of conservation of
energy, which they studied in the previous grade. As the teacher in the question drops a vertical distance ∆y, she gains an
amount of kinetic energy equal to the amount of gravitational potential energy she loses.
The first figure shows the system diagram after the teacher has swung downward and the rope is at an angle θ to the
horizontal. The second figure shows the corresponding FBD of the teacher at that instant.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 205
Considering the y-components of the forces:
ΣFy = ma y
mv 2
r
mv 2
2mg − mg sin θ =
r
2
v
2 g − g sin θ =
(Equation 1)
r
FT − mg sin θ =
Applying the law of conservation of energy:
EK = ∆EP
mv 2
= mgr sin θ
2
v2
= 2 g sin θ (Equation 2)
r
Substituting Equation 2 into Equation 1:
2 g − g sin θ = 2 g sin θ
2 − sin θ = 2 sin θ
2 = 3sin θ
sin θ =
2
3
Finally,
∆y = r sin θ
2
= (24 m)  
3
∆y = 16 m
Thus, the teacher has dropped by 16 m, and so is 8.0 m above the ground.
28. As the ball moves in its circular path, it has a centripetal acceleration toward the centre of the circle, P. This component of
the acceleration is toward the right at the instant shown in the text, Figure 6(b) on page 161. At the same instant, the ball
is experiencing tangential acceleration downward due to gravity. The net acceleration of these two components is in the
direction of vector D.
206 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
5.
6.
The shock absorber would be a spring with a large force constant to absorb lots of energy quickly, combined with a slow
air release to prevent too much bouncing. The size would be suited to fit into the length of the forks. The forces involved
would mean the choice of a strong material, such as steel, for strength.
Try This Activity: Which Ball Wins?
(Page 177)
(a) The ball on track Y will win because it converts its gravitational energy into kinetic energy more quickly, and will speed
up more quickly than the ball on track X.
(b) The total energy of the balls remains constant down the track. The energy is just transformed from one type to another.
4.1 WORK DONE BY A CONSTANT FORCE
PRACTICE
(Pages 181–182)
Understanding Concepts
1.
2.
F1 will do more work than F2 because the component of F2 in the direction of motion is smaller than F1.
No. The force of kinetic friction is always acting opposite to the direction of motion. Since negative work is always
opposite the direction of motion, the kinetic friction will always do negative work.
3. Yes. The force of gravity can move an object toward itself, and therefore does positive work on that object.
4. m = 2.75 kg
(a) ∆d = 1.37 m
W=?
W = ( F cosθ )∆d
= (mg cos θ )∆d
= (2.75 kg)(9.80 N/kg)(cos 0°)(1.37 m)
W = 36.9 J
The work done to move the plant 1.37 m up is 36.9 J.
(b) ∆y = 1.07 m
µK = 0.549
W=?
First we must calculate the normal force acting on the potted plant:
ΣFy = ma y = 0
FN − Fg = 0
FN = Fg
= mg
= (2.75 kg)(9.80 N/kg)
FN = 26.95 N
Copyright © 2003 Nelson
Chapter 4 Work and Energy 233
Let FA be the applied force to move the potted plant horizontally:
ΣFx = ma x = 0
FA − FK = 0
FA = FK
= µ FN
= (0.549)(26.95 N)
FA = 14.796 N
To calculate the work done on the potted plant:
W = ( FA cos θ )∆d
= (14.796 N)(cos 0°)(1.07 m)
5.
W = 15.8 J
The work done to move the plant 1.07 m across the shelf is 15.8 J.
m = 24.5 kg
G
F = 14.2 N [22.5° below the horizontal]
∆d = 14.8 m
W=?
We only need to consider the component of force in the direction of motion:
W = ( F cosθ )∆d
= (14.2 N)(cos 22.5°)(14.8 m)
6.
W = 194 J
The work done by the force is 194 J.
G
FT = 12.5 N [19.5° above the horizontal]
W = 225 J
∆d = ?
W = ( F cos θ )∆d
W
F cos θ
225 J
=
(12.5 N)(cos19.5°)
∆d = 19.1 m
The toboggan moves 19.1 m.
7. (a) We will calculate the area using the formula for a rectangle
A = lw
∆d =
= (4.0 N)(2.0 m)
A = 8.0 J
The area represents the work done on the object.
(b) First calculate the area of the second portion of the graph:
A = lw
= (−4.0 N)(6.0 m − 2.0 m)
A = −8.0 J
The total work is 8.0 + (–8.0) = 0.0 J
(c) One situation could be pushing a box across a table and pulling it back.
Applying Inquiry Skills
8.
You could set the pen on the paper and pull the paper across the desk. The force of static friction between the paper and
the pen is in the direction of motion, doing positive work on the pen.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 234
Understanding Concepts
9.
Four different situations are:
• a book sitting on a desk (∆d = 0)
• a student carrying a book at a constant speed (θ = 90º)
• a teacher whirling a putty pat in a circle at the end of a string
• a toy car travelling in a circular path
10. (a) A box being pulled by string at an angle involves the forward component doing positive work, and the vertical
component doing zero work.
(b) A box being pulled up a ramp involves the parallel component of gravity doing negative work, and the perpendicular
component of gravity doing zero work.
Section 4.1 Questions
(Page 183)
Understanding Concepts
1.
2.
3.
4.
The everyday use of the word “work” is different from the usage in physics when it is referring to employment, or a duty
to perform. “Working” as a teacher involves very little physical work. The physics definition of work means the transfer
of energy to an object to move it a certain distance. Many types of employment or daily activities involve physical work.
For example, the sentence “Loading the cement bags onto the truck was a lot of work,” uses the word “work” similar to
the physics definition of work.
The centripetal force is always directed toward the centre of the circle, and is by definition perpendicular to the motion of
the object. The 90º angle means that work is not done on the object by the centripetal force.
As you push on a wall, you are exerting a force, which involves the use of energy. Even though no physical work is being
done, your muscles are still burning your body’s fuel, causing you to become tired.
Assuming the classroom to be 4 m tall, and the student to have a mass of 70 kg:
W = ( F cosθ )∆d
= ( mg cos θ ) ∆d
= (70 N)(9.80 N/kg)(cos 0°)(4.0 m)
5.
W = 3.0 × 103 J
It would take about 3.0 ¯ 103 J, or 3.0 kJ of work to climb the ladder.
G
FA = 75 N [22° below the horizontal]
G
FT = 75 N [32° above the horizontal]
Copyright © 2003 Nelson
Chapter 4 Work and Energy 235
(a) The work done by the boy (WB):
WB = ( FT cos θ )∆d
= (75 N)(cos 32°)(13 m)
WB = 826.8 J
The work done by the girl (WG):
WG = ( FA cos θ )∆d
= (75 N)(cos 22°)(13 m)
WG = 904.0 J
The total work done:
Wtotal = WB + WG
= 828.8 J + 904.0 J
Wtotal = 1.73 × 103 J
The total amount of work done on the crate is 1.7 ¯ 10–3 J.
(b) The crate is moving at a constant speed, so it is not gaining any energy. This means that the crate will have the same
amount of energy before and after the move, so it must have work done on it opposite the direction of motion.
Therefore, the work done on the crate by the floor is –1.7 ¯ 103 J.
6. W = 9.65 ¯ 102 J
∆d = 45.3 m
G
F = 24.1 N [parallel to the handle of the sleigh]
θ=?
W = ( F cosθ )∆d
cos θ =
W
F ∆d
 W 
θ = cos −1 

 F ∆d 


965 J
= cos −1 

(24.1
N)(45.3
m)


θ = 27.9°
The angle between the snowy surface and the handle is 27.9º.
7. (a) ∆d = 38 m
m = 66 kg
G
FA = 58 N [18° above the horizontal]
FN = ?
µK = ?
First we must calculate the normal force:
ΣFy = ma y = 0
FN + FA sin18° − Fg = 0
FN = Fg − FA sin18°
= mg − FA sin18°
= (66 kg)(9.80 N/kg) − (58 N) sin18°
FN = 628.88 N
Copyright © 2003 Nelson
Chapter 4 Work and Energy 236
We then calculate the force of gravity:
ΣFx = ma x = 0
FA cos18° − FK = 0
FK = FA cos18°
= (58 N) cos18°
FK = 55.161 N
The coefficient of kinetic friction is:
F
µK = K
FN
55.161 N
628.88 N
µK = 0.088
The normal force is 6.3 ¯ 102 N, and the coefficient of kinetic friction between the toboggan and the snow is 0.088.
(b) W = ?
W = ( F cosθ )∆d
=
= (55.161 N)(cos180°)(38 m)
W = −2.1× 103 J
The work done by kinetic friction is –2.1 ¯ 103 J.
(c) The normal force, the gravitational force, and the vertical component of the applied force do no work on the toboggan.
(d) ∆d = 25 m
W=?
W = ( F cos θ )∆d
= (58 N)(cos18°)(25 m)
W = 1.4 × 103 J
The work done by the parent is 1.4 ¯ 103 J.
Applying Inquiry Skills
8.
As shown below, the graph indicates that the work done on a object is positive for angles less than 90º, zero for angles
equal to 90º, and negative for angles between 90º and 180º.
Making Connections
9.
Work done by friction has the effect of heating up the environment. Most forms of energy usually end up as thermal
energy.
4.2 KINETIC ENERGY AND THE WORK–ENERGY THEOREM
PRACTICE
(Pages 186–187)
Understanding Concepts
1.
The kinetic energy of a moving object is related to both the mass and the velocity. If the mass of the truck is large enough,
a slow moving truck can have more kinetic energy than a fast moving car.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 237
We then calculate the force of gravity:
ΣFx = ma x = 0
FA cos18° − FK = 0
FK = FA cos18°
= (58 N) cos18°
FK = 55.161 N
The coefficient of kinetic friction is:
F
µK = K
FN
55.161 N
628.88 N
µK = 0.088
The normal force is 6.3 ¯ 102 N, and the coefficient of kinetic friction between the toboggan and the snow is 0.088.
(b) W = ?
W = ( F cosθ )∆d
=
= (55.161 N)(cos180°)(38 m)
W = −2.1× 103 J
The work done by kinetic friction is –2.1 ¯ 103 J.
(c) The normal force, the gravitational force, and the vertical component of the applied force do no work on the toboggan.
(d) ∆d = 25 m
W=?
W = ( F cos θ )∆d
= (58 N)(cos18°)(25 m)
W = 1.4 × 103 J
The work done by the parent is 1.4 ¯ 103 J.
Applying Inquiry Skills
8.
As shown below, the graph indicates that the work done on a object is positive for angles less than 90º, zero for angles
equal to 90º, and negative for angles between 90º and 180º.
Making Connections
9.
Work done by friction has the effect of heating up the environment. Most forms of energy usually end up as thermal
energy.
4.2 KINETIC ENERGY AND THE WORK–ENERGY THEOREM
PRACTICE
(Pages 186–187)
Understanding Concepts
1.
The kinetic energy of a moving object is related to both the mass and the velocity. If the mass of the truck is large enough,
a slow moving truck can have more kinetic energy than a fast moving car.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 237
2.
The kinetic energy is proportional to the speed, so if the speed increases by
(a) 2, the kinetic energy will increase by a factor of 22, or 4
(b) 3, the kinetic energy will increase by a factor of 32, or 9
(c) 37%, the kinetic energy will increase by a factor of 1.372, or 1.9
3. Assume a 75-kg student running at 8.0 m/s:
1
EK = mv 2
2
1
= (75 kg)(8.0 m/s) 2
2
EK = 2.4 ×103 J
The kinetic energy at maximum speed is 2.4 kJ.
4. m = 45 g = 4.5 ¯ 10–2 kg
vi = 0 m/s
vf = 43 m/s
(a) W = ?
W = ∆E K
1
m(vf2 − vi2 )
2
1
2
2
= (4.5 × 10−2 kg) ( 43 m/s ) − (0 m/s )
2
W = 41.6025 J, or42 J
The work done by the club is 42 J.
(b) ∆d = 2.0 cm = 2.0 ¯ 10–2 m
F=?
W = F ∆d
=
(
)
W
∆d
41.6025 J
=
2.0 ×10 −2 m
F = 2.1 × 103 N
The average force exerted by the club is 2.1 × 103 N.
m = 27 g = 2.7 ¯ 10–1 kg
F = 95 N
∆d = 31 cm = 3.1 ¯ 10–1 m
vf = ?
F=
5.
W = ∆EK
1 2
mvf (since the initial speed is zero)
2
2 F ∆d
vf2 =
m
F ∆d =
vf =
=
6.
2 F ∆d
m
2(95 N)(3.1 ×10 −1 m)
2.7 × 10 −1 kg
vf = 47 m/s
The final speed of the arrow is 47 m/s.
m = 4.55 ¯ 104 kg
vi = 1.22 ¯ 104 m/s
F = 3.85 ¯ 105 N
∆d = 2.45 ¯ 106 m
vf = ?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 238
W = ∆EK
1 2 1 2
mvf − mvi
2
2
1 2
1 2
mvf = F ∆d + mvi
2
2
2 F ∆d
vf =
+ vi2
m
F ∆d =
=
7.
2(3.85 × 105 N)(2.45 × 106 m)
+ (1.22 × 10 4 m/s) 2
4.55 × 10 4 kg
vf = 1.38 × 10 4 m/s
The final speed of the probe is 1.38 ¯ 104 m/s.
m = 28.0 kg
G
FA = 95.6 N [35° above the horizontal]
FK = 75.5 N
vi = 0 m/s
∆d = 0.750 m
vf = ?
The total work done on the box will become kinetic energy. Since the initial speed is zero:
1
W = mvf2
2
1 2
FA cos 35.0°∆d + FK cos180°∆d = mvf
2
2∆d
2
vf =
( FA cos 35.0° + FK cos180° )
m
vf =
=
2 ∆d
( FA cos 35.0° + FK cos180° )
m
2(0.750 m)
((95.6 N)(0.81952) + (75.5 N)(−1) )
20.8 kg
vf = 0.45 m/s
8.
The final speed of the box is 0.45 m/s.
W = 1.47(cos 38º ) = 1.16
The toboggan would have increased its kinetic energy by 16%.
Applying Inquiry Skills
9.
W = F ∆d
= N⋅m
kg ⋅ m
⋅m
s2
kg ⋅ m 2
W=
s2
EK =
1 2
mv
2
m
= kg ⋅  
s 
=
EK =
2
kg ⋅ m 2
s2
The base units for both are the same.
Making Connections
10. m = 6.85 ¯ 103 kg
vA = 2.81 ¯ 103 m/s
vB = 8.38 ¯ 103 m/s
W=?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 239
(a) The work done is equal to the change in kinetic energy
W = ∆E K
1 2 1 2
mvB − 2 mvA
2
1
= m(vB2 − vA2 )
2
1
= (6.85 × 103 kg) (8.38 × 103 m/s) 2 − (2.81× 103 m/s) 2
2
W = 2.13 × 1011 J
(b) The work done by Earth to move the satellite from A to B is 2.13 × 1011 J.
W = ∆E K
=
(
1 2 1 2
mvA − mvB
2
2
1
= m(vA2 − vB2 )
2
1
= (6.85 × 103 kg) (2.81× 103 m/s) 2 − (8.38 × 103 m/s) 2
2
W = −2.13 × 1011 J
The work done by Earth to move the satellite from B to A is –2.13 × 1011 J.
)
=
(
)
Section 4.2 Questions
(Page 188)
Understanding Concepts
1.
The first doubling will require much less energy than the second doubling of the speed. This can clearly be shown using:
1
W = m(vf2 − vi2 )
2
W ∝ vf2 − vi2
To go from v to 2v:
W = vf2 − vi2
= (2v )2 − (v) 2
= 4v 2 − v 2
W = 3v 2
To go from 2v to 4v:
W = vf2 − vi2
= (4v )2 − (2v) 2
= 16v 2 − 4v 2
W = 11v 2
The first doubling of speed will require work proportional to 3 times the square of the original speed. The second
doubling will require work proportional to 11 times the square of the original speed.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 240
2.
m = 1.50 ¯ 103 kg
G
v = 18.0 m/s [E]
EK = ?
1 2
mv
2
1
= (1.5 × 103 kg)(18.0 m/s)2
2
EK = 2.43 ×105 J
The kinetic energy of the car is 2.43 ¯ 105 J.
3. (a) v = 1.150 × 18.0 = 20.7 m/s
EK = ?
1
EK = mv 2
2
1
= (1.5 × 103 kg)(20.7 m/s) 2
2
EK = 3.21× 105 J
The new kinetic energy of the car is 3.21 ¯ 105 J.
(b) The increase in EK is:
3.21× 105 J
= 1.32
2.43 × 105 J
We can verify this with (1.15)2 = 1.32
This represents an increase in the kinetic energy of 32%.
(c) W = ?
W = ∆E K
EK =
= 3.21× 105 J − 2.43 × 105 J
4.
W = 7.8 × 104 J
The work done to speed up the car was 7.8 ¯ 104 J.
m = 55 kg
EK = 3.3 ¯ 103 J
v=?
1
EK = mv 2
2
2 EK
v=
m
=
5.
2(3.3 ×103 J)
55 kg
v = 11 m/s
The speed of the sprinter is 11 m/s.
v = 12 m/s
EK = 43 J
m=?
1
EK = mv 2
2
2E
m = 2K
v
2(43 J)
=
(12 m/s)2
m = 0.60 kg
The mass of the basketball is 0.60 kg.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 241
6.
m = 0.353 kg
∆d = 89.3 cm = 0.893 m
(a) W = ?
W = (mg cos θ )∆d
= (0.353 kg)(9.80 N/kg)(cos 0°)(0.893 m)
W = 3.0892 J
The work done by gravity is 3.09 J.
(b) Using the work energy theorem, W = ∆EK .
Since vi = 0:
1
W = mvf2
2
2W
vf =
m
=
7.
2(3.0892 J)
0.353 kg
vf = 4.18 m/s
The speed of the plate just before it hits the floor is 4.18 m/s.
m = 61 kg
θ = 23°
FK = 72 N
vi = 3.5 m/s
∆d = 62 m
vf = ?
The component of gravity along the slope is mg sin 23º. Using the work energy theorem:
1
1
mg sin 23°(cos 0°) ∆d + FK (cos180°) ∆d = mvf2 − mvi2
2
2
1 2
1
mvf = mg sin 23°(1)∆d + FK (−1) ∆d + mvi2
2
2
2
F
∆
d
vf2 = 2 g sin 23°∆d − K
+ vi2
m
vf = 2 g sin 23°∆d −
2 FK ∆d
+ vi2
m
= 2(9.8 m/s 2 )(sin 23°)(62 m) −
8.
2(72 N)(62 m)
+ (3.5 m/s) 2
61 kg
vf = 18 m/s
The speed of the skier after travelling 62 m downhill is 18 m/s.
m = 55.2 kg
∆d = 4.18 m
µK = 0.27
vi = ?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 242
Using the FBD to calculate FN,
ΣFy = ma y = 0
FN − mg = 0
FN = mg
= (55.2 kg)(9.80 N/kg)
FN = 540.96 N
To calculate FK:
FK = µ K FN
= (0.27)(540.96 N)
FK = 146.06 N
Using the work-energy theorem:
W = ∆EK
FK (cos180°)∆d =
1
m(vf2 − vi2 )
2
Since vf = 0,
1
FK (cos180°)∆d = − mvi2
2
−2 FK (cos180°)∆d
vi =
m
=
−2(146.06 N)(−1)(4.18 m)
55.2 kg
vi = 4.7 m/s
The initial speed of the skater was 4.7 m/s.
Applying Inquiry Skills
9. (a)
Car Speed (m/s)
10.0
20.0
30.0
40.0
Car Energy (J)
4
6.0 × 10
5
2.4 × 10
5
5.4 × 10
5
9.6 × 10
Truck Speed (m/s)
Truck Energy (J)
10.0
20.0
30.0
40.0
2.5 × 10
8
1.0 × 10
8
2.2 × 10
8
4.0 × 10
To calculate the energy of the car and the truck, use the equation EK =
7
1 2
mv .
2
To convert tonnes to kilograms, multiply by 1000:
m = 1.2 t = 1.2 ¯ 103 kg (for the car)
m = 5.0 ¯ 102 t = 5.0 ¯ 105 kg (for the truck)
(b)
Copyright © 2003 Nelson
Chapter 4 Work and Energy 243
(c) A vehicle with a much larger mass has a much larger kinetic energy. As the speed of a vehicle increase, the kinetic
energy increases proportional to the square of the speed.
Making Connections
10. (a) The kinetic energy is used to permanently deform the object as energy is transformed into heat.
(b) The kinetic energy of a vehicle is transmitted (at least in part) to the occupants of the vehicle, and anything it contacts.
The high kinetic energy of a fast moving vehicle is more than sufficient to damage the human body beyond its limit,
causing death.
4.3 GRAVITATIONAL POTENTIAL ENERGY AT EARTH’S SURFACE
PRACTICE
(Page 191)
Understanding Concepts
1.
2.
The total work done by gravity is zero. The work on the way down is positive and is equal to the work done on the way
up, which is negative. At the end, the pen has not lost or gained gravitational potential energy. Alternatively, you could
argue that because the ∆d = 0 (25 cm – 25 cm), the work done must be equal to zero.
m = 62.5 kg
∆y = 346 m (using the ground as y = 0)
Eg = ?
∆Eg = mg ∆y
= (62.5 kg)(9.80 N/kg)(346 m)
∆Eg = 2.12 × 105 J
Relative to the ground, the gravitational potential energy is 2.12 ¯ 105 J.
3. m = 58.2 g = 5.82 ¯ 10–2 kg
∆y = 1.55 m
(a) ∆Eg = ?
At 1.55 m above the court:
∆Eg = mg ∆y
= (5.82 × 10−2 m)(9.80 N/kg)(1.55 m)
∆Eg = 0.884 J
At the court height:
∆Eg = mg ∆y
= (5.82 × 10−2 m)(9.80 N/kg)(0.00 m)
∆Eg = 0.00 J
The gravitational potential energy when the ball is above the court is 0.884 J, and as it strikes the court surface is
0.00 J.
(b) W = ?
W = ( F cosθ )∆d
= (mg cos θ )∆d
= (5.82 × 10−2 kg)(9.80 N/kg)(cos 0°)(1.55 m)
W = 0.884 J
At the instant the ball strikes the court surface, the force of gravity has done 0.884 J of work on the ball.
(c) The work done in (b) is equal to the change in kinetic energy of the ball.
4. m = 68.5 kg
m = 2.56 km = 2.56 ¯ 103 m
θ = 13.9°
Eg = ?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 244
(c) A vehicle with a much larger mass has a much larger kinetic energy. As the speed of a vehicle increase, the kinetic
energy increases proportional to the square of the speed.
Making Connections
10. (a) The kinetic energy is used to permanently deform the object as energy is transformed into heat.
(b) The kinetic energy of a vehicle is transmitted (at least in part) to the occupants of the vehicle, and anything it contacts.
The high kinetic energy of a fast moving vehicle is more than sufficient to damage the human body beyond its limit,
causing death.
4.3 GRAVITATIONAL POTENTIAL ENERGY AT EARTH’S SURFACE
PRACTICE
(Page 191)
Understanding Concepts
1.
2.
The total work done by gravity is zero. The work on the way down is positive and is equal to the work done on the way
up, which is negative. At the end, the pen has not lost or gained gravitational potential energy. Alternatively, you could
argue that because the ∆d = 0 (25 cm – 25 cm), the work done must be equal to zero.
m = 62.5 kg
∆y = 346 m (using the ground as y = 0)
Eg = ?
∆Eg = mg ∆y
= (62.5 kg)(9.80 N/kg)(346 m)
∆Eg = 2.12 × 105 J
Relative to the ground, the gravitational potential energy is 2.12 ¯ 105 J.
3. m = 58.2 g = 5.82 ¯ 10–2 kg
∆y = 1.55 m
(a) ∆Eg = ?
At 1.55 m above the court:
∆Eg = mg ∆y
= (5.82 × 10−2 m)(9.80 N/kg)(1.55 m)
∆Eg = 0.884 J
At the court height:
∆Eg = mg ∆y
= (5.82 × 10−2 m)(9.80 N/kg)(0.00 m)
∆Eg = 0.00 J
The gravitational potential energy when the ball is above the court is 0.884 J, and as it strikes the court surface is
0.00 J.
(b) W = ?
W = ( F cosθ )∆d
= (mg cos θ )∆d
= (5.82 × 10−2 kg)(9.80 N/kg)(cos 0°)(1.55 m)
W = 0.884 J
At the instant the ball strikes the court surface, the force of gravity has done 0.884 J of work on the ball.
(c) The work done in (b) is equal to the change in kinetic energy of the ball.
4. m = 68.5 kg
m = 2.56 km = 2.56 ¯ 103 m
θ = 13.9°
Eg = ?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 244
First, calculate the vertical lift:
∆y
2.56 × 103 m
∆y = 2.56 × 10−3 m(sin13.9°)
sin13.9° =
∆y = 614.98 m
Then calculate the gravitational potential energy:
∆Eg = mg ∆y
= (68.5 kg)(9.80 N/kg)(614.98 m)
∆Eg = 4.13 × 105 J
5.
The skier’s gravitational potential energy at the top of the mountain is 4.13 ¯ 105 J.
∆y = 2.36 m
∆Eg = –1.65 ¯ 103 J
m=?
The ground is 2.36 m down from the pole, therefore:
∆Eg = mg ∆y
m=
∆Eg
g ∆y
−1.65 × 103 J
(9.80 N/kg)(−2.36 m)
m = 71.3 kg
The mass of the jumper is 71.3 kg.
6. (a) The first coin does not need to be lifted, so no work is done on it. Each successive coin will need to be raised one more
y
coin thickness, t, than the previous. The thickness of each coin will be t = . The work done on each coin will be
N
equal to its increase in gravitational potential energy.
WT = W1 + W2 + " + WN
=
= mg ∆y1 + mg ∆y2 + " + mg ∆y N
= mg (∆y1 + ∆y2 + " + ∆yN )
= mg (0 + t + 2t + " Nt )
= mgt (0 + 1 + 2 + " N )
WT =
mgy
(1 + 2 + " + N )
N
The sum of an arithmetic series is:
t +t 
Sn = n  1 n 
 2 
For the series (1 + 2 + ··· + N):
1+ N 
Sn = N 

 2 
Copyright © 2003 Nelson
Chapter 4 Work and Energy 245
Substituting in the original equation to find the amount of work that must be done on the last coin:
mgy  1 + N 
WT =
N

N
 2 
1+ N 
WT = mgy 

 2 
(b) The energy stored is equal to the work done, therefore:
 1+ N 
∆Eg = mgy 

 2 
Applying Inquiry Skills
7.
The units for gravitational potential energy:
m
mg ∆y = ( kg )  2  m
s 
= kg ⋅ m 2 /s 2
The units for work:
F ∆d = ( N )( m )
 m
=  kg 2  ( m )
 s 
= kg ⋅ m 2 /s 2
The units for kinetic energy:
1 2
m
mv = ( kg )  
2
s 
2
= kg ⋅ m 2 /s 2
Therefore, all three units are the same.
Making Connections
8.
∆Eg = 6.1 ¯ 109 J
(a) ∆y = ?
Assume 920 students in the school. Assume an average mass of 70.0 kg per student.
m = 70.0 kg ¯ 920 = 6.44 ¯ 104 kg
∆Eg = mg ∆y
∆y =
=
∆Eg
mg
6.1× 109 J
(6.44 ×10 4 kg)(9.80 N/kg)
∆y = 9.472 ×10 4 m
The energy from one barrel of oil could raise the students 9.472 ¯ 104 m above ground level.
(b) There are 158.987 L in a barrel of oil
6.1× 109 J
= 3.8 × 107 J/L
158.987 L
There are 3.8 ¯ 107 J stored in each litre of oil.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 246
PRACTICE
(Page 193)
Understanding Concepts
9. (a) The Sun’s radiant energy is converted to thermal energy in the water, which is then converted into gravitational
potential energy as the water rises. The gravitational energy converts into kinetic energy as the water falls and turns the
turbine. The kinetic energy of the turbine is converted into electrical energy by the generator.
(b) A run-of-the-river generating station does not dam the water to create a large vertical drop in a short area, but rather
uses the natural drop of the land over a certain distance, and diverts part of the water to flow down this path.
Making Connections
(c) The main difference students will find is that most run-of-the-river generating stations in Canada are much smaller.
(d) (i) rainfall, rivers, and glacier/mountain snow melting
(ii) Bhutan relies heavily on its environment for exports from logging and energy.
(e) possible points: cost, limited suitable locations, low population density, long-term effects
Section 4.3 Questions
(Page 194)
Understanding Concepts
1.
2.
As the construction worker raises the wood, the wood’s gravitational energy increases.
m = 63 kg
∆y = 3.4 m
(a) On Earth, g = 9.80 N/kg
∆Eg = ?
∆Eg = mg ∆y
= (63 kg)(9.80 N/kg)(3.4 m)
∆Eg = 2.1× 103 J
The astronaut’s gravitational potential energy is 2.1 ¯ 103 J.
(b) On the Moon, g = 1.6 N/kg
∆Eg = ?
∆Eg = mg ∆y
= (63 kg)(1.6 N/kg)(3.4 m)
∆Eg = 3.4 × 102 J
The astronaut’s gravitational potential energy is 3.4 ¯ 102 J.
3. m = 125 g = 0.125 kg
∆y = 3.50 m
(a) ∆Eg = ? (of the pear relative to the ground)
The pear on the branch:
∆Eg = mg ∆y
= (0.125 kg)(9.80 N/kg)(3.50 m)
∆Eg = 4.29 J
The pear at ground level:
∆Eg = mg ∆y
= (0.125 kg)(9.80 N/kg)(0.00 m)
∆Eg = 0.00 J
The gravitational potential energy of the pear on the branch relative to the ground is 4.29 J. The gravitational potential
energy of the pear on the ground relative to the ground is 0.00 J.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 247
(b) ∆Eg = ? (of the pear relative to the branch)
The pear on the branch:
∆Eg = mg ∆y
= (0.125 kg)(9.80 N/kg)(0.00 m)
∆Eg = 0.00 J
The pear on the ground:
∆Eg = mg ∆y
= (0.125 kg)(9.80 N/kg)(−3.50 m)
∆Eg = −4.29 J
4.
The gravitational potential energy of the pear on the branch relative to the branch is 0.00 J. The gravitational potential
energy of the pear on the ground relative to the branch is –4.29 J.
m = 0.15 kg
∆Eg = 22 J
∆y = ?
∆Eg = mg ∆y
∆y =
∆Eg
mg
22 J
(0.15 kg)(9.80 N/kg)
∆y = 15 m
The ball’s maximum height is 15 m above the point where it was hit.
5. Let the subscript g represent gravity, and W represent the weightlifter.
m = 15 kg
∆y = 66 cm = 0.66
(a) Wg = ?
Wg = ( F cos θ )∆d
=
= (mg cos θ )∆d
= (15 kg)(−9.80 N/m)(cos180°)(0.66 m)
Wg = −97 J
The amount of work done by gravity on the mass is –97 J.
(b) WW = ?
WW = ( F cos θ )∆d
= (mg cos θ ) ∆d
= (15 kg)(9.80 N/kg)(cos 0°)(0.66 m)
WW = 97 J
The amount of work done by the weightlifter on the mass is 97 J.
(c) ∆Eg = ?
∆Eg = mg ∆y
= (15 kg)(9.80 N/kg)(0.66 m)
∆Eg = 97 J
The gravitational potential energy of the mass increases by 97 J.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 248
Applying Inquiry Skills
6.
Making Connections
7. (a) V = 32.8 km3 = 3.28 × 1010 m3
∆y = 23.1 m
ρ = 1.00 ¯ 103 kg/m3
∆Eg = ?
First we must determine the mass of the water:
m = ρV
= (1.00 × 103 kg/m3 )(3.28 × 1010 m3 )
m = 3.28 × 1013 kg
To calculate the gravitational potential energy:
∆Eg = mg ∆y
= (3.28 × 1013 kg)(9.80 N/kg)(23.1 m)
∆Eg = 7.43 × 1015 J
The gravitational potential energy of the lake relative to the turbines is 7.43 ¯ 1015 J.
7.43 × 1015 J
(b) This value is approximately
= 6.52 times the annual energy output of the Chukha plant in Bhutan.
1.14 × 1015 J
4.4 THE LAW OF CONSERVATION OF ENERGY
PRACTICE
(Page 197)
Understanding Concepts
1.
2.
3.
The force has done negative work on the ball. The speed of the ball has decreased, which corresponds to a decrease in
kinetic energy. This can only be done with negative work.
If the losses of energy are the same, the only source of kinetic energy is the conversion of gravitational potential. Mass
doesn’t matter (it cancels out in the equation), so they will both acquire the same final velocity.
∆y = 59.4 m
vi = 0.0 m/s
vf = ?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 249
Applying Inquiry Skills
6.
Making Connections
7. (a) V = 32.8 km3 = 3.28 × 1010 m3
∆y = 23.1 m
ρ = 1.00 ¯ 103 kg/m3
∆Eg = ?
First we must determine the mass of the water:
m = ρV
= (1.00 × 103 kg/m3 )(3.28 × 1010 m3 )
m = 3.28 × 1013 kg
To calculate the gravitational potential energy:
∆Eg = mg ∆y
= (3.28 × 1013 kg)(9.80 N/kg)(23.1 m)
∆Eg = 7.43 × 1015 J
The gravitational potential energy of the lake relative to the turbines is 7.43 ¯ 1015 J.
7.43 × 1015 J
(b) This value is approximately
= 6.52 times the annual energy output of the Chukha plant in Bhutan.
1.14 × 1015 J
4.4 THE LAW OF CONSERVATION OF ENERGY
PRACTICE
(Page 197)
Understanding Concepts
1.
2.
3.
The force has done negative work on the ball. The speed of the ball has decreased, which corresponds to a decrease in
kinetic energy. This can only be done with negative work.
If the losses of energy are the same, the only source of kinetic energy is the conversion of gravitational potential. Mass
doesn’t matter (it cancels out in the equation), so they will both acquire the same final velocity.
∆y = 59.4 m
vi = 0.0 m/s
vf = ?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 249
Using conservation of energy, we will have no kinetic energy at the top, and no gravitational energy at the bottom.
ET1 = ET2
1 2
mv
/ f
2
vf = 2 gy1
mgy
/ 1=
= 2(9.80 m/s 2 )(59.4 m)
vf = 34.1 m/s
To convert to km/h:
4.
m   3600 s   1 km 

