Unit 1 Forces and Motion: Dynamics
ARE YOU READY?
(Pages 2–3)
Knowledge and Understanding
1.
Scalar quantities include distance (metre, 5.0 m), time (second, 15 s), mass (kilogram, 65 kg), and frequency (hertz,
60 Hz). Vector quantities include velocity (metres per second, 15 m/s [E]), displacement (metre, 6.5 m [S]), acceleration
(metres per second squared, 9.8 m/s2 [down]), and force (newton, 25 N [forward]).
2. (a) Both masses will hit the floor at the same time since the speed at which an object falls is independent of mass, and is
related only to acceleration due to gravity (neglecting air resistance).
(b)
(c) m = 20 g = 0.02 kg
G
Fg = ?
G
G
Fg = mg
= (0.02 kg)(9.8 N/kg [down])
G
Fg = 0.2 N [down]
The weight of the 20-g mass is 0.2 N [down].
(d) One example is the force of Earth pulling downward on the 20-g mass and the force of the 20-g mass pulling upward
on Earth.
GM E mMoon
The magnitude of the force of gravity between Earth and the Moon depends linearly on the masses of
3. FG =
r2
Earth and the Moon, and depends inversely as the square of the distance between the centres of Earth and the Moon.
4. (a) Kinematics is the study of motion (e.g., analyzing motion with constant acceleration). Dynamics is the study of the
causes of motion (e.g., analyzing forces by applying Newton’s three laws of motion).
G
total distance
change of position
. Average velocity is a vector quantity, vav =
.
(b) Average speed is a scalar quantity, v =
av
total time
time interval
(c) Static friction is a force that acts to prevent a stationary object from starting to move. Kinetic friction is a force that acts
against a moving object. For a given situation, kinetic friction tends to be less than maximum static friction.
(d) Helpful friction is needed in many cases (e.g., turning a doorknob, walking, and travelling around a corner on a
highway). Unwanted friction usually increases the production of waste heat (e.g., friction in the moving parts of an
engine).
(e) Frequency is the number of cycles of a vibration per unit time; it is measured in hertz (Hz) or s−1. Period is the time for
one complete cycle of a vibration; it is measured in seconds (s).
(f) Rotation is the spinning of an object on its own axis (e.g., Earth rotates daily). Revolution is the motion of one body
around another (e.g., Earth revolves around the Sun once per year).
Inquiry and Communication
5. (a) The units of acceleration are m/s2, so the inspector could use a metre stick to measure the distance in metres (m), and a
stopwatch to measure the time in seconds (s).
(b) The acceleration is the dependent variable, and the distance and time are the independent variables. (Students will
discover in Chapter 3 that the distance is actually the radius of the circle and the time is the period of rotation of the
ride.)
Copyright © 2003 Nelson
Unit 1 Are You Ready?
1
6.
Error analysis can be reviewed by referring to page 755 of the text. Note that possible error is also called uncertainty.
(a) The possible error is ± half of the smallest division of the measurement, or ±0.05 m/s2.
possible error
× 100%
(b) % possible error =
measurement
=
±0.05 m/s
9.4 m/s
2
2
× 100%
% possible error = ±0.5%
The percent possible error is ±0.5%.
measured value − accepted value
× 100%
(c) % error =
accepted value
2
=
9.4 m/s − 9.8 m/s
9.8 m/s
2
2
× 100%
% error = − 4.0%
The percentage error is –4.0%.
difference in values
(d) % difference =
× 100%
average of values
2
=
9.7 m/s − 9.4 m/s
1
2
7.
(9.7 m/s
2
2
+ 9.4 m/s
2
)
× 100%
% difference = 3.0%
The percentage difference is 3.0%.
A prediction is a stated outcome expected from an experiment. A hypotenuse is a tentative explanation of what is
expected in an experiment.
Making Connections
8.
The rapid spinning in the centrifuge would cause the liquids to separate according to their densities, with the liquid of
greatest density moving farthest away, that is, toward the base of the tube.
G
9. (a) Let Fg = force of gravity on the truck
G
FN = normal force of the road on the truck
G
Fs = force of static friction on the truck
θ = angle of the banked curve in (a) and (c)
(a)
(b)
(c)
(b) Choice (c) would be the best because the force of the road on the truck (the normal force) will help a lot in forcing the
truck to the right.
2
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Math Skills
10. (a) The equation is rearranged as follows.
G G
1G
∆d = vi ∆t + a ∆t 2
2
G G
1G
∆d − vi ∆t = a ∆t 2
2
G G
 ∆d vi 
G
a = 2 2 − 
∆t 
 ∆t
(b) The quadratic formula is used to solve an equation in the form
ax 2 + bx + c = 0
As shown on page 750 of the text, the quadratic equation is
x=
−b ± b 2 − 4ac
2a
Solving for ∆t, we have:
G G
1G
∆d = vi ∆t + a ∆t 2
2
G
G
G
a 2
  ∆t = + (vi ) ∆t − ∆d
2
∆t =
G
−v ±
i
(vG )
G
G
− 2 ( a ) ∆d
G
a
2
i
( )
11. (a) Using the scale indicated in the question:
G
A = 29.0 m/s [35° N of E]
The north and east components of this vector are, respectively:
A (sin 35°) = 29.0 m/s (sin 35°) = 17 m/s
A (cos 35°) = 29.0 m/s (cos 35°) = 24 m/s
Notice that the answers are written to two significant digits because the angle is stated to two significant digits.