 34.1  

 = 123 km/h
s   h   1000 m 

The roller coaster reaches a maximum speed of 123 km/h at the bottom of the hill.
vi = v1 = 9.7 m/s
∆y = 4.2 m
vf = v2 = ?
We will use y = 0 at the point of contact on the hill.
ET1 = ET2
1 2
mv1 + mgy1 = 12 mv22 + mgy2
2
v12 + 2 gy1 = v22 + 2 gy2
v22 = v12 + 2 gy1 − 2 gy2
Since y2 = 0, 2gy2 = 0, therefore:
v2 = v12 + 2 gy1
= (9.7 m/s) 2 + 2(9.80 m/s 2 )(4.2 m)
5.
v2 = 13 m/s
The skier’s speed upon touching the hillside is 13 m/s.
∆y = 4.4 ¯ 102 m
v2 = 93 m/s
v1 = ?
ET1 = ET2
1 2
1
mv1 + mgy1 = mv22 + mgy2
2
2
2
v1 + 2 gy1 = v22 + 2 gy2
v12 = v22 + 2 gy2 − 2 gy1
v1 = v22 + 2 g ( y2 − y1 )
= (93 m/s)2 + 2(9.8 m/s 2 )(0 − 440 m)
6.
v1 = 5.0 m/s
The speed of the water at the top of the waterfall is 5.0 m/s.
v1 = 9.7 m/s
∆y = 4.7 m
v2 = ?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 250
ET1 = ET2
1 2
1
mv1 + mgy1 = mv22 + mgy2
2
2
v12 + 2 gy1 = v22 + 2 gy2
v22 = v12 + 2 gy1 − 2 gy2
v2 = v12 + 2 g ( y1 − y2 )
= (9.7 m/s)2 + 2(9.8 m/s 2 )(0 − 4.7 m)
7.
v2 = 1.4 m/s
The cyclist crests the hill at a speed of 1.4 m/s.
v2 = ?
First determine how high the pendulum is vertically raised:
h12 + 24.5 cm 2 = 85.5 cm 2
h1 = 85.5 cm 2 − 24.5 cm 2
h1 = 81.915 cm
h1 + h2 = 85.5 cm
h2 = 85.5 cm − h1
= 85.5 cm − 81.915 cm
h2 = 3.585 cm
Using the value h2 = 3.585 cm, or 0.03585 m, calculate the maximum speed:
ET1 = ET2
1 2
1
mv1 + mgy1 = mv22 + mgy2
2
2
since v1 = 0
2 gy1 = v22 + 2 gy2
v22 = 2 gy1 − 2 gy2
v2 = 2 g ( y1 − y2 )
= 2(9.80 m/s 2 )(0.03585 m − 0)
v2 = 0.838 m/s
The maximum speed of the pendulum bob is 0.838 m/s.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 251
Applying Inquiry Skills
8.
∆y (m)
∆Eg (J)
EK (J)
ET (J)
8.00
6.00
4.00
2.00
0.00
392
294
196
98.0
0.00
0.00
98.0
196
294
392
392
392
392
392
392
Making Connections
9.
A wrecking ball works as a pendulum that is slowly pulled back, increasing its gravitational energy. When it is released,
the gravitational potential energy is converted into kinetic energy which is used to destroy buildings.
PRACTICE
(Page 200)
Understanding Concepts
10. (a) The energy supplied becomes sound and thermal energy through friction.
(b) The energy supplied still produces sound and thermal energy, but some is also converted into kinetic energy.
11. FK = 67 N
∆d = 3.5 m
(a) W = ?
W = ( F cos θ )∆d
= (67 N)(cos180°)(3.5 m)
W = −2.3 × 10 2 J
The amount of work done by friction is –2.3 ¯ 102 J.
(b) Eth = ?
Eth = FK ∆d
= (67 N)(3.5 m)
Eth = 2.3 ×10 2 J
The amount of thermal energy produced is 2.3 ¯ 102 J.
12. Eth = 0.620 J
FK = 0.83 N
∆d = ?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 252
The 0.620 J of energy comes from the work done by friction, therefore:
Eth = FK ∆d
∆d =
Eth
FK
0.620 J
0.83 N
∆d = 0.75 m
The plate slides 0.75 m.
13. m = 22.0 kg
F = 98 N
FK = 87 N
vi = 0.0 m/s
∆d = 1.2 m
vf = ?
W = Eth + Ek
=
1
( F cos θ )∆d = FK ∆d + mv22
2
∆d − FK ∆d
F
θ
(
cos
)
v22 =
1
m
2
( F cos θ )∆d − FK ∆d
v2 =
0.5m
=
(98 N)(cos 0°)(1.2 m) − 87 N(1.2 m)
0.5(22.0 kg)
v2 = 1.1 m/s
The speed of the cabinet after it moves 1.2 m is 1.1 m/s.
14. m = 0.057 kg
∆d = 25 cm = 0.25 m
vf = 5.7 cm/s = 5.7 ¯ 10–2 m/s
FK = 0.15 N
vi = ?
EK1 = Eth + EK2
1 2
1
mv1 = FK ∆d + mv22
2
2
1
FK ∆d + mv22
2
v12 =
1
m
2
1
FK ∆d + mv22
2
v1 =
0.5m
1
(0.15 N)(0.25 m) + (0.057 kg)(5.7 ×10 −2 m/s) 2
2
=
0.5(0.057 kg)
v1 = 1.1 m/s
The initial speed of the pen is 1.1 m/s.
Applying Inquiry Skills
15. (a) The law of conservation of energy could be verified by observing how close to its original position the pendulum
would return after a swing.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 253
(b) Some sources of error would be loss of energy due to friction at the point of attachment and air friction of the moving
pendulum. Some error may be observed if the string is somewhat elastic.
Making Connections
16. The oil circulation system is an attempt to minimize the frictional forces between the moving parts of the engine, and
thereby reduce the loss (and damage) due to the thermal energy caused by friction. The water (coolant) circulation is used
to absorb thermal energy from the engine and dissipate it rapidly into the air to prevent damage from overheating.
Section 4.4 Questions
(Pages 201–202)
Understanding Concepts
1.
Roller coasters are gravity rides that have an initial input of gravitational potential energy that is converted to kinetic (and
back into gravitational) energy throughout the ride. To give them this initial energy, they must be pulled up the largest hill
at the beginning.
2. m = 0.052 kg
∆y = 11 cm = 0.11 m
y=0
(a) ∆Eg = ?
∆Eg = mg ∆y
= (0.052 kg)(9.80 m/s 2 )(0.11 m)
∆Eg = 0.056 J
The initial gravitational potential energy of the egg’s contents is 0.056 J.
(b) ∆Eg = ?
∆Eg = mg ∆y
= (0.052 kg)(9.80 m/s 2 )(0.0 m)
∆Eg = 0.0 J
The final gravitational potential energy of the egg’s contents is 0.0 J.
(c) ∆Eg = 0.0 – 0.056 = –0.056 J
The change in gravitational potential energy as the egg’s contents fall is –0.056 J.
(d) EK = ?
v=?
The kinetic energy will be equal to the loss of gravitational potential, so EK = 0.056 J.
1
EK = mv 2
2
2 EK
v=
m
=
3.
2(0.056 J)
0.052 kg
v = 1.5 m/s
The kinetic energy is 0.056 J. The speed of the egg’s contents just before hitting the pan is 1.5 m/s.
∆y = 1.2 m
vf = 9.9 m/s
vi = ?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 254
ET1 = ET2
1 2
1
mv1 + mgy1 = mv22 + mgy2
2
2
v12 + 2 gy1 = v22 + 2 gy2
v12 = v22 + 2 gy2 − 2 gy1
v1 = v22 + 2 g ( y2 − y1 )
= (9.9 m/s)2 + 2(9.8 m/s 2 )(1.2 m − 0 m)
v1 = 11 m/s
The initial speed of the ball was 11 m/s.
4. v1= 3.74 m/s
∆y = 8.74 m
m = 7.12 ¯ 104 kg
(a) ∆Eg = ?
∆Eg = mg ∆y
= (7.12 × 104 kg)(9.80 m/s 2 )(8.74 m)
∆Eg = 6.10 × 106 J
The gravitational potential energy of the mass of water at the top of the waterfall is 6.10 ¯ 106 J.
(b) v2 = ?
ET1 = ET2
1 2
1
mv1 + mgy1 = mv22 + mgy2
2
2
2
v1 + 2 gy1 = v22 + 2 gy2
v22 = v12 + 2 gy1 − 2 gy2
v2 = v12 + 2 g ( y1 − y2 )
= (3.74 m/s)2 + 2(9.80 m/s 2 )(8.74 m − 0 m)
v2 = 13.6 m/s
The speed of the water at the bottom of the waterfall is 13.6 m/s.
5. (a)
First, determine the vertical height above the bottom of the swing:
h
cos 48° = 1
3.7 m
h1 = 3.7 m (cos 48° )
h1 = 2.476 m
Copyright © 2003 Nelson
Chapter 4 Work and Energy 255
h1 + h2 = 3.7 m
h2 = 3.7 m − h1
= 3.7 m − 2.476 m
h2 = 1.224 m
Now we can calculate the acrobat’s speed:
ET1 = ET2
1 2
1
mv1 + mgy1 = mv22 + mgy2
2
2
Since v1 = 0:
2 gy1 = v22 + 2 gy2
v22 = 2 gy1 − 2 gy2
v2 = 2 g ( y1 − y2 )
= 2(9.80 m/s 2 )(1.224 m − 0 m)
v2 = 4.9 m/s
The acrobat’s speed at the bottom of the swing is 4.9 m/s.
(b) Due to conservation of energy, the maximum height on the other side is equal to the starting height of the acrobat.
6. m = 55 kg
∆d = 3.7 m
FK = 41.5 N
v1 = 65.7 cm/s = 0.657 m/s
v2 = 7.19 m/s
φ=?
Relating φ to h:
h
11.7
h = 11.7 sin φ
sin φ =
Using conservation of energy:
ET1 = ET2
EK1 + ∆Eg = EK 2 + Eth
1 2
1
mv1 + mgh = mv22 + FK ∆d
2
2
1
1
mgh = mv22 + FK ∆d − mv12
2
2
1
m(v22 − v12 ) + FK ∆d
2
h=
mg
1
2
2
(55.0 kg) ( 7.19 m/s ) − (0.657 m/s ) + (41.5 N)(11.7 m)
2
11.7 sin φ =
(55.0 kg)(9.80 m/s 2 )
sin φ = 0.30054
(
)
φ = 17.5°
The angle is 17.5°.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 256
7.
v1 = 0.0 m/s
v2 = 6.8 m/s
Since the skateboarder starts even with the height of the centre of the circle, the total vertical drop at the bottom will be
equal to the radius of the circle.
r = y1 = ?
ET1 = ET2
1 2
1
mv1 + mgy1 = mv22 + mgy2
2
2
v12 + 2 gy1 = v22 + 2 gy2
Since v1 = 0, and y2 = 0:
2 gy1 = v22
y1 =
v22
2g
6.8 m/s 2
2(9.8 m/s 2 )
y1 = 2.4 m
The radius of the half-pipe is 2.4 m.
8. m = 55 g = 5.5 ¯ 10–2 kg
v1 = 1.9 m/s
∆d = 54 cm = 0.54 m
v2 = 0.0 m/s
(a) µK = ?
First, solve for FK:
ET1 = ET2
=
EK = Eth
1 2
mv = FK ∆d
2
mv 2
FK =
2 ∆d
(5.5 × 10−2 kg)(1.9 m/s)2
=
2(0.54 m)
FK = 0.1838 N
Now calculate the normal force:
ΣFy = ma y = 0
FN − mg = 0
FN = mg
= (5.5 × 10−2 kg)(9.80 m/s 2 )
FN = 0.539 N
Solve for µK:
µK =
FK
FN
0.1838 N
0.539 N
µ K = 0.34
The coefficient of kinetic friction is 0.34.
=
Copyright © 2003 Nelson
Chapter 4 Work and Energy 257
(b) µK =?
First, solve for the acceleration:
vf2 = vi2 + 2a∆d
a=
=
vf2 − vi2
2 ∆d
(1.9 m/s )2 − (0 m/s )2
2(0.54 m)
a = 3.3426 m/s 2
Using the FBD, calculate the magnitude of kinetic friction:
ΣFx = max
FK = max
= (5.5 × 10−2 kg)(3.3426 m/s 2 )
FK = 0.1838 N
Calculate the normal force:
ΣFy = ma y = 0
FN − mg = 0
FN = mg
= (5.5 × 10−2 kg)(9.80 m/s 2 )
FN = 0.539 N
Calculate the coefficient of kinetic friction:
F
µK = K
FN
0.1838 N
0.539 N
µK = 0.34
The coefficient of kinetic friction is 0.34.
(c) The kinetic energy is converted into thermal energy.
9. vi = 85 km/h = 23.61 m/s
vf = 0.0 m/s
∆d = 47 m
FK = 7.4 ¯ 103 N
(a) Eth = ?
Eth = FK ∆d
=
= (7.4 × 103 N)(47 m)
Eth = 3.5 × 105 J
The amount of thermal energy produced is 3.5 ¯ 105 J.
(b) Before the skid, the thermal energy was kinetic energy.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 258
(c) m = ?
ET1 = ET2
EK = Eth
1 2
mv = FK ∆d
2
2 F ∆d
m = K2
v
2(7.4 × 103 N)(47 m)
=
2
( 23.61 m/s )
m = 1.2 ×103 kg
The mass of the car is 1.2 ¯ 103 kg.
(d) µK = ?
First calculate the normal force:
ΣFy = ma y = 0
FN − mg = 0
FN = mg
= (1.248 × 103 kg)(9.80 m/s 2 )
FN = 12 228
To calculate the coefficient of kinetic friction:
F
µK = K
FN
7.4 ×103 J
12228 J
µK = 0.61
The coefficient of kinetic friction is 0.61.
10. m = 22 kg
∆d = 2.5 m
θ = 44°
FK = 79 N
(a) W = ?
W = ( FK cos θ )∆d
=
= (79 N)(cos180°)(2.5 m)
W = −2.0 × 102 J
The work done by friction is –2.0 ¯ 102 J.
(b) EK = ?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 259
First, determine the vertical drop:
h
sin 44° =
2.5
h = 2.5sin 44°
h = 1.737 m
To calculate the final kinetic energy:
ET1 = ET2
∆Eg = EK + Eth
mg ∆y = EK + FK ∆d
EK = mg ∆y − FK ∆d
= (9.8 N/kg)(22 kg)(1.737 m) − (79 N)(2.5 m)
EK = 1.8 ×102 J
The box’s final kinetic energy is 1.8 ¯ 102 J.
(c) Eth = ?
Eth = FK ∆d
= (79 N)(2.5 m)
Eth = 2.0 ×10 2 J
The thermal energy produced is 2.0 ¯ 102 J.
Applying Inquiry Skills
11. (a) The loss of gravitational potential energy from the first lift to the second is equal to the amount of thermal energy
produced.
(b) The equation needed would be:
ET1 = ET2
∆Eg1 = ∆Eg2 + Eth
mgy1 = mgy2 + Eth
Eth = mgy1 − mgy2
Eth = mg ( y1 − y2 )
(c) You could determine the height at each point by standing a known distance from the base of the ride and measure the
angle up for each height. Using simple trigonometric ratios, you could estimate the amount of thermal energy
produced.
Making Connections
12. A trebuchet is very accurate. The raised mass supplies a fixed amount of gravitational potential energy. This potential
energy is converted into the kinetic energy of the projectile.
4.5 ELASTIC POTENTIAL ENERGY AND SIMPLE HARMONIC MOTION
PRACTICE
(Pages 206–207)
Understanding Concepts
1.
The higher the spring constant, the more force is needed to stretch the spring the same amount. For this reason, spring A
would be more difficult to stretch than spring B.
2. The spring would exert a southward force on you.
3. k = 25 N/m
(a) x = 16 cm = 0.16 m
Fx = kx
= (25 N/m)(0.16 m)
Fx = 4.0 N
Copyright © 2003 Nelson
Chapter 4 Work and Energy 260
First, determine the vertical drop:
h
sin 44° =
2.5
h = 2.5sin 44°
h = 1.737 m
To calculate the final kinetic energy:
ET1 = ET2
∆Eg = EK + Eth
mg ∆y = EK + FK ∆d
EK = mg ∆y − FK ∆d
= (9.8 N/kg)(22 kg)(1.737 m) − (79 N)(2.5 m)
EK = 1.8 ×102 J
The box’s final kinetic energy is 1.8 ¯ 102 J.
(c) Eth = ?
Eth = FK ∆d
= (79 N)(2.5 m)
Eth = 2.0 ×10 2 J
The thermal energy produced is 2.0 ¯ 102 J.
Applying Inquiry Skills
11. (a) The loss of gravitational potential energy from the first lift to the second is equal to the amount of thermal energy
produced.
(b) The equation needed would be:
ET1 = ET2
∆Eg1 = ∆Eg2 + Eth
mgy1 = mgy2 + Eth
Eth = mgy1 − mgy2
Eth = mg ( y1 − y2 )
(c) You could determine the height at each point by standing a known distance from the base of the ride and measure the
angle up for each height. Using simple trigonometric ratios, you could estimate the amount of thermal energy
produced.
Making Connections
12. A trebuchet is very accurate. The raised mass supplies a fixed amount of gravitational potential energy. This potential
energy is converted into the kinetic energy of the projectile.
4.5 ELASTIC POTENTIAL ENERGY AND SIMPLE HARMONIC MOTION
PRACTICE
(Pages 206–207)
Understanding Concepts
1.
The higher the spring constant, the more force is needed to stretch the spring the same amount. For this reason, spring A
would be more difficult to stretch than spring B.
2. The spring would exert a southward force on you.
3. k = 25 N/m
(a) x = 16 cm = 0.16 m
Fx = kx
= (25 N/m)(0.16 m)
Fx = 4.0 N
Copyright © 2003 Nelson
Chapter 4 Work and Energy 260
x = 32 cm = 0.32 m
Fx = kx
= (25 n/m)(0.32 m)
Fx = 8.0 N
The magnitude of force would be 4.0 N for a stretch of 16 cm, and 8.0 N for a stretch of 32 cm.
(b) x = 16 cm = 0.16m
Fx = −kx
= −(25 N/m)(0.16 m)
Fx = −4.0 N
x = 32 cm = 0.32 m
Fx = −kx
4.
= −(25 N/m)(0.32 m)
Fx = −8.0 N
The magnitudes of the forces are 4.0 N and 8.0 N, respectively.
k = 3.2 ¯ 102 N/m
x = 2.0 cm = 2.0 ¯ 10–2 m
Fx = −kx
= −(3.2 × 102 N/m)(2.0 × 10−2 m)
Fx = −6.4 N
The magnitude of the force applied by the air is 6.4 N.
5. m = 1.37 kg
k = 5.20 ¯ 102 N/m
(a) x = ?
ΣFy = ma = 0
Fx − mg = 0
kx = mg
mg
k
(1.37 kg)(9.80 N/kg)
=
5.20 × 102 N/m
x = 0.0258 m
x=
The spring stretches 0.0258 m.
(b) x = 1.59 cm = 0.0159 m
ΣFy = ?
ΣFy = Fx − mg
= kx − mg
= (5.20 × 102 N/m)(0.0159 m) − (1.37 kg)(9.80 N/kg)
ΣFy = −5.16 N
The net force on the fish is 5.16 N [down].
(c) x = 2.05 cm = 0.0205 m
a=?
ΣFy = ma
Fx − mg = ma
kx − mg
m
(5.20 × 102 N/m)(0.0205 m) − (1.37 kg)(9.80 N/kg)
=
1.37 kg
a=
a = −2.02 m/s 2
The acceleration of the fish is 2.02 m/s2 [down]
Copyright © 2003 Nelson
Chapter 4 Work and Energy 261
Applying Inquiry Skills
6. (a)
(b) The slope of the line is negative.
Making Connections
7. (a)
Mass (kg)
1.00
2.00
3.00
4.00
5.00
6.00
7.00
8.00
Stretch (m)
0.122
0.245
0.368
0.490
0.612
0.735
0.858
0.980
(b)
(c) The mass value may not be correct if the value of g is different then where it was calibrated. The weight value would
be accurate anywhere (even on the Moon).
Copyright © 2003 Nelson
Chapter 4 Work and Energy 262
PRACTICE
(Page 211)
Understanding Concepts
8. (a) The graph shown is the force applied by the spring. Since the spring is being stretched to the right (positive x), the force
exerted by the spring will be to the left (negative). This is what is shown on the graph.
(b) The force constant is the slope of the graph. Since the equation Fx = kx relates the force exerted by the spring, we must
change the direction (i.e., the sign) of the force,
− Fx
k=
x
−(−15 N)
=
0.40 m
k = 38 N/m
The force constant of the spring is 38 N/m.
(c) x = 35 cm = 0.35 m
The energy stored is the area between the curve and the x-axis.
1
Ee = A = bh
2
1
= (0.35 m)(13 N)
2
Ee = 2.3 J
The elastic potential energy is 2.3 J.
9. k = 9.0 ¯ 103 N/m
(a) x = 1.0 cm = 0.010 m
Ee = ?
1
Ee = kx 2
2
1
= (9.0 ×103 N/m)(0.010 m)2
2
Ee = 0.45 J
The elastic potential energy stored by the spring is 0.45 J.
(b) x = –2.0 cm = –0.020 m
Ee = ?
1
Ee = kx 2
2
1
= (9.0 ×103 N/m)(−0.020 m)2
2
Ee = 1.8 J
The elastic potential energy stored in the spring is 1.8 J.
10. m = 7.8 g = 0.0078 kg
k = 3.5 ¯ 102 N/m
x = –4.5 cm = –0.045 m
(a) Ee = ?
1
Ee = kx 2
2
1
= (3.5 × 102 N/m)( −0.045 kg) 2
2
Ee = 0.35 J
The elastic potential energy of the spring is 0.35 J.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 263
(b) v = ?
ET1 = ET2
Ee = EK
1 2 1 2
kx = mv
2
2
v=
=
kx 2
m
(3.5 × 102 N/m)(−0.045 m) 2
0.0078 kg
v = 9.5 m/s
The speed of the dart as it leaves the toy is 9.5 m/s.
11. m = 3.5 ¯ 10–3 kg
k = 9.5 N/m
(a) ∆y = 5.7 cm = 0.057 m
x=?
ET1 = ET2
Ee = ∆Eg
1 2
kx = mg ∆y
2
2mg ∆y
x=
k
2(3.5 × 10−3 kg)(9.8 N/kg)(0.057 m)
9.5 N/m
x = 0.020 m
The spring must be compressed by 0.020 m, or 2.0 cm.
(b) If friction was not negligible, the amount of compression would need to be increased. The energy supplied by the
compressed spring would now become both gravitational potential and thermal energy. In order to have the same final
gravitational energy, the spring would need to be compressed more.
12. m = 0.20 kg
k = 55 N/m
(a) ∆y = 1.5 cm = 0.015 m
v=?
ET1 = ET2
=
∆Eg = EK + Ee
1 2 1 2
mv + kx
2
2
mv 2 = 2 mg ∆y − kx 2
mg ∆y =
v=
=
2mg ∆y − kx 2
m
2(0.20 kg)(9.8 N/kg)(0.015 m) − (55 N/m)(0.015 m)2
0.20 kg
v = 0.48 m/s
The speed of the mass is 0.48 m/s.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 264
(b) For the fall, ∆y = x. At maximum fall, there will be no kinetic energy.
ET1 = ET2
∆Eg = Ee
1 2
kx
2
kx 2 − 2mg ∆y = 0
mg ∆y =
(55) x2 − 2(0.20)(9.8) x = 0
(55) x 2 − (3.92 ) x = 0
x(55 x − 3.92) = 0
55 x − 3.92 = 0
or
x=0
3.92
55
x = 0.071m or x = 0 m
The value x = 0 refers to the moment of release, so the maximum stretch will be 0.071 m.
13. k = 12 N/m
∆y = 93.0 cm = 0.93 m
m = 8.3 ¯ 10–3 kg
x = 4.0 cm = 0.040 m
∆d = ?
x=
Analyze as a projectile motion question. First, determine horizontal speed at launch:
ET1 = ET2
Ee = EK
1 2 1 2
kx = mv
2
2
v=
=
kx 2
m
(12 N/m)(−0.040 m)2
8.3 × 10 −3 kg
v = 1.521 m/s
Choosing down as positive, the time for the marble to drop 0.93 m vertically is:
1
∆y = vi ∆t + a(∆t ) 2
2
Since vi = 0:
1
a ( ∆t ) 2
2
2 ∆y
∆t =
a
∆y =
2(0.93 m)
9.8 m/s 2
∆t = 0.4357 s
=
The horizontal distance travelled during this time is:
∆d = v ∆t
= (1.521 m/s)(0.4357 s)
∆d = 0.66 m
The marble travels 0.66 m horizontally before hitting the floor.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 265
Applying Inquiry Skills
14. (a) The measurement needed would be the mass of the largest friend.
(b) Assuming a largest mass of 115 kg,
ΣFy = ma = 0
Fx − mg = 0
kx = mg
mg
x
(115 kg)(9.80 N/kg)
=
0.75 m
k = 1.5 × 103 N/m
The approximate force constant is 1.5 ¯ 103 N/m.
k=
Making Connections
15. m = 2.0 ¯ 102 µg = 2.0 ¯ 10–7 kg
∆y = 65 mm = 0.065 m
x = 75 cm = 0.75 m
Ee = ?
ET1 = ET2
Ee = x (∆Eg )
= x(mg ∆y )
= 0.75 m(2.0 ×10 −7 kg)(9.8 N/kg)(0.065 m)
Ee = 9.6 × 10−8 J
The initial quantity of elastic potential energy is 9.6 ¯ 10–8 J.
PRACTICE
(Pages 214–215)
Understanding Concepts
16. (a) The maximum displacement from the rest position will be at the top and bottom of the bounce.
(b) The speed is a maximum at the rest position.
(c) The speed will be zero at the top and the bottom of the bounce.
(d) The acceleration will be a maximum at the top and the bottom of the bounce.
(e) The acceleration will be zero at the rest position.
17. T = ?
f=?
(a) number of vibrations = 12
t = 48 s
total time
T=
number of complete vibrations
48
=
12
T = 4.0 s
number of complete vibrations
total time
12
=
48
f = 0.25 Hz
f =
Copyright © 2003 Nelson
Chapter 4 Work and Energy 266
Alternatively, you could calculate frequency by using the equation:
1
f =
T
1
=
4.0 s
f = 0.25 Hz
The period is 4.0 s, and the frequency is 0.25 Hz.
(b) number of vibrations = 210
t = 1 min = 60 s
total time
T=
number of complete vibrations
60
=
210
T = 0.29 s
number of complete vibrations
total time
210
=
60
f = 3.5 Hz
The period is 0.29 s, and the frequency is 3.5 Hz.
(c) number of vibrations = 2200
t = 5.0 s
total time
T=
number of complete vibrations
5.0
=
2200
T = 2.3 × 10 −3 s
f =
number of complete vibrations
total time
2200
=
5.0
f = 4.4 ×10 2 Hz
The period is 2.3 ¯ 10–3 s, and the frequency is 4.4 ¯ 102 Hz.
18. m = 0.25 kg
A = 8.5 cm
k = 1.4 ¯ 102 N/m
(a) d = ?
f =
During each cycle, the mass moves 4 amplitudes, or 4A. In 5 cycles, the mass will move:
d = 5× 4A
= 5 × 4(8.5 cm)
d = 1.7 × 10 2 cm
The mass moves 1.7 ¯ 102 cm in the first five cycles.
(b) T = ?
T = 2π
m
k
0.25 kg
1.4 ×10 2 N/m
T = 0.27 s
The period of vibration is 0.27 s.
= 2π
Copyright © 2003 Nelson
Chapter 4 Work and Energy 267
19. m = 0.10 kg
f = 2.5 Hz
k=?
f =
1
2π
k
m
k
m
2π f =
k = 4π 2 f 2 m
= 4π 2 (2.5 Hz)2 (0.10 kg)
k = 25 N/m
The force constant of the spring is 25 N/m.
20. k = 1.4 ¯ 102 N/m
T = 0.85 s
m=?
m
T = 2π
k
T
=
2π
m
k
T 2k
4π 2
(0.85 s) 2 (1.4 × 10 2 N/m)
=
4π 2
m = 2.6 kg
The mass would have to be 2.6 kg.
m=
Applying Inquiry Skills
21. Examining the base SI units for each:
x
m
=
a
m
 2
s 
= s
2
x
=s
a
m
kg
=
k
N
 
m
=
kg ⋅ m
 kg ⋅ m 
 2 
 s 
= s2
m
=s
k
Therefore, they are dimensionally equivalent.
Making Connections
22. number of vibrations = 6.0
t = 8.0 s
(a) k = ?
First we must calculate the frequency:
number of complete vibrations
f =
total time
6.0
=
8.0 s
f = 0.75 Hz
Copyright © 2003 Nelson
Chapter 4 Work and Energy 268
Using a mass of 75 kg:
f =
2π f =
1
2π
k
m
k
m
k = 4π 2 f 2 m
= 4π 2 (0.75 Hz) 2 (75 kg)
k = 1.7 × 103 N/m
The force constant is 1.7 ¯ 103 N/m.
(b) No, you are not undergoing SHM. When you leave the trampoline, there is a period of time when it is not exerting any
force on you. SHM requires that the force be proportional to the displacement.
PRACTICE
(Pages 217–218)
Understanding Concepts
23. (a) The speed is zero at lengths of 12 cm and 38 cm.
(b) The maximum speed will be at the rest position. The rest position will be halfway between the minimum and maximum
extensions:
12 + 38
= 25 cm
2
(c) The amplitude is 38 – 25 = 13 cm.
24. Ee = 5.64 J
m = 0.128 kg
k = 244 N/m
(a) x = ?
The maximum energy is constant, and all elastic potential at either end of the system, at the maximum amplitude.
1
Ee = kx 2
2
2 Ee
x=
k
2(5.64 J)
244 N/m
x = 0.215 m
The amplitude of the vibration is 0.215 m.
(b) v = ?
=
Approach 1: All of the energy will be kinetic as it passes through the rest position.
EK = mv 2
v=
=
2 EK
m
2(5.64 J)
0.128 kg
v = 9.39 m/s
Copyright © 2003 Nelson
Chapter 4 Work and Energy 269
Approach 2: All of the energy stored in the spring at maximum compression will be converted to kinetic energy.
ET1 = ET2
1 2 1 2
kx = mv
2
2
kx 2
m
v=
(244 N/m)(0.215 m)2
0.128 kg
=
v = 9.39 m/s
The speed is 9.39 m/s regardless of the approach used.
(c) x2 = 15.5 cm = 0.155 m
v=?
ET1 = ET2
1 2
1
1
kxmax = mv 2 + kx22
2
2
2
2
2
−
kx
kx
2
v 2 = max
m
v=
=
2
− x22 )
k ( xmax
m
(
(244 N/m) (0.215 m ) − (0.155 m )
2
2
)
0.128 kg
v = 6.51 m/s
The speed of the mass is 6.51 m/s.
25. x = 0.18 m
m = 58 g = 0.058 kg
k = 36 N/m
(a) Ee = ?
v=?
Maximum energy during maximum stretch/compression of the spring:
1
Emax = kx 2
2
1
= (36 N/m)(0.18 m) 2
2
Emax = 0.58 J
This will all be kinetic energy at the rest position.
1
EK = mv 2
2
2 EK
v=
m
=
2(0.58 J)
0.058 kg
v = 4.5 m/s
The maximum energy of the system is 0.58 J. The maximum speed of the mass is 4.5 m/s.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 270
(b) x = ?
Double the energy would be:
′ = 2 × Ee
Emax
1
= 2 × kx 2
2
= (36 N/m)(0.18 m) 2
′ = 1.1664 J
Emax
This is all stored as elastic potential energy at the full amplitude.
1
Ee = kx 2
2
2 Ee
x=
k
2(1.1664 J)
36 N/m
x = 0.25 m
The amplitude of vibration required would be 0.25 m.
(c) v = ?
1
EK = mv 2
2
2 EK
v=
m
=
=
2(1.1664 J)
0.058 kg
v = 6.3 m/s
The maximum speed of the mass is 6.3 m/s.
26. For the maximum speed, all of the elastic energy will be converted to kinetic:
ET1 = ET2
1 2 1 2
kx = mv
2
2
kx 2
m
v=
v=x
k
m
For a SHM system, x = A, therefore:
f =
2π f =
1
2π
k
m
k
m
Substituting above:
k
m
= A(2π f )
v=x
v = 2π fA
Copyright © 2003 Nelson
Chapter 4 Work and Energy 271
Applying Inquiry Skills
27. (a)
N
 
k 2
m
(A − x 2 ) =   (m 2 − m 2 )
m
kg
 kg ⋅ m 


m ⋅ s2  2
= 
(m )
kg
m2
s2
= m/s
The dimensions are m/s.
(b) This expression is to calculate the speed of an object at a location during SHM.
=
Making Connections
28. (a) Tuning fork prongs have slow damping to produce a long tone.
(b) The voltmeter needle has fast damping to stabilize the reading quickly.
(c) A guitar string has slow damping to play the note as long as possible.
(d) Saloon doors have medium damping to keep the doors from swinging too much, but they still swing back and forth.
(e) The string of the bow has fast damping to prevent dangerous vibration.
Section 4.5 Questions
(Pages 218–219)
Understanding Concepts
1.
When the two students pull on either end of the spring, it will not stretch as much as when it is pulled by both students
while attached to the wall. When they both pull on it from the same side, the wall pulls back with equal force. When they
pull from opposite ends, the force will be half as much, and the stretch will also be half as much.
2. The elastic potential energy is the same when the spring is stretched or compressed 2.0 cm. The amount of energy stored
only depends on the magnitude of the distortion, not the direction.
3. Harmonic means that it is regularly repeated, symmetrical motion.
1
4. (a) Period is inversely proportional to the frequency, T ∝ .
f
(b) The acceleration is directly proportional to the displacement, a ∝ x .
1
(c) The period is inversely proportional to square root of the force constant, T ∝
.
k
(d) The maximum speed is directly proportional to the amplitude, v ∝ A .
5. m = 62 kg
k = 2.4 ¯ 103 N/m
x=?
ΣFy = ma = 0
6 Fx − mg = 0
6kx = mg
mg
6k
(62 kg)(9.8 N/kg)
=
6(2.4 × 103 N/m)
x = 0.042 m
The compression of each spring is 0.042 m.
x=
Copyright © 2003 Nelson
Chapter 4 Work and Energy 272
6.
7.
k = 78 N/m
x = 2.3 cm = 0.023 m
Fx = ?
Fx = kx
= (78 N/m)(0.023 m)
Fx = 1.8 N
The magnitude of force is 1.8 N.
x1 = 1.85 cm
Fx1 = 85.5 N
x2 = 4.95 cm = 0.0495 m
Fx2 = ?
First we must calculate k:
Fx1 = kx1
k=
Fx1
x1
85.5 N
0.0185 m
k = 4621.6 N/m
=
Solve for x2:
8.
Fx 2 = kx2
= (4621.6 N/m)(0.0495 m)
Fx 2 = 229 N
The force required is 229 N.
m = 97 kg
k = 2.2 ¯ 103 N/m
a = 0.45 m/s2
x=?
Ignoring friction to calculate the applied force:
ΣFx = ma
FA = ma
= (97 kg)(0.45 m/s 2 )
FA = 43.65 N
This force is the force exerted on the spring:
Fx = kx
x=
Fx
k
43.65 N
2.2 × 103 N/m
x = 0.020 m
The spring stretches 0.020 m, or 2.0 ¯ 10–2 m.
9. m = 289 g = 0.289 kg
k = 18.7 N/m
(a) x = 10.0 cm = 0.100 m
ΣFy = ?
=
a=?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 273
To calculate the net force:
ΣFy = Fx − mg
= kx − mg
= (18.7 N/m)(0.100 m) − (0.289 kg)(9.80 N/kg)
ΣFy = −0.962 N
To calculate acceleration:
ΣFy = ma
a=
ΣFy
m
−0.962 N
=
0.289 kg
a = −3.33 m/s 2
The net force is 0.962 N [down], and the acceleration is 3.33 m/s2 [down].
(b) x = ?
ΣFy = ma = 0
Fx − mg = 0
kx = mg
mg
k
(0.289 kg)(9.80 N/kg)
=
18.7 N/m
x = 0.151 m
The spring will be stretched by 0.151 m.
10. m = 64.5 kg
∆y = 48.0 m –12.5 m = 35.5 m
k = 65.5 N/m
x = 35.5 m – 10.1 m = 25.4 m.
v=?
∆Eg = Ee + EK
x=
1 2 1 2
kx + mv
2
2
2
mv = 2 mg ∆y − kx 2
mg ∆y =
v=
=
2mg ∆y − kx 2
m
2(64.5 kg)(9.80 N/kg)(35.5 m) − (65.5 N/m)(25.4 m) 2
64.5 kg
v = 6.37 m/s
The jumper’s speed at a height of 12.5 m above the water is 6.37 m/s.
11. Fx = 8.6 N
x = 9.4 cm = 0.094 m
(a) k = ?
Fx = kx
Fx
x
8.6 N
=
0.094 m
k = 91 N/m
The force constant of the spring is 91 N/m.
k=
Copyright © 2003 Nelson
Chapter 4 Work and Energy 274
(b) Ee = ?
1 2
kx
2
1
= (91.49 N/m)(0.094 m) 2
2
Ee = 0.40 J
The maximum energy of the spring is 0.40 J.
12. x = 1.0 ¯ 10–7 m
Ee = 1.0 ¯ 10–13 J
k=?
1
Ee = kx 2
2
2E
k = 2e
x
2(1.0 × 10 −13 J)
=
(1.0 × 10−7 m) 2
Ee =
k = 2.0 × 101 N/m
The force constant is 2.0 ¯ 101 N/m.
13. m = 22 kg
θ = 29°
k = 8.9 ¯ 102 N/m
x = 0.30 m
d=?
To calculate the total vertical drop, ∆y:
ET1 = ET2
∆Eg = Ee
1 2
kx
2
kx 2
∆y =
2mg
mg ∆y =
(8.9 × 102 N/m)(0.30 m) 2
2(22 kg)(9.8 N/kg)
∆y = 0.1858 m
=
To calculate the distance:
∆y
d
∆y
d=
sin θ
0.1858 m
=
sin 29°
d = 0.38 m
The crate slides 0.38 m along the ramp.
14. m = 0.20 kg
k = 28 N/m
∆y = ?
sin θ =
Copyright © 2003 Nelson
Chapter 4 Work and Energy 275
For fall, ∆y = x. At maximum fall, there will be no kinetic energy.
ET1 = ET2
∆Eg = Ee
1 2
kx
2
kx 2 − 2mg ∆y = 0
mg ∆y =
28 x 2 − 2(0.20)(9.8) x = 0
28 x 2 − 3.92 x = 0
x(28 x − 3.92) = 0
28 x − 3.92 = 0
or
x=0
3.92
28
x = 0.14 m or x = 0
Since x = 0 refers to the moment of release, the maximum stretch will be 0.14 m.
x=
Applying Inquiry Skills
15. (a) (i) Line A (total energy is constant)
(ii) Line B (no kinetic energy at maximum stretch)
(iii) Line C (maximum elastic potential energy a maximum stretch)
(b) By observation from the graph, A = 10.0 cm = 0.100 m
(c) At the end, the total energy of 5.0 J is all elastic potential energy,
1
Ee = kx 2
2
2E
k = 2e
x
2(5.0 J)
=
(0.100 m)2
k = 1.0 ×103 N/m
The force constant of the spring is 1.0 ¯ 103 N/m.
(d) Maximum kinetic energy is 5.0 J when there is no elastic potential energy.
1
EK = mv 2
2
2 EK
v=
m
2(5.0 J)
0.12 m
v = 9.1 m/s
The maximum speed of the mass is 9.1 m/s.
16. (a) If the spring were cut in two, it would take more force to stretch the spring the same amount because each coil would
need to be moved twice as far. This will cause the force constant to double for the remaining two half springs. If a mass
hung from two identical springs attached together caused them to stretch 36 cm, than each of them would stretch half
of that amount. Since each bears the same weight, the force constant of each would only allow them to stretch 18 cm
under the same force, indicating a force constant that is twice as great.
(b) Answers will vary.
=
Making Connections
17. m = 5.5 ¯ 102 kg
number of vibrations = 6.0
t = 3.5 s
k=?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 276
Calculate the frequency:
number of complete vibrations
total time
6.0
=
3.5 s
f = 1.7143 Hz
f =
To calculate the force constant:
1
f =
2π
2π f =
k
m
k
m
k = 4π 2 f 2 m
= 4π 2 (1.7143 Hz)2 (5.5 × 102 kg)
k = 6.4 × 104 N/m
The force constant of each spring is 6.4 ¯ 104 N/m.
18. a = 25g
f = 8.9 Hz
A=?
T = 2π
A
a
1
= 2π
f
A
a
A
a
a
A= 2 2
4π f
1
=
2π f
25(9.8 m/s 2 )
4π 2 (8.9 Hz) 2
A = 0.078 m
The minimum amplitude is 0.078 m, or 7.8 ¯ 10–2 m.
=
CHAPTER 4 LAB ACTIVITIES
Activity 4.4.1: Applying the Law of Conservation of Energy
(Page 220)
(a) Energy conversions would be gravitational potential to kinetic to move the hands.
(b) Different types include water clocks, hour-glass clocks, and pendulum clocks.
(c) – (f) Answers will vary based on the students’ choice of design.
Investigation 4.5.1: Testing Real Springs
(Pages 220–221)
Questions
(i) By graphing the stretch as a function of the applied force we can learn the relationship is linear.
(ii) The force constant of two combined springs is always less than the force constant for either spring individually.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 277
CHAPTER 4 REVIEW
(Pages 226–229)
Understanding Concepts
1. (a) No work is done because the force is perpendicular to the displacement.
(b) The work is negative because the student is exerting a force opposite the direction of motion.
(c) The work is negative because gravity is exerting a force opposite the direction of motion.
(d) Assuming a level roadway, the work done is zero because the force is perpendicular to the displacement.
(e) No work is done because the electrical force is perpendicular to the displacement.
(f) No work is done because the tension is perpendicular to the displacement.
2. The force must be applied perpendicular to the object for it to do no work on the object.
3. The normal force can do work on an object. For example, when you jump, you push down on the ground and the normal
force pushes up on you and accelerates you up, giving you kinetic energy.
4. (a) No work is being done on the swimmer because the balancing forces forward and backward produce no motion.
(b) Work is done on the student to speed him up, but after that, there is no work being done on the student. Once the
student reaches the speed of the current, the only force is the upward buoyant force, which is perpendicular to the
waters surface. Technically, a small amount of work is being done by gravity as they whole river/student system is
pulled closer to the earth.
5. (a) The velocity will be changing because Newton’s second law states that if a net force acts on an object, it will accelerate
(i.e., change its velocity).
(b) It is possible the speed is constant if the particle is travelling in a circle.
(c) The kinetic energy is proportional to the square of the speed and the mass. Since both can remain constant under these
conditions, the kinetic energy may be constant.
6. Agree. In the absence of friction (including air resistance), all of the gravitational potential energy will be converted into
kinetic energy. The mass will cancel in each term.
7. (a) Damped vibrations are useful in the suspension of a vehicle.
(b) Damped vibrations are not useful in a pendulum clock.
8. It is not possible to have a motion that is not damped. Such a device would be a perpetual motion machine that cannot
exist. The force of friction within the system cannot be avoided.
9. m = 0.425 kg
∆y = 11.8 m
(a) W = ?
W = ( F cosθ )∆d
= (mg cos θ )∆y
= (0.425 kg)(9.80 m/s 2 )(cos180°)(11.8 m)
W = −49.1 J
The work gravity does on the ball on the way up is –49.1 J.
(b) W = ?
W = ( F cosθ )∆d
= (mg cos θ )∆y
= (0.425 kg)(9.80 m/s 2 )(cos 0°)(11.8 m)
W = 49.1 J
The work gravity does on the ball on the way down is 49.1 J.
10. F = 9.3 N
W = 87 J
∆d = 11 m
θ=?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 284
W = ( F cos θ )∆d
cos θ =
W
F ∆d
 W 
θ = cos −1 

 F ∆d 


87 J
= cos −1 

(9.3
N)(11
m)


θ = 32°
The angle between the applied force and the horizontal is 32°.
11. Let the subscript C represent the child, and TO represent the toboggan.
mC = 25.6 kg
mTO = 4.81 kg
∆y = 27.3 m
(a) WC = ?
First, find the actual distance:
27.3
sin φ =
∆d
27.3
∆d =
sin φ
The applied force, FA, will be equal to the component of gravity down the hill, mgsin φ
WC = ( F cos θ )∆d
= (mg sin φ )(cos θ ) ∆d
 27.3 
= (mg sin φ )(cos θ ) 

 sin φ 
= (mg cos θ )(27.3)
= (4.81 kg)(9.80 N/kg)(cos 0°)(27.3)
WC = 1.29 × 103 J
The total work done by the child is 1.29 ¯ 103 J.
(b) From part (a), the angle doesn’t matter, therefore W = 1.29 × 103 J.
(c) The total work on the child and toboggan during the slide will be equal to the work done to take them up the hill. Using
the equation derived in part (a),
WT = (mg cos θ )(27.3)
= (25.6 kg + 4.81 kg)(9.80 N/kg)(cos 0°)(27.3)
WT = 8.14 × 103 J
The total work on the child and the toboggan is 8.14 ¯ 103 J.
12. m = 73 kg
θ = 9.3°
vi = 4.2 m/s
∆d = ?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 285
The relation between the distance along the slope and the vertical height is:
∆y
sin 9.3° =
∆d
∆y = ∆d sin 9.3°
To calculate the change in distance:
ET1 = ET2
1 2
mv = mg ∆y
2
v 2 = 2 g ( ∆d sin 9.3°)
∆d =
v2
2 g (sin 9.3°)
(4.2 m/s)2
2(9.8 m/s)(sin 9.3°)
∆d = 5.6 m
The skier would travel 5.6 m along the hill before stopping.
13. An increase of 50% is equivalent of multiplying by 1.5:
1 2
mv
EK2 2 2
=
EK1 1 2
mv1
2
1.50 EK1 v22
= 2
EK1
v1
=
1.5 =
(v1 + 2.00) 2
v12
1.5v12 = v12 + 4v1 + 4
0.5v12 − 4v1 − 4 = 0
−(−4) ± (−4)2 − 4(0.5)(−4)
2(0.5)
4 ± 4.899
=
1
v1 = 8.90 m/s, or − 0.899 m/s
Since the negative value is not admissible, the original speed of the object was 8.90 m/s.
14. m = 7.0 ¯ 109 kg
∆y = 36 m
(a) ∆Eg = ?
∆Eg = mg ∆y
v1 =
= (7.0 × 109 kg)(9.8 N/kg)(36 m)
∆Eg = 2.5 × 1012 J
The gravitational potential energy is 2.5 ¯ 1012 J.
(b) The work done by on the pyramid by one person in 20 years is:
 5 × 106 J   40 d 
8
(0.20 ) 

 ( 20 a ) = 8 × 10 J/person
d
a




To calculate the total number of people:
2.5 × 1012
= 3 ×103 people
8 ×108
There were 3 ¯ 103 workers involved.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 286
15. m = 45 kg
∆d = 66 cm = 0.66 m
(a) W = ?
W = ( F cos θ )∆d
= (mg cos θ )∆d
= (45 kg)(9.80 N/kg)(cos180°)(0.66 m)
W = −2.9 × 102 J
The work done by gravity on the mass is –2.9 ¯ 102 J.
(b) W = ?
W = ( F cos θ )∆d
= (mg cos θ )∆d
= (45 kg)(9.80 N/kg)(cos 0°)(0.66 m)
W = 2.9 × 102 J
The work done by the weightlifter on the mass is 2.9 ¯ 102 J.
(c) ∆Eg = ?
∆Eg = mg ∆y
= (45 kg)(9.80 N/kg)(0.66 m)
∆Eg = 2.9 × 102 J
The change in gravitational potential energy is 2.9 ¯ 102 J.
16. m = 47 g = 0.047 kg
(a) ∆y = 4.3 m
v2 = ?
ET1 = ET2
1 2
1
mv1 + mgy1 = mv22 + mgy2
2
2
Since v1 = 0:
2 gy1 = v22 + 2 gy2
v22 = 2 gy1 − 2 gy2
v2 = 2 g ( y1 − y2 )
= 2(9.8 m/s 2 )(4.3 m)
v2 = 9.2 m/s
The speed of the stick just before it hits the ground is 9.2 m/s.
(b) If air resistance was included, the answer in part (a) would be slightly smaller. The energy from gravity would be
shared between the thermal energy and the kinetic energy.
17. y1 – 27 m
v1 = 18 m/s
θ = 37°
(a) v2 = ?
ET1 = ET2
1 2
1
mv1 + mgy1 = mv22 + mgy2
2
2
2
v1 + 2 gy1 = v22 + 2 gy2
v22 = v12 + 2 gy1 − 2 gy2
v2 = v12 + 2 g ( y1 − y2 )
= (18 m/s) 2 + 2(9.8 m/s 2 )(27 m − 0 m)
v2 = 29 m/s
The speed of the stick just before it hits the ground is 29 m/s.
(b) The angle the stick is thrown at does not appear in the equation, so the answer will still be 29 m/s.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 287
18. F = 1.5 ¯ 102 N [22° below the horizontal]
m = 18 kg
∆d = 1.6 m
µk = 0.55
(a) FN = ?
FK = ?
To calculate the normal force:
ΣFy = 0
FN − FA sin 22° − mg = 0
FN = FA sin 22° + mg
= (1.5 × 102 N)(sin 22°) + (18 kg)(9.8 N/kg)
= 232.6 N
FN = 2.3 ×10 2 N
To calculate the force of friction:
FK = µ K FN
= (0.55)(232.6 N)
= 127.9
FK = 1.3 × 102 N
The normal force on the box is 2.3 ¯ 102 N. The force of friction on the box is 1.3 ¯ 102 N.
(b) v = ?
If the box is starting from rest, there is no initial kinetic energy.
W = Eth + EK
1
( F cos θ )∆d = FK ∆d + mv 2
2
mv 2 = 2( F cos θ )∆d − 2 FK ∆d
v=
=
2( F cos θ )∆d − 2 FK ∆d
m
2(1.5 ×102 N)(cos 22°)(cos 0°)(1.6 m) − 2(127.9 N)(1.6 m)
18 kg
v = 1.4 m/s
The final speed of the box is 1.4 m/s.
(c) Eth = ?
Eth = FK ∆d
= (127.9 N)(1.6 m)
Eth = 2.0 × 102 J
The amount of thermal energy produced is 2.0 ¯ 102 J.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 288
19. m = 1.2 ¯ 103 kg
v1 = 9.5 ¯ 103 m/s
F = 9.2 ¯ 104 N
∆d = 86 km = 86 ¯ 103 m
v2 = ?
W = ∆EK
1 2 1 2
mvf − mvi
2
2
1 2
1 2
mvf = F ∆d + mvi
2
2
2 F ∆d
vf =
+ vi2
m
F ∆d =
=
2(9.2 × 104 N)(86 × 103 m)
+ (9.5 ×103 m/s) 2
1.2 × 103 kg
vf = 1.0 ×10 4 m/s
The final speed of the probe is 1.0 ¯ 104 m/s.
20. y1 = 1.15 m
y2 = 4.75 m
v1 = ?
ET1 = ET2
EK1 + Eg1 = Eg2
1 2
mv1 + mgy1 = mgy2
2
v12 = 2( gy2 − gy1 )
v1 = 2 g ( y2 − y1 )
= 2(9.80 m/s 2 )(4.75 m − 1.15 m)
v1 = 8.40 m/s
The speed with which the gymnast leaves the trampoline is 8.40 m/s.
21. The work in the area under the graph. Consider the area in three parts:
W = Atriangle 1 + Arectangle + Atriangle 2
1
1
b1h1 + lw + b2 h2
2
2
1
1
= (1.0 m)(12.0 N) + (2.0 m − 1.0 m)(12.0 N) + (6.0 m − 2.0 m)(12.0 N)
2
2
W = 42 J
The person does 42 J of work.
22. x = 0.418 m
F = 1.00 ¯ 102 N
(a) k = ?
Fx = kx
=
Fx
x
1.00 × 102 N
=
0.418 m
k = 239 N/m
The force constant of the spring is 239 N/m.
(b) x = 0.150 m
Fx = ?
k=
Copyright © 2003 Nelson
Chapter 4 Work and Energy 289
Fx = kx
= (239 N/m)(0.150 m)
Fx = 35.9 N
The force required to stretch the spring is 35.9 N.
(c) To stretch it 0.150 m:
1
Ee = kx 2
2
1
= (239 N/m)(0.150 m) 2
2
Ee = 2.69 J
To compress it 0.300 m:
1 2
kx
2
1
= (239 N/m)( −0.300 m) 2
2
Ee = 10.8 J
The work required is 2.69 J to stretch it, and 10.8 J to compress it.
23. k = 22 N/m
m = 7.5 ¯ 10–3 kg
FK = 4.2 ¯ 10–2 N
x = 3.5 cm = 0.035 m
∆d = ?
ET1 = ET2
Ee =
Ee = Eth
1 2
kx = FK ∆d
2
kx 2
∆d =
2 FK
(22 N/m)(0.035 m) 2
2(4.2 × 10 −2 N)
∆d = 0.32 m
The eraser will slide 0.32 m along the desk.
24. k = 75 N/m
A = 0.15 m
v = 1.7 m/s
x = 0.12 m
m=?
=
For SHM, all of the energy is Ee when at the maximum amplitude, A:
ET1 = ET2
Ee max = Ee + EK
1 2 1 2 1 2
kA = kx + mv
2
2
2
mv 2 = kA2 − kx 2
m=
=
k ( A2 − x 2 )
v2
(
75 N/m (0.15 m ) − (0.12 m )
2
(1.7 m/s)
2
2
)
m = 0.21 kg
The mass of the block is 0.21 kg.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 290
25. m = 0.42 kg
k = 38 N/m
A = 5.3 cm = 0.053 m
(a) Ee = ?
Maximum energy occurs at full amplitude.
1
Ee = kx 2
2
1
= (38 N/m)(0.053 m) 2
2
Ee = 0.053 J
The maximum energy of the mass-spring system is 0.053 J.
(b) v = ?
Maximum speed occurs when there is no elastic potential energy:
ET1 = ET2
Ee max = EK
1 2 1 2
kA = mv
2
2
2
mv = kA2
v=
=
kA2
m
38 N/m(0.053 m) 2
0.42 kg
v = 0.50 m/s
The maximum speed of the mass is 0.50 m/s.
(c) x = 4.0 cm = 0.040 m
v=?
ET1 = ET2
Ee max = Ee + EK
1 2 1 2 1 2
kA = kx + mv
2
2
2
2
2
mv = kA − kx 2
v=
=
k ( A2 − x 2 )
m
(
38 N/m ( 0.053 m ) − (0.040 m )
2
2
)
0.42 kg
v = 0.33 m/s
The speed of the mass is 0.33 m/s.
(d) x = 4.0 cm = 0.040 m
ET = ?
ET = Ee + EK
1 2 1 2
kx + mv
2
2
1
1
= (38 N/m)(0.040 m)2 + (0.42 kg)(0.33 m/s)2
2
2
ET = 0.053 J
The total energy is 0.053 J. The results are the same.
=
Copyright © 2003 Nelson
Chapter 4 Work and Energy 291
Applying Inquiry Skills
26. (a) One text weighs 2.0 kg and has a thickness of 3.6 cm. The first one doesn’t need to be raised at all.
W = Eg1 + Eg2 + Eg3 + Eg4 + Eg5
= mgy1 + mgy2 + mgy3 + mgy4 + mgy5
= mg ( y1 + y2 + y3 + y4 + y5 )
= (2.0 kg)(9.8 N/kg)(0 m + 0.036 m + 0.072 m + 0.108 m + 0.144 m)
W = 7.1 J
You would have to do 7.1 J of work.
(b) Some errors would be determining the mass of the text. Also, the thickness may compress the books on the bottom.
Not all texts may have the same mass.
27. The calculator would need to know the distance the box was pushed.
28.
29. The tractor seat would need to have a strong spring to absorb large bumps, and a shock absorber to prevent “launching”
the driver. There would have to be smaller springs on top of that to absorb small vibrations. Damping would be important
to prevent resonance.
Making Connections
30. Roller coasters are often shut down in high winds because of the loss of energy that may occur due to increased air
resistance. Cold can cause parts to shrink and increase frictional forces beyond safe limits.
31. (a) The ball on track B will arrive first. Shortly after the start, the vertical drop of the ball on track B causes an increase in
the speed, which it will enjoy for the majority of the race. At the end, it will slow down to the same speed that the ball
on track A has just accelerated to.
(b) Racing cyclists use this in a variety of ways. The sprinters stay high on the track until ready to make a break for it,
converting all of their stored gravitational energy into kinetic. Team racers use the change to minimize the work of the
rider. When the leader is ready to give up his spot, he rides up the hill a bit, converting some of his kinetic energy into
gravitational potential energy, reducing his speed. Once the last team mate has passed, he can drop down again, gaining
the gravitational potential energy back as kinetic energy without needing to supply it from his own body power.
32. (a) v1 = 0 m/s
v2 = ?
ET1 = ET2
1 2
1
mv1 + mgy1 = mv22 + mgy2
2
2
2
v1 + 2 gy1 = v22 + 2 gy2
v22 = 2 gy1 − 2 gy2
v2 = 2 g ( y1 − y2 )
= 2(9.80 m/s 2 )(37.8 m − 17.8 m)
v2 = 19.8 m/s
The speed of the coaster at position C is 19.8 m/s.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 292
(b) v1 = 5.00 m/s
v2 = ?
ET1 = ET2
1 2
1
mv1 + mgy1 = mv22 + mgy2
2
2
2
v1 + 2 gy1 = v22 + 2 gy2
v22 = v12 + 2 gy1 − 2 gy2
v2 = v12 + 2 g ( y1 − y2 )
= (5.00 m/s)2 + 2(9.80 m/s 2 )(37.8 m − 17.8 m)
v2 = 20.4 m/s
The speed of the coaster at position C is 20.4 m/s.
33. (a) The mass is not necessary because it cancels out of the mathematical equations.
(b) You might expect the speed would be 5.0 m/s more at the second point, but the reality is that the small kinetic energy
supplied by having a speed going over the first hill is negligible compared with the large amount of gravitational
potential energy.
34. a = 12 g [upward]
(a) FA = ?
ΣFy = ma
FA − mg = ma
FA = ma + mg
= m( a + g )
= m(0.12 g + g )
FA = 1.12mg
The force required is 1.12mg.
(b) W = ?
W = ( F cos θ )∆d
= (1.12mg )(cos 0°)(∆y )
W = 1.12 mg ∆y
The work done is 1.12mg∆y.
35. m = 1.5 kg
k = 2.1 ¯ 103 N/m
∆y = 0.37 m
x=?
Copyright © 2003 Nelson
Chapter 4 Work and Energy 293
ET1 = ET2
1 2
kx
2
1
(1.5)(9.8)(0.37 + x) = (2.1× 103 ) x 2
2
5.439 + 14.7 x = 1050 x 2
mg ∆y =
1050 x 2 − 14.7 x − 5.439 = 0
Using the quadratic equation:
−(−14.7) ± (−14.7) 2 − 4(1050)(−5.439)
2(1050)
14.7 ± 151.9
=
2100
= 0.079 m or x = −0.065 m (negative value inadmissable)
x=
x = 0.079 m
The maximum distance the spring is compressed is 0.079 m.
36. m = 0.55 g = 5.5 ¯ 10–4 kg
∆d = 95 cm = 0.95 m
∆dx = 3.7 m
Ee = ?
First we must calculate the time required for the vertical drop (vi = 0):
1
∆d = vi ∆t + a(∆t ) 2
2
1
∆d = a ( ∆t ) 2
2
2 ∆d
∆t =
a
2(−0.95 m)
−9.8 m/s 2
∆t = 0.44 s
=
To calculate the horizontal speed:
∆d x = v x ∆t
∆d x
∆t
3.7 m
=
0.44 s
vx = 8.4 m/s
vx =
We know that initial kinetic energy came from elastic potential energy, therefore:
ET1 = ET2
Ee = EK
1 2
mv
2
1
= (5.5 × 10−4 kg)(8.4 m/s)2
2
Ee = 0.019 J
The elastic potential energy stored was 0.019 J.
=
Copyright © 2003 Nelson
Chapter 4 Work and Energy 294
37. y1 = 16 m
y2 = 9.0 m
H=?
First we must solve for the launch speed if v1 = 0:
ET1 = ET2
1 2
1
mv1 + mgy1 = mv22 + mgy2
2
2
2 gy1 = v22 + 2 gy2
v22 = 2 gy1 − 2 gy2
v2 = 2 g ( y1 − y2 )
= 2(9.8 m/s 2 )(16 m − 9.0 m)
v2 = 11.71 m/s
Resolve vector components:
As just clearing the wall, the horizontal component will be the same, and the vertical component must produce a 30º angle
to the vertical.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 295
To calculate the vertical speed:
tan 30° =
vy =
=
v2 cos 45°
vy
v2 cos 45°
tan 30°
(11.71 m/s ) cos 45°
tan 30°
v y = 14.35 m/s [down]
Determine how far the skier has dropped from the launch point:
vf2 = vi2 + 2a∆d
∆d =
=
vf2 − vi2
2a
(
(−14.35 m/s)2 − (11.71 m/s )(sin 45° )
2
2
)
2(−9.8 m/s )
∆d = −7.0 m
Therefore, the wall must be 9.0 – 7.0 = 2.0 m tall.
38. y1 = 2.5 m
v1 = 9.0 m/s
y2 = 3.0 m
v2 = ?
ET1 = ET2
1 2
1
mv1 + mgy1 = mv22 + mgy2
2
2
v12 + 2 gy1 = v22 + 2 gy2
v22 = v12 + 2 gy1 − 2 gy2
v2 = v12 + 2 g ( y1 − y2 )
= (9.0 m/s)2 + 2(9.8 m/s 2 )(2.5 m − 3.0 m)
v2 = 8.4 m/s
The speed of the ball when it “swishes” through the hoop is 8.4 m/s.
39. g = 2.0
v1 = 6.0
y2 = 5.0
S=?
Solve for the launch speed at the top of the ramp where y1 = 0:
ET1 = ET2
1 2
1
mv1 + mgy1 = mv22 + mgy2
2
2
2
v1 + 0 = v22 + 2 gy2
v22 = v12 − 2 gy2
v2 = v12 − 2 gy2
= (6.0)2 − 2(2.0)(5.0)
v2 = 4.0 units
Copyright © 2003 Nelson
Chapter 4 Work and Energy 296
Analyze projectile motion to find the change in time:
1
∆ d y = v yi ∆ t + a ( ∆ t ) 2
2
1
−5.0 = (4.0sin 30°)∆t + (−2.0)(∆t )2
2
2
−5 = 2∆t − (∆t )
( ∆t ) 2 − 2 ∆t − 5 = 0
Using the quadratic equation:
−(−2) ± (−2)2 − 4(1)(−5)
2(1)
2 ± 4.9
=
2
= 3.45 units or − 1.45 units (dismiss negative answer)
∆t =
∆t = 3.45 units
To calculate horizontal range (S):
S = v x ∆t
= (4.0 cos 30°)(3.45)
S = 12 units
The shuttle lands a distance of 12 units from the ramp.
Copyright © 2003 Nelson
Chapter 4 Work and Energy 297
CHAPTER 5 MOMENTUM AND COLLISIONS
Reflect on Your Learning
(Page 230)
1. (a) The expression refers to once you start playing well, it is easier to keep playing well (or scoring, or winning).
(b) The physics meaning refers to a specific quantity. The everyday use of the word momentum can mean continuing to do
well and inertia.
2. (a) The momentum of car A is less than the momentum of car B.
(b) The momentum of bicycle and rider A is less than the momentum of bicycle and rider B.
(c) The momentum of the large truck A is greater than the momentum of car B.
3. (a) The phrase “follow-through” refers to continue to swing the racket/club even after contact with the ball is made.
This allows the racket/club to be in contact with the ball for a longer period of time.
(b) “Follow-through” affects the momentum (or change in it).
4. (a) One technique that could be used is to break the total area up into eight different rectangles. The height of each would
be an estimate. Each rectangle’s area could be determined and the sum would give the total area.
kg ⋅ m
kg ⋅ m
(b) F ⋅ t = N ⋅ s = 2 × s =
s
s
5. The car should be designed with crunch zones to absorb the energy. If elastic bumpers were used, the car would rebound
and be able to impact other objects.
Try This Activity: Predicting the Bounces
(Page 231)
(a) Prediction: The balls will bounce to the same height.
(b) Ball A and ball B both seem to be made from the same material. They possess approximately equal density and are
approximately the same size. Since the balls appear to be identical, they will possess the same amount of elasticity and
will bounce to the same height.
(c) The balls did not bounce to the same height. Our prediction and hypothesis were not supported by the evidence.
(d) Much of the kinetic energy of one ball is conserved in the bounce. The kinetic energy of the other ball is not conserved.
It is transformed into other forms. This ball loses kinetic energy in the bounce and does not bounce to the same height.
Bales of hay placed on the sides of ski-racing runs transform the kinetic energy of the skier into other forms, slowing the
skier down and reducing the chance for bodily harm.
5.1 MOMENTUM AND IMPULSE
PRACTICE
(Page 233)
Understanding Concepts
1. (a) m = 7.0 × 103 kg
G
v = 7.9 m/s
G
p=?
G
G
p = mv
= (7.0 ×103 kg)(7.9 m/s)
G
p = 5.5 × 104 kg ⋅ m/s [E]
The momentum of the African elephant is 5.5 × 104 kg ⋅ m/s [E] .
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 299
(b) m = 19 kg
G
v = 726 m/s
G
p=?
G
G
p = mv
= (19 kg)(26 m/s)
G
p = 4.9 × 102 kg ⋅ m/s [S]
The momentum of the mute swan is 4.9 ×102 kg ⋅ m/s [S] .
(c) m = 9.1 × 10–31 kg
G
v = 1.0 × 107 m/s
G
p=?
G
G
p = mv
= (9.1× 10−31 kg)(1.0 ×107 m/s)
G
p = 9.1× 10 −24 kg ⋅ m/s [forward]
2.
3.
The momentum of the electron is 9.1× 10−24 kg ⋅ m/s [forward] .
m = 405 kg
G
p = 5.02 × 103 kg ⋅ m/s
G
v =?
G
G
p = mv
G
G p
v=
m
5.02 × 103 kg ⋅ m/s
=
405 kg
G
v = 12.4 m/s [W]
The velocity of the craft is 12.4 m/s [W] .
G
p = 4.5 kg ⋅ m/s
G
v = 9.0 × 102 m/s
m=?
G
G
p = mv
G
p
m= G
v
4.5 kg ⋅ m/s
=
9.0 ×10 2 m/s
m = 5.0 × 10−3 kg
The mass of the bullet is 5.0 × 10−3 kg .
4. (a) Assuming a mass of 58 kg and a top speed of 8.0 m/s.
p = mv
= (58 kg)(8.0 m/s)
p = 4.6 ×10 2 kg ⋅ m/s
The magnitude of the momentum of an average student would be 4.6 ×102 kg ⋅ m/s .
(b) Assume a typical compact car to have a mass of 1200 kg.
p = mv
p
m
4.6 ×10 2 kg ⋅ m/s
=
1200 kg
v = 0.39 m/s
A typical compact car would have to travel at a velocity of 0.39 m/s to achieve the same momentum.
v=
300 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
PRACTICE
(Page 237)
Understanding Concepts
5.
impulse
F ∆t = N ⋅ s
change in momentum
∆p = m ( ∆v )
kg ⋅ m
m
= kg  
×s
2
s
s 
kg ⋅ m
kg ⋅ m
F ∆t =
∆p =
s
s
Therefore, the units of impulse and change in momentum are equivalent.
6. (a) m = 0.065 kg
viy = –3.8 m/s
piy = ?
=
piy = mviy
= (0.065 kg)(−3.8 m/s)
piy = −0.25 kg ⋅ m/s
The momentum of the snowball before hitting the ground is –0.25 kg⋅m/s.
(b) vfy = 0.0 m/s
pfy = ?
pfy = mvfy
= (0.065 kg)(0.0 m/s)
pfy = 0.0 kg ⋅ m/s
The momentum of the snowball after hitting the ground is 0.0 kg⋅m/s.
(c) ∆p y = pfy − piy
= 0.0 kg ⋅ m/s − (−0.25 kg ⋅ m/s)
∆p y = 0.25 kg ⋅ m/s
The change in momentum is 0.25 kg⋅m/s.
7. (a) ΣFx = 4.8 × 103 N
∆t = 3.5 s
impulse = ΣFx ∆t = ?
ΣFx ∆t = (4.8 × 103 N)(3.5 s)
ΣFx ∆t = 1.7 × 104 N ⋅ s [W]
The impulse on the truck over this time interval is 1.7 × 104 N ⋅ s [W] .
(b) pix = 5.8 × 104 kg ⋅ m/s
pfx = ?
ΣFx ∆t = ∆px
ΣFx ∆t = pfx − pix
pfx = ΣFx ∆t + pix
= 1.7 × 104 N ⋅ s + 5.8 × 104 kg ⋅ m/s
pfx = 7.5 ×104 kg ⋅ m/s [W]
The final momentum of the truck is 7.5 × 104 kg ⋅ m/s [W] .
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 301
8.
m = 0.27 kg
vix = 2.7 m/s
vfx = 0 m/s
ΣFx = 33 N [W]
∆t = ?
Choosing the initial direction of the ball (east) as positive:
ΣFx ∆t = ∆px
ΣFx ∆t = m(vfx − vix )
∆t =
m(vfx − vix )
ΣFx
0.27 kg(0 m/s − 2.7 m/s)
−33 N
∆t = 0.022 s, or 22 ms
The ball is in contact with the net for 22 ms.
v = 75.5 m/s [11.1° below the horizontal]
m = 1.24 × 105 kg
px = ?
py = ?
G
p =p
=
9.
= mv
= (1.24 × 105 kg)(75.5 m/s)
p = 9.362 × 106 kg ⋅ m/s
px
p
p x = p cos11.1°
cos11.1° =
= (9.362 × 106 kg ⋅ m/s)(cos11.1°)
p x = 9.19 × 10 6 kg ⋅ m/s
sin11.1° =
py
p
p y = p sin11.1°
= (9.362 × 106 kg ⋅ m/s)(sin11.1°)
p y = 1.80 ×106 kg ⋅ m/s
The horizontal component of the plane’s momentum is 9.19 × 106 kg ⋅ m/s and the vertical component is
1.80 ×106 kg ⋅ m/s .
Applying Inquiry Skills
10. (a) Using the area of a triangle:
impulse = A =
1
bh
2
1
(0.5 s)(4.0 N)
2
impulse = 1.0 N ⋅ s [E]
The impulse imparted is 1.0 N⋅s [E].
=
302 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
(b) Estimate the shape from 0 s – 0.2 s to be a triangle.
Estimate the shape from 0.2 s – 0.4 s to be a trapezoid.
Estimate the shape from 0.4 s – 1.0 s to be a trapezoid.
The sum of all three is the total impulse:
1
h +h
impulse = A = bh +  1 2
2
 2
  h1′ + h2′ 
 b +  2  b′
 

1
 30 N + 40 N 
 40 N + 60 N 
= (0.2 s)(30 N) + 
 (0.4 s − 0.2 s) + 
 (1.0 s − 0.4 s)
2
2
2




impulse = 4.0 × 101 N ⋅ s [S]
The impulse imparted is 4.0 × 101 N⋅s [S].
Making Connections
11. (a) Padded gloves increase the duration of time that the moving hand (punch) comes to rest. This causes a decrease in the
size of the force applied to the head (and the hand), reducing fractures.
(b) By “rolling with a punch,” a boxer increases the contact time, and reduces the applied force as the punch is applied.
Section 5.1 Questions
(Page 238)
Understanding Concepts
1.
The net force on an object is the rate of change in momentum.
G ∆pG
ΣF =
∆t
2. Impulse is most useful when there is only a single net force acting on an object.
G
3. (a) ΣF = 24 N [E]
∆t = 3.2 s
G
Σ F ∆t = ?
G
ΣF ∆t = (24 N)(3.2 s)
G
ΣF ∆t = 77 N ⋅ s [E]
The impulse exerted is 77 N ⋅ s [E] .
G
(b) ΣF = 1.2 × 102 N [forward]
∆t = 9.1 ms
G
Σ F ∆t = ?
G
ΣF ∆t = (1.2 × 102 N)(9.1× 10−3 s)
G
ΣF ∆t = 1.1 N ⋅ s [forward]
The impulse exerted is 1.1 N ⋅ s [forward] .
(c) m = 12 kg
G
g = 9.8 m/s2 [down]
∆t = 3.0 s
G
Σ F ∆t = ?
G
ΣF ∆t = mg ∆t
= (12 kg)(9.8 m/s 2 )(3.0 s)
G
ΣF ∆t = 3.5 × 102 N ⋅ s [down]
The impulse exerted is 3.5 × 102 N ⋅ s [down] .
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 303
(d) Solve for the area under the graph. Estimate the geometric shapes shown.
G
ΣF ∆t = A1 + A2 + A3
=
4.
1
 1.0 N + 3.7 N 
 3.9 N + 5.0 N 
(0.010 s)(0.81 N) + 
 (0.020 s − 0.010 s) + 
 (0.040 s − 0.020 s)
2
2
2




G
ΣF ∆t = 0.12 N ⋅ s [S]
The impulse exerted is 0.12 N ⋅ s [S] .
msystem = 41 kg + 21 kg = 62 kg
∆t = 2.0 s
ΣFx = 75 N [W]
vix = 0 m/s
vfx = ?
ΣFx ∆t = ∆px
ΣFx ∆t = mvfx − mvix
mvfx = ΣFx ∆t + mvix
ΣFx ∆t + mvix
m
ΣFx ∆t
=
+ vix
m
(75 N)(2.0 s)
=
+ 0 m/s
(41 kg + 21 kg)
vfx = 2.4 m/s [W]
The final velocity of the cart and the child will be 2.4 m/s [W].
m = 1.1× 103 kg
vfx =
5.
vix = 22 m/s
vfx = 0 m/s
∆t = 1.5 s
ΣFx = ?
304 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
ΣFx ∆t = ∆px
ΣFx ∆t = m(vfx − vix )
m(vfx − vix )
∆t
(1.1× 103 kg)(0 m/s − 22 m/s)
=
1.5 s
4
ΣFx = −1.6 × 10 N [E], or 1.6 ×10 4 N [W]
ΣFx =
The average force required to stop the car is 1.6 ×104 N [W] .
6. (a) m = 0.17 kg
vix = 2.1 m/s [right]
vfx = 1.8 m/s [left]
∆p x = ?
Choose right as the positive direction.
∆p x = m(vfx − vix )
= (0.17 kg)(−1.8 m/s − 2.1 m/s)
= −0.66 kg ⋅ m/s [right]
∆p x = 0.66 kg ⋅ m/s [left]
The change in momentum of the ball is 0.66 kg ⋅ m/s [left] .
(b) ΣFx ∆t = ?
The impulse is equal to the change in momentum, but the units are written in a different form.
ΣFx ∆t = m(vfx − vix )
= (0.17 kg)(−1.8 m/s − 2.1 m/s)
= −0.66 N ⋅ s [right]
ΣFx ∆t = 0.66 N ⋅ s [left]
The impulse given to the ball by the cushion is 0.66 N ⋅ s [left] .
7. (a) m = 0.16 kg
vfx = 11 m/s
vix = 18 m/s
∆p x = ?
∆p x = m(vfx − vix )
= (0.16 kg)(11 m/s − 18 m/s)
= −1.1 kg ⋅ m/s [forward]
∆p x = 1.1 kg ⋅ m/s [backward]
The change in momentum of the puck is 1.1 kg ⋅ m/s [backward] .
(b) ΣFx ∆t = ?
ΣFx ∆t = m(vfx − vix )
= (0.16 kg)(11 m/s − 18 m/s)
= −1.1 N ⋅ s [forward]
ΣFx ∆t = 1.1 N ⋅ s [backward]
The impulse exerted by the snow on the puck is 1.1 N ⋅ s [backward] .
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 305
(c) ∆t = 2.5 s
FK = ?
ΣFx ∆t = m(vfx − vix )
m(vfx − vix )
∆t
(0.16 kg)(11 m/s − 18 m/s)
=
2.5 s
= −0.45 N [forward]
FK =
8.
FK = 0.45 N [backward]
The average frictional force exerted by the snow on the puck is 0.45 N [backward].
m = 0.50 kg
∆t = 1.5 s
ΣFx = 1.4 N [W]
vix = 2.4 m/s
vfx = ?
Choose east as the positive direction.
ΣFx ∆t = ∆px
ΣFx ∆t = mvfx − mvix
mvfx = ΣFx ∆t + mvix
ΣFx ∆t + mvix
m
ΣFx ∆t
=
+ vix
m
(−1.4 N)(1.5 s)
=
+ 2.4 m/s
(0.50 kg)
= −1.8 m/s [E]
vfx =
9.
vfx = 1.8 m/s [W]
The final velocity of the disk is 1.8 m/s [W].
m = 2.0 kg
∆t = 0.50 s
ΣFx = 6.0 N [N]
vfx = 4.5 m/s
vix = ?
Choose north as the positive direction.
ΣFx ∆t = ∆px
ΣFx ∆t = mvfx − mvix
mvix = mvfx − ΣFx ∆t
mvfx − ΣFx ∆t
m
ΣFx ∆t
= vfx −
m
(6.0 N)(0.50 s)
= 4.5 m/s −
(2.0 kg)
vix = 3.0 m/s [N]
The initial velocity of the skateboard is 3.0 m/s [N].
vix =
306 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
10. (a) m = 0.61 kg
vix = 9.6 m/s [down]
vfx = 8.5 m/s [up]
∆p x = ?
Choose up as positive.
∆p x = m(vfx − vix )
= (0.61 kg)(8.5 m/s − (−9.6 m/s))
∆p x = 11 kg ⋅ m/s [up]
The change in momentum of the basketball is 11 kg ⋅ m/s [up] .
(b) ∆t = 6.5 ms = 6.5 × 10−3 s
Ffloor = ?
Choose up as positive.
ΣFx ∆t = m(vfx − vix )
m(vfx − vix )
∆t
(0.61 kg)(8.5 m/s − (−9.6 m/s))
=
6.5 × 10−3 s
= 1.7 × 103 N [up]
Ffloor =
Ffloor
The average force exerted on the basketball by the floor is 1.7 × 103 N [up] .
11. The force of gravity can be ignored in question 10 because it is so small compared to the force exerted by the floor
(approximately 6.0 N compared with 1700 N).
Applying Inquiry Skills
12. By using a high-speed digital camera, take rapid photos during the swing. For each athlete, compare the distance that the
racket is in contact with the ball. The longer the distance, the greater the follow-through.
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 307
Making Connections
13. The change in momentum when you land is the same for either type of landing. When you bend your knees, the time that
the force is applied is longer, so the required force is less. This smaller force is less likely to cause pain.
14. (a) The analyst could determine the initial speed of the bumper component from its skid marks. To estimate the initial
speeds of each vehicle, you would need to know the mass of each vehicle and how far each one skidded after the
collision. Using this information, you can also determine approximate coefficients of friction and solve for speeds
immediately after the collision. This would provide enough information to calculate the change in momenta of each
vehicle.
(b) Answers will vary.
5.2 CONSERVATION OF MOMENTUM IN ONE DIMENSION
PRACTICE
(Pages 243–244)
Understanding Concepts
1.
2.
The net force on the system must be zero for momentum to be conserved.
For the centre of mass of the system, the statement is equivalent to Newton’s first law. With no net force acting on the
system, the centre of mass of the system will not experience any change in velocity. However, the individual parts of the
system will undergo various changes in speed and direction.
3. (a) Earth will exert a downward force on the hairbrush.
(b) The hairbrush will exert an upward force on Earth.
(c) The forces in (a) and (b) are the same in magnitude.
(d) The net force of the system containing Earth and the hairbrush is zero.
(e) Momentum of this system will be conserved.
(f) Earth will move up as the hairbrush falls down.
(g) Choose up as positive.
m1 = 0.0598 kg
m2 = 5.98 × 10 24 kg
v1 y = 10 m/s [down]
v2 y = ?
p y = p ′y
0 = m1v1 y + m2 v2 y
v2 y = −
=−
m1v1 y
m2
(0.0598 kg)(−10 m/s)
5.98 × 1024 kg
v2 y = 1× 10−25 m/s [up]
Earth’s speed at this time is 1×10−25 m/s [up] .
4.
m1 = 45 kg
m2 = 33 kg
v1x = 1.9 m/s [E]
v2 x = ?
308 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
Making Connections
13. The change in momentum when you land is the same for either type of landing. When you bend your knees, the time that
the force is applied is longer, so the required force is less. This smaller force is less likely to cause pain.
14. (a) The analyst could determine the initial speed of the bumper component from its skid marks. To estimate the initial
speeds of each vehicle, you would need to know the mass of each vehicle and how far each one skidded after the
collision. Using this information, you can also determine approximate coefficients of friction and solve for speeds
immediately after the collision. This would provide enough information to calculate the change in momenta of each
vehicle.
(b) Answers will vary.
5.2 CONSERVATION OF MOMENTUM IN ONE DIMENSION
PRACTICE
(Pages 243–244)
Understanding Concepts
1.
2.
The net force on the system must be zero for momentum to be conserved.
For the centre of mass of the system, the statement is equivalent to Newton’s first law. With no net force acting on the
system, the centre of mass of the system will not experience any change in velocity. However, the individual parts of the
system will undergo various changes in speed and direction.
3. (a) Earth will exert a downward force on the hairbrush.
(b) The hairbrush will exert an upward force on Earth.
(c) The forces in (a) and (b) are the same in magnitude.
(d) The net force of the system containing Earth and the hairbrush is zero.
(e) Momentum of this system will be conserved.
(f) Earth will move up as the hairbrush falls down.
(g) Choose up as positive.
m1 = 0.0598 kg
m2 = 5.98 × 10 24 kg
v1 y = 10 m/s [down]
v2 y = ?
p y = p ′y
0 = m1v1 y + m2 v2 y
v2 y = −
=−
m1v1 y
m2
(0.0598 kg)(−10 m/s)
5.98 × 1024 kg
v2 y = 1× 10−25 m/s [up]
Earth’s speed at this time is 1×10−25 m/s [up] .
4.
m1 = 45 kg
m2 = 33 kg
v1x = 1.9 m/s [E]
v2 x = ?
308 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
Choose east as positive.
p x = p ′x
0 = m1v1x + m2 v2 x
v2 x = −
m1v1x
m2
(45 kg)(1.9 m/s)
33 kg
= −2.6 m/s [E]
=−
5.
v2 x = 2.6 m/s [W]
The velocity of the raft relative to the water is 2.6 m/s [W].
m1 = 56.9 kg
v1x = 3.28 m/s
v2 x = −3.69 m/s
m2 = ?
p x = p ′x
0 = m1v1x + m2 v2 x
m2 = −
m1v1x
v2 x
(56.9 kg)(3.28 m/s)
−3.69 m/s
m2 = 50.6 kg
The mass of the other skater is 50.6 kg.
m1 = 11 kg
m2 = 24 kg
=−
6.
v1x = 95 m/s
v2 x = ?
p x = p x′
0 = m1v1x + m2 v2 x
v2 x = −
m1v1x
m2
(11 kg)(95 m/s)
24 kg
v2 x = −44 m/s
The speed of the 24-kg piece is 44 m/s in the opposite direction.
m1 = 1.37 × 104 kg
=−
7.
m2 = 1.12 × 10 4 kg
v1 = 20.0 km/h [N]
v1′ = v2′ = v′ = 18.3 m/s
(when coupled)
v2 = ?
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 309
Since v1′ = v2′ = v ′,
m1v1 + m2 v2 = m1v1′ + m2 v2′
m1v1 + m2 v2 = (m1 + m2 )v ′
m2 v2 = (m1 + m2 )v′ − m1v1
(m + m2 )v′ − m1v1
v2 = 1
m2
(1.37 × 104 kg + 1.12 × 104 kg)(18.3 km/h) − (1.37 × 104 kg)(20.0 km/h)
1.12 × 104 kg
v2 = 16.2 km/h [N]
The initial velocity of the second car is 16.2 km/h [N].
m1 = 0.15 kg
=
8.
m2 = 0.045 kg
v1 = 56 m/s
v2 = 0
v2′ = 67 m/s
v1′ = ?
m1v1 + m2 v2 = m1v1′ + m2 v2′
m1v1 + 0 = m1v1′ + m2 v2′
m1v1′ = m1v1 − m2 v2′
m v − m2 v2′
v1′ = 1 1
m1
m2 v2′
= v1 −
m1
= 56 m/s −
(0.045 kg)(67 m/s)
0.15 kg
v1′ = 36 m/s
The head of the driver has a speed of 36 m/s immediately after the collision.
Applying Inquiry Skills
9. (a) Cart A moves toward the stationary cart B. This causes the two carts to collide, imparting speed to cart B. There is no
data to tell what happens to cart A during the last part of the collision and after.
(b) Cart B has no speed, so the system momentum will be equal to the momentum of cart A.
psystem = pA
= mA vA
= (0.40 kg)(0.60 m/s)
psystem = 0.24 kg ⋅ m/s [E]
The momentum of the system of carts before the collision is 0.24 kg ⋅ m/s [E] .
(c) mA = 0.40 kg
mB = 0.80 kg
vA = 0.60 m/s [E]
vB = 0
vB′ = 0.35 m/s [E]
vA′ = ?
310 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
Choosing east as positive.
p = p′
mA vA + mB vB = mA vA′ + mB vB′
mA vA + 0 = mA vA′ + mB vB′
mA vA′ = mA vA − mB vB′
mA vA − mB vB′
mA
m v′
= vA − B B
mA
vA′ =
(0.80 kg)(0.35 m/s)
0.40 kg
′
vA = −0.10 m/s [E], or 0.10 m/s [W]
The velocity of cart A after the collision is 0.10 m/s [W].
= 0.60 m/s −
(d)
Making Connections
10. Assuming the astronaut has something to throw, he could throw it away from the spaceship. This would cause him to
move toward the ship.
Section 5.2 Questions
(Pages 244–245)
Understanding Concepts
1. (a) F The forces are the same (Newton’s third law).
(b) F The magnitude of the changes in momenta will be equal.
(c) T
(d) F The magnitude of the changes in momenta will be equal.
2. Yes. An example would be two dynamics carts on a track that have had a spring bumper explosion.
3. (a) Momentum is conserved. There is no net external force of the system of objects.
(b) Momentum is conserved. There is no net external force of the system of objects.
(c) Momentum is not conserved. There is an external force applied by the stove on the pan.
4. It is not possible to react against the vacuum of space. A rocket ship works by pushing exhaust gasses out the back, and
the exhaust gases push back on the rocket ship. There is not involvement of the air (or not).
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 311
5.
m1 = 57 kg
m2 = 27 kg
v1 = v2 = v = 3.2 m/s [forward] (when together)
v1′ = 3.8 m/s [forward]
v2′ = ?
Choose forward as positive.
m1v1 + m2 v2 = m1v1′ + m2 v2′
(m1 + m2 )v = m1v1′ + m2 v2′
m2 v2′ = ( m1 + m2 )v − m1v1′
(m + m2 )v − m1v1′
v2′ = 1
m2
(57 kg + 27 kg)(3.2 m/s) − (57 kg)(3.8 m/s)
27 kg
v2′ = 1.9 m/s [forward]
The final velocity of the cart is 1.9 m/s [forward].
m1 = 65 kg + 35 kg = 100 kg (treating the hiker and raft as a single object)
=
6.
m2 = 19 kg
v1 = 0
v2 = 0
v1′ = 1.1 m/s [S]
v2′ = ?
Choose south as positive.
p = p′
m1v1 + m2 v2 = m1v1′ + m2 v2′
0 = m1v1′ + m2 v2′
v2′ = −
m1v1′
m2
(100 kg)(1.1 m/s)
19 kg
v2′ = −5.8 m/s [S], or 5.8 m/s [N]
The hiker threw the backpack with a velocity of 5.8 m/s [N] relative to the water.
=−
7.
m1 = 1.13 × 103 kg
m2 = 1.25 × 103 kg
v1 = 25.7 m/s [E]
v2 = 13.8 m/s [W]
v1′ = v2′ = v′ = ? (when coupled)
Choose east as positive.
m1v1 + m2 v2 = m1v1′ + m2 v2′
m1v1 + m2 v2 = (m1 + m2 )v ′
v′ =
m1v1 + m2 v2
m1 + m2
(1.13 × 103 kg)(25.7 m/s) + (1.25 × 103 kg)( −13.8 m/s)
1.13 × 103 kg + 1.25 × 103 kg
v ′ = 4.95 m/s [E]
The velocity of the system after the collision is 4.95 m/s [E].
=
312 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
8. (a) For the first vehicle:
m1 = 1.13 × 103 kg
v1 = 25.7 m/s [E]
v1′ = v2′ = v′ = 4.95 m/s [E] (when coupled)
G
∆p = ?
G
∆p = p2 − p1
= m1v ′ − m1v1
= m1 (v′ − v1 )
= (1.13 × 103 kg)(4.95 m/s − 25.7 m/s)
G
∆p = −2.34 × 104 kg ⋅ m/s [E]
For the second vehicle:
m2 = 1.25 × 103 kg
v2 = 13.8 m/s [W]
v1′ = v2′ = v′ = 4.95 m/s [E] (when coupled)
G
∆p = ?
G
∆p = p2 − p1
= m2 v′ − m2 v2
= m2 (v′ − v2 )
= (1.25 × 103 kg)(4.95 m/s − ( −13.8 m/s))
G
∆p = 2.34 × 104 kg ⋅ m/s [E]
(b) These two quantities are equal in magnitude and opposite in direction.
(c) The total change in momentum of the two-automobile system is zero.
9. mL = 89 kg
vQ = 0
vL = 5.2 m/s
vQ′ = vL′ = v′ = 2.7 m/s
mQ = ?
mQ vQ + mL vL = mQ vQ′ + mL vL′
0 + mL vL = mQ v ′ + mL v ′
mQ v′ = mL vL − mL v ′
mL (vL − v ′)
v′
(89 kg)(5.2 m/s − 2.7 m/s)
=
2.7 m/s
mQ = 82 kg
mQ =
The mass of the quarterback is 82 kg.
10. Choose original direction of 0.25-kg ball as positive.
m1 = 0.18 kg
m2 = 0.25 kg
v1 = −2.5 m/s
v2 = 1.7 m/s
v2′ = −0.10 m/s
v1′ = ?
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 313
m1v1 + m2 v2 = m1v1′ + m2 v2′
m1v1′ = m1v1 + m2 v2 − m2 v2′
v1′ =
m1v1 + m2 (v2 − v2′ )
m1
(0.18 kg)(−2.5 m/s) + (0.25 kg)(1.7 m/s − (−0.10 m/s))
0.18 kg
′
v1 = 0 m/s
The velocity of the 0.18-kg ball after the collision is 0 m/s.
=
Applying Inquiry Skills
11. Set up the carts as shown in the text on a track. Attach a spark timing recorder to each cart to record location for each time
interval. Turn on the timers, and release the carts. Analyze each tape close to the collision to determine the speed of each
cart. Mass each cart and calculate the momentum of each cart after the collision. Check to see if each one adds up to zero
(as it should).
12.(a) Since the car and SUV came to an immediate halt at the location of the crash, the total momentum of the system was
zero, before and after the collision. That would mean that the total momentum before and after would have the same
magnitude.
mC vC = mS vS
(but mS = 2mC )
mC vC = (2mC )vS
mC vC = 2mC vS
vC = 2vS
This data indicates the car was travelling at twice the speed of the SUV.
(b) If both drivers had the same speed, there would have been momentum after the collision in the direction of the original
motion of the SUV.
5.3 ELASTIC AND INELASTIC COLLISIONS
Try This Activity: Newton’s Cradle
(Page 248)
(a) Each collision is successive. Assuming an elastic collision,
m1v1 + m2 v2 = m1v1′ + m2 v2′ (but v2 = 0 and all masses are equal)
mv1 + 0 = mv1′ + mv2′
v1 = v1′ + v2′
v2′ = v1 − v1′
(Equation 1)
and
1
2
m1v12 + 12 m2 v22 = 12 m1v1′2 + 12 m2 v2′2
(but v2 = 0 and all masses are equal)
mv12 + 0 = mv1′2 + mv2′2
v12 = v1′2 + v2′2
v2′2 = v12 − v1′2
(Equation 2)
Substitute Equation 1 into Equation 2:
(v1 − v1′ )2 = v12 − v1′2
v12 − 2v1v1′ + v1′2 = v12 − v1′2
2/ v1′2 − 2/ v1v1′ = 0
v1′ (v1′ − v1 ) = 0
v1′ = 0
or v1′ − v1 = 0
If v1′ − v1 = 0, then v1′ = v1 (i.e., the speed of the ball after the collision is unchanged). This is not possible, so v1′ = 0.
The final speed of the first ball is zero after the collision.
314 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
m1v1 + m2 v2 = m1v1′ + m2 v2′
m1v1′ = m1v1 + m2 v2 − m2 v2′
v1′ =
m1v1 + m2 (v2 − v2′ )
m1
(0.18 kg)(−2.5 m/s) + (0.25 kg)(1.7 m/s − (−0.10 m/s))
0.18 kg
′
v1 = 0 m/s
The velocity of the 0.18-kg ball after the collision is 0 m/s.
=
Applying Inquiry Skills
11. Set up the carts as shown in the text on a track. Attach a spark timing recorder to each cart to record location for each time
interval. Turn on the timers, and release the carts. Analyze each tape close to the collision to determine the speed of each
cart. Mass each cart and calculate the momentum of each cart after the collision. Check to see if each one adds up to zero
(as it should).
12.(a) Since the car and SUV came to an immediate halt at the location of the crash, the total momentum of the system was
zero, before and after the collision. That would mean that the total momentum before and after would have the same
magnitude.
mC vC = mS vS
(but mS = 2mC )
mC vC = (2mC )vS
mC vC = 2mC vS
vC = 2vS
This data indicates the car was travelling at twice the speed of the SUV.
(b) If both drivers had the same speed, there would have been momentum after the collision in the direction of the original
motion of the SUV.
5.3 ELASTIC AND INELASTIC COLLISIONS
Try This Activity: Newton’s Cradle
(Page 248)
(a) Each collision is successive. Assuming an elastic collision,
m1v1 + m2 v2 = m1v1′ + m2 v2′ (but v2 = 0 and all masses are equal)
mv1 + 0 = mv1′ + mv2′
v1 = v1′ + v2′
v2′ = v1 − v1′
(Equation 1)
and
1
2
m1v12 + 12 m2 v22 = 12 m1v1′2 + 12 m2 v2′2
(but v2 = 0 and all masses are equal)
mv12 + 0 = mv1′2 + mv2′2
v12 = v1′2 + v2′2
v2′2 = v12 − v1′2
(Equation 2)
Substitute Equation 1 into Equation 2:
(v1 − v1′ )2 = v12 − v1′2
v12 − 2v1v1′ + v1′2 = v12 − v1′2
2/ v1′2 − 2/ v1v1′ = 0
v1′ (v1′ − v1 ) = 0
v1′ = 0
or v1′ − v1 = 0
If v1′ − v1 = 0, then v1′ = v1 (i.e., the speed of the ball after the collision is unchanged). This is not possible, so v1′ = 0.
The final speed of the first ball is zero after the collision.
314 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
Substituting back into Equation 1 gives:
v2′ = v1 − v1′
= v1 − 0
v2′ = v1
Each sphere collision will leave the previous one stationary as the next one moves on. Only one sphere can move out from
the end.
(b) Similarly, only one sphere can end up with all of the kinetic energy.
(c) According to the above calculations, each sphere will “pass” its momentum onto the next. When the last sphere receives
the momentum, and therefore, speed, it will rise to the same vertical height as the original sphere was dropped.
(d) If two spheres are used, then two spheres will move out from the other side. If three spheres are used, then three spheres
will move out from the other side.
PRACTICE
(Page 248)
Understanding Concepts
1.
2.
3.
All of the original kinetic energy is transformed to other forms if both objects come to rest after the collision.
Yes, we can conclude the collision is completely inelastic. After a “hit-and-stick” collision, no energy is stored as elastic
potential to be returned to either object at the end of the collision.
A head on collision is very dangerous because of the high relative velocity between the vehicles and the large (and rapid)
change in speed for each one. These large accelerations produce large forces that are capable of inflicting serious damage
on a human body.
Applying Inquiry Skills
4.
If the ball tends to return to shape quickly when squeezed, it will have a more elastic collision than one that returns to its
original shape more slowly.
5. (a)
(b)
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 315
Making Connections
6. (a) The impact force is reduced by the soft interior because it takes the same impulse, and causes the interaction to last
longer. The larger time interval of collision means that a smaller force is applied. A hard interior would have a short
duration of collision, and a higher force.
(b) If a helmet does not fit properly, the force applied to the head is not evenly distributed to the whole head. The smaller
distribution area means the force would be applied to a smaller area, increasing the pressure to that portion of the head.
(c) Once a helmet is involved in a collision, it should be replaced. One way a helmet is designed to reduce injury is to
absorb some of the impact by breaking. This is only able to help the wearer one time. In future collisions, the safety of
the wearer could be compromised.
7. Some possible answers:
Flexible
• absorb energy
• reduce deceleration of train (and passengers)
Rigid
• prevent car from collapsing and injuring passengers located at either end
• less likely to have debris flying about the interior (due to crumpling car)
PRACTICE
(Pages 251–252)
Understanding Concepts
8.
The larger truck would have the larger momentum.
mv 2
EK =
2
m2v 2
=
2m
(mv)2
=
2m
p2
EK =
2m
p = 2mEK
psmall = 2msmall EK small
and
plarge = 2mlarge EK large
Since EK small = EK large ,
psmall = 2msmall EK
and
plarge = 2mlarge EK
The mass of the larger vehicle is larger, and the momentum will be too.
9. (a) All objects have mass. If it also has a velocity, then it will have both momentum and kinetic energy.
(b) For an isolated system of two objects, it is possible to have a momentum of zero. One example would be two carts
released from an internal spring “explosion”. The total momentum is zero, but both of the individual parts have kinetic
energy. If the system of objects has momentum, then at least one of them is moving, and there will be kinetic energy. It
is not possible to have momentum and zero kinetic energy at the same time.
10. m1 = 0.15 kg
m2 = 0.15 kg
v1 = 22 m/s [N]
v2 = 22 m/s [S]
v1′ = v2′ = v′ = ? (when coupled)
316 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
Choose north as positive.
m1v1 + m2 v2 = m1v1′ + m2 v2′
m1v1 + m2 v2 = (m1 + m2 )v ′
v′ =
m1v1 + m2 v2
m1 + m2
(0.15 kg)(22 m/s) + (0.15 kg)(−22 m/s)
0.15 kg + 0.15 m/s
v′ = 0 m/s
The velocity of the two-ball system after the collision is 0 m/s.
11. Conservation of Momentum
m1v1 + m2 v2 = m1v1′ + m2 v2′ (but v2 = 0 and the masses are equal)
mv1 + 0 = mv1′ + mv2′
=
v1 = v1′ + v2′
v2′ = v1 − v1′
(Equation 1)
Conservation of Energy
1
2
m1v12 + 12 m2 v22 = 12 m1v1′2 + 12 m2 v2′2
mv12
2
+ 0 = mv1′ +
(but v2 = 0 and the masses are equal)
mv2′2
v12 = v1′2 + v2′2
v2′2 = v12 − v1′2
(Equation 2)
Substitute Equation 1 into Equation 2:
(v1 − v1′ )2 = v12 − v1′2
v12 − 2v1v1′ + v1′2 = v12 − v1′2
2/ v1′2 − 2/ v1v1′ = 0
v1′ (v1′ − v1 ) = 0
v1′ = 0
or v1′ − v1 = 0
If v1′ − v1 = 0, then v1′ = v1 (i.e., the speed of the proton after the collision is unchanged). This is not possible, so v1′ = 0.
The final speed of the first proton is zero after the collision.
Substituting back into Equation 1 gives:
v2′ = v1 − v1′
= v1 − 0
= v1
′
v2 = 815 m/s
The final velocity of the second proton after the collision is 815 m/s in the direction of the initial velocity.
12. For an inelastic collision, the vehicles stick together.
m1 = 1.3 × 104 kg
m2 = 1.1× 103 kg
v1 = 90 km/h [N]
v2 = 30 km/h [N]
v1′ = v2′ = v′ = ? (when coupled)
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 317
Choose north as positive.
m1v1 + m2 v2 = m1v1′ + m2 v2′
when coupled, v1′ = v2′ = v′
m1v1 + m2 v2 = (m1 + m2 )v ′
v′ =
m1v1 + m2 v2
m1 + m2
(1.3 ×10 4 kg)(90 km/h) + (1.1× 103 kg)(30 km/h)
1.3 × 104 kg + 1.1× 103 kg
v′ = 85 km/h [N]
The velocity of the vehicles is 85 km/h [N] just after the collision.
13. mt = 1.3 ×104 kg
=
mc = 1.1 ×103 kg
vt = 90 km/h [N] = 25 m/s [N]
vc = 30 km/h [N] = 8.333 m/s [N]
ET = EK truck + E K car
1
1
mt vt2 + mc vc2
2
2
1
1
= (1.3 × 104 kg)(25 m/s)2 + (1.1× 103 kg)(8.333 m/s) 2
2
2
6
ET = 4.1× 10 J
=
The final velocity of the vehicles after the collision is 85.319 km/h [N] = 23.7 m/s [N].
ET′ = EK′ truck + EK′ car
1
1
mt vt′2 + mc vc′2
2
2
1
1
= (1.3 × 104 kg)(23.7 m/s) 2 + (1.1× 103 kg)(23.7 m/s) 2
2
2
6
ET′ = 4.0 × 10 J
The decrease in kinetic energy is 4.1 × 106 – 4.1 × 106 = 1 × 105 J.
14. Choose the original direction of motion as the positive direction.
mO = 5.31 ×10−26 kg
=
mN = 4.65 × 10−26 kg
vO = 0
vO′ = 4.81 ×10 2 m/s
vN′ = −34.1 m/s
vN = ?
mN vN + mO vO = mN vN′ + mO vO′
mN vN + 0 = mN vN′ + mO vO′
m v′ + mO vO′
vN = N N
mN
m v′
= vN′ + O O
mN
= −34.1 m/s +
(5.31× 10−26 kg)(4.81× 10 2 m/s)
4.65 × 10 −26 kg
vN = 5.15 × 102 m/s
The initial speed of the nitrogen molecule was 5.15 × 102 m/s .
318 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
Applying Inquiry Skills
15. (a) Line A represents the total momentum because the total momentum in the system is constant.
(b) The collision is an inelastic one. If the collision were elastic, the total kinetic energy before and after would have been
the same.
Making Connections
16. (a) The rubber bullet would have the elastic collision and the lead bullet would have the inelastic collision.
G
G
(b)
∆pT = −∆pR
G
G
G
G
G
G
G
mT vT2 − mT vT1 = −(mR vR2 − mR vR1 )
vT1 = 0, and vR2 = −vR1
G
G
G
mT vT2 − 0 = mR vR1 − mR (−vR1 )
G
G
G
G
G
mT vT2 = mR vR2 + mR vR1
vR1 = vR2
G
G
G
mT vT2 = mR vR2 + mR vR1
G
G
G
pT′ = pR′ + pR
G
G
∆pT = −∆pL
G
G
G
G
mT vT2 − mT vT1 = −mL vL2 + mL vL1
G
G
G
mT vT2 − 0 = mL vL2 − mL vL2
G
G
G
mT vT2 = mL vL1 − mL vL1
G
G
G
pT′ = pL − pL′
G
vT1 = 0
The rubber bullet transfers more momentum to the target.
(c) Rubbers bullets are preferred in crowd control because they are less likely to kill or permanently injure any of the
crowd and they impart a larger backward impulse on the crowd.
Section 5.3 Questions
(Page 253)
Understanding Concepts
1. (a) It is not possible for both objects to be at rest. If they were both at rest, the initial momentum of the first object would
have violated the law of conservation of t momentum.
(b) It is possible for the first object to be at rest after the collision. One example is a curling stone that strikes another and
then stops moving.
2. The does not violate the law of conservation of momentum for the system which contains the earth and the snowball.
The earth exerts a net external force on the tree/snowball system. This external force negates the conservation of
momentum for that system.
3. There momentums will only be the same if they have the same mass. The relationship between momentum and kinetic
energy is
mv 2
EK =
2
m2v 2
=
2m
(mv)2
=
2m
p2
EK =
2m
p = 2mEK
When two objects have the same kinetic energy, the object with the larger mass will always have a larger momentum.
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 319
4.
Conservation of Momentum
m1v1 + m2 v2 = m1v1′ + m2 v2′
m1v1 + 0 = m1v1′ + m2 v2′
m1v1 − m1v1′ = m2 v2′
m1 (v1 − v1′ ) = m2 v2′
(but v2 = 0)
(Equation 1)
Conservation of Energy
1
2
m1v12 + 12 m2 v22 = 12 m1v1′2 + 12 m2 v2′2
m1v12
m1v12
2
+ 0 = m1v1′ +
(but v2 = 0)
m2 v2′2
− m1v1′2 = m2 v2′2
m1 (v12 − v1′2 ) = m2 v2′2 (Equation 2)
Divide Equation 2 by Equation 1:
m1 (v12 − v1′2 ) m2 v2′2
=
m1 (v1 − v1′ )
m2 v2′
(v1 + v1′ )(v1 − v1′ )
= v2′
(v1 − v1′ )
v2′ = v1 + v1′ (Equation 3)
or
v1′ = v2′ − v1 (Equation 4)
Substitute Equations 3 and 4 back into the original conservation of momentum equation:
m1v1 + 0 = m1v1′ + m2 v2′
m1v1 = m1v1′ + m2 (v1 + v1′ )
m1v1 = m1v1′ + m2 v1 + m2 v1′
m1v1′ + m2 v1′ = m1v1 − m2 v1
v1′ (m1 + m2 ) = (m1 − m2 )v1
 m − m2
v1′ =  1
 m1 + m2