(b) The vectors can be added by using a vector scale diagram (adding the vectors head-to-tail), by using components, or by
applying trigonometry (the laws of sines and cosines).
(c) Scale: 1.0 cm = 5.0 cm
G G
B + A = 3.9 cm × 5.0 cm = 20 cm [65° N of E]
G G
A − B = 3.3 cm × 5.0 cm = 17 cm [2° S of E]
Copyright © 2003 Nelson
Unit 1 Are You Ready?
3
Technical Skills and Safety
12. (a) The total time is 6(0.10 s) = 0.60 s.
(b)
As shown in the illustration, the x-component of each displacement vector is about 1.0 cm in the diagram, or 5.0 cm
using the scale indicated. We can conclude that the motion in the x direction is constant-speed motion.
(c) The y-component of the displacements constantly changes, becoming smaller as the puck rises, and then becoming
larger as the puck descends.
(d) The displacement (or change of position) from the initial position to the final position is:
G
∆d = 6.15 cm × 5.0 cm/cm = 31 cm [right]
∆t = 0.60 s
G
vav = ?
G
∆d
G
vav =
∆t
31 cm [right]
=
0.60 s
G
vav = 52 cm/s [right]
The average velocity of the puck is 52 cm/s [right].
13. (a) Using a stopwatch, determine the total time for a certain number of complete revolutions of the stopper (e.g., 20
cycles). Then apply the following relationships:
number of cycles
frequency: f =
total time
period: T =
total time
number of cycles
(b) The string should be strong, the stopper should be securely attached to the string, and the lab partner should hold the
string securely while twirling the stopper a safe distance away from objects or people.
(c) Typical sources of error are:
• starting and stopping the stopwatch at the precise instant required (also called human reaction time error.)
• choosing the same position to indicate both the starting and finishing locations of the motion
• keeping the stopper moving at a constant speed and/or in a horizontal circle
• measuring the radius of the circle of motion
4
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
CHAPTER 1 KINEMATICS
Reflect on Your Learning
(Page 4)
1. (a)
(b) Ball B will land first because it has an initial downward component of velocity. Balls A and C will land at the same
instant because, as students will discover in the Try This Activity, page 5, the horizontal component of the velocity of
ball C does not affect its downward acceleration. Ball D will land last because its initial velocity has an upward vertical
component.
2. (a)
(b) The canoeists arrive at the north shore at the same time. The motion of the canoeist in the river is perpendicular to the
flow of the river, so the motion is not affected by the flow of the river.
(c) The figure below shows that the canoeist must aim upstream in order to arrive directly north of the starting position.
This trip will take longer because the component of the velocity perpendicular to the shore is less than it was
previously.
Try This Activity: Choose the Winner
It is important in setting up this demonstration that the device be fixed horizontally. For an alternative suggestion to the
apparatus shown in the text, refer to Section 1.4 Questions, page 51, question 10, and the corresponding solution.
(a) Predictions may vary. Refer to the Reflect on your learning questions 1(b) above.
(b) Balls A and B will land simultaneously. The horizontal motion of the ball projected horizontally is independent of its
downward acceleration.
Copyright © 2003 Nelson
Chapter 1 Kinematics
5
1.1 SPEED AND VELOCITY IN ONE AND TWO DIMENSIONS
PRACTICE
(Pages 7–8)
Understanding Concepts
1. (a) The motion of a tennis ball that falls vertically downward is in one dimension.
(b) The motion of a tennis ball that falls vertically downward and then bounces is in one dimension.
(c) The motion of a basketball moving through the air toward the hoop is in two dimensions.
(d) The motion of a curve ball is in three dimensions.
(e) The motion of a passenger seat of a Ferris wheel is in two dimensions.
(f) The motion of a roller coaster is in three dimensions.
2. (a) The force exerted by an elevator cable is a vector measurement.
(b) The reading on a car’s odometer is a scalar measurement.
(c) The gravitational force of Earth on you is a vector measurement.
(d) The number of physics students in your class is a scalar measurement.
(e) Your age is a scalar measurement.
3. A car’s speedometer indicates instantaneous speed, a scalar quantity. It does not indicate any direction.
4. (a) ∆t = 6.69 h
d = 4.02 km × 200 laps = 8.04 × 102 km
vav = ?
d
vav =
∆t
8.04 × 10 2 km
=
6.69 h
vav = 1.20 × 10 2 km/h
The average speed in 1911 was 1.20 × 102 km/h.
(b) ∆t = 3.32 h
vav = ?
d
vav =
∆t
8.04 × 10 2 km
=
3.32 h
vav = 2.42 × 10 2 km/h
The average speed in 1965 was 2.42 × 102 km/h.
(c) ∆t = 2.69 h
vav = ?
d
vav =
∆t
8.04 × 10 2 km
=
2.69 h
vav = 2.99 × 10 2 km/h
The average speed in 1990 was 2.99 × 102 km/h.
5. (a) d = 16 m
∆t = 21 s
vav = ?
d
vav =
∆t
16 m
=
21s
vav = 0.76 m/s
The swimmer’s average speed is 0.76 m/s.
6
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b) circumference = πD = π (16 m) = 50.26 m
∆t = ?
d
vav =
∆t
d
∆t =
vav
=
50.26 m
0.76 m/s
∆t = 66s
It would take the swimmer 66 s to swim around the edge of the pool.
6. (a) vav = 342 m/s
∆t = 3.54 × 10–2 s
d=?
d = vav ∆t
= (342 m/s) (3.54 × 10−2 s)
d = 12.1m
The distance travelled is 12.1 m.
(b) vav = 1.74 