 v1

 0.022 kg − 0.027 kg 
=
 (3.5 m/s)
 0.022 kg + 0.027 kg 
v1′ = −0.36 m/s [forward], or 0.36 m/s [backward]
The velocity of the 22-g superball after the collision is 0.36 m/s [backward] .
m1v1 + 0 = m1v1′ + m2 v2′
m1v1 = m1 (v2′ − v1 ) + m2 v2′
m1v1 = m1v2′ − m1v1 + m2 v2′
m1v2′ + m2 v2′ = 2m1v1
v2′ (m1 + m2 ) = 2m1v1
 2m1 
v2′ = 
 v1
 m1 + m2 


2(0.022 kg)
=
 (3.5 m/s)
 0.022 kg + 0.027 kg 
v2′ = 3.1 m/s [forward]
The velocity of the 27-g superball after the collision is 3.1 m/s [forward] .
320 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
5.
Using the equations derived in question 4 ,
 m − m2 
v1′ =  1
 v1
 m1 + m2 
 m − m2 
1
v1 = 
 v1
3
 m + m2 
m + m2 = 3(m − m2 )
m + m2 = 3m − 3m2
4m2 = 2m
m2 =
6.
1
m
2
m = 66 kg
g = 9.8 m/s 2
∆y = 25 m
v1 = ?
Use conservation of energy to solve for the speed of the moving skier at the bottom of the hill.
ET = ET′
1
mv12
2
v1 = 2 g ∆y
mg ∆y =
= 2(9.8 m/s 2 )(25 m)
v1 = 22.14 m/s
m2 = 72 kg
v1 = 22.14 m/s
v2 = 0
v1′ = v2′ = v′ = ?
(when coupled)
m1v1 + m2 v2 = m1v1′ + m2 v2′
v2 = 0, and when coupled, v1′ = v2′ = v′
m1v1 + 0 = (m1 + m2 )v′
v′ =
m1v1
m1 + m2
(66 kg)(22.14 m/s)
66 kg + 72 kg
v′ = 11 m/s
The speed of the two-skier system immediately after the collision is 11 m/s.
=
Applying Inquiry Skills
7. (a) The collision takes place very quickly. The duration of time that the bullet and block are interacting, the strings are
vertical, and only balance the gravitational forces. During the collision (not during the swing), momentum is conserved.
(b) Let V be the speed of the bullet and block combination after the collision.
m1v1 + m2 v2 = m1v1′ + m2 v2′
(but v2 = 0 and when coupled, v1′ = v2′ = V )
mv + 0 = (m + M )V
mv
m+M
(c) The law of conservation of energy can be used to relate the maximum vertical height to the speed just after collision.
V=
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 321
(d)
ET = ET′
1 2
mv
2
1
gh = V 2
2
V2
h=
2g
mg ∆y =
 mv 


m+ M 
=
2g
(e)
2
h=
m2v 2
2 g (m + M ) 2
h=
m2v 2
2 g (m + M ) 2
m 2 v 2 = 2 gh (m + M )2
v=
2 gh(m + M )2
m2
m+ M 
= 2 gh 

 m 
2
m+M 
v = 2 gh 

 m 
(f) m = 87 g = 0.0087 kg
M = 5.212 kg
g = 9.8 m/s 2
h = 6.2 cm = 0.062 m
v=?
m+M 
v = 2 gh 

 m 
 0.0087 kg + 5.212 kg 
= 2(9.8 m/s 2 )(0.062 m) 

0.0087 kg


v = 6.6 ×10 2 m/s
The initial speed of the bullet was 6.6 × 102 m/s .
(g) Some of the sources of error would be friction of the moving parts of the gun and loss of energy to thermal energy in
the spring. The frictional draw of the catch mechanism would use some of the energy that should be converted into
gravitational potential.
8. There are a number of advantages of a crumple zone. One is that it converts a significant part of the kinetic energy of the
vehicle into thermal energy as the steel become permanently deformed. The crumple zone also causes the vehicle to slow
down over a greater distance. This greater distance increases the length of time the vehicle is slowing down, and the
average force to stop the vehicle (and its passengers) decreases.
9. Most meteorites burn up in the atmosphere. Larges known one is about 60 metric tons, and the next known one is about
30 metric tons. Large collisions have rarely (ever 300 million years or so) and can cause devastating climactic change.
322 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
5.4 CONSERVATION OF MOMENTUM IN TWO DIMENSIONS
PRACTICE
(Pages 257–258)
Understanding Concepts
1.
2. (a) ms = 52 kg
mc = 26 kg
G
vs = 1.2 m/s [W]
G
vc = 1.2 m/s [S]
G
vs′ = ?
G
vc′ = ?
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 323
Calculate the momenta:
G
G
ps = ms vs
= (52 kg)(1.2 m/s)
G
ps = 62.4 kg ⋅ m/s [S]
G
G
pc = mc vc
= (26 kg)(1.2 m/s)
G
pc = 31.2 kg ⋅ m/s [S]
G G
ps + pc = 93.6 kg ⋅ m/s [S]
G
G
ps′ = ms vs′
= (52 kg)(1.0 m/s)
G
ps′ = 52 kg ⋅ m/s [W]
Measure length and angle to get:
G
pc′ = 1.1× 102 kg ⋅ m/s [61° S of E]
G
G
pc′ = mc vc′
G
pc′
G
′
vc =
mc
1.1 × 10 2 kg ⋅ m/s
26 kg
G
vc′ = 4.1 m/s [61° S of E]
The approximate final velocity of the cart is 4.1 m/s [61° S of E] .
(b) Using the diagram from part (a), the total momentum after is
G
G 2 G G 2
pc′ = ps′ + ps + pc
=
= (52 kg ⋅ m/s) 2 + (93.6 kg ⋅ m/s)2
G
pc′ = 1.1 × 10 2 kg ⋅ m/s
tan θ =
3.
G G
ps + pc
G
ps′
G G
 p +p 
θ = tan −1  s G c 
 ps′ 
 93.6 
= tan −1 

 52 
θ = 61°
G
2
So, pc′ = 1.1× 10 kg ⋅ m/s [61° S of E].
Diagram is not to scale. Dotted line represents the direction “after.”
A completely inelastic collision means they will stick together.
m1 = 1.4 × 103 kg
m2 = 1.3 × 103 kg
G
v1 = 45 km/h [S] = 12.5 m/s [S]
G
v2 = 39 km/h [E] = 10.83 m/s [E]
′ =?
v12
324 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
G
G
p1 = m1 v1
= (1.4 ×103 kg)(12.5 m/s)
G
p1 = 1.75 × 10 4 kg ⋅ m/s
G
G
p2 = m2 v2
= (1.3 ×103 kg)(10.83 m/s)
G
p1 = 1.408 ×10 4 kg ⋅ m/s
G
′ =
p12
G 2 G
p1 + p2
2
= (1.75 ×104 kg ⋅ m/s)2 + (1.408 × 104 kg ⋅ m/s)2
G
′ = 2.246 × 104 kg ⋅ m/s
p12
G
′ = m12 v12
′
p12
G
p′
′ = 12
v12
m12
=
2.246 ×104 kg ⋅ m/s
1.3 × 103 kg + 1.4 × 103 kg
′ = 8.3 m/s
v12
G
p
tan θ = G 1
p2
G
 p 
θ = tan −1  G 1 
 p2 
 1.75 × 10 4 kg ⋅ m/s 
= tan −1 

4
 1.408 ×10 kg ⋅ m/s 
θ = 51°
The final speed of the cars is 8.3 m/s [51º S of E], or 3.0 × 101 km/h [51º S of E].
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 325
4. (a) Diagram not to scale.
(b)
0 = mvBy − mvAy
0 = vBy − vAy
vBy = vAy
vB sin 48.9° = vA sin 31.1°
vB =
sin 31.1°
vA
sin 48.9°
mvA1 = mvAx + mvBx
vA1 = vAx + vBx
2.25 = vA cos 31.1° + vB cos 48.9°
 sin 31.1° 
vA  cos 48.9°
2.25 = vA cos 31.1° + 
 sin 48.9° 
2.25 = 1.307vA
vA = 1.72 m/s
sin 31.1°
vA
sin 48.9°
sin 31.1°
(1.72 m/s)
=
sin 48.9°
vB = 1.18 m/s
vB =
The velocities of the balls are 1.18 m/s at 48.9° and 1.72 m/s at 31.1°.
326 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
(c)
′ = 12 mvA2 + 12 mvB2
EKT
(d)
2
EKT = 12 mvA1
= 12 m(2.25) 2
= 12 m(vA2 + vB2 )
= 12 m(1.722 + 1.182 )
′ = 12 m(4.35)
EKT
EKT = 12 m(5.06)
The total kinetic energy after is less than the total kinetic energy before, so the collision is not elastic.
5. (a) The residual nucleus will move in the opposite direction of the combined momentums of the two particles so the total
momentum is still zero. (Note: Diagram is not to scale.)
G
pn = 4.8 × 10−21 kg ⋅ m/s [S]
G
pe = 9.0 × 10−21 kg ⋅ m/s [E]
θ =?
tan θ =
pn
pe
p 
θ = tan −1  n 
 pe 
 4.8 × 10 −21 kg ⋅ m/s 
= tan −1 

−21
 9.0 × 10 kg ⋅ m/s 
θ = 28°
The nucleus will move in the direction 28º N of W.
(b) pnucleus = ?
pnucleus =
pn2 + pe2
= (4.8 ×10−21 kg ⋅ m/s) 2 + (9.0 ×10−21 kg ⋅ m/s) 2
pnucleus = 1.0 × 10−20 kg ⋅ m/s
The magnitude of the nucleus’s momentum is 1.0 ×10−20 kg ⋅ m/s .
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 327
(c) mnucleus = 3.6 ×10−25 kg
G
vnucleus = ?
pnucleus = mnucleus vnucleus
vnucleus =
=
pnucleus
mnucleus
1.0 ×10 −20 kg ⋅ m/s
3.6 ×10 −25 kg
vnucleus = 2.8 × 104 m/s
G
vnucleus = 2.8 × 104 m/s [28° N of W]
The recoil velocity of the nucleus is 2.8 ×104 m/s [28° N of W] .
Applying Inquiry Skills
6. (a) The black car was travelling faster at the moment of impact. There was a larger component of momentum in the
original direction of the motion of the black car because there is less deviation in its path than the path of the white car.
(b) The investigator could determine more precise information by measuring the angle of deviation for the cars and the
total distance that the combined cars slid.
Making Connections
7.
Answers will vary depending on the type of equipment chosen. Most likely there will be some padding included in it that
will reduce the force required by increasing the time of collision.
Section 5.4 Questions
(Pages 258–239)
Understanding Concepts
1.
328 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
2.
mLi = 1.2 × 10−26 kg
′ = 0.40 km/s [54° to original direction of motion of neutron]
vLi
Diagram is not to scale.
vn′ = 2.5 km/s
′ =?
pLi
pn′ = ?
θ =?
′ = mLi vLi
′
pLi
= (1.2 ×10 −26 kg)(0.40 km/s)
′ = 4.8 × 10−27 kg ⋅ km/s
pLi
pn′ = mn vn′
= (1.7 × 10−27 kg)(2.5 km/s)
pn′ = 4.25 × 10 −27 kg ⋅ km/s
In the y-direction:
p y = p ′y
′ y − pn′ y
0 = pLi
′ sin 54° − pn′ sin θ
0 = pLi
p ′ sin 54°
sin θ = Li
pn′
3.
 p ′ sin 54° 
θ = sin −1  Li

pn′


−27
 (4.8 ×10 kg ⋅ km/s)(sin 54°) 
= sin −1 

4.25 × 10 −27 kg ⋅ km/s


θ = 66° [from original direction of motion of the neutron]
The neutron is now travelling at 66° from its original direction of motion.
Diagram is not to scale.
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 329
m1 = 71 kg
G
v1 = 2.3 m/s [12° N of E]
v2 = 1.9 kg ⋅ m/s [52° S of W]
m2 = ?
p1 = m1v1
= (71 kg)(2.3 m/s)
p1 = 163.3 kg ⋅ m/s
p2 = m2 v2
= m2 (1.9 kg ⋅ m/s)
p2 = 1.9 m2 kg ⋅ m/s
Using the sine law:
sin101° sin 39°
=
163.3
1.9m2
4.
 sin 39°   163.3 kg ⋅ m/s 
m2 = 


 1.9 kg ⋅ m/s   sin101° 
m2 = 55 kg
The mass of the second skater is 55 kg.
Diagram not to scale.
p1x = m1v1x
p1′ = m1v1′
p2′ = m2 v2′
= (0.50 kg)(2.0 m/s)
p1x = 1.0 kg ⋅ m/s
= (0.50 kg)(1.5 m/s)
p1′ = 0.75 kg ⋅ m/s
p2′ = 0.30v2′ kg ⋅ m/s
= (0.30 kg)v2′
p1x = p1′x + p2′ x
p1x = p1′ cos 30° + p2′ cos θ
0 = p1′y − p2′ y
1.0 = (0.75) cos 30° + (0.30v2′ ) cos θ
0 = (0.75) sin 30° − (0.30v2′ ) sin θ
v2′ cos θ = 1.168
(Equation 1)
0 = p1′ sin 30° − p2′ sin θ
v2′ sin θ = 1.25
(Equation 2)
Divide Equation 2 by Equation 1:
v2′ sin θ
1.25
=
v2′ cos θ 1.168
tan θ = 1.070
θ = tan −1 (1.070)
θ = 47°
330 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
Substitute back into Equation 2:
v2′ sin θ = 1.25
1.25
sin θ
1.25
=
sin 47°
v2′ = 1.7 m/s
The velocity of the second ball after collision is 1.7 m/s [47º S of E].
v2′ =
Applying Inquiry Skills
5. (a) Referring to the diagram:
Ignore points A3 and B3 for calculations because they are during the collision.
Measure between points A1 and A2, A4 and A5, and divide by 0.50 s to calculate velocities.
Repeat for Puck B (using two values for the after calculation because you have them).
Measure angles before and after for velocities.
2.00
G
vA =
= 4.0 cm/s [45° N of E]
0.50
1.80
G
v′A =
= 3.6 cm/s [N]
0.50
1.9
G
vB =
= 3.8 cm/s [38° W of N]
0.50
vB′ =
Copyright © 2003 Nelson
1.3 + 1.4
= 2.7 cm/s [43° E of N]
0.50 + 0.50
Chapter 5 Momentum and Collisions 331
Let the x direction be the east-west direction.
p x = p ′x
mA vAx − mB vBx = mA vA′ x + mB vB′ x
vA′ x = 0
mA vAx − mB vBx = 0 + mB vB′ x
mB vB′ x + mB vBx = mA vAx
mB (vB′ x + vBx ) = mA vAx
mB =
mA vAx
vB′ x + vBx
(0.32 kg)(4.0 cos 45°)
2.7 sin 43° + 3.8sin 38°
= 0.2165 kg
=
mB = 0.22 kg
(b) mA = 0.32 kg
mB = 0.2165 kg
vA = 0.040 m/s
vB = 0.038 m/s
EK lost = ?
1
1
mA vA2 + mB vB2
2
2
1
1
= (0.32 kg)(0.040 m/s) 2 + (0.2165 kg)(0.038 m/s) 2
2
2
−4
ET = 4.123 × 10 J
ET =
1
1
mA vA′2 + mB vB′2
2
2
1
1
= (0.32 kg)(0.036 m/s) 2 + (0.2165 kg)(0.027 m/s)2
2
2
ET′ = 2.863 × 10−4 J
ET′ =
EK lost = 4.123 × 10−4 J − 2.863 × 10−4 J
EK lost = 1.3 × 10 −4 J
The total kinetic energy lost is 1.3 × 10 −4 J .
(c) This is an inelastic collision.
(d) The most likely source of error is measuring the angles from the diagram.
Making Connections
6. (a) Some possible points:
• The perception of a safe vehicle is a common advertising feature of most vehicles.
• The idea of protecting loved ones and yourself in a collision is socially desirable.
• Insurance companies offer better rates for vehicles with high safety performance in collisions.
(b) Some points are:
• seat belts
• roll bars (roll safety)
• bumpers
• crumple zones
• air bag
• anti-lock braking systems
• traction control (4WD, AWD, tires)
332 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
(c) Any of the ones listed in (b).
(d) Could compare safety ratings and insurance prices, cost of vehicle (often safer vehicles or options are more expensive),
and fuel economy.
CHAPTER 5 LAB ACTIVITIES
Investigation 5.2.1: Analyzing One-Dimensional Collisions
(Pages 260–262)
Question
Observing two objects before, during, and after a collision allows the verification of conservation of momentum and kinetic
energy theory.
Prediction
(a) For all three categories, the total momentum of the system will always be the same.
(b) Category I: The total kinetic energy before and after will be the same. During the collision there will be some loss of
kinetic energy as it is stored as some other form in the repulsive device.
Category II: The total kinetic energy before the collision will be zero. It will be a maximum at the end of the explosion,
increasing as the explosion takes place.
Category III: The total kinetic energy will be a maximum before, decreasing throughout the collision and a minimum at
the end.
Hypothesis
(c) The velocity of both carts before and after the collision (an adhesive collision would be the easiest) can be determined
using the ticker-tape timer. Using the known mass of the carts, the conservation of momentum can be used to calculate the
unknown mass.
Before the Collision
Collision
I (a)
I (b)
I (c)
II (a)
II (b)
II (c)
III (a)
III (b)
III (c)
IV
After the Collision
Total p
(kg⋅⋅m/s)
Total EK
(J)
m1
(kg)
v1
(m/s)
m2
(kg)
v2
(m/s)
m1
(kg)
v1’
(m/s)
m2
(kg)
v2’
(m/s)
before
after
before
after
Loss
in EK
(%)
0.50
1.0
0.50
0.50
1.0
0.50
0.50
1.0
0.50
0.50
0.19
0.18
0.15
0.0
0.0
0.0
0.24
0.20
0.18
0.22
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
m2
0.0
0.0
–0.13
0.0
0.0
0.0
0.0
0.0
–0.17
0.0
0.50
1.0
0.50
0.50
1.0
0.50
0.50
1.0
0.50
0.50
0.0
0.060
–0.12
–0.22
–0.091
–0.10
0.11
0.12
0.0
0.088
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
m2
0.17
0.22
0.14
0.21
0.27
0.12
0.11
0.12
0.0
0.088
0.095
0.18
0.010
0.0
0.0
0.0
0.12
0.20
0.0050
—
0.085
0.17
0.010
–0.0050
0.044
0.010
0.11
0.18
0.0
—
0.0090
0.016
0.0098
0.0
0.0
0.0
0.014
0.020
0.015
—
0.0072
0.014
0.0085
0.023
0.022
0.0061
0.0060
0.011
0.0
—
20
12
13
—
—
—
57
45
100
—
Analysis
(d) For each collision, the total momentum before and after was essentially the same.
(e) For Category I collisions, there was very little loss of kinetic energy during the collisions. Category II collisions did not
start with any, but received the kinetic energy from the spring bumper. Category III collisions lost a significant amount of
kinetic energy during the collisions.
(f) Category I collisions were all elastic (within experimental error). Category III collisions were all completely inelastic.
(g) During Category I collisions, there was a temporary loss of kinetic energy that reappeared after the collision.
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 333
Multiple Choice
11. (e)
12. (d) Momentum will increase by 2 × 2, and kinetic enegy by 2 × 22.
13. (d)
14. (d)
15. (e)
16. (d) Some energy is lost to thermal energy of the collision.
17. (b) The momentum is p = 2mEK so p ∝ EK .
18. (c) M only stops, while T must continue to act on R to give it momentum in the opposite direction.
19. (a) They have both stopped, so the total momentum is zero, but the initial kinetic energy is stored as electric potential.
20. (d)
1

21. (a) mv + (2m)  (−v )  = (m + 2m)v ′
2

mv − mv
v′ =
3m
v′ = 0
22.(d) 0 = mv − ( M − m)V
V=
mv
M −m
CHAPTER 5 REVIEW
(Pages 269–271)
Understanding Concepts
1.
2.
3.
4.
Impulse is the product of the force and the time. A smaller force exerted over a long period of time can impart a larger
impulse than a large force for a short time.
As the meteor comes to a stop, the kinetic energy is converted into thermal energy which melts the material at the impact
site.
This observation does not contradict the law of conservation of momentum. The momentum of the earth increases toward
the falling object. The amount of change in the earth’s velocity is so small it is imperceptible, so it appears the law of
conservation is being contradicted, but it is not.
The change in momentum of the ball that bounces is greater than the putty that sticks to the floor. Assuming both have the
same mass and are dropped from the same height:
For the putty,
∆p = m(v2 − v1 )
= m(0 − v1 )
∆p = −mv1
5.
6.
For the ball
∆p = m(v2 − v1 )
= m((−v1 ) − v1 )
∆p = −2mv1
mv 2
2
m2 v 2
=
2m
p2
EK =
2m
The momentum of the two players is the same, but the person with the larger mass will have a smaller kinetic energy. It
would be better to avoid the faster moving lighter player, p2 = m2v2.
A car crashing into a tree is an example. Momentum is not conserved in the tree-car system because the tree is attached to
the earth. The earth supplies a net external force to the system of objects so that momentum is not conserved.
EK =
336 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
7.
Diagram is not to scale.
m = 1.3 × 102 kg
G
v = 8.7 m/s [44° E of N]
pN = ?
pE = ?
G
G
p = mv
= (1.3 ×10 2 kg)(8.7 m/s)
G
p = 1131 kg ⋅ m/s [44° E of N]
pN = p cos 44°
= (1131 kg ⋅ m/s)(cos 44°)
pN = 8.1× 102 kg ⋅ m/s
pE = p sin 44°
= (1131 kg ⋅ m/s)(sin 44°)
pE = 7.9 × 102 kg ⋅ m/s
8.
The northward component of the boat’s momentum is 8.1× 102 kg ⋅ m/s . The eastward component is 7.9 × 102 kg ⋅ m/s .
Diagram is not to scale.
pE = 2.6 × 104 kg ⋅ m/s
m = 1.1× 103 kg
θ = 22°
v=?
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 337
pE = mv cos 22°
pE
m cos 22°
2.6 × 104 kg ⋅ m/s
=
(1.1× 103 kg)(cos 22°)
v = 25 m/s
The car is travelling at a speed of 25 m/s.
Choose west as positive.
m = 1.2 × 103 kg
v=
9.
v1 = 53 km/h [W] = 14.72 m/s [W]
v2 = 0
∆t = 55 ms = 5.5 × 10−2 s
ΣFx = ?
ΣFx ∆t = ∆p
m(v2 − v1 )
∆t
(1.2 × 103 kg)(0 m/s − 14.72 m/s)
=
5.5 × 10−2 s
= −3.2 ×105 N [W]
ΣFx =
ΣFx = 3.2 × 105 N [E]
The average force exerted on the car by the pole is 3.2 × 105 N [E] .
10. (a) ΣFx = 324 N
∆t = 5.1 ms = 5.1× 10−3 s
ΣFx ∆t = ?
ΣFx ∆t = (324 N)(5.1 × 10 −3 s)
ΣFx ∆t = 1.7 N ⋅ s [fwd]
The impulse on the ball is 1.7 N⋅s [fwd].
(b) m = 0.059 kg
v1 = 0
v2 = ?
ΣFx ∆t = ∆p
ΣFx ∆t = m(v2 − v1 )
ΣFx ∆t = mv2
ΣFx ∆t
m
(324 N)(5.1× 10−3 s)
=
0.059 kg
v2 = 28 m/s [fwd]
The velocity of the ball just as it leaves the racket is 28 m/s [fwd].
11. m1 = 0.112 kg
m2 = 0.154 kg
v1′ = 1.38 m/s
v2 =
v2′ = ?
338 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
p = p′
0 = m1v1′ + m2 v2′
m v′
v2′ = − 1 1
m2
(0.112 kg)(1.38 m/s)
0.154 kg
′
v2 = −1.00 m/s
The speed of the second car is 1.00 m/s.
12. mP = 1.67 × 10−27 kg
=−
mα = 6.64 × 10 −27 kg
vP = 1.57 km/s
vα = 0
vP′ = −0.893 km/s
vα′ = ?
mP vP + mα vα = mP vP′ + mα vα′
mP vP + 0 = mP vP′ + mα vα′
m v − mP vP′
vα′ = P P
mα
m (v − v ′ )
= P P P
mα
(1.67 ×10 −27 kg)(1.57 km/s − ( −0.893 km/s))
6.64 × 10−27 kg
vα′ = 0.620 km/s
The speed of the alpha particle is 0.620 km/s.
13. m1 = 2.67 kg
m2 = 5.83 kg
G
v1 = 1.70 × 102 m/s [toward Jupiter]
G
v1′ = 185 m/s [toward Jupiter]
G
v2′ = 183 m/s [toward Jupiter]
G
v2 = ?
=
Choose the direction toward Jupiter as positive.
m1v1 + m2 v2 = m1v1′ + m2 v2′
m2 v2 = m1v1′ − m1v1 + m2 v2′
m (v′ − v ) + m2 v2′
v2 = 1 1 1
m2
m1 (v1′ − v1 )
=
+ v2′
m2
=
(2.67 kg)(185 m/s − 1.70 × 10 2 m/s)
+ 183 m/s
5.83 kg
v2 = 1.90 × 10 2 m/s
G
v2 = 1.90 × 10 2 m/s [toward Jupiter]
The velocity of the more massive rock is 1.90 × 102 m/s [toward Jupiter] .
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 339
14. (a) v1i = 6.0 m/s
v2i = 0
v1f = v2f = 2.0 m/s
The two objects have the same final velocity. Having the same final velocity is an indication of a ‘hit-and-stick’
collision. All collisions that have the objects with the same speed (in the same direction) after are completely inelastic
collisions.
(b) v1i = 24 m/s
v2i = 0
v1f = −4.0 m/s
v2f = 14 m/s
1
1
1
1
m1v1i2 + m2 v2i2
EKf = m1v1f2 + m2 v2f2
2
2
2
2
1
1
1
1
2
2
= (2.0 kg)(24 m/s) + (4.0 kg)(0 m/s)
= (2.0 kg)(−4.0 m/s)2 + (4.0 kg)(14 m/s) 2
2
2
2
2
EKi = 576 J
EKf = 408 J
There is some loss of kinetic energy. This is an inelastic collision.
(c) v1i = 12 m/s
v2i = 0
EKi =
v1f = −4.0 m/s
v2f = 8.0 m/s
1
1
1
1
2
m1v1i2 + m2 v2i2
EKf = m1v1f2 + m2 v2f
2
2
2
2
1
1
1
1
= (2.0 kg)(12 m/s)2 + (4.0 kg)(0 m/s)2
= (2.0 kg)(−4.0 m/s)2 + (4.0 kg)(8.0 m/s) 2
2
2
2
2
EKi = 144 J
EKf = 144 J
There is no loss of kinetic energy, so this collision is elastic.
15. Choose north as positive.
EKi =
Conservation of Energy
1
m v 2 = 12 m1v1′2 + 12 m2 v2′2
2 1 1
m1v12
2
= m1v1′ +
Conservation of Momentum
m1v1 = m1v1′ + m2 v2′
(0.253)(1.80) = 0.253v1′ + 0.232v2′
m2 v2′2
2
0.253(1.80) = 0.253v1′2 + 0.232v2′2
3.24 = v1′
2
+ 0.917v2′2
(Equation 1)
1.80 = v1′ + 0.917v2′
v1′ = 1.80 − 0.917v2′
(Equation 2)
Substitute Equation 2 into Equation 1:
3.24 = (1.80 − 0.917v2′ )2 + 0.917v2′2
3.24 = (3.24 − 3.30v2′ + 0.841v2′2 ) + 0.917v2′2
0 = −3.30v2′ + 1.758v2′2
0 = v2′ (−3.30 + 1.758v2′ )
v2′ = 0
or
− 3.30 + 1.758v2′ = 0
Since v′2 = 0 is not valid (no change in speed), then
1.758v2′ = 3.30
v2′ = 1.88 m/s [N]
340 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
Substitute back into Equation 2:
v1′ = 1.80 − 0.917(1.88)
v1′ = 0.08 m/s [N]
The velocity of the first cart is 0.08 m/s [N]. The velocity of the second cart is 1.88 m/s [N].
16. (a) Diagram not to scale.
(b)
0 = mvBy − mvAy
0 = vBy − vAy
vBy = vAy
vB sin 46° = vA sin 33°
vB =
sin 33°
vA
sin 46°
mvA1 = mvAx + mvBx
vA1 = vAx + vBx
5.4 = vA cos 33° + vB cos 46°
 sin 33° 
vA  cos 46°
5.4 = vA cos 33° + 
 sin 46° 
5.4 = 1.365vA
vA = 3.957 m/s, or 4.0 m/s
Substitute back into equation for vB :
sin 33°
vA
sin 46°
sin 33°
=
(3.957 m/s)
sin 46°
vB = 3.0 m/s
The speed of the first puck is 4.0 m/s after the collision. The speed of the second puck is 3.0 m/s after the collision.
vB =
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 341
17. θ A = 67.8°
θ B = 30.0°
vB
= 3.30
vA
mB
=?
mA
0 = pA′ y − pB′ y
0 = mA vA sin 67.8° − mB vB sin 30.0°
mB vB sin 30.0° = mA vA sin 67.8°
mB vA sin 67.8°
=
mA vB sin 30.0°
=
=
sin 67.8°
vB
(sin 30.0°)
vA
sin 67.8°
(3.30)(sin 30.0°)
mB
= 0.561
mA
The ratio of the masses of the particles is 0.561.
18. Diagrams not to scale.
342 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
Conserve momentum in the y direction.
p1 y + p2 y = − p1′y + p2′ y
0 + mv2 y = − mv1′y + mv2′ y
v2 y = −v1′y + v2′ y
= −v1′ sin 69.2° + v2′ sin 62.8°
= −(2.37 m/s)(sin 69.2°) + (2.49 m/s)(sin 62.8°)
v2 y = 0.00 m/s
Conserve momentum in the x direction.
p1x + p2 x = p1′x − p2′ x
mv1x + mv2 x = mv1′x − mv2′ x
v1x + v2 x = v1′x − v2′ x
v2 x = v1′x − v2′ x − v1x
= v1′ cos 69.2° − v2′ cos 62.8° − v1
= (2.37 m/s)(cos 69.2°) − (2.49 m/s)(cos 62.8°) − 2.70 m/s
= −3.00 m/s [E]
v2 x = 3.00 m/s [W]
There is no y component for velocity, so the unknown initial velocity is 3.00 m/s [W]
19. (a) g = 9.8 m/s 2
∆y = 0.26 m
v=?
Use conservation of mechanical energy for the swing.
1
mg ∆y = mv 2
2
v = 2 g ∆y
= 2(9.8 m/s 2 )(0.26 m)
= 2.2574 m/s
v = 2.3 m/s
The speed of the ball just before the collision is 2.3 m/s.
(b) Conservation of Energy
Conservation of Momentum
1
1
1
2
2
2
m1v1 = m1v1′ + m2 v2′
m1v1 = m1v1′ + m2 v2′
2
2
2
(0.25)(2.2574) = 0.25v1′ + 0.21v2′
m1v12 = m1v1′2 + m2 v2′2
2.2574 = v1′ + 0.84v2′
0.25(2.2574)2 = 0.25v1′2 + 0.21v2′2
v1′ = 2.2574 − 0.84v2′
5.096 = v1′2 + 0.84v2′2 (Equation 1)
(Equation 2)
Substitute Equation 2 into Equation 1:
5.096 = (2.2574 − 0.84v2′ )2 + 0.84v2′2
5.096 = (5.096 − 3.792v2′ + 0.7056v2′2 ) + 0.84v2′2
0 = −3.792v2′ + 1.5456v2′2
0 = v2′ (−3.792 + 1.5456v2′ )
v2′ = 0
Copyright © 2003 Nelson
or − 3.792 + 1.5456v2′ = 0
Chapter 5 Momentum and Collisions 343
Since v2′ = 0 is not valid (no change in speed), then:
1.5456v2′ = 3.792
v2′ = 2.5 m/s [fwd]
The speed of the 0.21-kg ball just after the collision is 2.5 m/s [fwd].
20. (a) m1 = 0.45 kg
m2 = 0.79 kg
v1 = 2.2 m/s
′ =?
v12
′
m1v1 = (m1 + m2 )v12
′ =
v12
m1v1
m1 + m2
(0.45 kg)(2.2 m/s)
0.45 kg + 0.79 kg
′ = 0.80 m/s
v12
The speed of the ball and the box just after the collision is 0.80 m/s.
(b) ∆d = 0.051 m
vi = 0.80 m/s
=
vf = 0
ΣF = ?
v +v 
∆d =  i f  ∆t
 2 
 2 
∆t = 
 ∆d
 vi + vf 
2


=
 (0.051 m)
 0.80 m/s + 0 m/s 
∆t = 0.1275 s
Σ F ∆t = ∆p
m(v2 − v1 )
∆t
(0.79 kg + 0.45 kg)(0 m/s − 0.80 m/s)
=
0.1275 s
ΣF = −7.8 N
The magnitude of the friction force acting on the ball and the box is 7.8 N.
21. m1 = 1.9 × 104 kg
ΣF =
m2 = 1.7 × 104 kg
v1′ = 3.5 × 103 km/h
v2′ = 3.4 × 103 km/h
θ = 180.0º – 5.1º – 5.9º = 169.0º
v=?
344 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
Using the sine law:
sin 5.1°
sin θ
=
m2 v2′
(m1 + m2 )v
m2 v2′ sin θ
v=
(m1 + m2 ) sin 5.1°
=
(1.7 × 104 kg)(3.4 × 103 km/h) sin169.0°
(1.9 × 104 kg + 1.7 × 104 kg) sin 5.1°
v = 3.4 × 103 km/h
The original speed of the spacecrafts was 3.4 × 103 km/h .
22. The total momentum before is zero, so the total momentum after must also be zero.
Diagram not to scale.
m1 = 2.0 kg
m2 = 3.0 kg
m3 = 4.0 kg
v1 = 1.5 m/s [N]
v2 = 2.5 m/s [E]
v3 = ?
p32 = p12 + p22
(m3v3 )2 = (m1v1 )2 + (m2 v2 )2
m32 v32 = (m1v1 ) 2 + (m2 v2 )2
v3 =
=
(m1v1 ) 2 + (m2 v2 ) 2
m32
(2.0 kg(1.5 m/s)) 2 + (3.0 kg(2.5 m/s)) 2
(4.0 kg) 2
v3 = 2.0 m/s
tan θ =
m1v1
m2 v2
 (2.0 kg)(1.5 m/s) 
θ = tan −1 

 (3.0 kg)(2.5 m/s) 
θ = 22°
The final velocity of the third piece is 2.0 m/s [22º S of W].
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 345
Applying Inquiry Skills
23.
24.
25. (a) The coefficient of restitution is a measure of how quickly an object returns to its shape after it has been compressed.
The apparatus shown could help determine the coefficient of restitution because the speed the sphere returns to its
original shape corresponds to the height the ball bounces to.
26. (a) The bed sheet spreads the force applied over a greater distance (and therefore time) than if the egg was to strike the
ground directly. The same impulse is imparted to the egg (i.e. it is brought to rest), so the force required is much
smaller.
(b) This can be used to cushion the landing of a person who was falling from a high position by reducing the average force
needed to stop them. You could test several heights with a sack of potatoes and plot the results on a graph. For a
human, you could extrapolate from the graph.
Making Connections
27. It is better for safety to have telephone poles that collapse upon impact. In this way, they absorb some energy and increase
the time of collision. Both of these reduce the force imparted to the vehicle and occupants in a collision.
28. High speed photography and spark timers both show the location of an object at fixed time intervals. High speed
photography gives more thorough information because you can see the state of the objects during each stage of the
collision.
29. Many arrester cables are connected to a water squeezer that dampens the motion and converts the kinetic energy of the
aircraft to kinetic energy of ejected water.
30. The main purpose of the ablation shield is to protect the shuttle during re-entry into the earth’s atmosphere.
346 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
Extension
31. (a) Conservation of Momentum
m1v1 + m2 v2 = m1v1′ + m2 v2′
m1v1 + 0 = m1v1′ + m2 v2′
m1v1 − m1v1′ = m2 v2′
m1 (v1 − v1′ ) = m2 v2′
(but v2 = 0)
and
(Equation 1)
Conservation of Energy
1
1
1
1
m1v12 + m2 v22 = m1v1′2 + m2 v2′2
2
2
2
2
2
2
′
m1v1 + 0 = m1v1 + m2 v2′2
(but v2 = 0)
m1v12 − m1v1′2 = m2 v2′2
m1 (v12 − v1′2 ) = m2 v2′2
(Equation 2)
Divide Equation 2 by Equation 1:
m1 (v12 − v1′2 ) m2 v2′2
=
m1 (v1 − v1′ )
m2 v2′
′
′
(v1 + v1 )(v1 − v1 )
= v2′
(v1 − v1′ )
v2′ = v1 + v1′
(Equation 3) or v1′ = v2′ − v1
(Equation 4)
Substitute Equations 3 and Equation 4 back into the original conservation of momentum equation:
m1v1 + 0 = m1v1′ + m2 v2′
m1v1 + 0 = m1v1′ + m2 v2′
m1v1 = m1v1′ + m2 (v1 + v1′ )
m1v1 = m1 (v2′ − v1 ) + m2 v2′
m1v1 = m1v1′ + m2 v1 + m2 v1′
m1v1′ + m2 v1′ = m1v1 − m2 v1
v1′ (m1 + m2 ) = (m1 − m2 )v1
 m − m2 
v1′ =  1
 v1
 m1 + m2 
 m − m2 
(b) v1′ =  1
 v1
 m1 + m2 
m−m
=
 v1
m+m
v1′ = 0
 2m1 
v2′ = 
 v1
 m1 + m2 
 2m 
=
 v1
m+m
v2′ = v1
 m − m2 
(c) v1′ =  1
 v1
 m1 + m2 
 m −0
= 1
 v1
 m1 + 0 
 2m1 
v2′ = 
 v1
 m1 + m2 
 2m1 
=
 v1
 m1 + 0 
 m − m2 
(d) v1′ =  1
 v1
 m1 + m2 
 2m1 
v2′ = 
 v1
 m1 + m2 
v1′ = v1
 0 − m2 
=
 v1
 0 + m2 
v1′ = −v1
m1v1 = m1v2′ − m1v1 + m2 v2′
m1v2′ + m2 v2′ = 2m1v1
v2′ (m1 + m2 ) = 2m1v1
 2m1 
v2′ = 
 v1
 m1 + m2 
v2′ = 2v1
 2m1 
=
 v1
 0 + m2 
2m
v2′ = 1 v1
m2
32. mP = 1.0 × 103 kg
mB = 2.0 × 103 kg
vP = 50 m/s
vB = 0
∆d = ?
Copyright © 2003 Nelson
Chapter 5 Momentum and Collisions 347
For the collision:
mP vP + mB vB = (mP + mB )vPB
vPB =
mP vP + mB vB
mP + mB
(1.0 ×103 kg)(50 m/s) + (2.0 ×103 kg)(0)
1.0 × 103 kg + 2.0 × 103 kg
= 16.67 m/s
=
vPB
Calculate the frictional force on the plane and on the barge:
For the plane
1
FK = mg
4
1
= (1.0 × 103 kg)(9.8 m/s 2 )
4
FK = 2450 N [backward]
Acceleration of the plane
ΣF = ma
ΣF
m
−2450 N
=
1.0 × 103 kg
For the barge
By Newton’s third law, FK = 2450 N [forward]
Acceleration of the barge
ΣF = ma
ΣF
m
2450 N
=
2.0 ×103 kg
a=
a=
a = −2.45 m/s 2
a = 1.225 m/s 2
Distance plane will travel during collision
vf2 = vi2 + 2a∆d
vf2 − vi2
2a
(16.67 m/s)2 − (50 m/s)2
=
2(−2.45 m/s 2 )
∆d = 453.5 m
∆d =
Distance barge will travel during collision
vf2 = vi2 + 2a∆d
vf2 − vi2
2a
(16.67 m/s)2 − (0 m/s)2
=
2(1.225 m/s 2 )
∆d = 113.4 m
∆d =
The required length of the barge is 453.5 – 113.4 = 3.4 × 102 m.
348 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
6.1 GRAVITATIONAL FIELDS
PRACTICE
(Pages 276–277)
Understanding Concepts
1.
As they are moving at a high speed in their orbits, space vehicles are attracted to Earth by the force of gravity, which
keeps them in their orbits. Also, the vehicles have small booster rockets to control the location of the orbit and to
counteract low amounts of friction in the upper atmosphere.
2. (a) G = 6.67 × 10−11 N ⋅ m 2 /kg 2
M E = 5.98 ×1024 kg
mMoon = 7.35 ×10 22 kg
rMoon = 3.84 × 108 m
FG =
=
GM E mMoon
rMoon 2
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(7.35 × 1022 kg)
(3.84 ×108 m) 2
FG = 1.99 × 1020 N [toward Earth's centre]
The gravitational force exerted on the Moon by Earth is 1.99× 1020 N [toward Earth’s centre].
GM E mMoon
(b) FG =
rMoon 2
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(7.35 × 1022 kg)
(3.84 ×108 m) 2
FG = 1.99 × 1020 N [toward Moon's centre]
3.
The gravitational force exerted on Earth by the Moon is 1.99× 1020 N [toward Moon’s centre].
GM E
GM E
Let gr represent the gravitational field strength at the desired radius r. Thus, g r =
.
and g =
2
(r + rE )
rE 2
(a) r = rE
gr =
=
gr =
GM E
(rE + rE ) 2
GM E
4rE 2
g
4
The magnitude of the gravitational field strength (in terms of g) at 1.0 Earth radius is
g
.
4
(b) r = 3rE
gr =
=
gr =
GM E
(3rE + rE ) 2
GM E
16rE 2
g
16
The magnitude of the gravitational field strength (in terms of g) at 3.0 Earth radii is
350 Unit 2 Energy and Momentum
g
.
16
Copyright © 2003 Nelson
(c) r = 4.2rE
gr =
GM E
(4.2 rE + rE ) 2
GM E
=
27 rE 2
gr =
g
27
The magnitude of the gravitational field strength (in terms of g) at 4.2 Earth radii is
4.
g
.
27
Let gP represent the magnitude of the gravitational field strength of the planet. Thus, g P =
gP =
GM E
(0.50rE )
2
and g =
GM E
rE 2
.
4.0GM E
rE 2
g P = 4.0 g
The planet has a surface gravitational field strength of magnitude 4.0g.
5. (a) g = 1.6 N/kg
r = 1.74 ×106 m
g=
GM Moon
r2
gr 2
G
(1.6 N/kg)(1.74 × 106 m) 2
=
6.67 ×10−11 N ⋅ m 2 /kg 2
M Moon =
M Moon = 7.3 × 1022 kg
The mass of the Moon is 7.3 × 1022 kg.
(b) M Moon = 7.35 × 1022 kg
mstudent = 55 kg (example mass of student)
FG =
GM Moon mstudent
r2
(6.67 ×10 −11 N ⋅ m 2 /kg 2 )(7.35 × 10 22 kg)(55 kg)
=
(1.74 × 106 m)2
FG = 89 N
A 55-kg student would have a weight of 89 N on the Moon. (Students can also determine the weight by applying the
fact that the ratio of the Moon’s gravitational field strength to Earth’s gravitational field strength is 1.6:9.8.)
6. (a) g = 5.42 × 10−9 N/kg
m = 1.00 kg
FG = mg
= (1.00 kg)(5.42 × 10−9 N/kg)
FG = 5.42 × 10−9 N
The magnitude of the gravitational force is 5.42 × 10−9 N.
(b) m = 8.91× 105 kg
FG = mg
= (8.91× 105 kg)(5.42 × 10−9 N/kg)
FG = 4.83 × 10−3 N
The magnitude of the gravitational force is 4.83 × 10−3 N.
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 351
Applying Inquiry Skills
7.
It is an inverse square relationship. The graphs for (a) and (b) are the same except that the vertical axes are labelled
differently.
Making Connections
8. (a) If Earth’s density were greater, then its mass would also be greater. Therefore, Earth’s surface gravitational field
strength would also be much greater since it is proportional to Earth’s mass.
(b) Since the surface gravitational field strength of Earth would be much greater if its density were greater, a human’s
weight would also be greater. To support the increased weight, bones would need to be very sturdy (and heavier).
Shorter, thicker bones would be more efficient at supporting the larger weight.
(c) Answers will vary. Some effects in nature could be shorter maximum heights of plants and more powerful winds.
Examples of effects on human activities include different sizes of equipment used in sports (balls, bats, rackets, etc.)
and small sizes of some transportation vehicles, especially aircrafts.
Section 6.1 Questions
(Page 277)
Understanding Concepts
1.
The weight of a space probe decreases as the probe travels away from Earth because the gravitational field strength of
Earth, gE, decreases. However, as it approaches the Moon, the space probe experiences the gravitational field strength of
the Moon, gM. There is a location where gE and gM are equal, but in opposite directions. At this location, the weight of the
probe is zero, although the mass of the space probe does not change. Since Earth is almost 100 times more massive than
the Moon, this location is much closer to the Moon.
2. (a) Let the subscript s represent the satellite.
m = 225 kg
M E = 5.98 ×1024 kg
rs = 8.62 ×106 m
rE = 6.38 ×106 m
r = rs + rE
FG =
GM E m
r2
(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(5.98 × 10 24 kg)(225 kg)
=
(8.62 × 106 m + 6.38 × 106 m)2
FG = 3.99 × 102 N
The magnitude and direction of the gravitational force on the satellite is 3.99 × 102 N [toward Earth’s centre].
G
G
(b) ∑ F = ma
G
G F
a= G
m
3.99 × 102 N [toward Earth's centre]
=
225 kg
G
2
a = 1.77 m/s [toward Earth's centre]
The resulting acceleration of the satellite is 1.77 m/s2 [toward Earth’s centre].
352 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
3.
r = 7.4 ×107 m
M E = 5.98 × 1024 kg
GM
g = 2E
r
(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
=
(7.4 × 107 m) 2
g = 7.3 × 10−2 m
The magnitude and direction of the gravitational field strength is 7.3 × 10–2 N/kg [toward Earth’s centre].
4. (a) Let the subscript s represent the satellite.
g s = 4.5 N/kg
rE = 6.38 × 106 m
g = 9.8 N/kg
Since gs =
GM E
r
2
and g =
g  GM E
=
g s  rE 2
GM E
rE 2
:
 r 2 

 GM 
E 

g
r2
= 2
g s rE
r = rE
g
gs
= 6.38 × 106 m
9.8 N/kg
4.5 N/kg
r = 9.42 × 106 m
The distance the satellite is above Earth = r – rE
= 9.42 × 106 m – 6.38 × 106 m
= 3.0 × 106 m
6
Thus, the satellite is 3.0 × 10 m above Earth’s surface.
(b) m = 6.2 × 102 kg
FG = mg
= (6.2 × 102 kg)(4.5 N/kg)
FG = 2.8 × 103 N
5.
The gravitational force on the satellite is 2.8 × 103 N.
rN = 2.48 × 107 m
M N = 1.03 × 1026 kg
GM
g = 2N
r
(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(1.03 × 1026 kg)
=
(2.48 ×107 m) 2
g = 11.2 N/kg
The magnitude of Neptune’s surface gravitational field strength is 11.2 N/kg. The value given in Table 1 is
1.14 × 9.8 N/kg = 11.2 N/kg. Therefore, the values are the same.
6. (a) r = 2.5 ×10 7 m
m = 456 kg
v = 3.9 km/s = 3.9 × 103 m/s
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 353
v2
r
(3.9 × 103 m/s) 2
=
2.5 × 107 m
ac = 0.61m/s 2
The acceleration of the satellite is 0.61 m/s2 [toward Earth’s centre].
GM E m
(b) FG =
r2
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(456 kg)(5.98 × 1024 kg)
=
(2.5 × 107 m)2
ac =
FG = 2.9 × 102 N
The gravitational force on the satellite is 2.9 × 102 N [toward Earth’s centre].
7. (a) g = 1.3 N/kg
M = 1.3 × 1023 kg
GM
g= 2
r
GM
r=
g
(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(1.3 ×1023 kg)
1.3 N/kg
=
= 2.6 × 106 m
r = 2.6 × 103 km
Titan’s radius is 2.6 × 103 km.
(b) m = 0.181kg
FG = mg
= (0.181kg)(1.3 N/kg)
8.
FG = 0.24 N
The force of gravity on a 0.181-kg rock on Titan is 0.24 N.
Let r2 be the distance from Earth’s centre at which the gravitational field strength has a magnitude, g2, of 3.20 N/kg.
g E = 9.80 N/kg at Earth's surface
g 2 = 3.20 N/kg at distance r2
Using ratio and proportion:
g 2 GmE GmE
= 2 ÷ 2
gE
r2
rE
=
r2 2 =
r2 =
=
rE 2
r2 2
g E rE 2
g2
g E rE 2
g2
(9.80 N/kg)(rE 2 )
(3.20 N/m)
r2 = 1.75rE
354 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
d = 1.75rE − 1.00rE
d = 0.75rE
The gravitational field strength is 3.20 N/kg at a distance of 0.75rE above Earth’s surface.
Applying Inquiry Skills
9.
The diagram below shows examples of the required FBDs.
Making Connections
10. Based on the data in Table 1, some astronomers might argue that Pluto should not be considered to be a planet because
the gravitational field strength of Pluto is so much smaller than the other planets. On average, Pluto’s gravitational field
strength is less than one-tenth the gravitational field strength of the other planets.
6.2 ORBITS AND KEPLER’S LAWS
PRACTICE
(Page 279)
Understanding Concepts
1.
The Moon does not fall into Earth because it travels at a specific speed around Earth. This keeps the Moon at an
approximately constant distance from Earth’s centre called the orbital radius.
2. Since the space probe is in a circular orbit, the direction of the gravitational force is perpendicular to the direction of the
instantaneous velocity. Thus, the force of gravity does not do any work on the probe, and there is no change in the kinetic
energy (or speed) of the probe.
3. Let the subscript s represent the satellite.
ME = 5.98 × 1024 kg
rE = 6.38 × 106 m
rs = 525 km = 5.25 × 105 m
r = rE + rs
GM E
(a) vs =
r
=
(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(5.98 × 10 24 kg)
(6.38 ×106 m) + (5.25 ×105 m)
vs = 7.60 × 103 m/s
The speed of the satellite is 7.60 × 103 m/s.
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 355
d = 1.75rE − 1.00rE
d = 0.75rE
The gravitational field strength is 3.20 N/kg at a distance of 0.75rE above Earth’s surface.
Applying Inquiry Skills
9.
The diagram below shows examples of the required FBDs.
Making Connections
10. Based on the data in Table 1, some astronomers might argue that Pluto should not be considered to be a planet because
the gravitational field strength of Pluto is so much smaller than the other planets. On average, Pluto’s gravitational field
strength is less than one-tenth the gravitational field strength of the other planets.
6.2 ORBITS AND KEPLER’S LAWS
PRACTICE
(Page 279)
Understanding Concepts
1.
The Moon does not fall into Earth because it travels at a specific speed around Earth. This keeps the Moon at an
approximately constant distance from Earth’s centre called the orbital radius.
2. Since the space probe is in a circular orbit, the direction of the gravitational force is perpendicular to the direction of the
instantaneous velocity. Thus, the force of gravity does not do any work on the probe, and there is no change in the kinetic
energy (or speed) of the probe.
3. Let the subscript s represent the satellite.
ME = 5.98 × 1024 kg
rE = 6.38 × 106 m
rs = 525 km = 5.25 × 105 m
r = rE + rs
GM E
(a) vs =
r
=
(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(5.98 × 10 24 kg)
(6.38 ×106 m) + (5.25 ×105 m)
vs = 7.60 × 103 m/s
The speed of the satellite is 7.60 × 103 m/s.
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 355
2π r
T
2π r
T=
v
(b) v =
=
(
2π 6.38 × 106 m + 5.25 × 105 m
)
3
7.60 × 10 m/s
3
T = 5.71× 10 s
4.
 1h 
The period of revolution of the satellite is 5.71 × 103 s or (5.71× 103 s) 
 = 1.59 h.
 3600 s 
Let the subscript s represent the satellite.
MMoon = 7.35 × 1022 kg
rMoon = 1.74 × 106 m
vs =
=
GM Moon
r
(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(7.35 × 1022 kg)
1.74 × 106 m
vs = 1.68 × 103 m/s
The speed of the satellite is 1.68 × 103 m/s.
Applying Inquiry Skills
5. (a) The speed of a satellite around a central body is inversely proportional to the square root of the radius of the satellite’s
1
orbit. Thus, vs ∝
.
rs
(b)
Making Connections
6.
Recommended web sites are:
www.space.com/spacewatch/space_junk.html
http://earthwatch.unep.net/solidwaste/spacejunk.html
Students are likely to encounter the following points about the problem of space junk:
• a study done in 1999 estimated 4 million pounds of space junk in low-Earth orbit
• objects that are baseball-size and bigger may threaten the safety of astronauts in space; collisions involving even the
smallest of objects may be damaging due to the high speeds of the objects
• the U.S. Space Command agency counted the number of objects in space as of June 21, 2000 and found 2671 satellites,
90 space probes, and 6096 chunks of debris
• some objects re-enter Earth’s atmosphere, but most burn up on re-entry, or land in water or uninhabited land
• NASA calculates that if the amount of debris equal to or larger than 1 centimeter exceeds 150 000, it could make space
flight impossible
356 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
• as technology improves, more and more satellites and space objects are being launched into space, not only by space
•
agencies, but other industries including telecommunications and organizations interested in astronomy
possible solutions include better tracking of space junk, improved methods of bringing down satellites to Earth, and
launching satellites to sweep up the space junk that is already in orbit
PRACTICE
(Page 283)
Understanding Concepts
7.
Relative to the rest of the solar system, Earth’s frame of reference is accelerating, so the geocentric model is the
noninertial frame of reference. The heliocentric model is an inertial frame of reference if the solar system is considered to
be isolated. However, it is a noninertial frame with respect to the Milky Way Galaxy and the rest of the universe.
8. Tycho Brahe (1546–1601) made precise, comprehensive astronomical measurements of the solar system and more than 700
stars. For 20 years, he made countless naked-eye observations using large instruments he made himself. He was able to
collect data for Mercury, Venus, Earth, Mars, and Saturn, because those were planets that he could see. The other planets
were beyond his scope of vision and the telescope was not invented until the early 17th century.
9. Using Kepler’s second law, Earth sweeps out equal areas in equal time intervals. Therefore, when Earth is closest to the Sun,
it is moving fastest. Conversely, if the orbit is divided into 180° “halves,” the portion closest to the Sun will have a smaller
area, and therefore a shorter time. Since there are three fewer days between September 21 and March 21, Earth must be
closest to the Sun at that time.
r3
10. The ratio 2 is calculated for each planet in Table 1.
T
Table 1
Object
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Pluto
Mean
Radius of
Orbit (m)
10
5.79 × 10
11
1.08 × 10
11
1.49 × 10
11
2.28 × 10
11
7.78 × 10
12
1.43 × 10
12
2.87 × 10
12
4.50 × 10
12
5.91 × 10
Period of
Revolution
of Orbit (s)
6
7.60 × 10
7
1.94 × 10
7
3.16 × 10
7
5.94 × 10
8
3.75 × 10
8
9.30 × 10
9
2.65 × 10
9
5.20 × 10
9
7.82 × 10
3
r ∝T
2
r3
3 2
(m /s )
T2
18
3.36 × 10
18
3.35 × 10
18
3.31 × 10
18
3.36 × 10
18
3.35 × 10
18
3.38 × 10
18
3.37 × 10
18
3.37 × 10
18
3.38 × 10
C=
All proportionality constants calculated in Table 1 are within 1.5% of the average value. This verifies Kepler’s third law.
11. (a) The average value (in SI base units) of the constant of proportionality in r3 ∝ T2 is 3.36 × 1018 m3/s2.
(b) CS = 3.36 × 1018 m3/s2 (from (a))
GM S
CS =
4π 2
C 4π 2
MS = S
G
(3.36 ×1018 m3 /s 2 )4π 2
=
6.67 ×10 −11 N ⋅ m 2 / kg 2
M S = 1.99 × 1030 kg
The mass of the Sun is 1.99 × 1030 kg.
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 357
12. (a) rMoon = 3.84 × 108 m
TMoon = 2.36 × 106 s
3
r
CE = Moon 2
TMoon
(3.84 ×108 m)3
(2.36 × 106 s)2
=
CE = 1.02 × 1013 m3 /s 2
Thus, Kepler’s third law constant, CE, is 1.02 × 1013 m3/s2 for objects orbiting Earth.
(b) Let the subscript s represent the satellite.
ME = 5.98 × 1024 kg
 3600 s 
4
Ts = 4.0 h = (4.0 h) 
 = 1.4 × 10 s
1
h


Since CE =
GM E
4π 2
rs 3 =
rs =
=
rs 3
Ts 2
, then
GM E
4π 2
=
rs3
Ts 2
.
GM ETs 2
4π 2
3
(6.67 × 10−11 N ⋅ m 2 / kg 2 )(5.98 × 1024 kg)(1.4 × 104 s)2
4π 2
rs = 1.3 × 107 m or 1.3 × 104 km
The satellite must be 1.3 × 104 km above the centre of Earth.
2π r
vs =
T
2π (1.26 × 10 7 m )
=
1.4 × 10 4 s
vs = 5.6 × 103 m/s
The speed of the satellite is 5.6 × 103 m/s.
Applying Inquiry Skills
13.
In the sample diagram above, the shape of each figure is approximately that of a triangle, so the approximate areas are:
1
1
A1 = b1h1
A2 = b2 h2
2
2
1
1
= (5.0 mm)(54 mm)
= (12 mm)(22 mm)
2
2
A1 = 1.3 × 10 2 mm 2
A2 = 1.3 × 10 2 mm 2
In the diagram, d2 > d1, but t2 = t1, so v2 > v1.
Making Connections
14. (a) Let the subscript C represent the central body around which another body (subscript B) revolves in an orbit of known
GM C rB3
4π 2 rB3
period and average radius. Since
,
the
equation
for
the
mass
of
the
central
body
is
.
M
=
=
C
4π 2
TB 2
GTB2
358 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
(b) According to the equation derived in (a) above, G is known and if the mass of the star can be estimated, the period of
revolution of the planet must be found in order to solve for the radius of the orbit. Astronomers have discovered that
the mass of a star can be estimated by determining its luminosity. (This applies to “main-sequence” stars, by far the
majority of stars.) The period of revolution is determined by measuring the period of the wobble of the star as the
planet tugs on the star. The radius of the orbit is found using the equation r =
3
GM Star (TPlanet )2
4π 2
.
Section 6.2 Questions
(Page 284)
Understanding Concepts
1.
2.
3.
4.
According to Kepler’s first law, a comet travels in an elongated elliptical orbit. Kepler’s second law implies that the
portion of the comet’s orbit closet to the Sun will have a much smaller area than the rest of its orbit. Thus, the time spent
in this region (where the comet may be visible to observers on Earth) will be far less compared to the total orbital period.
Kepler’s second law states that Earth sweeps out equal areas in equal time intervals. Therefore, on January 4 when Earth
is closest to the Sun, it is moving most rapidly because the distance travelled is greatest for equal time intervals. The Earth is
moving least rapidly on July 5 when it is farthest from the Sun.
Although the nonrotating frame of reference is placed at the centre of the Sun, the Sun is in orbit around the centre of the
Milky Way Galaxy, so it is constantly accelerating. This means it is a noninertial frame of reference within the galaxy.
r = 4.8 × 1011 m
MS = 1.99 × 1030 kg
GM S
CS =
4π 2
(6.67 × 10−11 N ⋅ m 2 / kg 2 )(1.99 × 1030 kg)
=
4π 2
CS = 3.36 × 1018 m3 /s 2
r3
T2
r3
T2 =
CS
CS =
(4.8 × 1011 m)3
3.36 × 1018 m3 /s 2
T=
5.
T = 1.8 ×108 s
The orbital period of the asteroid is 1.8 × 108 s.
Let the subscript P represent the unknown planet.
r3
CS = E 2
TE
CS =
rP 3
TP 2
=
rP 3
TP 2
rE 3
TE 2
TP = 2TE
rP
3
(2TE )2
=
rP =
rE 3
TE 2
3
4TE 2 rE 3
TE 2
= 3 4rE
rP = 1.6rE
The small planet would be 1.6 times farther from the Sun than Earth.
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 359
6.
ME = 5.98 × 1024 kg
rE = 6.38 × 1024 m
rs = 2rE
GM E
2
4π
GM E
4π 2
=
=
T=
=
7.
(2rE )3
T2
8rE 3
T2
32π 2 rE 3
GM E
32π 2 (6.38 × 106 m)3
(6.67 × 10 N ⋅ m 2 / kg 2 )(5.98 × 1024 kg)
−11
 1h 
= 1.43 ×10 4 s 

 3600s 
T = 4.0 h
The period of revolution is 4.0 h.
Let the subscript P represent Phobos, the subscript D represent Deimos, and the subscript M represent Mars.
 3600 s 
 60 s 
5
TD = 30 h 18 min = 30 h 
 + 18 min 
 = 1.09 × 10 s
1
h
1
min




rD = 2.3 × 104 km
 3600 s 
 60 s 
4
TP = 7 h 39 min = 7 h 
 + 39 min 
 = 2.75 × 10 s
 1h 
 1 min 
r 3
CM = D 2
TD
=
(2.3 × 107 m)3
(1.09 × 105 s)2
CM = 1.02 × 1012 s
Substitute the value of CM into the equation for Phobos:
r3
CM = P 2
TP
rP 3 = CM TP 2
rP = 3 (1.02 × 1012 s)(2.75 × 10 4 s) 2
rP = 9.2 ×106 m
Phobos is 9.2 × 106 m from the centre of Mars.
Applying Inquiry Skills
8.
GM
(N ⋅ m 2 / kg 2 )(kg)
=
r
m
=
(kg ⋅ m/s 2 )(m2 / kg 2 )(kg)
m
=
m3 /s 2
m
= m 2 /s 2
= m/s
Therefore, the SI base units are metres per second.
360 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
9.
Since r3 = CT2 (from Kepler’s third law), the line on the graph is straight and the slope is Kepler’s third law constant for
the Sun.
Making Connections
10. (a) Galileo was born near Pisa, on February 15, 1564. In 1609, after learning that a telescope had been invented in
Holland, he built his own telescope of 20 times magnification. The strength of this magnification allowed Galileo to see
mountains and craters on the Moon, and to discover the four largest satellites of Jupiter.
Tycho’s work was done between 1581 and 1601, during which he made numerous naked-eye observations with
large instruments. Kepler began analyzing Tycho’s data in 1601. In 1609, Kepler published his first two laws. He
published his third law in 1619.
(b) Relating Kepler’s third law and the law of universal gravitation, we find that the mass of Jupiter is given by
4π 2 rM 2
, where rM and TM are the radius of the orbit and period of revolution, respectively, of any moon around
MJ =
GTM 2
Jupiter. Thus, Galileo would need to know the values of rM and TM for at least one moon, as well as the universal
gravitation constant, G.
(c) Calculating the mass of Jupiter was not possible until the value of G was determined, which was not possible until
Kepler’s third law was formed. (As mentioned in Section 3.3, it was Cavendish who first determined that value in
1798.)
6.3 GRAVITATIONAL POTENTIAL ENERGY IN GENERAL
PRACTICE
(Pages 287–288)
Understanding Concepts
1.
ME = 5.98 × 1024 kg
m = 0.0123ME
r = 3.84 × 105 km
GM E m
r
GM E 2 0.0123
=−
r
Eg = −
(6.67 ×10
=−
−11
)(
N ⋅ m 2 /kg 2 5.98 × 1024 kg
3.84 × 108 m
) (0.0123)
2
28
Eg = −7.64 × 10 J
The gravitational potential energy of the Earth-Moon system is –7.64 × 1028 J.
2. (a) m = 1.0 kg
rE = 6.38 × 106 m
r = 1.0 × 102 km = 1.0 × 105 m
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 361
9.
Since r3 = CT2 (from Kepler’s third law), the line on the graph is straight and the slope is Kepler’s third law constant for
the Sun.
Making Connections
10. (a) Galileo was born near Pisa, on February 15, 1564. In 1609, after learning that a telescope had been invented in
Holland, he built his own telescope of 20 times magnification. The strength of this magnification allowed Galileo to see
mountains and craters on the Moon, and to discover the four largest satellites of Jupiter.
Tycho’s work was done between 1581 and 1601, during which he made numerous naked-eye observations with
large instruments. Kepler began analyzing Tycho’s data in 1601. In 1609, Kepler published his first two laws. He
published his third law in 1619.
(b) Relating Kepler’s third law and the law of universal gravitation, we find that the mass of Jupiter is given by
4π 2 rM 2
, where rM and TM are the radius of the orbit and period of revolution, respectively, of any moon around
MJ =
GTM 2
Jupiter. Thus, Galileo would need to know the values of rM and TM for at least one moon, as well as the universal
gravitation constant, G.
(c) Calculating the mass of Jupiter was not possible until the value of G was determined, which was not possible until
Kepler’s third law was formed. (As mentioned in Section 3.3, it was Cavendish who first determined that value in
1798.)
6.3 GRAVITATIONAL POTENTIAL ENERGY IN GENERAL
PRACTICE
(Pages 287–288)
Understanding Concepts
1.
ME = 5.98 × 1024 kg
m = 0.0123ME
r = 3.84 × 105 km
GM E m
r
GM E 2 0.0123
=−
r
Eg = −
(6.67 ×10
=−
−11
)(
N ⋅ m 2 /kg 2 5.98 × 1024 kg
3.84 × 108 m
) (0.0123)
2
28
Eg = −7.64 × 10 J
The gravitational potential energy of the Earth-Moon system is –7.64 × 1028 J.
2. (a) m = 1.0 kg
rE = 6.38 × 106 m
r = 1.0 × 102 km = 1.0 × 105 m
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 361
On Earth’s surface,
GM E m
Eg1 = −
rE
(6.67 ×10
=−
−11
)(
)
N ⋅ m 2 /kg 2 5.98 × 10 24 kg (1.0 kg )
6
6.38 × 10 m
Eg1 = −6.25 × 107 J
In orbit,
Eg2 = −
GM E m
r + rE
(6.67 ×10
=−
−11
)(
)
N ⋅ m 2 /kg 2 5.98 ×10 24 kg (1.0 kg )
5
6
1.0 × 10 m + 6.38 × 10 m
7
Eg2 = −6.15 ×10 J
∆Eg = Eg2 − Eg1
= −6.15 × 107 J − (−6.25 × 107 J)
∆Eg = 1.0 × 107 J
The change in gravitational potential energy for a 1.0-kg mass lifted 1.0 × 102 km above Earth’s surface is 1.0 × 106 J
(b) g = 9.80 N/kg
∆Eg = mg ∆y
= (1.0 kg)(9.80 N/kg)(1.0 × 105 m)
∆Eg = 9.8 × 105 J
The percentage error can be calculated as:
 measured value − accepted value 
% error = 
 × 100%
accepted value


 9.8 × 105 J − 1.0 × 106 J 
= 
 × 100%
1.0 × 106 J


% error = − 2.0%
Thus, the answer calculated using the equation ∆Eg = mg∆y would be 2.0% less than the answer calculated in (a).
(c) The low percentage error (less than 2.0%) tells us that there is little need for more exact treatment in most normal
Earth-bound problems. The approximation assuming a constant value of g is quite good, even for an altitude of
1.0 × 102 km.
3. ME = 5.98 × 1024 kg
rE = 6.38 × 106 m
Since EK = ∆Eg, EK = Eg2 –Eg1 where Eg1 is the gravitational potential energy on Earth’s surface and Eg2 is the
gravitational potential energy in orbit. Thus,
GM E m  GM E m 
1 2
−−
mv = −

2
2rE
rE 

1
1 
v = 2GM E  −

 rE 2rE 
(
= 2 6.67 × 10 −11 N ⋅ m 2 /kg 2
)(5.98 ×10
24


1
1
−
kg 

6
6
 6.38 × 10 m 2(6.38 ×10 m) 
)
v = 7.91× 103 m/s
The object must be projected with an initial velocity of 7.91 × 103 m/s.
362 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
4.
MS = 1.99 × 1030 kg
ME = 5.98 × 1024 kg
rp = 1.47 × 1011 m; the distance from the Sun to Earth at perihelion
ra = 1.52 × 1011 m; the distance from the Sun to Earth at aphelion
(a) The maximum change in Earth’s gravitational potential energy during one orbit of the Sun is ∆Eg = Ea –Ep where Ea is
the gravitational potential energy at aphelion and Ep is the gravitational potential energy at perihelion.
GM S M E  GM S M E 
∆Eg = −
−−



ra
rp


1 1
= GM S M E  − 
 rp ra 


1
1


= 6.67 ×10−11 N ⋅ m 2 /kg 2 5.98 × 10 24 kg 1.99 × 1030 kg 
−

11
11
 1.47 × 10 m 1.52 × 10 m 
(
)(
)(
)
∆Eg = 1.8 × 1032 J
The maximum change in Earth’s gravitational potential energy during one orbit of the Sun is 1.8 × 1032 J.
(b) Earth is moving the fastest at perihelion, which is its closest approach to the Sun. The maximum change in kinetic
energy during one orbit is ∆EK = ∆Eg = 1.8 × 1032 J.
Making Connections
5.
m = 5.00 × 102 kg
r1= 2rE
r2 = 3rE
ME = 5.98 × 1024 kg
rE = 6.38 × 106 m
(a) In the satellite’s initial orbit:
GM E m
Eg1 = −
r1
(6.67 ×10
=−
−11
)(
)(
)
)(
)
N ⋅ m 2 /kg 2 5.98 × 1024 kg 5.00 ×10 2 kg
6
2(6.38 × 10 m)
10
Eg1 = −1.56 × 10 J
In the satellite’s final orbit:
GM E m
Eg2 = −
r2
(6.67 ×10
=−
−11
)(
N ⋅ m 2 /kg 2 5.98 × 1024 kg 5.00 × 10 2 kg
6
3(6.38 ×10 m)
10
Eg2 = −1.04 × 10 J
The gravitational potential energy in the satellite’s initial orbit is −1.56 × 1010 J. The gravitational potential energy in
the satellite’s final orbit is −1.04 × 1010 J.
(b) ∆Eg = Eg2 − Eg1
= −1.04 × 1010 J − (−1.56 × 1010 J)
∆Eg = 5.2 ×109 J
The change in gravitational potential energy from the first orbit to the second orbit is 5.2 × 109 J.
(c) W = ∆Eg
W = 5.2 × 109 J
The work done in raising the satellite as it moves from the first orbit to the second orbit is 5.2 × 109 J.
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 363
PRACTICE
(Page 293)
Understanding Concepts
6.
7.
The escape speed of a space probe depends on the mass of Earth, not on the mass of the probe, according to the equation
2GM E
v=
.
r
MJ = 318 ME
rJ = 10.9 rE
2GM J
2GM E
vJ =
and vE =
rJ
rE
2G (318M E )
10.9rE
vJ
=
vE
2GM E
rE
 2G (318M E )   rE 
= 
 

 10.9r
E

  2GM E 
318
10.9
=
vJ
= 5.40
vE
The ratio of the escape speed from Jupiter to the escape speed from Earth is 5.40:1.
8. mMoon = 7.35 × 1022 kg
r = 3.84 × 108 m (from Earth’s centre)
GM E m
(a) Eg = −
r
(6.67 ×10
=−
−11
)(
)(
N ⋅ m 2 /kg 2 5.98 × 1024 kg 7.35 × 1022 kg
8
)
3.84 × 10 m
28
Eg = −7.63 ×10 J
The gravitational potential energy of the Moon with respect to Earth is –7.63 × 1028 J.
(b) EK = ET – Eg
1
EK = Eg − Eg
2
1
= − Eg
2
EK = 3.82 × 1028 J
1 2
mv
2
2 EK
v=
m
Since EK =
=
2(3.82 × 1028 J)
7.35 ×10 22 kg
v = 1.02 ×103 m/s
The Moon’s kinetic energy is 3.82 × 1028 J and its speed in orbit is 1.02 × 103 m/s.
(c) The Moon’s binding energy to Earth is ET =
364 Unit 2 Energy and Momentum
1
Eg = 3.82 × 1028 J.
2
Copyright © 2003 Nelson
9.
m = 2.0 × 103
r = 5.0 × 102 km = 5.0 × 105 m
ME = 5.98 × 1024 kg
rE = 6.38 × 106 m
W = ∆E = ET (in orbit) − ET (on Earth)
1
Eg2 − Eg1
2
1 GM E m  GM E m 
=−
−−

2 ( r + rE ) 
rE 
∆E =
1

1
= GM E m  −

r
2(
r
r
)
+
E 
 E


1
1
= 6.67 ×10−11 N ⋅ m 2 /kg 2 5.98 × 1024 kg 2.0 ×103 kg 
−

6
5
6
 6.38 × 10 m 2(5.0 × 10 m + 6.38 × 10 m) 
(
)(
)(
)
∆E = 6.7 ×1010 J
The total amount of energy needed to place the satellite in circular Earth orbit is 6.7 × 1010 J.
1 GM E m
10. The binding energy of the satellite is
. Thus, the additional energy that would have to be supplied to the satellite
2 r + rE
would be:
1 GM E m
E=
2 r + rE
=
=
1 GM E m
2 (r + rE )
(6.67 ×10
−11
)(
)(
N ⋅ m 2 /kg 2 5.98 × 10 24 kg 2.0 ×103 kg
5
6
)
2(5.0 ×10 m + 6.38 ×10 m)
10
E = 5.8 ×10 J
Therefore, 5.8 × 1010 J of additional energy would have to be supplied to the satellite to allow it to escape from Earth’s
gravitational field.
11. T = 24 h = 24 h × 3600 s/h = 8.64 × 104 s
ME = 5.98 × 1024 kg
GM E
CE =
4π 2
=
(6.67 ×10
−11
)(
N ⋅ m 2 /kg 2 5.98 ×10 24 kg
4π
2
)
CE = 1.01×1013 m3 /s 2
r 3 = CE T 2
r = 3 CE T 2
= 3 (1.01× 1013 m3 /s 2 )(8.64 × 104 s) 2
r = 4.23 × 107 m
2π r
T
2π (4.23 × 107 m)
=
8.64 × 10 4 s
v = 3.1 × 103 m/s
The satellite’s speed in orbit is 3.1 × 103 m/s.
v=
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 365
1
Eg2 − Eg1
2
1 2
1 GM E m  GM E m 
mv = −
−−

rE 
2
2 r

EK = ∆E =
1 1 
v 2 = 2GM E  − 
 rE 2r 


1
1
v = 2 6.67 ×10−11 N ⋅ m 2 /kg 2 5.98 ×10 24 kg 
−

6
7
×
×
6.38
10
m
2(4.23
10
m)


(
)(
)
v = 1.08 ×104 m/s
The satellite must reach a speed of 1.1 × 104 m/s during launch to get into a geosynchronous orbit.
12. m = 4.00 MS
MS = 1.99 × 1030 kg
c = 3.00 × 108 m/s
2GM
r= 2
c
2(6.67 ×10 −11 N ⋅ m 2 /kg 2 )(4.00 ×1.99 ×1030 kg)
=
(3.00 ×108 m/s) 2
r = 1.18 × 10 4 m, or 11.8 km
The Schwartzschild radius of a black hole is 11.8 km.
Applying Inquiry Skills
13. As shown in Figure 4 on page 288 of the text, a graph of gravitational potential energy as a function of distance from the
centre of the body yields a potential well graph.
Making Connections
14. (a) m = 65.0 kg
ME = 5.98 × 1024 kg
rE = 6.38 × 106 m
On Earth’s surface:
GM E m
E=
rE
(6.67 ×10
=
−11
)(
)
N ⋅ m 2 /kg 2 5.98 × 1024 kg (65.0 kg )
6
6.38 × 10 m
9
E = 4.06 ×10 J
The binding energy of a 65.0-kg person on Earth’s surface is 4.06 × 109 J.
366 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
(b) To just escape from Earth, EK = binding energy. Thus, the person would need 4.06 × 109 J of kinetic energy.
(c) g = 9.80 N/kg
∆y = 1.00 m
W = mg ∆y
= (65 kg)(9.80 N/kg)(1.00 m)
W = 6.37 × 102 J
To raise the person 1.00 m at Earth’s surface would require 6.37 × 102 J of work.
(d) One of NASA’s objectives in designing launches into space is to minimize the mass of the payload. Since the binding

GM E m 
energy is proportional to the mass of the object  binding energy =
 , the lower the mass, the lower the binding
rE 

energy. Minimizing the mass of the payload would therefore require less energy and work to launch the vehicle into
space, reduce the amount of fuel needed, and be more cost effective.
Section 6.3 Questions
(Page 294)
Understanding Concepts
1.
2.
GM E m
. Therefore, the escape energy is proportional to the mass of the object. Since the mass of a
rE
1500-kg rocket is three times the mass of a 500-kg rocket, the escape energy of the 1500-kg rocket will also be three times
as great as the escape energy of the 500-kg rocket.
The statement, “No satellite can orbit Earth in less than about 80 min” is correct. The minimum period of orbit for a
satellite corresponds to the radius of Earth.
GM E
CE =
4π 2
(6.67 × 10−11 N ⋅ m2 / kg 2 )(5.98 × 1024 kg)
=
4π 2
13 3
CE = 1.01× 10 m
Escape energy =
Cs =
TE 2 =
TE =
rs3
Ts 2
rE 3
CE
(6.38 ×106 m)3
(1.01× 1013 m3 / s 2 )
= (5.07 × 103 s)(1 min/60 s)
TE = 85 min
Thus, the minimum period of orbit corresponds to 85 min.
3. (a) ME = 5.98 × 1024 kg
mt = 1.2 × 103 kg (mass of the tank)
rE = 6.38 × 106 m
rt = 6.38 × 106 m + 2.0 × 106 m = 8.28 × 106 m
W=?
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 367
Applying the law of conservation of energy:
W = ∆Eg
=−
GM E mt  GM E mt 
−−

rE
rt


1 1
= GM E mt  − 
 rt rE 
1
1


= (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.2 × 103 kg) 
−

6
6
×
×
8.28
10
m
6.38
10
m


W = −1.7 × 1010 J
The work done by gravity on the tank is −1.7 × 1010 J. (The gravitational potential energy at Earth’s surface is more
negative than it is at higher altitudes.)
(b) v = 0
v′ = ?
Since the loss in gravitational potential energy (found above) equals the gain in kinetic energy,
∆EK = −∆Eg = 1.7 × 107 J.
m (v′ )
2
2
m (v )
2
−
2
= 1.7 ×107 J
(v′ )2 =
(
2 1.7 × 107 J
)
m
(
2 1.72 × 107 J
v′ =
)
m
(
2 1.7 ×107 J
=
)
3
1.2 ×10 kg
v′ = 5.4 ×103 m/s
The speed of impact at Earth’s surface is 5.4 × 103 m/s.
4. EK = 5.0 × 109 J
Eg = −6.4 × 109 J
ET = EK + Eg
= (5.0 × 109 J) + (−6.4 × 109 J)
ET = −1.4 × 109 J
The binding energy of the space vehicle is 1.4 × 109 J.
5. m = 2.00 × 103 kg
r = 4.00 × 102 km = 4.00 × 105 m
ME = 5.98 × 1024 kg
rE = 6.38 × 106 m
GM E m
(a) Eg = −
r + rE
(6.67 ×10
=−
−11
)(
)(
N ⋅ m 2 /kg 2 5.98 × 1024 kg 2.00 × 103 kg
5
6
)
4.00 × 10 m + 6.38 × 10 m
11
Eg = −1.18 × 10 J
Thus, the average value of gravitational potential energy of the satellite in orbit is −1.18 × 1011 J.
(b) EK = ET – Eg
1
= Eg –Eg
2
1
= − Eg
2
EK = 5.88 × 1010 J
Thus, the average value of orbital kinetic energy is 5.88 × 1010 J.
368 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
1
Eg
2
ET = –5.88 × 1010 J
Thus, the total energy of the satellite while in orbit is –5.88 × 1010 J.
(d) Let the subscript p represent the satellite at perigee.
rp = 2.80 × 102 km = 2.80 × 105 m
GM E m
Egp = −
rp + rE
(c) ET =
(6.67 ×10
=−
−11
)(
)(
N ⋅ m 2 /kg 2 5.98 × 10 24 kg 2.00 × 103 kg
5
6
)
2.80 × 10 m + 6.38 × 10 m
11
Egp = −1.20 × 10 J
To determine the satellite’s orbital speed at perigee:
1
EK = − Egp
2
1 2
1
mv = − (−1.20 × 1011 J)
2
2
v=
1.20 ×1011 J
2.00 ×103 kg
v = 7.74 ×103 m/s
The satellite’s orbital speed at perigee is 7.74 × 103 m/s.
6. m = 5.00 × 102 kg
r = 2.00 × 102 km
ME = 5.98 × 1024 kg
rE = 6.38 × 106 m
GM E m
(a) Eg = −
r + rE
(6.67 ×10
=−
−11
)(
)(
N ⋅ m 2 /kg 2 5.98 × 1024 kg 5.00 × 102 kg
5
6
)
2.00 × 10 m + 6.38 × 10 m
Eg = −3.03 ×1010 J
The gravitational potential energy of the satellite is −3.03 × 1010 J.
(b) EK = ET – Eg
1
= Eg –Eg
2
1
= − (−3.03 × 1010 J)
2
EK = 1.52 × 1010 J
Thus, the average value of kinetic energy of the satellite is 1.52 × 1010 J.
1
(c) binding energy = Eg = –1.52 × 1010 J
2
Thus, the binding energy of the satellite while in orbit is –1.52 × 1010 J.
(d) To launch the satellite into orbit:
∆E = ET (in orbit) – ET (on Earth)
1
= Eg2 − Eg1
2
1 GM E m  GM E m 
=−
−−

2 ( r + rE ) 
rE 
1

1
∆E = GM E m  −

 rE 2(r + rE ) 
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 369
To escape from Earth:
GM E m
ET =
rE
ET
=
∆E
=
 GM E m 


 rE 
1

1
GM E m  −

 rE 2(r + rE ) 
1
 
 rE 
1

1
 −

r
2(
r
+
r
)
E 
 E
1




6.38 × 106 m 

=


1
1
−


6
5
6
 6.38 × 10 m 2(2.00 × 10 m + 6.38 × 10 m) 
ET
= 1.94
∆E
The percentage increase in launching energy required for the satellite to escape from Earth is 94%.
7. (a) MS = 1.99 × 1030 kg
rS = 6.96 × 108 m
v = ? (escape speed)
2GM S
v=
rS
=
2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
6.96 ×108 m
v = 6.18 × 105 m/s
The escape speed from our solar system is 6.18 × 105 m/s.
(b) The energy needed to escape from the solar system is the addition of the escape energy from Earth and the escape
energy from the Sun at the location of Earth’s orbit. The escape energy can be used to determine the escape speed.
ME = 5.98 × 1024 kg
rE = 6.38 × 106 m
rE-S = 1.49 × 1011 m (radius of Earth’s orbit)
Et = ? (the total escape energy)
v=?
GM S m GM E m
+
Et =
rE-S
rE
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)(m) (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(m)
+
1.49 × 1011 m
6.38 × 106 m
Et = 9.53 ×108 (m) J
=
1 2
mv
2
2E
v2 = K
m
Now, EK = Et =
v=
=
2 EK
m
2(9.53 ×108 (m) J
m
v = 4.37 ×10 4 m/s
The escape speed is 4.37 × 104 m/s.
370 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
8.
mMoon = 7.35 × 1022 kg
rMoon = 1.74 × 106 m
GM Moon m  GM Moon m 
1 2
−−
mv = −

2
2rMoon
rMoon 

 1
1 
−
v = 2GM Moon 

2
r
r
Moon 
 Moon
(
= 2 6.67 × 10 −11 N ⋅ m 2 /kg 2
9.
)(7.35 × 10
22


1
1
−
kg 

6
6
 1.74 × 10 m 2(1.74 × 10 m) 
)
v = 1.68 × 103 m/s
An object projected from the Moon must have a speed of 1.68 × 103 m/s to reach an altitude equal to the Moon’s radius.
MS = 1.99 × 1030 kg
r = 15.4 km = 1.54 × 104 m
c = 3.00 × 108 m/s
2GM
r= 2
c
rc 2
M =
2G
(1.54 × 10 4 m)(3.00 × 108 m/s) 2
=
2(6.67 × 10−11 N ⋅ m 2 /kg 2 )
M = 1.04 × 1031 kg
In terms of the Sun’s mass, the mass of the black hole is:
M
1.04 × 1031 kg
=
M S 1.99 × 1030 kg
M = 5.22 M S
Thus, the mass of the black hole is 5.22 MS.
Applying Inquiry Skills
10. The data for the required graph, generated by using the equation Eg =
−GM M m
, are shown in the table below.
r
9
Position
× 10 J)
Eg (×
rM
2rM
3rM
4rM
5rM
−25.0
−12.5
−8.33
−6.25
−5.00
11. (a) ME = 5.98 × 1024 kg
c = 3.00 × 108 m/s
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 371
r=
2GM E
c2
2(6.67 ×10 −11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
=
(3.00 ×108 m/s) 2
= 8.86 × 10−3 m
r = 8.86 mm
The theoretical Schwartzschild radius of the black hole is 8.86 mm.
(b) The very small radius of the black hole implies that it is extremely dense, according to ρ ∝
1
. Thus,
r3
r3
ρ
= E3
ρE r
=
(6.38 ×106 m)3
(8.86 × 10 −3 m)3
ρ
= 3.73 ×10 26
ρE
Therefore, the black hole is 3.73 × 1026 times more dense than Earth.
Making Connections
12. The escape energy of an object from the Moon is much less than the escape energy of the same object from Earth. (In fact,
calculations show that the ratio of the escape energy from Earth is more than 22 times as great as that from the Moon.)
Thus, less fuel is required to send the spacecraft from the Moon to Earth than from Earth to the Moon.
CHAPTER 6 LAB ACTIVITIES
Lab Exercise 6.3.1: Graphical Analysis of Energies
(Page 295)
Procedure
1.
Changing the numerical values in the given data to megametres and gigajoules is suggested just to make data entry in a
software program easier. The results of the graphing and the analysis are not affected by this suggestion. The data are:
r (Mm)
Eg (GJ)
1.6
3.2
4.8
6.4
8.0
9.6
1.12
1.28
1.44
1.60
−3.0
−1.5
−1.0
−7.5
−6.0
−5.0
−4.3
−3.8
−3.3
−3.0
2 – 5. The four lines required on the graph are shown below.
372 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
23. (c) Let the subscript 1 represent the mass of the Sun at its current value and the subscript 2 represent the mass of the Sun at
1
1
half its current value. Since MS2 = MS1, CS2 = CS1. Thus,
2
2
r3
r3
CS1 = 2 and CS2 = 2 .
T
T
 r3 
 2
CS1  T1 
=
CS2  r 3 
 2
 T2 
T2 2
T12
=2
T2
= 2
T1
24. (c) A satellite in geosynchronous orbit has a period of revolution of 24 h.
25. (a) The speed of the comet increases as it comes closer to the Sun. Position A is closet to the Sun, and therefore has the
greatest speed. Position C is furthest from the Sun, and so has the slowest speed. Positions B and D are equidistant
from the Sun and are between positions A and C. Thus, vA > vB = vD > vC.
26. (e) MP = ME
1
rP = rE
4
2GM E
2GM E
8GM E
vP =
=
vE =
and
1
rE
rE
rE
4
 8GM E

rE
vP 
=
vE  2GM E

rE

vP
= 4
vE






vP = 2vE
CHAPTER 6 REVIEW
(Pages 300–301)
Understanding Concepts
1.
2.
3.
The escape energy (and thus the escape speed) from the Sun is much greater than that from Earth, so the rocket given the
speed needed to escape from Earth would not have enough speed to escape from the solar system. Space vehicles sent to
explore distant planets have a much lower binding energy by the time they reach those distant locations, and could acquire
enough energy to escape from the solar system by taking advantage of the force of gravity of the distant planet.
Since Earth rotates eastward, an eastward orientation of the rocket as it is being launched means that the rocket already
has a component of the required velocity before blasting off. This means that less energy will be needed to launch the
rocket eastward than would be required to launch it westward in order to achieve the same speed.
gU = 1.0 N/kg
MU = 8.80× 1025 kg
rU = 2.56× 107 m
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 375
gU =
r=
=
GM U
(r + rU ) 2
GM U
− rU
gU
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(8.80 × 1025 kg)
− 2.56 × 107 m
1.0 N/kg
= 5.1 × 10 7 m
4.
5.
r = 5.1 × 10 4 km
Uranus has a gravitational field strength of 1.0 N/kg at an elevation of 5.1 × 104 km above the surface.
M = 1.48 × 1023 kg
r = 5.55 × 103 km = 5.55 × 106 m
GM
g= 2
r
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.48 × 1023 kg)
=
(5.55 × 106 m)2
g = 0.318 N/kg
The magnitude of Ganymede’s gravitational field strength at a point in space 5.55 × 103 km from its centre is 0.318 N/kg.
Use the spacecraft-to-Earth line as the reference for the coordinate system.
ME = 5.98 × 1024 kg
MMoon = 7.35 × 1022 kg
rE-spacecraft = 3.07 × 108 m
rMoon-spacecraft = 2.30 × 108 m
G
G
G
g T = g E + g Moon
G
g T,x = g E
G
g T,y = g Moon
GM
G
gE = 2 E
r
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 10 24 kg)
=
(3.07 × 108 m)2
G
g E = 4.23 ×10 −3 N/kg
GM Moon
G
g Moon =
r2
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.35 × 1022 kg)
=
(2.30 × 108 m)2
G
g M = 9.26 × 10−5 N/kg
G
g T = g T,x 2 + g T,y 2
=
(4.23×10
−3
N/kg
G
g T = 4.23 × 10−3 N/kg
376 Unit 2 Energy and Momentum
) + (9.26 ×10
2
−5
N/kg
)
2
Copyright © 2003 Nelson
tan θ =
6.
g T,y
g T,x
 g T,y 
θ = tan −1 


 g T,x 
θ = 1.26°
The total gravitational field strength (magnitude and direction) of the Earth-Moon-spacecraft system is 4.23 × 10−3 N/kg
[1.26° from the spacecraft-to-Earth line].
rE = 6.38 × 106 m
dM = 0.38dE
gE = 9.8 N/kg
gM = 0.38gE
Since Mercury’s diameter is 0.38 times that of Earth’s, Mercury’s radius is also 0.38 times that of Earth. Therefore,
rM = 0.38 rE.
GM M
gM =
rM 2
gM =
GM M
(0.38rE )2
Substituting gM = 0.38 gE:
GM M
0.38 g E =
2
(0.38rE )
(0.38 g E )(0.38rE )
2
MM =
G
(0.38 ) (9.80 N/kg) ((0.38)(6.38 × 106 m) )
2
=
6.67 × 10−11 N ⋅ m 2 /kg 2
M M = 3.3 × 10 23 kg
Therefore, Mercury’s mass is 3.3 × 1023 kg.
7. v = 7.15 × 103 m/s
ME = 5.98 × 1024 kg
rE = 6.38 × 106 m
GM E
(a) v =
r
GM
r = 2E
v
(6.67 ×10
=
−11
)(
N ⋅ m 2 /kg 2 5.98 × 1024 kg
(7.15 ×10
3
m/s
)
2
)
6
r = 7.80 × 10 m
In terms of Earth’s radius, the satellite’s distance from Earth’s centre is
r 7.80 × 106 m
=
rE 6.38 × 106 m
r = 1.22 rE
Thus, the satellite is 1.22 rE from Earth’s centre.
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 377
(b) altitude = r − rE
= 1.22 rE − rE
altitude = 0.22 rE
The satellite has an altitude of 0.22 rE.
8. vTethys = 1.1 × 104 m/s
MS = 5.67 × 1026 kg
GM S
(a) v =
r
GM S
r= 2
v
=
(6.67 ×10
−11
)(
m/s )
N ⋅ m 2 /kg 2 5.67 × 1026 kg
(1.1×10
4
2
)
r = 3.1 × 108 m, or 3.1× 105 km
The orbital radius of Tethys is 3.1 × 105 km.
2π r
(b) T =
v
=
9.
(
2π 3.1 ×108 m
4
)
1.1× 10 m/s
 1 h  1 d 
= (1.8 × 105 s) 


 3600 s  24 h 
T = 2.1 d
The orbital period of Tethys is 2.1 d.
MS = 1.99 × 1030 kg
TV = 1.94 × 107 m
GM S
CS =
4π 2
r3
= CS
TV 2
r3 =
TV 2 GM S
4π 2
r=
3
=
3
TV 2 GM S
4π 2
(1.94 × 107 s) 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
4π 2
r = 1.08 × 1011 m
The average Sun-Venus distance is 1.08 × 1011 m.
10. ME = 5.98 × 1024 kg
rE = 6.38 × 106 m
v = 9.00 km/s = 9.00 × 103 m/s
mR = 4.60 kg
378 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
(a) r′ = ?
Applying the law of conservation of energy:
Eg + EK = Eg′ + EK′
−
GM E mR 1
GM E mR
+ mR v 2 = −
+0
rE
r′
2
−
GM E 1 2
GM E
+ v =−
rE
2
r′
GM E GM E 1 2
=
− v
2
r′
rE
r′ =
=
GM E
GM E 1 2
− v
2
rE
(6.67 ×10
(6.67 ×10
−11
2
−11
N ⋅ m /s
2
)(
N ⋅ m 2 /s 2 5.98 × 1024 kg
)(5.98 ×10
24
6
kg
)−1
2
6.38 × 10 m
)
(9.00 ×10
3
m/s
)
2
r ′ = 1.85 × 107 m
Let the altitude be A.
A = r ′ − rE
= 1.85 × 107 m − 6.38 × 106 m
A = 1.21× 107 m
The altitude above Earth’s surface is 1.21 × 107 m or 1.21 × 104 km.
(b) At the altitude found in (a), the gravitational potential energy is negative, the kinetic energy is zero (because the speed
is zero), and the binding energy, EB′, is the extra energy needed to give the rocket a total energy of zero.
EB′ + Eg′ + EK′ = 0
EB′ = − Eg′
 GM E mR 
= −−

r′


GM E mR
=
r′
=
(6.67 ×10
−11
)(
)
N ⋅ m 2 /s 2 5.98 × 10 24 kg ( 4.60 kg )
7
1.85 × 10 m
EB′ = 9.92 × 10 J
The binding energy is 9.92 × 107 J.
11. MT = 1.35 × 1023 kg (mass of Titan)
rT = 2.58 × 103 km = 2.58 × 106 m
mR = 2.34 × 103 kg (mass of rocket)
(a) vesc = ? (escape speed)
2GM T
vesc =
rT
7
=
(
)(
2 6.67 × 10 −11 N ⋅ m 2 /s 2 1.35 × 10 23 kg
6
)
2.58 × 10 m
3
vesc = 2.64 × 10 m/s
The escape speed from Titan is 2.64 × 103 m/s.
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 379
(b) Eesc = ? (escape speed)
At the surface of Titan, the rocket is at rest, so its kinetic energy is zero. Thus, its total energy is Eg and the escape
energy is the extra energy needed to give the rocket a total energy of zero.
Eg + Eesc = 0
Eesc = − Eg
 GM T mR 
= − −

rT


GM T mR
=
rT
=
(6.67 ×10
−11
)(
)(
N ⋅ m2 /s 2 1.35 × 1023 kg 2.34 × 103 kg
)
6
2.58 × 10 m
Eesc = 8.17 × 109 J
The escape energy is 8.17 × 109 J. This value can also be found by using the escape speed of the rocket in the equation
1
2
EK = mR (vesc ) .
2
12. ME = 5.98 × 1024 kg
rE = 6.38 × 106 m
mR = 1.00 × 104 kg
rR = 1.00 × 1010 m
(a) Eg = ?
GM E mR
Eg = −
rR
=−
(6.67 ×10
−11
)(
)(
N ⋅ m 2 /s 2 5.98 × 10 24 kg 1.00 × 10 4 kg
10
1.00 ×10
)
m
8
Eg = −3.99 × 10 J
The gravitational potential energy is −3.99 × 108 J.
(b) Since the total energy, EK + Eg, must be at least zero, the kinetic energy needed to escape is +3.99 × 108 J.
(c) vesc = ?
2GM E
vesc =
rR
=
(6.67 ×10
−11
)(
N ⋅ m 2 /s 2 5.98 ×10 24 kg
10
1.00 × 10
)
m
2
vesc = 2.82 × 10 m/s
The escape speed from this position is 2.82 × 102 m/s. The escape speed can also be found by applying the escape
2EK
, where EK = Eesc.
energy found in (b) to the equation involving the kinetic energy, vesc =
m
13. MS = 1.99 × 1030 kg
ME = 5.98 × 1024 kg
rE = 1.49 × 1011 m
GM S M E
Eg = −
rE
=−
(6.67 ×10
−11
)(
)(
N ⋅ m 2 /kg 2 1.99 × 1030 kg 5.98 × 1024 kg
11
)
1.49 × 10 m
33
Eg = −5.33 × 10 J
The gravitational potential energy of the Sun-Earth system is −5.33 × 1033 J.
380 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
14. (a) MM = 3.28 × 1023 kg
rM = 2.44 × 106 m
2GM M
rM
v=
(
)(
2 6.67 × 10 −11 N ⋅ m 2 /kg 2 3.28 ×1023 kg
=
6
)
2.44 × 10 m
v = 4.23 × 103 m/s, or 4.23 km/s
The escape speed from Mercury is 4.23 km/s.
(b) MMoon = 7.35 × 1022 kg
rMoon = 1.74 × 106 m
2GM Moon
rMoon
v=
(
2 6.67 × 10 −11 N ⋅ m 2 /kg 2
=
6
)(7.35 ×10
22
kg
)
1.74 × 10 m
3
v = 2.37 × 10 m/s, or 2.37 km/s
The escape speed from Earth’s Moon is 2.37 km/s.
15. (a) Mstar = 3.4 × 1030 kg
1.7 × 104 m
= 8.5 × 103 m
rstar =
2
2GM star
v=
rstar
(
)(
2 6.67 × 10 −11 N ⋅ m 2 /kg 2 3.4 × 1030 kg
=
3
)
8.5 × 10 m
8
v = 2.3 × 10 m/s
The escape speed from a neutron star is 2.3 × 108 m/s.
(b) c = 3.00 × 108 m/s
v
2.3 × 108 m/s
=
c 3.00 × 108 m/s
v
= 0.77
c
v
Thus, the percentage equals   ×100% = 77% .
c
The escape speed from a neutron star is 77% the speed of light.
5.06 × 107 m
16. (a) r =
= 2.53 ×107 m
2
v = 24 km/s = 2.4 × 104 m/s
v=
M =
2GM
r
rv 2
2G
(2.53 ×10 m )(2.4 ×10 m )
=
2 (6.67 × 10 N ⋅ m /kg )
7
2
4
−11
2
2
M = 1.1 × 10 26 kg
The planet’s mass is 1.1 × 1026 kg.
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 381
(b) According to the data in Appendix C, the planet is Neptune. (There is a 6% difference in mass from Appendix C.)
17. MS = 1.99 × 1030 kg
MP = 1.67 × 10−27 kg
r = 1.4 × 109 m (initial position)
v = 3.5 × 105 m/s (initial speed)
(a) r′ = 2.8 × 109 m (final position)
v′ = ? (final speed)
Applying the law of conservation of energy:
Eg + EK = Eg′ + EK′
−
GM S mP 1
GM S mP 1
2
+ mP v 2 = −
+ mP (v′ )
2
2
r
r′
GM S 1 2
GM S 1
2
−
+ v =−
+ (v ′ )
2
2
r
r′
2
 1 1 2
(v′ ) = 2GM S  −  + v
 r′ r 
 1 1
v′ = 2GM S  −  + v 2
 r′ r 
=
(
2 6.67 × 10−11 N ⋅ m 2 /s 2
)(1.99 × 10
30
1
)  2.8 × 10
kg 
8

5
 + 3.5 × 10 m/s
m 1.4 × 10 m 
−
1
9
(
)
2
v′ = 1.7 × 105 m/s
The proton’s speed is 1.7 × 105 m/s.
(b) vesc′ = ? (escape speed at the final position)
vesc′ =
=
2GM S
r′
2(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
2.8 × 109 m
vesc′ = 3.1× 105 m/s
The escape speed is 3.1 × 105 m/s at the location indicated; this is greater than the speed found in (a), so the proton will
not escape.
18. When light strikes a piece of black paper, a small portion of the light is reflected. However, when light strikes a black
hole, the light is absorbed, making the black hole even blacker than black paper.
19. m = 1.1 × 1011 MS
MS = 1.99 × 1030 kg
c = 3.00 × 108 m/s
2GM
r= 2
c
2(6.67 ×10 −11 N ⋅ m 2 /kg 2 )(1.1× 1011 (1.99 × 1030 kg))
=
(3.00 × 108 m/s)2
r = 3.2 ×1014 m
The Schwartzschild radius of the black hole is 3.2 × 1014 m.
Applying Inquiry Skills
20. Table 1 provides the missing answers concerning some of the moons of Uranus.
(a) Kepler’s third law constant for Uranus (CU) can be calculated using the ratio
(b) The average of the CU values of the calculations in (a) is 1.48 × 1014 m3/s2.
382 Unit 2 Energy and Momentum
r3
.
T2
Copyright © 2003 Nelson
(c) MU = 8.80 × 1025 kg
GM U
CU =
4π 2
(6.67 × 10−11 N ⋅ m 2 / kg 2 )(8.80 × 1025 kg)
=
4π 2
CU = 1.49 × 1014 m3 /s 2
The values agree (0.7% difference).
(d) Table completed using equations T =
r3
and r = 3 CUT 2 .
CU
(e) Students who speculate that only the larger moons can be observed using Earth-based telescopes are right. Thus, only
the larger moons were discovered hundreds of years ago. Students who research the physical data of the moons that
orbit Uranus will find that Titania and Oberon have diameters greater than 1500 km, whereas all the other moons listed
are less than 100 km in diameter.
Table 1 Data of Several Moons of the Planet Uranus for Question 20
Moon
Discovery
raverage (km)
Ophelia
Desdemona
Juliet
Portia
Rosalind
Belinda
Titania
Oberon
Voyager 2 (1986)
Voyager 2 (1986)
Voyager 2 (1986)
Voyager 2 (1986)
Voyager 2 (1986)
Voyager 2 (1986)
Herschel (1787)
Herschel (1787)
3
2
T (Earth days)
CU (m /s )
0.375
0.475
0.492
0.512
0.556
0.621
8.66
13.46
1.48 × 10
14
1.46 × 10
14
1.48 × 10
14
1.48 × 10
14
1.48 × 10
14
1.48 × 10
14
1.48 × 10
14
1.48 × 10
4
5.38 × 10
4
6.27 × 10
4
6.44 × 10
4
6.61 × 10
4
6.99 × 10
4
7.52 × 10
5
4.36 × 10
5
5.85 × 10
14
21. (a) Some students may think the problem makes sense. However, many students will realize that the (theoretical) radius of
an orbit that has a period of 65 min would be less than Earth’s radius. (Students may recall that the typical orbital
period of a satellite in low-altitude orbit is about 80 min. For example, see question 22 on page 168 of the text.)
(b) ME = 5.98 × 1024 kg
T = 65 min = (65 min)(60 s/min) = 3.90 × 103 s
r=?
r 3 GM E
=
T2
4π 2
 GM E  2
r3 = 
T
2 
 4π 
r=
3
GM ET 2
=
3
(6.67 ×10
4π 2
−11
)(
)(
N ⋅ m 2 /s 2 5.98 × 1024 kg 3.90 ×103 s
)
2
4π 2
r = 5.36 × 106 m
The theoretical radius of the orbit is 5.36 × 106 m.
(c) Earth’s radius (6.38 × 106 m) is larger than the theoretical radius found in (b), so the calculated orbit cannot exist.
(d) The skill of analyzing a situation is valuable in order to reduce the chances of wasting time on calculations that don’t
make sense and to increase the chances of being able to estimate whether or not a solution to a problem is logical.
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 383
22. (a) The rocket’s mass can be calculated from the gravitational potential energy at rest (on Earth’s surface at rE). From the
graph Eg = –10 × 1010 J = –1.0× 1011 J.
−GM E m
Eg =
r
−rEg
m=
GM E
=
(
)(
− 6.38 ×106 m −1.0 × 1011 J
(6.67 ×10
−11
2
N ⋅ m /kg
2
)
)(5.98 ×10
24
kg
)
3
m = 1.6 × 10 kg
The rocket’s mass is 1.6 × 103 kg.
(b) The escape energy can be determined using the value of gravitational potential energy at rest (1.0 × 1011 J).
(c) The launch speed of the rocket can be calculated using the value of the initial kinetic energy EK (on Earth’s surface at
rE). From the graph EK = –12 × 1010 J = –1.2× 1011 J.
1
EK = mv 2
2
2 EK
v=
m
=
2(1.2 × 1011 J)
1.6 × 103 kg
v = 1.2 ×10 4 m/s
The launch speed is 1.2 × 104 m/s.
(d) Extrapolating from the graph, the kinetic energy EK, approaches 2.0 × 1010 J as the distance approaches infinity, where
Eg would approach zero. This can be approximated: at 5rE, the kinetic energy is 4.0 × 1010 J and Eg is –2.0 × 1010 J.
1
EK = mv 2
2
2 EK
v=
m
=
2(2.0 ×1010 J)
1.6 × 103 kg
v = 5.0 ×103 m/s
The speed is 5.0 × 103 m/s.
Making Connections
23. (a) Turning the high-speed craft around would require a fairly large amount of energy, so mission control decided to have
the craft continue on toward the Moon. The idea was to take advantage of the Moon’s gravity to act as a sort of slingshot to help the craft accelerate in turning around and begin its return journey at the highest speed possible.
(b) One major risk was the chance that there would not be enough electrical power available to guide the craft around the
Moon at the most crucial times.
Extension
24. Let L represent the large planet and S represent the small planet.
rL = 2rS
DL = DS (densities)
384 Unit 2 Energy and Momentum
Copyright © 2003 Nelson
Thus, using V for volume, the ratio of the masses is:
mS DSVS
=
mL DLVL
=
VS
VL
4 3
  π rS
3
= 
4 3
  π rL
3
r3 
=  S 3 
 rL 
mS  1 
= 
mL  2 
3
The centripetal acceleration of the satellite is caused by the force of gravity in each case. Thus, using magnitudes:
mv 2 4π 2 mr
ΣF = mac =
=
r
T2
GMm 4π 2 mr
=
r2
T2
3
r
GM
=
2
T
4π 2
T 2 4π 2
=
r 3 GM
 TL 2   4π 2 
 3 

 rL  =  GmL 
 TS2   4π 2 
 3 

 rS   GmS 
TL 2
rL
3
×
rS3
TS
2
TL 2
TS
2
=
=
mS
mL
rL3
rS
3
×
mS
mL
 r 3  m 
TL 2 = TS2  L3   S 
 rS   mL 
3
2 1
TL 2 = TS2    
1 2
3
TL 2 = TS2
TL = TS
The shortest possible period is 40 min.
25. Since the radius of the path is 2.0 × 1011 m, the distance between the stars is 2(2.0 × 1011 m) = 4.0 × 1011 m,
MS = 3.0 × 1030 kg (mass of each star). The only force acting on each star is the force of gravity of the other star, which
causes the circular motion of one star around the other.
Copyright © 2003 Nelson
Chapter 6 Gravitation and Celestial Mechanics 385
ΣF = mac
GM S M S
( 2r )2
GM S
=
M Sv 2
r
=
v2
r
4r 2
GM S
= v2
4r
GM S  2π r 
=

4r
 T 
2
GM S 4π 2 r 2
=
4r
T2
16π 2 r 3
T2 =
GM S
16π 2 r 3
GM S
T=
(
16π 2 2.0 × 1011 m
=
(6.67 ×10
−11
)(
)
3
N ⋅ m 2 /s 2 3.0 × 1030 kg
)
T = 7.9 × 107 s
The period of one complete cycle is 7.9 × 107 s.
26. E is the amount of energy per unit area, and that area is proportional to
1
, where r is the distance from the Sun to the
r2
planet, so:
1
r2
k
E= 2
r
E∝
From Kepler’s third law:
r3
=C
T2
r 3 = CT 2 where C is Kepler’s third law constant for the Sun
2
4
r 2 = C 3T 3
Substituting into the first equation:
k
E= 2 4
C 3T 3
−2
E = kC 3 T
−4
3
E = (constant)T
Thus, E is proportional to T
386 Unit 2 Energy and Momentum
−4
3
−4
3
. (Solving for the “constant” is unnecessary.)
Copyright © 2003 Nelson
7.1 ELECTRIC CHARGE AND THE ELECTRICAL STRUCTURE OF MATTER
Try This Activity: Charging Objects
(Page 323)
The smoke particles are attracted to the charged object due to induction.
PRACTICE
(Pages 324–325)
Understanding Concepts
1.
2.
3.
4.
5.
6.
7.
If the base layer is positive it attracts the negatively charged beads and the microcapsule will appear dark when viewed
from the top. If the base layer is negative it repels the negatively charged beads that float to the top, making the
microcapsule appear white when viewed from the top.
(a) The charge is the same but opposite in sign.
(b) You achieve this transfer of charge by friction. Both objects are neutral at the start, but walking involved some rubbing
motion between the carpet and your feet. This resulted in an opposite charge on each object.
(c) If the air is not dry then the water vapour and dissolved substances that ionize in the air can act as conductors and
reduce the charge on your body. This makes it more difficult to build up and retain any excess charge.
(a) Some of the excess electrons from the charged object pass into the electroscope and spread out over the surface of the
conducting material. The two leaves become negatively charged and repel each other.
(b) The further apart the leaves separate, the greater the charge on the object.
(c) No, if the charged object comes close to the electroscope without touching it, the object will cause the electrons in the
electroscope to redistribute either towards the leaves or away from them. In either case, the leaves will develop similar
charges and repel.
Touching the metal casing neutralizes your body so that any excess charge does not damage the memory when you handle
the chip. The memory is stored magnetically and the charge could change the magnetization on the disk and cause a loss
of data.
Heat from pressure rollers is used to melt the toner and fix it to the paper. The paper sticks together because the toner
particles are given a negative charge and the paper is given a positive charge. The interaction of these charges causes the
papers to stick together.
The toner and the paper have to be replenished from time to time.
Selenium is a photoconductor. It acts as an insulator in the dark and as a conductor when exposed to light. When light is
reflected from the copy through lenses and onto selenium, it loses its charge. Where the copy is dark no light is reflected
and the selenium retains its charge. In these places it can attract negatively charged toner particles.
Making Connections
8.
Using warm water will tend to melt the toner particles causing them to stick more readily to your hands. Using cold water
will help to remove the particles without additional melting.
Section 7.1 Questions
(Page 326)
Understanding Concepts
1. (a) The glass rod will develop a positive charge where it is rubbed and the plastic bag will have a negative charge.
Electrons will move from the glass into the plastic and stay in that area.
(b) The ebonite rod will develop a negative charge where it is rubbed and the fur will develop a positive charge. Electrons
will move from the fur into the ebonite and stay in that area.
(c) Both objects will end up with positive charges spread over the surface of the objects. Electrons will move from the
neutral rod into the positively charged rod.
(d) Both objects will end up with negative charges spread over the surface of the objects. Electrons will move from the
negatively charged sphere into the neutral rod.
420 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
2.
Bring the negatively charged object close to the neutral grounded object. Remove the ground and then the charged object.
The object will end up with a positive charge. In this process electrons in the neutral object are repelled by the negatively
charged object and flow out of the object into the ground.
3. The charged object causes the electrons to shift in position inside the object, which causes the centre of negative charge to
move closer to the positively charged object or further from the negatively charged object. The centre of positive charge
does not move. This causes a charge separation within the object. The neutral object is attracted to the charged object
because the centre of the opposite charge on the neutral object is closer than the centre of similar charge to the charging
object.
4. (a) Water vapour and ions within the air act as conductors causing the charged object to lose charge over time.
(b) (i) More water vapour in the air means there are more charge carriers in the air to act as conductors and promote the
loss of charge in the object.
(ii) X-ray radiation and other types of radiation are more likely to interact with a neutral gas at greater altitudes,
changing the gas into a conductor which reduces the charge on the object.
5. (a) Aluminum is a conductor. It is used in the selenium-coated drum to conduct charge away from the selenium that has
been exposed to light.
(b) Paper is an insulator. Positive charge is spread uniformly over its surface to attract the negatively charged toner
particles away from the selenium.
Applying Inquiry Skills
6.
Diagrams and designs may vary but most will involve using charged conductors to attract the charged particles in the air
ducts. Maintenance is required to clean these conductors from the particles stuck to the surface.
7. If we use the negatively charged electroscope first and the object is repelled then the object has a negative charge. If it is
attracted to the electroscope it could be either neutral or positively charged. If this happened, we could further determine
the charge by using the positively charged electroscope. If the object is attracted then the object is neutral, and if it is
repelled then the object is positively charged.
8. (a) The paper will be attracted to the charged objects to varying degrees.
(b) The paper is an insulator and when the charged object is brought close, it causes an induced charge separation in the
object, which is then attracted to the charged object.
(c) The charged object can lose its charge over time and loses the ability to hold up the paper.
(d) Charge from the sphere is transferred to the paper. The similarly charged objects now repel one another and jump apart.
Making Connections
9. (a) Instead of reflecting light from a copy, the copy is stored digitally and a laser is aimed at the selenium drum to release
the toner in places on the copy where it is not needed.
(b) The laser beam can be very thin, making the possible detail of the copy greater.
10. (a) The clothes rub up against each other in the dryer and transfer charge in the process. The oppositely charged clothes
then attract each other.
(b) The fabric softener sheet is used to help transfer charge back and forth between the clothes after the charge develops to
help reduce the buildup of charge on the clothes.
11. (a) There is friction between the plane and the air as it moves through the air resulting in a buildup of charge on the plane.
(b) There is water vapour and ions in the air that act as conductors to remove the charge from the plane.
(c) Charge will concentrate on longer and thinner areas of a conductor. The charge that gathers on these strips is more
likely to attract ions which will then reduce the charge on the plane.
7.2 ELECTRIC FORCES: COULOMB’S LAW
PRACTICE
(Pages 330–331)
Understanding Concepts
1.
d = 10.0 cm
F1 = 3.0 ¯ 10–6 N
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 421
2.
Bring the negatively charged object close to the neutral grounded object. Remove the ground and then the charged object.
The object will end up with a positive charge. In this process electrons in the neutral object are repelled by the negatively
charged object and flow out of the object into the ground.
3. The charged object causes the electrons to shift in position inside the object, which causes the centre of negative charge to
move closer to the positively charged object or further from the negatively charged object. The centre of positive charge
does not move. This causes a charge separation within the object. The neutral object is attracted to the charged object
because the centre of the opposite charge on the neutral object is closer than the centre of similar charge to the charging
object.
4. (a) Water vapour and ions within the air act as conductors causing the charged object to lose charge over time.
(b) (i) More water vapour in the air means there are more charge carriers in the air to act as conductors and promote the
loss of charge in the object.
(ii) X-ray radiation and other types of radiation are more likely to interact with a neutral gas at greater altitudes,
changing the gas into a conductor which reduces the charge on the object.
5. (a) Aluminum is a conductor. It is used in the selenium-coated drum to conduct charge away from the selenium that has
been exposed to light.
(b) Paper is an insulator. Positive charge is spread uniformly over its surface to attract the negatively charged toner
particles away from the selenium.
Applying Inquiry Skills
6.
Diagrams and designs may vary but most will involve using charged conductors to attract the charged particles in the air
ducts. Maintenance is required to clean these conductors from the particles stuck to the surface.
7. If we use the negatively charged electroscope first and the object is repelled then the object has a negative charge. If it is
attracted to the electroscope it could be either neutral or positively charged. If this happened, we could further determine
the charge by using the positively charged electroscope. If the object is attracted then the object is neutral, and if it is
repelled then the object is positively charged.
8. (a) The paper will be attracted to the charged objects to varying degrees.
(b) The paper is an insulator and when the charged object is brought close, it causes an induced charge separation in the
object, which is then attracted to the charged object.
(c) The charged object can lose its charge over time and loses the ability to hold up the paper.
(d) Charge from the sphere is transferred to the paper. The similarly charged objects now repel one another and jump apart.
Making Connections
9. (a) Instead of reflecting light from a copy, the copy is stored digitally and a laser is aimed at the selenium drum to release
the toner in places on the copy where it is not needed.
(b) The laser beam can be very thin, making the possible detail of the copy greater.
10. (a) The clothes rub up against each other in the dryer and transfer charge in the process. The oppositely charged clothes
then attract each other.
(b) The fabric softener sheet is used to help transfer charge back and forth between the clothes after the charge develops to
help reduce the buildup of charge on the clothes.
11. (a) There is friction between the plane and the air as it moves through the air resulting in a buildup of charge on the plane.
(b) There is water vapour and ions in the air that act as conductors to remove the charge from the plane.
(c) Charge will concentrate on longer and thinner areas of a conductor. The charge that gathers on these strips is more
likely to attract ions which will then reduce the charge on the plane.
7.2 ELECTRIC FORCES: COULOMB’S LAW
PRACTICE
(Pages 330–331)
Understanding Concepts
1.
d = 10.0 cm
F1 = 3.0 ¯ 10–6 N
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 421
(a) F2 = ?
F ∝ q1q2
F2 ( 2q1 )( 2q2 )
=
F1
q1q2
F2 = 4 F1
= 4(3.0 × 10−6 N)
F2 = 1.2 × 10−5 N
The resulting force would be 1.2 ¯ 10–5 N.
(b) F2 = ?
F ∝ q1q2
1 
q (q )
F2  2 1  2
=
F1
q1q2
1
F1
2
1
= (3.0 × 10−6 N)
2
F2 = 1.5 × 10 −6 N
F2 =
1
F1
2
The resulting force would be 1.5 ¯ 10–6 N.
(c) F2 = ?
1
F∝ 2
r
F2 d12
=
F1 d 2 2
F2 =
F2 = F1
d12
d22
(10.0 cm )
2
(30.0 cm )
2
= (3.0 × 10−6 N)
F2 = 3.3 × 10 −7 N
The resulting force would be 3.3 ¯ 10–7 N.
2. F1 = 3.6 ¯ 10–5 N
d1 = 0.12 m
(a) d2 = 0.24 m
F2 = ?
1
F∝ 2
r
F2 d12
=
F1 d 2 2
F2 = F1
d12
d22
 0.12 m 
= (3.6 ×10 N) 

 0.24 m 
F2 = 9.0 × 10 −6 N
The force of repulsion is 9.0 ¯ 10–6 N.
(b) d2 = 0.30 m
F2 = ?
2
−5
422 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
F2 = F1
d12
d22
 0.12 m 
= (3.6 × 10 −5 N) 

 0.30 m 
F2 = 5.8 × 10 −6 N
The force of repulsion is 5.8 ¯ 10–6 N.
(c) d2 = 0.36 m
F2 = ?
d2
F1 = F2 1 2
d2
2
2
3.
4.
5.
 0.12 m 
= (3.6 × 10 −5 N) 

 0.36 m 
F2 = 4.0 × 10−6 N
The force of repulsion is 4.0 ¯ 10–6 N.
q1 = 5.0 ¯ 10–8 C
q2 = 1.0 ¯ 10–7 C
r = 5.0 cm = 5.0 ¯ 10–2 m
F=?
kq q
F = 12 2
r
(9.0 ×109 N ⋅ m2 /C2 )(5.0 ×10−8 C )(1.0 ×10−7 C )
=
2
(5.0 ×10−2 m )
F = 1.8 ×10 −2 N
The force between the charges is 1.8 ¯ 10–2 N.
q1 = 1.5 × 10–8 C
q2 = 3.2 × 10–7 C
r = 1.5 m
F=?
kq q
F = 12 2
r
×109 N ⋅ m 2 /C 2 )(1.5 ×10 −8 C )(3.2 ×10 −7 C )
9.0
(
=
2
(1.5 m )
F = 1.9 N
The force between the charges is 1.9 N.
k = 9.0 ¯ 109 N·m2/C2
r = 4.0 ¯ 10–2 m
F = 1.2 ¯ 10–2 N
q=?
Let the charges be q and 2q in magnitude.
kq q
F = 12 2
r
9.0 × 109 N ⋅ m 2 /C 2 ) ( q )( 2q )
(
−2
1.2 ×10 N =
2
(4.0 ×10−2 m )
q 2 = (1.07 × 10−22 C )
2
q = 1.0 × 10−11 C
Therefore, one charge is 1.0 ¯ 10–11 C, and the other charge is 2(1.0 ¯ 10–11 C) = 2.0 ¯ 10–11 C. It doesn’t matter which
charge is positive.
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 423
6.
q1 = q2 = 1.1 × 10–7 C
F = 4.2 × 10–4 N
r=?
kq1 q2
r2
kq1q2
r=
F
F=
(9.0 ×10 N ⋅ m /C )(1.1×10 C )
9
=
2
2
−7
2
4.2 ×10 −4 N
= 0.259 m 2
7.
r = 0.51 m
The centres of the two charges are 0.51 m apart.
Resolving the tension in the x and y directions and using the fact that the object is in equilibrium:
T cos 15º = mg
T sin 15º = FE
FE
sin15°
=
= tan15°
mg cos15°
FE = mg tan15°
= ( 2.0 × 10−3 kg ) (9.8 N/kg )( 0.268 )
FE = 5.25 × 10 −3 N
The free body diagram for the sphere on the left:
424 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
For the distance between the two charges (r):
r
2 = sin15°
0.30 m
r = 2(0.30 m)(sin15°)
r = 0.155 m
Then FE =
kq1q2
, but q1 = q2 = q
r2
Fd 2
k
q=
(5.25 ×10
=
−2
N ) (0.155 m )
2
9.0 ×109 N ⋅ m 2 /C 2
= 1.40 × 10−14 C 2
q = 1.2 × 10 −7 C
The magnitude of the charge on each sphere is 1.2 ¯ 10–7 C.
PRACTICE
(Page 334)
Understanding Concepts
8.
qA = –4.0 ¯ 10–6 C
qB = –6.0 ¯ 10–6 C
qC = +9.0 ¯ 10–6 C
rAB = rBC = 10.0 cm
F1 , F2 , F3 = ?
For the force exerted on sphere 1 by sphere 2:
kq q
F12 = 1 2 2
r12
(9.0 ×10 N ⋅ m /C )( 4.0 ×10 C )(6.0 ×10 C )
9
=
2
−6
2
−6
(0.50 m )
2
F12 = 0.864 N [left]
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 425
For the force exerted on sphere 1 by sphere 3:
kq q
F13 = 1 2 3
r13
(9.0 ×10 N ⋅ m /C )(4.0 ×10 C )(9.0 ×10 C )
9
=
2
−6
2
−6
(1.0 m )
2
F13 = 0.324 N [right]
The total force acting on sphere 1 is:
F1 = F12 + F13
= 0.864 N [left] + 0.324 N [right]
F1 = 0.54 N [left]
For the force exerted on sphere 2 by sphere 3:
kq q
F23 = 2 2 3
r23
(9.0 ×10 N ⋅ m /C )(6.0 ×10 C )(9.0 ×10 C )
9
=
2
−6
2
−6
(0.50 m )
2
F23 = 1.944 N [right]
The total force acting on sphere 2 is:
F2 = F21 + F23
= 0.864 N [right] + 1.944 N [right]
F2 = 2.8 N [right]
The total force acting on sphere 3 is:
F3 = F31 + F32
= 0.324 N [left] + 1.944 N [left]
9.
F3 = 2.3 N [left]
The net forces acting on spheres 1, 2, and 3 are 0.54 N [left], 2.8 N [right], and 2.3 N [left].
q1 = q2 = q3 = –4.0 ¯ 10–6 C
F1 , F2 , F3 = ?
426 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
For the magnitude of each repulsive force:
kq q
F = 12 2
r
(9.0 ×10 N ⋅ m /C )(4.0 × 10 C )
=
9
2
−6
2
2
(0.20 m )
2
F = 3.6 N
For sphere 1:
Using components in the x – y plane:
F3 x = − F3 cos 60° = −(3.6 N)(0.50) = −1.8 N
F3 y = F3 sin 60° = (3.6 N)(0.867) = 3.1 N
F2 x = F2 cos 60° = (3.6 N)(0.50) = 1.8 N
F2 y = F2 sin 60° = (3.6 N)(0.867) = 3.1 N
∑F
∑F
x
= −1.8 N + 1.8 N = 0
y
= 3.1 N + 3.1 N = 6.2 N
The net force on sphere 1 is:
∑ F1 = 6.2 N [up]
∑F
1
= 6.2 N [outward, 150° away from the side]
The net force acting on sphere 1 is 6.2 N [outward, 150° away from the side]. The net forces on spheres 2 and 3 have the
same magnitude as sphere 1, and act along lines away from the triangle at an angle of 150º from the sides.
Section 7.2 Questions
(Pages 335–336)
Understanding Concepts
1. (a) The electric force is very strong, is directly proportional to the product of the charges, and is inversely proportional to
the distance between them squared. The gravitational force is much weaker and is directly proportional to the product
of the masses instead of the charge, but is also an inverse square law.
(b) Coulomb’s law: The force between two point charges is inversely proportional to the square of the distance between
the charges and directly proportional to the product of the charges.
Newton’s law of universal gravitation: The force of gravitational attraction between any two objects is directly
proportional to the product of the masses of the objects, and inversely proportional to the square of the distance
between their centres.
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 427
Information to be entered into Venn diagram on page 335:
Newton’s law differences: Directly proportional to the product of the two masses. The force is relatively weak. Can
only attract.
Common: Both are inverse square laws. Forces act along the line between the centers of mass or charge. Distance is
measured from the centers of the spheres.
Coulomb’s law differences: Directly proportional to the product of the two charges. The force is relatively strong.
Can attract and repel.
2. (a) F2 = ?
1
F∝ 2
r
F2 d12
=
F1 d 2 2
F2 = F1
d12
d22
=F
(1)
2
(3 )
2
F
9
The magnitude of the electric force would be divided by a factor of 9.
(b) F2 = ?
1
F∝ 2
r
F2 d12
=
F1 d 2 2
F2 =
F2 = F1
d12
d22
(1)
2
=F
2
1
 
2
F2 = 4 F
The magnitude of the electric force would be multiplied by a factor of 4.
(c) F2 = ?
F ∝ q1q2
F2 ( 2q1 )( 2q2 )
=
F
q1q2
F2 = 4 F
The magnitude of the electric force would be multiplied by a factor of 4.
(d) The magnitude will not change, but the direction will change. They will now attract instead of repel.
(e) F2 = ?
We first consider the effect of the neutral sphere:
F ∝ q1q2
1 
q (q )
F2  2 1  2
=
F
q1q2
F2 =
1
F
2
428 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
Now the effect of decreasing the distance:
1
F∝ 2
r
F2
d12
= 2
1
F d2
2
1 d2
F2 = F 1 2
2 d2
1 (1)
F
2  2 2
 
3
 1  9 
=    F
 2  4 
9
F2 = F
8
2
=
9
.
8
The magnitude of the sphere would increase by a factor of
3.
q1 = 5.0 × 10–6 C
q2 = 4.0 × 10–6 C
r = 2.0 m
F=?
F=
=
kq1 q2
r2
(9.0 ×109 N ⋅ m2 /C2 )(5.0 ×10−6 C )( 4.0 ×10−6 C )
( 2.0 m )
2
F = 4.5 ×10 −2 N
The magnitude of the force is 4.5 ¯ 10–2 N.
4. (a) m1 = m2 = 10.0 kg
r = 5.00 × 102 m
FG = ?
Gm1m2
FG =
r2
(6.67 ×10
=
−11
N ⋅ m 2 /kg 2 ) (10.0 kg )
2
(5.00 ×10 m )
2
2
FG = 2.67 × 10−14 N
The force of gravity between the two objects is 2.67 ¯ 10–14 N.
(b) q1 = q2 = 1.0 C
r = 5.00 × 102 m
F=?
kq q
F = 12 2
r
(9.0 ×10 N ⋅ m /C ) (1.0 C )
=
(5.00 ×10 m )
9
2
2
2
2
2
F = 3.6 ×104 N
The electric force between the two objects is 3.6 ¯ 104 N.
(c)
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 429
(d) FG = –2.67 ¯ 10–14 N
FE = 3.6 ¯ 104 N
m = 10.0 kg
a=?
The force of gravity will pull the objects together and the electric force will force them apart.
∑ F = FG + FE
= −2.67 × 10−14 N + 3.6 × 104 N
∑ F = 3.6 ×10
4
N
For the acceleration we have:
∑F
a=
m
3.6 ×10 4 N
=
10.0 kg
a = 3.6 × 103 m/s 2
The net force acting on each object is 3.6 ¯ 104 N. The acceleration is 3.6 × 103 m/s2 away from the other object.
(e) The answers are the same (3.6 ¯ 104 N, 3.6 ¯ 103 m/s2). The force of gravity does not make any significant
contribution to the acceleration because it is so small when compared with the electric force.
5. F = 36 N
r = 2.0 m
q=?
kq q
F = 12 2
r
(9.0 ×10 N ⋅ m /C ) (q )
36 N =
9
2
2
2
( 2.0 m )
2
q 2 = 1.6 ×10 −8 C 2
6.
q = 1.3 × 10 −4 C
The charge on each sphere is 1.3 ¯ 10–4 C. The charges could either be positive or negative.
mA = mB = 0.10 kg
q=?
1
The charge on each sphere is now − q . The free body diagram for sphere A is:
2
430 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
Resolving the tension in the x and y directions and using the fact that the object is in equilibrium:
T cos 12º = mg
T sin 12º = FE
FE
sin12°
=
= tan12°
mg cos12°
FE = mg tan12°
= (1.0 × 10−1 kg ) (9.8 N/kg )( tan12° )
FE = 2.08 × 10 −1 N
For the distance between the two charges (r):
r
= sin12°
2.0 m
r = (2.0 m)(sin12°)
r = 0.416 m
Then FE =
kq1q2
q
but q1 = q2 =
2
r2
4 Fr 2
k
q=
4 ( 2.0 × 10−1 N ) (0.416 m )
2
=
7.
9.0 ×109 N ⋅ m 2 /C 2
q = 3.9 ×10 −6 C
The initial charge on B is 3.9 ¯ 10–6 C.
q1 = +5.0 µC
q2 = –6.0 µC
q3 = +7.0 µC
r12 = 0.80 m
r23 = 0.40 m
F1 , F2 , F3 = ?
For the force exerted on sphere 1 by sphere 2:
kq q
F12 = 1 2 2
r12
(9.0 ×10 N ⋅ m /C )(5.0 ×10 C )(6.0 ×10 C )
9
=
2
−6
2
−6
( 0.80 m )
2
F12 = 0.422 N [right]
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 431
For the force exerted on sphere 1 by sphere 3:
kq q
F13 = 1 2 3
r13
(9.0 ×10 N ⋅ m /C )(5.0 ×10 C )(7.0 ×10 C )
9
=
2
−6
2
−6
(1.2 m )
2
F13 = 0.219 N [left]
The total force on sphere 1 is:
F1 = F12 + F13
= 0.422 N [right] + 0.219 N [left]
F1 = 0.203 N [right]
For the force exerted on sphere 2 by sphere 3:
kq q
F23 = 2 2 3
r23
(9.0 ×10 N ⋅ m /C )(6.0 ×10 C )(7.0 ×10 C )
9
=
2
−6
2
−6
(0.40 m )
2
F23 = 2.363 N [right]
The total force acting on sphere 2 is:
F2 = F21 + F23
= 0.422 + 2.363
F2 = 1.94 N [right]
The total force acting on sphere 3 is:
F3 = F31 + F32
= 0.219 N [right] + 2.363 N [left]
8.
F3 = 2.14 N [left]
The net force acting on sphere 1 is 0.20 N [right]; the net force acting on sphere 2 is 1.94 N [right]; the net force acting on
sphere 3 is 2.14 N [left].
Let the subscript S represent the sides, d represent the smaller diagonal, and D represent the larger diagonal.
q1 = q2 = q3 = q4 = 1.0 ¯ 10–5 C
r = 1.0 m
F on all sides = ?
We must calculate forces along the sides and the two on the diagonals. Along the sides we have:
kq q
FS = 12 2
r
(9.0 ×10 N ⋅ m /C )(1.0 ×10 C )
=
9
2
−6
2
2
(1.0 m )
2
FS = 9.0 × 10−3 N
The smaller diagonal d can be found using the cosine law:
d 2 = a 2 + b 2 − 2ab cos D
= (1.0 m)2 + (1.0 m)2 − 2(1.0 m)(1.0 m) cos 45°
d = 0.765 m
432 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
To calculate the force along the smaller diagonal:
kq q
Fd = 12 2
r
(9.0 ×10 N ⋅ m /C )(1.0 ×10 C )
=
9
2
−6
2
2
(0.765 m )
2
Fd = 1.54 × 10−2 N
The larger diagonal D is:
D2 = (1.0 m)2 + (1.0 m)2 – 2(1.0 m) (1.0 m)cos135º
D = 1.848 m
The force along the larger diagonal D is:
kq q
FD = 12 2
r
(9.0 ×10 N ⋅ m /C )(1.0 ×10 C )
=
9
2
−6
2
2
(1.848 m )
2
FD = 2.64 ×10 −3 N
For the sphere in the bottom left corner, the angle between adjacent forces is 22.5º:
Components in the x direction:
x = −9.0 ×10−3 N − 2.64 ×10−3 N(cos 22.5°) − 9.0 ×10−3 N(cos 45°) = −1.78 × 10−2 N
Components in the y direction:
y = −2.64 × 10−3 N(sin 22.5°) − 9.0 × 10−3 N(sin 45°) = −7.37 × 10−3 N
Therefore,
∑F
∑F
θ=
+ ( Fy )
=
( Fx )
=
( −1.78 ×10
2
2
−2
N ) + ( −7.37 ×10 −3 N )
2
2
= 1.9 ×10 −2 N
Fy
Fx
 −7.37 × 10−3 N 
= tan −1 

−2
 −1.78 × 10 N 
θ = 22°
The force on the sphere in the bottom left corner is 1.9 ¯ 10–2 N [158° from the side]. From symmetry the force on the
sphere in the upper right corner is also 1.9 ¯ 10–2 N [158° from the side].
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 433
For the top left charge we have:
Components in the x direction:
x = −9.0 ×10−3 N − 1.54 ×10−3 N(cos 67.5°) + 9.0 ×10−3 N(cos 45°) = −3.22 × 10−3 N
Components in the y direction:
y = 1.54 × 10−3 N(sin 67.5°) + 9.0 × 10−3 N(sin 45°) = 7.79 × 10−3 N
Therefore,
∑F
∑F
θ=
9.
+ ( Fy )
=
( Fx )
=
( −3.22 ×10
2
2
−3
N ) + (7.79 ×10−2 N )
2
2
= 8.4 ×10 −2 N
Fy
Fx
 7.79 × 10−3 N 
= tan −1 

−3
 3.22 × 10 N 
θ = 68°
The force on the top left charge is 8.4 ¯ 10–3 N [112° from the side]. From symmetry the force on the sphere in the
bottom right corner is also is 8.4 ¯ 10–3 N [112° from the side].
q1 = 1.6 ¯ 10–5 C
q2 = 6.4 ¯ 10–5 C
x = 2.0 m
q3 = 3.0 ¯ 10–6 C
x=?
If we place the charge anywhere other than on the line joining the two charges, the forces from the two charges on the
third charge will not be parallel and will not cancel. Therefore we must place the charge on the line joining the two
charges. If the charge is placed to the left of the smaller charge then both forces will point in the same direction. The same
is true if we place it to the right of the larger charge. Either case will not produce a net force of 0. The third force must be
placed between the two charges q1 and q2, a distance x from q1, as shown.
434 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
If there is no net force on q3,
F31 = F32
kq3 q1
kq3 q2
=
2
2
x
( 2.0 m − x )
kq3 (1.6 ×10 −5 C )
x2
=
kq3 (6.4 × 10 −5 C )
( 2.0 m − x )
2
Omitting units for simplicity, multiplying both sides by 105 and cancelling q3:
(1.6)(2 − x) 2 = 6.4 x 2
6.4 − 6.4 x + 1.6 x 2 = 6.4 x 2 (divide by a factor of 1.6)
4 − 4 x + x2 = 4 x2
3x 2 + 4 x − 4 = 0
Using the quadratic formula:
x=
−b ± b 2 − 4ac
2a
−4 ± 16 − 4(3)(−4)
6
−4 ± 8
=
6
4
−12
x=
, or
6
6
The negative answer is inadmissible and so the third charge is located 0.67 m from the 1.6 × 10–5-C charge.
The third charge divides out of the original equation. We do not need to know its value or even the sign.
10. r = 8.0 cm, 2r = 16.0 cm
q1 = q2 = 2.5 ¯ 10–6 C
k=?
=
The distance between the charges doubles meaning each spring must compress 4.0 cm. The force of the spring must be
equal to the repulsion from the other charge:
Fs = Fe
kx =
kq 2
r2
(9.0 ×10 N ⋅ m /C )( 2.5 ×10 C )
=
9
k ( 0.04 m )
2
−6
2
2
(0.16 m )
2
k = 55 N/m
The force constant of the springs is 55 N/m.
Applying Inquiry Skills
11. Determine the spring constant if it is not known. Use the technique of charging one sphere and then touching it to the
other, which will divide the charge in half. Grounding and charging by contact will help to reduce the charge by factors of
one half again. The charges are similar and so they will repel and compress the spring until the force of the spring is equal
to the electric force. The compression of the spring is directly related to the product of the charge and inversely related to
the square of the distance between the charges.
Making Connections
12. (a) The universal gravitation constant is very small and often the charges on objects are very small.
(b) One set of examples are the force of Earth on objects, the force of gravity on the Moon, and its effect on the tides. In all
cases one of the objects has a very large mass. Another set of examples are static cling between clothes from a dryer,
paper sticking together from a copier, charge stored on the surface of a TV screen. In all cases the charge can be large
and/or the distance between the charges is small.
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 435
13. (a) The Moon is in circular orbit around Earth.
Fc = FE
4π 2 mm ro kq 2
= 2
T2
ro
q=
4π 2 mm ro 3
kT 2
4π 2 (7.35 ×10 22 kg )(3.84 ×108 m )
3
=
(9.0 ×10 N ⋅ m /C )(2.36 ×10 s )
9
2
2
6
2
q = 5.7 × 1013 C
The magnitude of the charge required on each to hold the Moon in orbit is 5.7 ¯ 1013 C.
(b) The charge would decrease over time since the charge is so large it might be lost to the atmosphere and then into space.
The attraction will decrease and the Moon will escape its orbit around Earth.
7.3 ELECTRIC FIELDS
PRACTICE
(Pages 343–344)
Understanding Concepts
1. (a) q = 2.4 ¯ 10–6 C
FE = 3.2 N [left]
FE
q
3.2 N [left]
=
−2.4 × 10 −6 C
= −1.3 × 10 6 N/C [left]
ε =
ε = 1.3 × 106 N/C [right]
The magnitude of the electric field at that point is 1.3 ¯ 106 N/C [right].
(b) The electric field at that point is independent of the charge placed there. The answer is the same.
2. ε = 12 N/C
q = 2.5 ¯ 10–7 C
FE = ?
FE = qε
= ( 2.5 ×10−7 C ) (12 N/C )
3.
FE = 3.0 ×10 −6 N
The magnitude of the electric force is 3.0 ¯ 10–6 N.
r = 3.0 m
q = 5.4 ¯ 10–4 C
ε =?
F
ε= E
q
kq
= 2
r
(9.0 ×109 N ⋅ m2 /C2 )(5.4 ×10−4 C )
=
2
(3.0 m )
ε = 5.4 ×105 N/C [right]
The magnitude of the electric field is 5.4 ¯ 105 N/C [right].
436 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
13. (a) The Moon is in circular orbit around Earth.
Fc = FE
4π 2 mm ro kq 2
= 2
T2
ro
q=
4π 2 mm ro 3
kT 2
4π 2 (7.35 ×10 22 kg )(3.84 ×108 m )
3
=
(9.0 ×10 N ⋅ m /C )(2.36 ×10 s )
9
2
2
6
2
q = 5.7 × 1013 C
The magnitude of the charge required on each to hold the Moon in orbit is 5.7 ¯ 1013 C.
(b) The charge would decrease over time since the charge is so large it might be lost to the atmosphere and then into space.
The attraction will decrease and the Moon will escape its orbit around Earth.
7.3 ELECTRIC FIELDS
PRACTICE
(Pages 343–344)
Understanding Concepts
1. (a) q = 2.4 ¯ 10–6 C
FE = 3.2 N [left]
FE
q
3.2 N [left]
=
−2.4 × 10 −6 C
= −1.3 × 10 6 N/C [left]
ε =
ε = 1.3 × 106 N/C [right]
The magnitude of the electric field at that point is 1.3 ¯ 106 N/C [right].
(b) The electric field at that point is independent of the charge placed there. The answer is the same.
2. ε = 12 N/C
q = 2.5 ¯ 10–7 C
FE = ?
FE = qε
= ( 2.5 ×10−7 C ) (12 N/C )
3.
FE = 3.0 ×10 −6 N
The magnitude of the electric force is 3.0 ¯ 10–6 N.
r = 3.0 m
q = 5.4 ¯ 10–4 C
ε =?
F
ε= E
q
kq
= 2
r
(9.0 ×109 N ⋅ m2 /C2 )(5.4 ×10−4 C )
=
2
(3.0 m )
ε = 5.4 ×105 N/C [right]
The magnitude of the electric field is 5.4 ¯ 105 N/C [right].
436 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
4.
k = 9.0 ¯ 109 N⋅m2/C2
q1 = 50.0 µC = 5.0 ¯ 10–5 C
q2 = –10.0 µC = –1.0 ¯ 10–5 C
rXY = 0.45 m
rYZ = 0.30 m
ε =?
First we must calculate the electric field exerted on point Z by point X:
kq
ε ZX = 2
rZX
(9.0 ×10 Nim /C )(5.0 ×10 C )
9
=
2
−5
2
( 0.75 m )
2
ε ZX = 8.0 × 105 N/C [right]
Now we must calculate the electric field exerted on point Z by point Y:
kq
ε ZY = 2
rZY
(9.0 ×10 N ⋅ m /C )(1.0 ×10 C )
9
=
2
−5
2
(0.30 m )
2
ε ZY = 1.0 × 106 N/C [left]
Therefore,
ε Z = ε ZX + ε ZY
= 8.0 × 105 N/C [right] + 1.0 × 106 N/C [left]
5.
ε Z = 2.0 × 105 N/C [left]
The electric field at point Z is 2.0 ¯ 105 N/C [left].
qX = +2.0 ¯ 10–8 C
qY = +2.0 ¯ 10–8 C
ε =?
First we must determine the distance between point X and point Z. This will be equal to the distance between point Y and
point Z, since X and Y are both 3.0 cm from the centre of the base of the triangle.
Using the Pythagorean theorem:
rXZ = rYZ =
(3.0 cm ) + (4.0 cm )
2
2
rXZ = rYZ = 5.0 cm
 4.0 
θ = tan −1 

 3.0 
θ = 53°
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 437
We can now calculate the electric field exerted on point Z by point X:
kq
ε ZX = X2
rZX
(9.0 ×10 N ⋅ m /C )( 2.0 ×10 C )
(5.0 ×10 m )
9
=
2
−8
2
−2
2
ε ZX = 7.2 × 104 N/C [53° U from R]
Similarly, ε ZY = 7.2 × 104 N/C [53° U from L].
Components in the x direction:
(ε ZX )X = (7.2 ×104 N/C ) cos 53° = 4.3 ×104 N/C
(ε ZY )X = − ( 7.2 × 104 N/C ) cos 53° = −4.3 × 104 N/C
Components in the y direction:
(ε ZX )Y = (7.2 ×104 N/C ) sin 53° = 5.75 × 104 N/C
(ε ZY )Y = (7.2 ×104 N/C ) sin 53° = 5.75 × 104 N/C
Therefore,
6.
7.
(ε Znet )X = 0
(ε Znet )Y = 1.15 × 105 N/C
ε Znet = 1.2 × 105 N/C [Up]
The electric field at point Z is 1.2 ¯ 105 N/C [up].
Since the electric field is constant between the plates, ε1 = ε 2 = 3.0 × 103 N/C .
ε1 = 3.0 ¯ 103 N/C
ε2 = ?
ε ∝q
ε 2 q2
=
ε1 q1
q 
ε 2 = ε1  2 
 q1 
1
= (3.0 × 103 N/C )  
2
3
ε 2 = 1.5 × 10 N/C
The electric field strength would now be 1.5 ¯ 103 N/C.
PRACTICE
(Page 347)
Understanding Concepts
8. (a) Since the conductor is in electrostatic equilibrium the charges experience no net force. This means the electric field
inside the conductor must be zero.
(b) A Faraday cage is a container made of a conductor. Any charge on the cage spreads out over the surface and quickly
reaches electrostatic equilibrium. Therefore, there is no electric field inside the container.
(c) If the field is not perpendicular to the surface of the conductor then the component of the field parallel to the surface of
the conductor causes free electrons to move. This is not the case (since the conductor is in electrostatic equilibrium)
and so the electric field is perpendicular to the surface.
438 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
9. (a) No, the electric field from the charge will cause an induced charge separation between the inner and outer surfaces of
the sphere. The field lines will start and end on the charge inside the sphere and the inner surface of the sphere.
The excess charge on the surface of the sphere will cause an electric field outside the sphere.
(b) If the charge is not inside the sphere then the sphere will shield the inside of the sphere. If we leave the charge inside
the sphere then we can shield the outside world by placing an equal but opposite charge on the sphere. This charge will
be attracted to the inner surface of the sphere and the electric field lines will end there. There will be no external field
lines since there is no net charge on the surface of the sphere.
10. A coaxial cable consists of a solid wire in the middle of the cable surrounded by an insulating sleeve, a metallic
cylindrical braid, and an outer insulating jacket. Stray electric fields outside the cable stop at the surface of the metallic
braid shielding the current inside the central solid wire.
Section 7.3 Questions
(Pages 347–348)
Understanding Concepts
1. (a) If the test charge is not small then its field will significantly affect the nature of the field in the area, changing its nature
and what we intended to measure. For a smaller charge the effect is minimal, therefore providing a clearer picture of
the nature of the field when no test charge is present.
(b) If a negative test charge is used then the direction of the field is opposite to the direction of the net force on the charge.
We can easily account for this and so it does not have to be positive.
2.
3. (a) Diagram is not to scale. The distance between the charges should be 15 cm and the dotted circle should have a radius of
3 cm.
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 439
(b) The electric field is strongest at C. It is the only point that has both electric fields in the same direction. For the other
points the distances are larger for at least one of the charges (reducing the force) and the forces are not in the same
direction. This results in a smaller net electric field.
(c) Since the distance from any point on the circle to the positive charge is 3.0 cm, the field due to the positive charge is
the same in magnitude. Therefore we will concentrate on the field from the negative charge. The point on the circle
closest to the negative charge is at S and the field is in the same direction so this is the place where the field is
strongest. At W the fields are in opposite directions and so it is reasonable to assume the field is weakest here.
(See diagram for 3(a).)
4. (a)
5.
6.
7.
8.
(b) The electric field is strongest at D because both electric fields point in the same direction.
(c) The electric field is strongest at S because the fields are in the same direction. It is weakest at W where the fields are in
opposite directions.
Electric field lines indicate the direction of the net force on a small positive test charge. This test charge cannot have two
net forces in different directions because then they are not net forces. Therefore, electric field lines cannot cross.
(a) The electric field lines indicate the direction of the force on the charge.
(b) The electric field lines indicate the direction of the acceleration.
(c) The electric field lines are not related to the velocity of the charge.
F
Magnitude of the electric field is defined as the electric force per unit positive charge, or
, therefore they are the same.
q
The negative charge causes a charge separation between the inner and outer surfaces of the sphere. Electric field lines
extend from the inner surface of the sphere (starting from positive charge) and end on the negative charge. Since the outer
surface has a negative charge, electric field lines end on the surface of the sphere directed towards the centre of the
sphere.
Applying Inquiry Skills
9.
First students must decide on a method to measure or indicate the electric field of the plates. A computer probe or rayon
fibers in mineral oil are good choices. Then they must decide what properties they will investigate. Examples include the
effect of the amount of charge on the plates, and the distance between the plates. They can then investigate the effect on
440 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
the field between the plates, the field near the end of the plates (curved), and the field outside the plates (zero). They must
ensure that they change only one factor at a time.
10. (a) The field lines from the small charge must end on an equal amount of negative charge, which then induces a charge
separation on the inner and outer surfaces of the sphere. Since the outside of the sphere now has a net positive charge
we get a reading on the electrometer.
(b) The negative charge on the inside of the large sphere is equal to the positive charge on the small sphere. When they
come into contact, electrons move into the small sphere, making it neutral.
(c) The positive charge on the outside of the large sphere stays where it is. Students should remember that the excess
charge on a conductor spreads out over the outer surface.
(d) We cannot shield an electric field from the outside by placing it in a neutral conductor.
11. By definition, a field of force exists in a region of space when an appropriate object (in this case a charge) placed in the
field experiences a force. Since the test charge experiences a force then there is a field.
12. (a) The planet experiences a force of attraction towards the star (and vice versa) and so there is a field. In this case the
appropriate object is anything with mass (the planet) and it is located in the field of the star.
(b) The electron experiences a force of attraction towards the proton (and vice versa) and so there is a field. In this case the
appropriate object is anything with charge (the electron) and it is located in the field of the electron. However, students
should remember that the field of the electron is similar in magnitude to the field of the proton (unlike the planet and
star) and so we can reverse the roles here easily. The mass of the electron is a lot less than the proton and so it is
usually described this way. (The electron orbits the proton.)
13. (a) Student answers may include the following concepts:
Like gravity this is an action-at-a-distance force, the main reason for introducing fields. Since charges attract and repel
when they are not in contact it seems reasonable to describe them with a field because we are familiar with the concept
and it has worked well in the past. Both laws are inverse square laws and so it also seems reasonable to assume they
both involve fields. Test charges placed anywhere in the area of another charge experience a force. This alone implies a
field in the area around the field.
(b) It must be an action-at-a-distance force involving an appropriate object that experiences a force anywhere in the field.
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 441
Making Connections
14. One of the devices could be causing an electric field in the area, and if the other device is close enough its operation may
be affected if it is not shielded properly. Two possible solutions are to increase the distance between the objects since the
field strength obeys an inverse square law, or check the covering of the improperly working device to make sure it is
properly in place.
15. Some eels use electric fields to stun fish while most other organisms use electric fields to detect prey. In many cases, the
ability to detect these fields is similar to that of the hammerhead.
7.4 ELECTRIC POTENTIAL
PRACTICE
(Page 354)
Understanding Concepts
1.
r = 25 cm
V = –6.4 ¯ 104 V
k = 9.0 ¯ 109 N⋅m2/C2
q=?
kq
r
rV
q=
k
(0.25 m ) ( −6.4 × 104 V )
=
(9.0 ×109 N ⋅ m 2 /C2 )
V=
2.
q = −1.8 × 10 −6 C
The point charge is –1.8 ¯ 10–6 C.
W = 4.2 ¯ 10–3 J
q = 1.2 ¯ 10–6 C
V=?
W
V=
q
4.2 × 10−3 J
1.2 × 10−6 C
V = 3.5 × 103 V
The potential difference between X and Y is 3.5 ¯ 103 V.
r = 5.0 mm = 5.0 ¯ 10–3 m
V = 3.0 ¯ 102 V
V
ε=
r
3.0 ×10 2 V
=
5.0 × 10−3 m
ε = 6.0 ×104 N/C
The electric field strength is 6.0 ¯ 104 N/C.
r = 1.2 cm = 1.2 ¯ 10–2 m
ε = 1.5 ¯ 104 N/C
V=?
V
ε=
r
V = εr
=
3.
4.
= (1.5 × 105 N/C )(1.2 × 10 −2 m )
V = 1.8 × 102 V
A potential difference of 1.8 ¯ 102 V would have to be maintained.
442 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
Making Connections
14. One of the devices could be causing an electric field in the area, and if the other device is close enough its operation may
be affected if it is not shielded properly. Two possible solutions are to increase the distance between the objects since the
field strength obeys an inverse square law, or check the covering of the improperly working device to make sure it is
properly in place.
15. Some eels use electric fields to stun fish while most other organisms use electric fields to detect prey. In many cases, the
ability to detect these fields is similar to that of the hammerhead.
7.4 ELECTRIC POTENTIAL
PRACTICE
(Page 354)
Understanding Concepts
1.
r = 25 cm
V = –6.4 ¯ 104 V
k = 9.0 ¯ 109 N⋅m2/C2
q=?
kq
r
rV
q=
k
(0.25 m ) ( −6.4 × 104 V )
=
(9.0 ×109 N ⋅ m 2 /C2 )
V=
2.
q = −1.8 × 10 −6 C
The point charge is –1.8 ¯ 10–6 C.
W = 4.2 ¯ 10–3 J
q = 1.2 ¯ 10–6 C
V=?
W
V=
q
4.2 × 10−3 J
1.2 × 10−6 C
V = 3.5 × 103 V
The potential difference between X and Y is 3.5 ¯ 103 V.
r = 5.0 mm = 5.0 ¯ 10–3 m
V = 3.0 ¯ 102 V
V
ε=
r
3.0 ×10 2 V
=
5.0 × 10−3 m
ε = 6.0 ×104 N/C
The electric field strength is 6.0 ¯ 104 N/C.
r = 1.2 cm = 1.2 ¯ 10–2 m
ε = 1.5 ¯ 104 N/C
V=?
V
ε=
r
V = εr
=
3.
4.
= (1.5 × 105 N/C )(1.2 × 10 −2 m )
V = 1.8 × 102 V
A potential difference of 1.8 ¯ 102 V would have to be maintained.
442 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
PRACTICE
(Page 358)
Understanding Concepts
5. (a) Inside the clouds there is a charge separation, with a negative charge on the bottom. The negatively charged bottom of
the cloud repels the electrons in the ground, inducing a positive charge on the surface of the ground. Since the surfaces
are oppositely charged there is an attraction between the two.
(b) If the bottom of the cloud is negatively charged then the lightning rod will be positively charged because the electrons
in the lightning rod are repelled.
(c) The electric field near the tip of the lightning rod will be strong enough to ionize the air (because of the small radius of
curvature), changing the air from an insulator to a conductor. Lightning will more readily follow this path of ionized air
towards the lightning rod striking it instead of the roof of the building.
6. (a) Figure 13 on page 356 shows a positively charged layer of extracellular fluid parallel to a negatively charged layer of
intracellular fluid. These two parallel layers are separated by a cell membrane giving the appearance of a set of parallel
plates.
(b) ∆V = 0.070 V
d = 5.0 nm = 5.0 ¯ 10–9 m
ε=?
If we assume this acts like a set of parallel plates then the electric field is uniform:
∆V
ε=
d
0.070 V
=
5.0 × 10−9 m
ε = 1.4 × 107 N/C
The magnitude of the electric field is 1.4 ¯ 107 N/C.
(c) q = 1.6 ¯ 10–19 C
W=?
W = Vq
= (0.070 V)(1.6 × 10−19 C)
W = 1.1× 10 −20 J
You would have to do 1.1 ¯ 10–20 J of work on the sodium ion.
Section 7.4 Questions
(Pages 358–359)
Understanding Concepts
1.
r = 0.35 m
V = 110 V = 1.1 ¯ 102 V
k = 9.0 ¯ 109 N·m2/C2
q=?
kq
r
rV
q=
k
(0.35 m ) (1.1× 102 V )
=
(9.0 ×109 Nm 2 /C2 )
V=
q = 4.3 × 10 −9 C
The electric charge is +4.3 ¯ 10–9 C.
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 443
2.
r1 = 0.16 m
r2 = 0.40 m
q1
=?
q2
Assuming both charges do not equal zero, one charge must be negative and the other positive. (Otherwise they could not
add to zero.)
kq
V =
r
V1 + V2 = 0
kq1 kq2
+
=0
r1
r2
kq1
kq
=− 2
r1
r2
q1
r
=− 1
q2
r2
=−
0.16 m
0.40 m
q1
= −0.40
q2
q1
is –0.40.
q2
3. (a) The electric field is zero at X, the point midway between the two charges.
The ratio of
(b) No. Electric potential is a scalar quantity and in this case both potentials are positive therefore the sum of the two
potentials is always greater than zero.
(c) Electric field is a vector quantity that can equal zero when the two fields from each charge are equal in magnitude but
opposite in direction (at X). Electric potential is a scalar quantity, and since the electric potential from each positive
charge is positive they cannot add to zero.
4. (a) Electric potential is a scalar measurement and electric field is a vector measurement. The magnitude of the electric field
is equal to the change in the potential difference over the distance.
(b) Electric potential is the electric potential energy stored per unit charge. Electric potential is a characteristic associated
with a single charge while electric potential energy is the energy stored in a system of two or more charges.
5. A trivial example is if there are no charges present then the electric potential is zero and so is the electric field. If there are
charges present then there must be at least two charges of the opposite signs otherwise the electric potential cannot be
zero. If the charges are equal in magnitude (to simplify the argument) then the electric potential is zero anywhere on the
perpendicular bisector of the line joining the two charges, but the electric field can only be zero on the line joining the two
charges and it is not zero between the two charges. Therefore, if there are charges present the electric field is not zero.
Similar arguments can be given by examining the equations and the vector nature of electric fields.
6. Yes, if the particle has the same sign of charge as the electric potential then the product of the electric potential with the
charge will be positive, and the electric potential energy will increase from a lower value to a higher value.
444 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
7.
V = 120 V
ε = 3.0 ¯ 106 N/C
r=?
V
r
V
r=
ε
ε=
120 V
3.0 × 106 N/C
r = 4.0 × 10−5 m
The smallest possible gap between the plates is 4.0 ¯ 10–5 m.
8. (a) q1 = +4.0 µm = +4.0 ¯ 10–6 C
q2 = +2.0 µm = +2.0 ¯ 10–6 C
q3 = –4.0 µm = –4.0 ¯ 10–6 C
r = 2.0 m
ET = ?
=
Since electric potential energy is scalar, we simply add all the energy associated with the different combinations of
charges.
kq q
E= 1 2
r
(9.0 ×109 N ⋅ m2 /C2 ) ( 4.0 ×10−6 C )(2.0 ×10−6 C ) + (4.0 ×10−6 C )(−4.0 ×10−6 C ) + ( −4.0 ×10−6 C )(2.0 ×10−6 C )
ET =
2.0 m
ET = 3.6 ×10−2 J
The total electric potential energy is 3.62 ¯ 10–2 J.
(b) VT = ?
The distance from a charge to the middle of the opposite side is 3 m for triangles with angles 30º, 60º, 90º. (Students
may also use the Pythagorean theorem.) Otherwise the distance is 1.0 m.
For the bottom side:
V=
kq
r
 −4.0 ×10 −6 C 2.0 ×10 −6 C 4.0 ×10 −6 C 
+
+
VT = (9.0 × 109 N ⋅ m 2 /C 2 ) 

1.0 m
3m 
 1.0 m
VT = 2.8 ×103 V
For the top left side:
 −4.0 ×10 −6 C 4.0 ×10 −6 C 2.0 ×10 −6 C 
+
+
VT = (9.0 × 109 N ⋅ m 2 /C 2 ) 

1.0 m
3m 
 1.0 m
VT = 1.0 ×10 4 V
For the top right side:
 4.0 × 10−6 C 2.0 × 10−6 C 4.0 × 10−6 C 
+
−
VT = (9.0 ×109 N ⋅ m 2 /C 2 ) 

1.0 m
3m 
 1.0 m
VT = 3.3 × 10 4 V
The electric potential for the top left side is 1.0 ¯ 104 V, for the top right side is 3.3 ¯ 104 V, and at the midpoint of
the bottom side is 2.8 ¯ 103 V.
9. (a) r = 12 cm = 0.12 m
V = 85 kV = 8.5 ¯ 104 V
q=?
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 445
kq
r
rV
q=
k
(0.12 m ) (8.5 × 104 V )
=
(9.0 ×109 N ⋅ m2 /C2 )
V=
q = 1.1× 10 −6 C
The charge on the sphere is 1.1 ¯ 10–6 C.
(b) ε = ?
V
ε=
r
8.5 ×10 4 V
=
0.12 m
ε = 7.1× 105 N/C
The electric field is 7.1 ¯ 105 N/C.
(c) No, the electric field near the surface would have to be at least 3.0 × 106 N/C at the surface to ionize air.
Applying Inquiry Skills
10. Place a charge on a sphere of radius R and measure the potential difference between the sphere and the ground. Measure
the potential difference between the charged sphere and another neutral sphere of radius r. Connect the two spheres using
a long wire ensuring the spheres are far apart. Measure the potential difference between the two spheres again. Repeat for
different amounts of charge on the spheres.
11. The negatively charged bottom of the clouds will induce a positive charge on the lightning rod. Because the radius of
curvature of the lightning rod is small it will easily produce large enough electric fields to ionize the air and form positive
ions. The positively charged ions will be attracted to the negatively charged plate and the negatively charged sphere
reducing the charge on the spheres. The ions will be repelled from the positively charged spheres.
Making Connections
12. (a) The gap is small to maximize the electric field between the conductors and help ionize the gases so that a spark will
jump between the conductors.
(b) The breakdown value is higher because there are gases around the conductors other than air. The predominant gas is
octane and the mixture of these gases and air has a higher breakdown value than air.
(c) When a spark jumps from one conductor to the other they become neutral (for an instant) and the potential difference
between the two drops to zero and then quickly increases again until the breakdown value is reached. Another spark is
created and the process continues.
13. If the bottom of the cloud is negatively charged then the tip of lightning rods will be positively charged. The electric field
near the tip of the lightning rod will reach the breakdown value and create positively charged ions in the air. It is believed
that these ions might serve to help neutralize the cloud because they are now attracted to the negatively charged cloud.
The problem is in the delay (the ions have to reach the cloud before lightning starts) and the shear number of ions
required. This implies that if this could work, we would need a high number of lightning rods in the area.
7.5 THE MILLIKAN EXPERIMENT: DETERMINING THE ELEMENTARY CHARGE
PRACTICE
(Pages 362–363)
Understanding Concepts
1.
q = 8.0 ¯ 10–8 C
e = 1.6 ¯ 10–19 C
N=?
446 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
kq
r
rV
q=
k
(0.12 m ) (8.5 × 104 V )
=
(9.0 ×109 N ⋅ m2 /C2 )
V=
q = 1.1× 10 −6 C
The charge on the sphere is 1.1 ¯ 10–6 C.
(b) ε = ?
V
ε=
r
8.5 ×10 4 V
=
0.12 m
ε = 7.1× 105 N/C
The electric field is 7.1 ¯ 105 N/C.
(c) No, the electric field near the surface would have to be at least 3.0 × 106 N/C at the surface to ionize air.
Applying Inquiry Skills
10. Place a charge on a sphere of radius R and measure the potential difference between the sphere and the ground. Measure
the potential difference between the charged sphere and another neutral sphere of radius r. Connect the two spheres using
a long wire ensuring the spheres are far apart. Measure the potential difference between the two spheres again. Repeat for
different amounts of charge on the spheres.
11. The negatively charged bottom of the clouds will induce a positive charge on the lightning rod. Because the radius of
curvature of the lightning rod is small it will easily produce large enough electric fields to ionize the air and form positive
ions. The positively charged ions will be attracted to the negatively charged plate and the negatively charged sphere
reducing the charge on the spheres. The ions will be repelled from the positively charged spheres.
Making Connections
12. (a) The gap is small to maximize the electric field between the conductors and help ionize the gases so that a spark will
jump between the conductors.
(b) The breakdown value is higher because there are gases around the conductors other than air. The predominant gas is
octane and the mixture of these gases and air has a higher breakdown value than air.
(c) When a spark jumps from one conductor to the other they become neutral (for an instant) and the potential difference
between the two drops to zero and then quickly increases again until the breakdown value is reached. Another spark is
created and the process continues.
13. If the bottom of the cloud is negatively charged then the tip of lightning rods will be positively charged. The electric field
near the tip of the lightning rod will reach the breakdown value and create positively charged ions in the air. It is believed
that these ions might serve to help neutralize the cloud because they are now attracted to the negatively charged cloud.
The problem is in the delay (the ions have to reach the cloud before lightning starts) and the shear number of ions
required. This implies that if this could work, we would need a high number of lightning rods in the area.
7.5 THE MILLIKAN EXPERIMENT: DETERMINING THE ELEMENTARY CHARGE
PRACTICE
(Pages 362–363)
Understanding Concepts
1.
q = 8.0 ¯ 10–8 C
e = 1.6 ¯ 10–19 C
N=?
446 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
q
e
8.0 ×10 −8 C
=
1.6 ×10 −19 C
N = 5.0 × 1011 electrons
The number of electrons that must be removed is 5.0 ¯ 1011.
N = 1.0 ¯ 108 electrons
F=?
N=
2.
First we must calculate the charge on each sphere:
q = N ( −e)
= (1.0 × 108 )(−1.6 × 10−19 C)
q = 1.6 × 10−11 C
We then calculate the magnitude of the force:
kq q
F = 12 2
r
(9.0 ×10 N ⋅ m /C )(1.6 ×10
=
9
2
2
(1.0 m )
−11
C)
2
2
3.
F = 2.3 ×10 −12 N
The force of electric repulsion is 2.3 ¯ 10–12 N.
N = 5.00 ¯ 109 electrons
r = 0.500 m
ε=?
V=?
First we must calculate the charge of the object:
q = Ne
= (5.00 × 109 )(−1.6 × 10−19 C)
q = −8.0 × 10−10 C
To calculate the electric field intensity:
kq
ε= 2
r
9.0 ×109 N ⋅ m 2 /C 2 )(8.0 × 10 −10 C )
(
=
2
( 0.50 m )
ε = 29 N/C
To calculate the electric potential:
kq
V=
r
9.0 ×109 N ⋅ m 2 /C 2 )( −8.00 ×10 −10 C )
(
=
0.50 m
V = −14 V
The magnitude of the electric field is 29 N/C. The electric potential is –14 V.
4. r = 0.050 m
FE = 4.5 ¯ 10–15 N
(a) q = 6.4 ¯ 10–19 C
V=?
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 447
First we must calculate the electric field intensity:
F
ε= E
q
4.5 × 10 −15 N
6.4 × 10−19 C
ε = 7.0 ×103 N/C
=
For the potential difference:
V = εr
= (7.0 ×103 N/C ) (0.050 m )
V = 3.5 × 10 2 V
The potential difference between the plates is 3.5 ¯ 102 V.
(b) m = ?
Fg
m=
g
F
= E
g
=
5.
4.5 × 10−15 N
9.8 N/kg
m = 4.6 × 10 −16 kg
The mass of the sphere is 4.6 ¯ 10–16 kg.
m = 4.95 ¯ 10–15 kg
r = 1.0 cm = 1.0 ¯ 10–2 m
V = 510 V
q=?
First we must calculate the force of electric repulsion:
FE = Fg
= mg
= (4.95 × 10−15 kg)(9.8 N/kg)
FE = 4.9 × 10−14 N
We must also calculate the electric field intensity:
V
ε=
r
510 V
=
1.0 ×10 −2 m
ε = 5.1× 104 N/C
Therefore, to calculate the charge:
F
ε= E
q
FE
q=
ε
4.9 × 10−14 N
=
5.1× 10 4 N/C
q = 9.6 × 10−19 C
448 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
6.
The charge on the drop is 9.6 ¯ 10–19 C, which can be expressed as a multiple of elementary charge:
9.6 × 10−19 C
q=
1.6 × 10−19 C
q = 6e
Since the upper plate is positive, the charge must be negative due to an excess of electrons.
ε = 1.0 ¯ 102 N/C
m = 2.0 ¯ 10–15 kg
q=?
F
ε= E
q
FE = qε = mg
q=
=
mg
ε
(2.0 ×10−15 kg ) (9.8 N/kg )
−1.0 × 10 2 N/C
q = −2.0 ×10 −16 C
The charge on the oil drop is –2.0 ¯ 10–16 C, or a multiple of –1.2 ¯ 103 e.
Section 7.5 Questions
(Page 364)
Understanding Concepts
1.
qA = –3q
qB = +5q
r = 1.5 m
FE = 8.1 ¯ 10–2 N
(a) N = ?
Since the spheres are identical they will share the total charge equally. The charge on each sphere is
FE =
( −3q + 5q )
2
=q.
kq 2
r2
q=
=
FE r 2
k
(8.1×10
−2
N ) (1.5 m )
2
9.0 ×109 N ⋅ m 2 /C 2
q = 4.5 × 10 −6 C
For this to occur a charge of – 4q must transfer from the negative to the positive charge:
− 4 q = N ( − e)
N=
=
−4 q
−e
−4 ( 4.5 × 10−6 C )
−1.6 × 10 −19 C
N = 1.1× 1014
The number of transferred electrons is 1.1 ¯ 1014.
(b) Since the charges are equal, the distance from each charge is equal and the fields are in opposite directions. Therefore,
the electric field midway between the two charges is zero.
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 449
For the electric potential, r =
1.5 m
= 0.75 m, since we are determining the electric potential midway between the two
2
spheres:
VT = 2V
=2
=2
kq
r
(9.0 ×109 N ⋅ m2 /C2 )(4.5 ×10−6 C )
0.75 m
5
VT = 1.1 ×10 V
The electric potential is 1.1 ¯ 105 V.
(c) FE = ?
kq q
FE = 12 2
r
k (3q )(5q )
=
r2
kq 2
= 15 2
r
= 15 (8.1× 10−2 N )
FE = 1.2 N
The initial electric force is 1.2 N.
2. m = 4.3 ¯ 10–9 kg
ε = 9.2 ¯ 102 N/C [up]
(a) The charge on the drop must be positive so that the electric force is up to cancel the force of gravity.
(b) N = ?
FE = Fg
qε = mg
mg
ε
mg
Ne =
ε
mg
N=
εe
q=
(4.3 ×10 kg ) (9.8 N/kg )
(9.2 ×10 N/C )(1.6 ×10 N ⋅ m /C )
−9
=
2
−19
2
2
N = 2.9 ×108
The number of extra electrons or protons is 2.9 ¯ 108.
3. m = 4.7 ¯ 10–15 kg
(a) q = ?
FE = Fg
qε = mg
∆V
= mg
r
mgr
q=
∆V
(4.7 ×10−15 kg ) (9.8 N/kg ) (5.0 ×10−3 m )
=
120 V
q = 1.92 × 10−18 C, or 1.9 × 10−18 C
The charge on the oil drop is 1.9 ¯ 10–18 C.
q
450 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
(b) N = ?
q = Ne
q
e
1.92 × 10−18 C
=
1.6 × 10−19 C
N = 12
The number of elementary charges required is 12.
(c) The charge must have a deficit of electrons since the top plate is negative.
4. m = 2.0 ¯ 10–5 kg
q=?
N=
The charges must be similar so that the electric force pushes the objects apart to cancel the force of gravity.
FE = Fg
kq 2 Gm1m2
=
r2
r2
Gm 2
q2 =
k
q=±
(6.67 ×10
−11
Nm 2 /kg 2 )( 2.0 × 10−5 kg )
2
9.0 ×109 Nm 2 /C 2
q = ±1.7 × 10 −15 C
The charge on one object will be +1.7 ¯ 10–15 C, and the charge on the other object will be –1.7 ¯ 10–15 C.
5. m = 5.0 ¯ 10–2 kg
NA = 1.0 ¯ 1012 electrons (excess)
NB = 4.5 ¯ 1012 electrons (deficit)
r = 0.12 m
(a) θ = ?
To determine the charge on sphere A:
qA = N ( −e )
= (1.0 ×1012 )( −1.6 × 10−19 C)
qA = −1.6 ×10 −7 C
To determine the charge on sphere B:
q B = N ( e)
= (4.5 × 1012 )(1.6 ×10−19 C)
qB = 7.2 ×10−7 C
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 451
We can calculate the electric force:
kq q
FE = A2 B
r
×109 N ⋅ m 2 /C 2 )(1.6 × 10−7 C )(7.2 × 10−7 C )
9.0
(
=
2
(0.12 m )
FE = 7.2 ×10 −2 N
Also,
Fg = mg
= (5.0 × 10−2 kg)(9.8 N/kg)
Fg = 0.49 N
From the free body diagram for A:
Equating the components:
T sin θ = FE
T cos θ = Fg
Dividing the components:
T sin θ FE
=
T cos θ Fg
tan θ =
FE
Fg
 7.2 ×10 −2 N 
θ = tan −1 

 0.49 N 
θ = 8.4°
The angle between the thread and the vertical is 8.4°.
(b) T = ?
T cos θ = Fg
T=
Fg
cos θ
0.49 N
=
cos 8.4°
T = 0.50 N
The tension in the thread is 0.50 N.
Applying Inquiry Skills
6. (a) Water evaporates over time and, therefore, while trying to suspend the water droplet the mass is decreasing, which
changes the amount of electric force required to suspend the drop.
(b) A droplet initially suspended starts to rise as the water evaporates and the mass of the drop decreases.
(c) The scientist could not be certain for a variety of reasons. Possible answers include the charge on the drop was not
constant, the equipment was not working properly or lacked the ability to suspend the drop for an extended period of
time, there is something wrong with the procedure, and the electric field between the plates is not constant.
7. (a) If each charge is a multiple of some elementary charge then each charge can be expressed as an integer times this
elementary charge. Dividing any two of the charges will yield a ratio of integers that can be written as a fraction in
lowest terms. The numerator and denominator represent the required integers and can be used to find the elementary
charge. Other charges should be checked to make sure the others are multiples of this charge.
28125 45
(b) Dividing the charge on oil drop 2 by the charge on oil drop 1 yields 2.8125 or
. Therefore the first charge
=
10000 16
is 16 times the fundamental charge and the second is 45 times the fundamental charge. Dividing these charges by the
number of fundamental charges yields a fundamental charge of 4.0 × 10–20 C. Since this number divides evenly into all
the other charges there is no need to make any adjustments.
452 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
(c) No, the fundamental charge could be any positive integer multiple of this charge. We can increase our certainty if we
had more experimental data.
(d) This charge is actually smaller than the actual fundamental charge and so we must conclude that there is some error in
the experiment. If it was a positive integer multiple of the actual fundamental charge then we might conclude that more
data is required.
Making Connections
8.
ε = 1.0 ¯ 102 N/C
(a) q = ?
ε=
kq
rE 2
q=
ε rE 2
k
(1.0 ×10
=
2
N/C )(6.4 ×10 6 m )
2
9.0 ×109 N ⋅ m 2 /C 2
q = 4.5 × 105 C
The charge on Earth is 4.5 ¯ 105 C. Since the field lines are directed towards Earth the charge is negative.
(b) (i) Both fields can be represented by straight field lines directed towards the centre of the earth.
(ii) Gravity always attracts objects while the electric field can either attract or repel objects.
(iii) Both fields are inverse square laws meaning they decrease with the square of the distance from the centre of Earth.
(c) m = ?
FE = Fg
eε = mg
eε
g
m=
(1.6 ×10
=
−19
C )(1.0 × 10 2 N/C )
9.8 N/kg
−18
m = 1.6 × 10 kg
The largest mass that could be suspended is 1.6 ¯ 10–18 kg.
(d) Both particles have small enough masses but the electric field points towards the centre of Earth meaning Earth has a
negative charge. It will attract the proton (which cannot be suspended) and repel the electron. The electron must be at a
higher altitude since its mass is so small.
9. The dust particles could have a negative charge and the electric field of the Earth would then suspend them above the
surface causing them to float. If this is true a negatively charged ebonite rod will repel them.
7.6 THE MOTION OF CHARGED PARTICLES IN ELECTRIC FIELDS
PRACTICE
(Page 368)
Understanding Concepts
1.
V = 1.5 ¯ 102 V
W = 0.24 J
q=?
W = Vq
q=
W
V
0.24 J
1.5 ×10 2 V
q = 1.6 × 10 −3 C
The magnitude of the charge is 1.6 ¯ 10–3 C.
=
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 453
(c) No, the fundamental charge could be any positive integer multiple of this charge. We can increase our certainty if we
had more experimental data.
(d) This charge is actually smaller than the actual fundamental charge and so we must conclude that there is some error in
the experiment. If it was a positive integer multiple of the actual fundamental charge then we might conclude that more
data is required.
Making Connections
8.
ε = 1.0 ¯ 102 N/C
(a) q = ?
ε=
kq
rE 2
q=
ε rE 2
k
(1.0 ×10
=
2
N/C )(6.4 ×10 6 m )
2
9.0 ×109 N ⋅ m 2 /C 2
q = 4.5 × 105 C
The charge on Earth is 4.5 ¯ 105 C. Since the field lines are directed towards Earth the charge is negative.
(b) (i) Both fields can be represented by straight field lines directed towards the centre of the earth.
(ii) Gravity always attracts objects while the electric field can either attract or repel objects.
(iii) Both fields are inverse square laws meaning they decrease with the square of the distance from the centre of Earth.
(c) m = ?
FE = Fg
eε = mg
eε
g
m=
(1.6 ×10
=
−19
C )(1.0 × 10 2 N/C )
9.8 N/kg
−18
m = 1.6 × 10 kg
The largest mass that could be suspended is 1.6 ¯ 10–18 kg.
(d) Both particles have small enough masses but the electric field points towards the centre of Earth meaning Earth has a
negative charge. It will attract the proton (which cannot be suspended) and repel the electron. The electron must be at a
higher altitude since its mass is so small.
9. The dust particles could have a negative charge and the electric field of the Earth would then suspend them above the
surface causing them to float. If this is true a negatively charged ebonite rod will repel them.
7.6 THE MOTION OF CHARGED PARTICLES IN ELECTRIC FIELDS
PRACTICE
(Page 368)
Understanding Concepts
1.
V = 1.5 ¯ 102 V
W = 0.24 J
q=?
W = Vq
q=
W
V
0.24 J
1.5 ×10 2 V
q = 1.6 × 10 −3 C
The magnitude of the charge is 1.6 ¯ 10–3 C.
=
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 453
2.
q = 2e
m = 6.6 ¯ 10–27 kg
V = 2.0 ¯ 103 V
v=?
(a) Using the equation:
W = qV =
v=
=
1 2
mv
2
2Vq
m
2 ( 2.0 × 103 V )(3.2 × 10−19 C )
6.6 × 10−27 kg
v = 4.4 × 105 m/s
The α particle would reach the plate at a velocity of 4.4 ¯ 105 m/s.
1
(b) V = ( 2.0 × 103 V ) = 1.0 × 103 V
2
2Vq
v=
m
=
2 (1.0 × 103 V )(3.2 × 10−19 C )
6.6 ×10 −27 kg
v = 3.1× 105 m/s
The α particle would reach the plate at a velocity of 3.1 ¯ 105 m/s.
3. m = 1.0 × 10–5 kg
q = 4.0 × 10–7 C
d = 50.0 cm = 0.50 m
V = 8.0 × 102 V
(a) W = ?
W = qV
= ( 4.0 × 10−7 C )(8.0 × 102 V )
W = 3.2 ×10 −4 J
The work done by the string is 3.2 ¯ 10–4 J.
(b) F = ?
W
F=
d
3.2 ×10 −4 J
=
0.50 m
F = 6.4 ×10 −4 N
The average force required is 6.4 ¯ 10–4 N.
(c) EK = ?
EK = W
EK = 3.2 × 10 −4 J
The kinetic energy of the pith ball is 3.2 ¯ 10–4 J.
454 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
(d) v = ?
1 2
mv = EK
2
2 EK
v=
m
=
4.
2 (3.2 ×10 −4 J )
1.0 × 10 −5 kg
v = 8.0 m/s
The speed of the pith ball is 8.0 m/s.
r = 1.0 ¯ 10–12 m
EK = ?
v=?
kq q
EE = 1 2
r
(9.0 ×10 N ⋅ m /C )(1.6 ×10
=
9
2
2
−19
1.0 ×10 −12 m
C)
2
EE = 2.3 × 10 −16 J
Etotal = EK + EE = 2.3 × 10 −16 J
′ = EK′ + EE′
Etotal
′
Etotal
(since EK = 0)
1

= 2  mv 2  + 0
2


= mv 2
′ = 2.3 ×10 −16 J
But Etotal = Etotal
v=
=
5.
Etotal
m
2.3 × 10 −16 J
9.1× 10 −31 kg
v = 1.6 × 107 m/s
The kinetic energy of the each electron is 2.3 ¯ 10–16 J. The speed of each electron is 1.6 ¯ 107 m/s.
m = 3.3 ¯ 10–27 kg
q = 1.6 ¯ 10–19 C
v = 5.0 ¯ 106 m/s
V=?
1
W = qV = mv 2
2
2
mv
V=
2q
(3.3 ×10
=
−27
kg )(5.0 × 106 m/s )
2
2 (1.6 ×10 −19 C )
V = 2.6 × 105 V
The potential difference required is 2.6 ¯ 105 V.
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 455
Section 7.6 Questions
(Page 371)
Understanding Concepts
1.
V = 1.2 ¯ 103 V
d = 0.12 m
(a) v = ?
W = qV =
v=
=
1 2
mv
2
2Vq
m
2 (1.2 × 103 V )(1.6 × 10−19 C )
9.11 ×10 −31 kg
v = 2.1 × 107 m/s
The electron will pass through the hole in the positive plate at a speed of 2.1 ¯ 107 m/s.
(b) The electron is not pulled back because there is no electric field outside of the parallel plate apparatus.
(c) Place a small hole in the negative plate instead. Some source of protons would be required. (A hot filament will not
suffice.)
(d) v = ?
2Vq
v=
m
=
2 (1.2 × 103 V )(1.6 × 10−19 C )
1.67 ×10 −27 kg
v = 4.8 × 105 m/s
The speed of the emerging proton is 4.8 ¯ 105 m/s.
2. ε = 2.5 ¯ 102 N/C
v = 1.2 ¯ 106 m/s
(a) r = 2.5 cm = 2.5 ¯ 10–2 m
W=?
W = qV
= qε r
= (1.6 × 10−19 C )( 2.5 × 10 2 N/C )( 2.5 × 10 −2 m )
W = 1.0 × 10−18 J
The work done is 1.0 ¯ 10–18 J.
(b) v΄ = ?
Knowing the relationship between work and energy, W = ∆EK = EK′ − EK , we can calculate the kinetic energy:
EK′ = W + EK
1
= W + mv 2
2
= 1.0 × 10 −18 J +
2
1
9.11× 10 −31 kg )(1.2 × 106 m/s )
(
2
EK′ = 1.7 × 10 −18 J
456 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
We can then calculate the speed:
EK′ = 12 mv′2
2 EK
m
v′ =
2 (1.7 ×10 −18 J )
=
9.11× 10−31 kg
v′ = 1.9 × 106 m/s
The speed of the electron at this point is 1.9 ¯ 106 m/s.
(c) r = ?
1
W = qV = mv 2
2
1 2
qε r = mv
2
1 2
mv
r= 2
qε
2
1
(9.11×10−31 kg )(1.2 ×106 m/s )
2
=
(1.6 ×10−19 C )( 2.5 ×102 N/C )
r = 1.6 × 10−2 m
The electron will move 1.6 ¯ 10–2 m, or1.6 cm into the field before it comes to rest.
3. v = 3.5 ¯ 106 m/s
(a) r = ?
ET = ET′
2 EK = EE′
mv 2 =
r=
=
ke 2
r2
ke 2
mv 2
k e
 
mv
9.0 × 109 N ⋅ m 2 /C 2  1.6 × 10−19 C 


9.11× 10 −31 kg  3.5 × 106 m/s 
r = 4.5 × 10 −6 m
The smallest possible distance between the two electrons is 4.5 ¯ 10–6 m.
(b) No, this would only occur during a head on collision, which is unlikely to occur for two electrons fired at each other
from far apart.
4. m = 6.64 ¯ 10–27 kg
q1 = +2e
r = 4.7 ¯ 10–15 m
q2 = +79e
EE = ?
kq q
EE = 1 2
r
=
(9.0 ×10 N ⋅ m /C ) (2 (1.6 ×10
9
=
2
2
4.7 ×10
−19
−15
)(
C ) 79 (1.6 × 10−19 C )
)
m
−12
EE = 7.7 × 10 J
The α particle must have an initial energy of 7.7 ¯ 10–12 J.
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 457
5.
v = 3.00 ¯ 106 m/s
ε = 120 N/C
∆l = 4.0 cm = 4.0 ¯ 10–2 m
(a) ∆dy = ?
We treat this like a projectile motion problem. The horizontal motion is uniform.
The time to cross the plates is:
∆l
∆t =
vx
4.0 ×10 −2 m
3.00 × 106 m/s
∆t = 1.3 × 10 −8 s
=
The acceleration of the electron is directed up with magnitude:
F
a= E
m
eε
=
m
(1.6 ×10−19 C ) (120 N/C )
=
9.11× 10−31 kg
a = 2.1× 1013 m/s 2
The initial velocity up is zero, therefore:
1
2
∆d y = a ( ∆t )
2
2
1
= ( 2.1 ×1013 m/s 2 )(1.3 ×10 −8 s )
2
∆d y = 1.8 × 10 −3 m
The vertical deflection of the electron is 1.8 ¯ 10–3 m.
(b) v′y = ?
v′y = 0 + a∆t
= ( 2.1 × 1013 m/s 2 )(1.3 × 10−8 s )
v′y = 2.7 ×105 m/s
The vertical component of the final velocity is 2.7 ¯ 105 m/s.
(c) θ = ?
v′y
tan θ =
vx
 2.7 × 105 m/s 
θ = tan −1 

6
 3.00 × 10 m/s 
θ = 5.1°
The angle at which the electron emerges is 5.1°.
Making Connections
6. (a) Have the beam of electrons pass through a pair of horizontal parallel plates. The electrons will accelerate towards the
positive plate. We can accelerate the electrons up or down by changing the amounts of charge on each plate and
changing which plate is negative.
(b) Have the beam pass through a pair of vertical parallel plates and deflect them in a similar fashion as explained above.
We can increase the deflection from left to right by gradually increasing the potential difference between the plates that
will increase the horizontal deflection. We can make the beam jump from right to left by reversing the charge on the
plates.
458 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
(c) The potential difference measured on the surface of the skin of the patient (which is used to indicate potential changes
from the heart) is used to control the potential changes on the horizontal plates causing the beam to deflect vertically.
The vertical plates change their potential independent of measurements to ensure the beam moves across the screen
horizontally at a regular rate.
(d) Answers may vary depending on the speed of the electrons, the length of the plates, and the electric field between the
plates. For example, in question 5 the deflection is 5.1º, which is close enough. If we double the electric field we will
double the acceleration of the electron meaning we will also double the final vertical and so the angle will be about 10º.
tan10°
This is true even though tangent is not a linear function since
= 2 (to two significant digits).
tan 5°
7. (a) The objects will accelerate towards each other and as they do so the acceleration and velocity will both increase, but
energy and momentum will be conserved. Electric potential energy and gravitational potential energy will be
transformed into kinetic energy.
(b) The motion will be similar but only gravitational potential energy will be available to convert into kinetic energy.
CHAPTER 7 LAB ACTIVITIES
Investigation 7.2.1: Factors Affecting Electric Force between Charges
(Pages 372–373)
Questions
(i) The force is larger when the distance between the charges is smaller and the charges are larger. The force is directly
proportional to the product of the charges, and inversely proportional to the distance between them squared.
kq q
(ii) FE = 12 2
r
Hypothesis/Prediction
(a) The force is directly proportional to the product of the charges and inversely proportional to the distance between them
kq q
squared: FE = 12 2 .
r
Analysis
(b)
T cos θ = mg
T sin θ = FE
F
T sin θ
= E
T cos θ mg
FE = mg tan θ
(c) Answers may vary. Students will plot a curve (an inverse square).
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 459
CHAPTER 7 SUMMARY
Make a Summary
(Page 376)
kq1q2
. Force is a vector and this applies to point charges (not parallel plates
r2
or other more complicated charge distributions.) This equation is used to derive many of the other equations in this chapter
such as the electric field, electric potential energy, and electric potential either directly or indirectly. For the electric field of a
kq
point charge we have ε = 21 which is a vector. It can be justified by eliminating the second charge in Coulomb’s law.
r
The electric potential energy between two point charges is related to the electric force in a similar manner as
kq q
gravitational potential energy is related to the gravitational force. The equation EE = 1 2 applies to point charges only and
r
kq1
represents a scalar. The energy per unit charge or electric potential V =
is also a scalar for point charges derived from the
r
electric potential energy. As a result we can write E = qV and for charged parallel plates we can find the electric field between
From Coulomb’s law we have the equation FE =
the plates using the equation ε =
V
.
r
CHAPTER 7 SELF QUIZ
(Page 377)
True/False
1.
2.
3.
4.
5.
6.
F The forces are the same in magnitude.
F The electric force can also repel and depends on charge not mass, etc.
T
F The electric field lines indicate the direction of the net electric force on a positive charge.
F The charge will stay on the outer surface of the car.
T
Multiple Choice
7.
(b)
8.
(e) F =
F=
9.
10.
11.
12.
13.
k (2q1 )q2
(3r ) 2
2 kq1 q2
9 r2
 32 
(e) ε 2 = ε1  2 
6 
(e)
(b)
(b)
(b)
CHAPTER 7 REVIEW
(Pages 378–379)
Understanding Concepts
1. (a) Charge transmitted to the child’s hair causes the individual strands of hair to repel each other and as a result the hair
stands on end.
(b) Charge is also transmitted to this child having the same effect on the hair.
(c) If the children were grounded the would lose the excess charge and the hair would return to normal. Therefore they are
not grounded.
462 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
2. Answers may vary. Most will include the following:
Law
Newton’s law of
gravitation
Coulomb’s law
3.
4.
Factors Affecting
Magnitude
the product of the masses
and the square of the
distance between the
objects
the product of the masses
and the square of the
distance between the
objects
Relative Strength
Direction
weak
always towards the
other charge
strong
can repel or attract
The conditions are when we are dealing with small uniformly charged spheres with a small charge, and they are separated
by a distance that is large compared to the radius of the sphere. If the charge is uniformly distributed over the surface of
the charge then we can treat it as a point charge centered at the centre of the sphere. If the charge or force is large, or the
distance is small, the distribution of charge might shift and the distance cannot be measured from the center. Otherwise
the objects must be so far apart compared to their own size that they may be considered point charges.
F1 = 1.6 ¯ 10–2 N
F2 = ?
kq q
F = 12 2
r
 1  1 
k  q1   q2 
 2  2 
2
( 2r )
F2
=
kq1q2
F1
r2
2
 1  1   1 
F2 = F1     
 2  2   2 
1
= 1.6 × 10 −2 N  
 16 
5.
F2 = 1.0 × 10 −3 N
The magnitude will become 1.0 ¯ 10–3 N.
e = 1.602 ¯ 10–19 C
F = 4.0 ¯ 10–11 N
r=?
kq q
F = 12 2
r
kq1q2
r=
F
(9.0 ×10 N ⋅ m /C )(1.6 ×10
9
=
6.
2
2
4.0 ×10 −11 N
−19
C)
2
r = 2.4 ×10 −9 m
The distance between the two protons is 2.4 ¯ 10–9 m.
r = 5.3 ¯ 10–11 m
m1 = 1.67 ¯ 10–27 kg
m2 = 9.1 ¯ 10–31 kg
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 463
(a) F = ?
F=
kq1 q2
r2
(9.0 ×10 N ⋅ m /C )(1.6 ×10
=
(5.3 × 10 m )
9
2
2
−11
−19
C)
2
2
F = 8.2 ×10 −8 N
The electrostatic force is 8.2 ¯ 10–8 N.
(b) FG = ?
Gm1m2
FG =
r2
(6.67 ×10−11 N ⋅ m2 /kg 2 )(1.67 × 10−27 kg )(9.11×10−31 kg )
=
2
(5.3 ×10−11 m )
FG = 3.6 ×10 −47 N
The gravitational force is 3.6 ¯ 10–47 N.
(c) The electrostatic force is responsible for the electron’s centripetal motion around the proton.
(d) v = ?
T=?
To calculate velocity:
mv 2
r
Fr
v=
m
F=
=
(8.2 ×10
−2
N )(5.3 × 10−11 m )
9.11× 10 −31 kg
v = 2.2 ×106 m/s
To calculate the period:
2π r
v
2π (5.3 ×10 −11 m)
=
2.2 × 106 m/s
T = 1.5 × 10−16 s
The speed of the electron is 2.2 ¯ 106 m/s. The period is 1.5 ¯ 10–16 s.
q1 = 4.0 ¯ 10–5 C
q2 = –1.8 ¯ 10–5 C
r12 = 24 cm = 0.24 m
q3 = –2.5 ¯ 10–6 C
T=
7.
464 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
(a) r23 = 12 cm = 0.12 m
F3 = ?
Since the third charge is placed on the side of the negative charge:
r13 = r12 + r23 = 0.36 m
First, we must calculate the force exerted by the first charge on the third charge:
kq q
F13 = 1 2 3
r13
(9.0 ×10 N ⋅ m /C )(4.0 ×10 C )( 2.5 ×10 C )
9
=
2
−5
2
−6
(0.36 m )
2
F13 = 6.9 N
F13 = 6.9 N [left]
We must then calculate the force exerted by the second charge on the third charge:
kq q
F23 = 2 2 3
r23
(9.0 ×10 N ⋅ m /C )(1.8 ×10 C )( 2.5 ×10 C )
9
=
2
−5
2
−6
(0.12m )
2
F23 = 28.1 N
F23 = 28.1 N [right]
Therefore, for the total force exerted on the third charge:
F3 = 28.1 N [right] + 6.9 N [left]
F3 = 21 N [right]
The force on the third charge is 21 N [right].
(b) r23 = 12 cm = 0.12 m
F3 = ?
Since the third charge is placed on the side of the positive charge:
r13 = r12 – r23 = 0.12 m
First we must calculate the force exerted by the first charge on the third charge:
kq q
F13 = 1 2 3
r13
(9.0 ×10 N ⋅ m /C )(4.0 ×10 C )( 2.5 ×10 C )
=
9
2
−5
2
−6
(0.12 m )
2
F13 = 62.5 N
F13 = 62.5 N [right]
We then calculate the force exerted by the second charge on the third charge:
kq q
F23 = 2 2 3
r23
(9.0 ×10 N ⋅ m /C )(1.8 ×10 C )( 2.5 ×10 C )
9
=
2
−5
2
−6
( 0.12 m )
2
F23 = 3.1 N
F23 = 3.1 N [left]
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 465
Therefore, for the total force exerted on the third charge:
F3 = 62.5 N [right] + 3.1 N [left]
F3 = 59 N [right]
The force exerted on the third charge is 59 N [right].
(c) F3 = ?
Since the original distance between the two point charges was 0.24 m, F13 = F23 = 0.12 m.
First we must calculate the force exerted by the first charge on the third charge:
kq q
F13 = 1 2 3
r13
(9.0 ×10 N ⋅ m /C )(4.0 ×10 C )( 2.5 ×10 C )
9
=
2
−5
2
−6
(0.12 m )
2
F13 = 62.5 N
F13 = 62.5 N [left]
We then calculate the force exerted by the second charge on the third charge:
kq q
F23 = 2 2 3
r23
(9.0 ×10 N ⋅ m /C )(1.8 ×10 C )( 2.5 ×10 C )
9
=
2
−5
2
−6
(0.12 m )
2
F23 = 28.1 N
F23 = 28.1 N [left]
Therefore, the total force on the third charge:
F3 = 62.5 N [left] + 28.1 N [left]
F3 = 91 N [left]
The total force on the third charge is 91 N [left].
8. The test charge is small so that the effect of the test charge’s electric field on the electric field is minimal. The electric
field at any point is the vector sum of all the individual fields present. A ‘small’ test charge would have such a negligible
electric field around it that it would not disturb the electric field being measured.
9. It will begin to travel in the direction tangential to the electric field since the electric field is in the direction of the force
on the positive test charge.
10. The automobile acts like a Faraday cage. The charge stays on the outside of the car and the electric field inside the car is
zero.
11.
466 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
12. q = 1.0 µC = 1.0 ¯ 10–6 C
FE = 6.0 ¯ 10–6 N
(a) ε = ?
ε=
FE
q
6.0 × 10−6 N [right]
1.0 × 10−6 C
ε = 6.0 N/C [right]
The electric field strength is 6.0 N/C [right].
(b) q = –7.2 ¯ 10–4 C
FE = ?
=
FE = qε
= ( −7.2 × 10−4 C ) ( 6.0 N/C [right])
= −4.3 × 10 −3 N [right]
FE = 4.3 × 10 −3 N [left]
A force of 4.3 ¯ 10–3 N [left] would be exerted on the charge.
13. r = 1.5 m
q = 8.0 ¯ 10–3 C
ε =?
kq
ε= 2
r
(9.0 ×109 N ⋅ m 2 /C2 )(8.0 ×10−3 C )
=
2
(1.5 m )
ε = 3.2 ×107 N/C [right]
The electric field strength is 3.2 ¯ 107 N/C [right].
14. q1 = 2.0 ¯ 10–5 C
q2 = 8.0 ¯ 10–6 C
rXZ = 90.0 cm = 0.900 m
rYZ = 30.0 cm = 0.300 m
ε =?
We define the positive direction to the right:
ε = ε X +ε Y
kq
kq
ε = 2X + 2Y
rX
rY
(9.0 ×10 N ⋅ m /C )(−2.0 ×10 C ) + (9.0 ×10 N ⋅ m /C )(8.0 ×10 C )
=
9
2
−5
2
(0.900 m )
2
9
2
−5
2
(0.300 m )
2
= −2.22 × 105 N/C + 8.00 × 105 N/C
ε = 5.8 × 105 N/C [right]
The electric field strength is 5.8 ¯ 105 N/C [right].
15. m = 3.0 ¯ 10–4 kg
r = 10.0 cm = 0.100 m
V = 420 V
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 467
(a) ε = ?
V
r
420 V
=
0.10 m
ε = 4.2 × 103 N/C
The electric field strength is 4.2 ¯ 103 N/C.
(b) T = ?
ε=
Drawing a free body diagram for the ball:
Resolving T into horizontal and vertical components:
T cos θ = mg
T sin θ = FE
For θ we have:
 0.010 m 
θ = sin −1 

 1.0 m 
θ = 0.57°
T cos 0.57º = mg
T=
(3.0 ×10
−4
)(
kg 9.8 m/s 2
)
cos 0.57°
T = 2.9 × 10 N
The tension in the thread is 2.9 ¯ 10–3 N.
(c) FE = ?
T sin θ = FE
−3
FE = (2.9 ×10 −3 N)(sin 0.57º )
FE = 2.9 × 10−5 N
The magnitude of the electric force is 2.9 ¯ 10–5 N.
(d) q = ?
F
q= E
ε
2.9 × 10−5 N
=
4.2 × 103 N/C
q = 6.9 × 10 −9 C
The charge on the ball is 6.9 ¯ 10–9 C.
468 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
16. No net work is required to move a charged particle between two points of the same electric potential. Although there
could be paths of zero force, there could also be paths over which the force changes over a great range of values, as long
as the net wok is zero.
17. No work is required to move a charged particle through an electric field if it always moves perpendicular to electric field
lines. The electric potential remains constant along such a path, by definition.
18. q = 1.2 ¯ 10–3 C
w = 30.0 cm
l = 40.0 cm
FE = ?
ε=?
V=?
For the length of the diagonal:
d = ( w)2 + (l )2
=
(30.0 cm ) + ( 40.0 cm )
2
2
d = 50.0 cm
For the angle of the diagonal:
30.0 cm
40.0 cm
θ = 37°
sin θ =
Each charge experiences three forces: one along each side, and one along the diagonal.
For the force on q4:
kq q
F1 = 1 2 4
r14
(9.0 ×10 N ⋅ m /C )(1.2 ×10 C )
=
9
2
−3
2
2
(0.40 m )
2
F1 = 8.1 × 10 4 N
F2 =
kq2 q4
r24 2
(9.0 ×10 N ⋅ m /C )(1.2 ×10 C )
=
9
2
−3
2
2
(0.30 m )
2
F2 = 1.44 × 105 N
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 469
F3 =
kq3 q4
r34 2
(9.0 ×10 N ⋅ m /C )(1.2 ×10 C )
=
9
2
2
−3
2
(0.50 m )
2
F3 = 5.2 ×10 4 N
Therefore, Fnet = F1 + F2 + F3 .
Taking components in the x – y plane:
x
F1
–8.1 × 104 N
0
F2
F3
(–5.2 × 104 N)(cos 37º) = –4.2 × 104 N
–1.23 × 105 N
Fnet
Fnet =
=
2
Fx + Fy
y
0
1.44 × 105 N
(5.2 × 104 N)(sin 37º) = 3.1 × 104 N
1.75 × 105 N
2
(−1.23 ×10 N ) + (1.75 ×10 N )
5
2
5
2
Fnet = 2.1× 105 N
 Fy 

φ = tan −1 
 Fx 


 1.75 × 105 N 
= tan −1 

5
 1.23 × 10 N 
φ = 55°
Therefore, Fnet = 2.1× 105 N [55° up from the left] and the force is symmetrically the same at each of the other corners.
F
= 0 . This is because the forces due to each of the four charges are equal and
q
opposite, in pairs, so that the net force is 0.
For the electric field at the centre, ε =
For the electric potential:
V=
=
kq
r
(9.0 ×109 N ⋅ m2 /C2 )(1.2 ×10−3 C )
0.25 m
V = 4.32 × 107 V
Therefore, for each charge:
Vtotal = 4 ( 4.32 ×107 V )
Vtotal = 1.7 ×108 V
The electric force on each charge is 2.1 ¯ 105 N [55° up from the left]. The electric field in the centre is zero. The electric
potential at the centre is 1.7 ¯ 108 V.
470 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
19. q = 4.5 ¯ 10–4 C
r = 0.50 m
V=?
V=
=
kq
r
(9.0 ×109 N ⋅ m2 /C2 )( 4.5 × 10−4 C )
0.50 m
6
V = 8.1× 10 V
The electric potential is 8.1 ¯ 106 V.
20. q1 = 1.0 ¯ 10–6 C
q2 = 3.2 ¯ 10–3 C
r1 = 1.0 ¯ 102 cm
r2 = 40.0 cm = 0.400 m
W=?
First we must calculate the electric potential before the charge moved:
kq
V2 =
r2
(9.0 ×10 N ⋅ m /C )(3.2 ×10 C )
9
=
2
2
−3
0.40 m
7
V2 = 7.2 ×10 V
Now we must calculate the electric potential after the charge moved
kq
V1 =
r1
(9.0 ×10 N ⋅ m /C )(3.2 ×10 C )
9
=
2
2
−3
1.0 m
7
V1 = 2.9 ×10 V
To calculate the work:
W = q∆V
= q (V2 − V1 )
= (1.0 × 10−6 C )(7.2 × 107 V − 2.9 × 107 V )
W = 43 J
The amount of work required was 43 J.
21. V = 2.5 ¯ 104 V
∆EK = ?
∆E K = q ∆V
= (1.6 ×10 −19 C )( 2.5 × 10 4 V )
∆EK = 4.0 × 10 −15 J
The amount of kinetic energy gained is 4.0 ¯ 10–15 J.
22. r = 1.0 ¯ 10–15 m
W=?
kq
W=
r
(9.0 ×10 N ⋅ m /C )(1.6 ×10
=
9
2
2
1.0 × 10−15 m
−19
C)
2
W = 2.3 × 10 −13 J
The amount of work done is 2.3 ¯ 10–13 J.
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 471
23. r = 2.0 cm = 2.0 ¯ 10–2 m
V = 450 V
ε=?
ε=
V
r
450 V
2.0 ×10 −2 m
ε = 2.3 × 104 N/C
The electric field between the two plates is 2.3 ¯ 104 N/C.
24. r = 8.0 cm = 8.0 ¯ 10–2 m
ε = 2.5 ¯ 103 N/C
V=?
V = εr
=
= ( 2.5 × 103 N/C )(8.0 × 10−2 m )
V = 2.0 × 102 V
The potential difference would be 2.0 ¯ 102 V.
25. The Millikan experiment would not work in a vacuum. The drop could be balanced but when the balancing voltage was
removed, it would accelerate under gravity, allowing no measurement of its terminal velocity and hence no calculation of
its mass. Also, the oil drops would move very quickly, making them difficult to balance, and they would evaporate.
26. r = ?
Fe = Fg
kq1q2
= mg
r2
r=
ke 2
mg
(9.0 ×10 N ⋅ m /C )(1.6 ×10 C )
(9.11×10 kg ) (9.8 N/kg )
9
=
2
2
−19
2
−31
r = 5.1 m
The other electron would have to be 5.1 m vertically above the first electron.
27. m = 2.6 ¯ 10–15 kg
r = 0.50 cm = 5.0 ¯ 10–3 m
V = 270 V
q=?
N=?
Since q =
FE
, for the electric potential energy:
ε
FE = Fg
FE = mg
= ( 2.6 ×10 −15 kg ) (9.8 N/kg )
FE = 2.55 × 10−14 N
For the electric field:
ε=
V
r
270 V
5.0 × 10−3 m
ε = 5.4 × 104 N/C
=
472 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
We can now calculate the charge:
F
q= E
ε
2.55 × 10 −14 N
=
5.4 ×10 4 N/C
q = 4.7 × 10 −19 C
To find the number of electrons:
q
N=
e
4.7 × 10 −19 C
=
1.6 × 10−19 C
N =3
The charge on the oil droplet is 4.7 ¯ 10–19 C. We cannot tell if there is an excess or a deficit unless we know which plate
is positive.
28. m = 0.10 g = 1.0 ¯ 10–4 kg
q = 5.0 ¯ 10–6 C
r = 25 cm = 2.5 ¯ 10–1 m
V=?
To calculate the electric potential:
FE = Fg
FE = mg
= (1.0 ×10 −4 kg ) (9.8 N/kg )
FE = 9.8 × 10−4 N
We now calculate the electric field:
F
ε= E
q
9.8 × 10−4 N
5.0 ×10 −6 C
ε = 1.96 × 102 N/C
=
Using this value, we can calculate the potential difference:
V = εr
= (1.96 × 10 2 N/C ) (0.25 m )
V = 49 V
The potential difference required is 49 V.
29. r = 0.40 m
N = 1.0 ¯ 1012 electrons
V=?
ε=?
First we must calculate the charge:
q = N ( −e )
= (1× 1012 )( −1.6 × 10 −19 C )
q = −1.6 × 10−7 C
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 473
To calculate the electric potential:
kq
V=
r
9.0 ×109 N ⋅ m 2 /C 2 )( −1.6 ×10 −7 C )
(
=
0.40m
V = −3.6 ×103 V
To calculate the electric field:
kq
ε= 2
r
9.0 × 109 N ⋅ m 2 /C 2 )(1.6 × 10−7 C )
(
=
2
(0.40 m )
ε = 9.0 ×103 N/C
The electric potential is –3.6 ¯ 103 V. The electric field is 9.0 ¯ 103 N/C.
30. V = 5.0 ¯ 102 V
v=?
Using the relationships:
W = qV = ∆EK =
1 2
mv
2
Therefore,
v=
=
2qV
m
2 (1.6 × 10−19 C )(5.0 × 102 V )
9.11× 10−31 kg
v = 1.3 × 107 m/s
The speed of the electron is 1.3 ¯ 107 m/s.
31. EK = 1.9 ¯ 10–15 J
q = +2e
V=?
∆EK = qV
V =
∆E K
q
1.9 ×10 −15 J
3.2 ×10 −19 C
V = 5.9 ×103 V
The potential difference would be 5.9 ¯ 103 V.
32. vi = 5.0 ¯ 106 m/s
vf = 1.0 ¯ 106 m/s
V=?
=
Using the relationships:
W = qV = ∆EK = EK′ − EK
We must first calculate the work required:
1
1
W = mvf 2 − mvi 2
2
2
2
2
1
= (9.11× 10 −19 kg ) (1.0 × 106 m/s ) − (5.0 × 106 m/s )
2
W = −1.09 × 10 −17 J
(
474 Unit 3 Electric, Gravitational, and Magnetic Fields
)
Copyright © 2003 Nelson
We now calculate the potential difference:
∆E K
V=
q
1.09 × 10−17 J
1.6 × 10 −19 C
V = 68 V
The potential difference is 68 V.
33. rXW = rWY = rYZ = 4.0 cm
V1 = 3.0 ¯ 102 V
V2 = 5.0 ¯ 102 V
(a) v = ?
1 2
mv = qV
2
2qV
v=
m
=
=
2 (1.6 × 10−19 C )(3.0 × 102 V )
9.11× 10−31 kg
v = 1.0 × 107 m/s
The speed of the electron at hole W is 1.0 ¯ 107 m/s.
(b) Since plates W and Y are connected together, there is no field between them, so the charge drifts with a constant speed
through hole Y. Since 3.0 ¯ 102 V was used to accelerate the electron then the same amount is required to stop it. The
distance from the plate Y is:
3.0 × 102 V
d
=
4.0 cm 5.0 × 102 V
d = 2.4 cm
Therefore the electron is 4.0 cm – 2.4 cm = 1.6 cm from plate Z when it stops.
(c) The path is reversible, so that the electron arrives back at X with v = 0 m/s. Since this 3.0 × 102 V was initially used to
accelerate the electron then it will also stop it.
34. v = 3.0 ¯ 106 m/s
r=?
First we must calculate the kinetic energy:
1

EK = 2  mv 2 
2

= (6.6 × 10−27 kg )(3.0 × 106 m/s )
2
EK = 5.94 × 10−14 J
At minimum separation:
∆EE = −∆EK
= − (0 J − 5.94 × 10 −14 J )
∆EE = 5.94 × 10−14 J
Therefore,
kq1 q2
r
kq q
r= 1 2
EE
EE =
(9.0 ×10 N ⋅ m /C )(3.2 ×10
=
9
2
2
5.94 × 10−14 J
r = 1.6 ×10 −14 m
The minimum separation is 1.6 ¯ 10–14 m.
Copyright © 2003 Nelson
−19
C)
2
Chapter 7 Electric Charges and Electric Fields 475
35. ∆dx = 10 cm = 0.10 m
vx = 8.0 ¯ 107 m/s
V = 6.0 ¯ 102 V
(a) ∆dy = ?
For the time the electron takes to move through the apparatus:
∆d x
∆t =
vx
0.10 m
8.0 × 107 m/s
∆t = 1.25 × 10 −9 s
=
For its vertical acceleration:
F
a= E
m
qε
a=
m
where ε =
V
r
Therefore,
qV
mr
(1.6 ×10−19 C )(6.0 ×102 V )
a=
=
(9.11×10
−31
kg )( 2.0 × 10−2 m )
a = 5.27 × 1015 m/s 2
For its vertical deflection:
1
a ∆t 2
2
2
1
= (5.27 × 1015 m/s 2 )(1.25 × 10−9 s )
2
∆d y = 4.1× 10 −3 m
∆d y =
The vertical deflection of the electron is 4.1 ¯ 10–3 m, or 0.41 cm.
(b) v2 = ?
For its vertical velocity:
v2 y = a ∆ t
= (5.27 × 1015 m/s 2 )(1.25 × 10−9 s )
v2 y = 6.6 × 106 m/s
To calculate the velocity when the electron leaves the plate apparatus:
v2 = v x 2 + v2 y 2
=
(8.0 ×10 m/s ) + (6.6 ×10 m/s )
7
2
6
2
v2 = 8.0 × 107 m/s
To find the angle at which the electron leaves:
 6.6 × 106 m/s 
θ = tan −1 

7
 8.0 × 10 m/s 
θ = 4.7°
Therefore the electron leaves the plate apparatus at a velocity of 8.0 ¯ 107 m/s [4.7° up from the right].
476 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
Applying Inquiry Skills
36. Students will find that the television screen is not charged if it has been off for a long time. It develops a charge after
turning it on and it retains this charge for a while after it has been turned off.
37. Students should investigate the nature of the field near the conductor noticing that it is perpendicular to the surface and
zero inside the conductor. This works well if a large metal ring is available.
38. (a) Earth and the plates are negatively charged since the field lines are directed towards them. Beneath the plates the
surface of Earth is neutral.
(b) The top plate will shield the lower plate from the field lines. The lower plate will lose its charge since the electrons will
move into Earth. The ammeter will measure a current.
(c) The lower plate will no longer be shielded from the electric field lines. Electrons will flow back into the lower plate
and the ammeter will measure a current in the opposite direction.
(d) The ammeter will measure an AC current since the charge will repeatedly flow out of and back into the lower plate due
to the periodic shielding effect of the upper plate.
(e) If the electric field is larger the charge flowing into and out of the lower plate is larger meaning we could calibrate the
ammeter to indicate the magnitude of the electric field present.
39. (a) Electric field lines are always perpendicular to the surface of the conductor. The field is strongest at the surface of the
conductor closest to the charged plate and very weak on the opposite side.
(b) The electric field inside the conductor is zero. The conductor shields the inside from the external charge
40. Electric potential differences across the fluorescent light bulbs cause them to glow. Placing charges on the bulbs causes
the necessary potential differences which make the light bulbs glow, as well as bringing a charged object close. The closer
the charge and the larger the charge the greater the electric potential difference across the bulb and the brighter the glow.
The electric potential difference from the ground to a rural power line is high enough to cause a light bulb to glow as well,
as demonstrated in Figure 8 on page 381.
Making Connections
41. Negatively charged dust particles will be attracted to the positively charged film. Introducing positively charged particles
into the air will help reduce the number of negatively charged dust particles because these particles will attach themselves
to the dust, making them electrically neutral before they reach the film.
42. Pressure on these crystals cause a charge separation and induces an electric potential difference. This electric potential
difference is used to help wrist watches keep time since the vibrating crystal has a regular frequency and it loses very little
energy.
43. The arc starts when the electric field between the plates is at the breakdown value. This causes the air to ionize, changing
it from an insulator to a conductor. The largest electric field will occur between the plates when the distance between the
V
plates is smallest since ε = . Once the air is ionized in an area it will promote charge to flow from one plate to the other
r
in that area, in turn causing the spark to climb up the plates.
Extension
44. Gauss’s law relates the electric field through a surface area surrounding a charge (or flux) to the amount of charge inside
the surface area. It is used to determine the nature of electric fields near charge distributions or around wires etc. It gives a
clearer picture of the electric field produced by charge distributions rather than restricting ourselves to just point charges.
45. v = 2.4 ¯ 106 m/s [45° up from the horizontal]
r = 2.5 mm = 2.5 ¯ 10–3 m
V = 1.0 ¯ 102 V
(a) Approach this like a projectile motion problem except use the electric force instead of gravity. For the vertical
acceleration:
Fnet = FE
ma = qε
a=
=
qV
mr
(1.6 ×10−19 C )(1.0 × 102 V )
(9.11×10
−31
kg )( 2.5 × 10−3 m )
a = 7.0 ×1015 m/s 2 [down]
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 477
For the maximum height above the bottom plate the velocity in the y direction is zero:
v y′ = v y + a∆t
0 = ( 2.4 ×106 m/s ) sin 45° + ( −7.0 × 1015 m/s 2 ) ∆t
∆t = 2.4 × 10 −10 s
Therefore,
∆d y = v y ∆t + 12 a∆t 2
= ( 2.4 ×106 m/s (sin 45° ))( 2.4 × 10−10 s ) +
2
1
−7.0 × 1015 m/s 2 )( 2.4 × 10−10 s )
(
2
∆d y = 2.1× 10 −4 m
If the electron enters in the middle of the plate then it comes 1.25 × 10-3 m – 2.1 × 10-4 m = 1.0 × 10-3 m or about 1.0
mm to the top plate.
(b) We must first determine how much time it takes for the electron to reach the bottom plate:
1
∆d y = v y ∆t + a ∆t 2
2
1
−3
−1.25 × 10 m = ( 2.4 × 106 m/s (sin 45° )) ∆t + ( −7.0 × 1015 m/s 2 ) ∆t 2
2
15
2
6
−3
3.5 × 10 ∆t − 1.7 × 10 ∆t − 1.25 × 10 = 0
Using the quadratic formula:
∆t =
− ( −1.7 × 106 ) ±
( −1.7 ×10 ) − 4 (3.5 ×10 )(−1.25 ×10 )
2 (3.5 ×10 )
6 2
15
−3
15
∆t = 8.9 × 10 −10 s
To find where it strikes the bottom plate:
∆d x = v x ∆t
= ( 2.4 ×106 m/s ) (cos 45° ) (8.9 ×10−10 s )
∆d x = 1.5 × 10 −3 m
The electron strikes the bottom plate at a distance of 1.5 ¯ 10–3 m.
478 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson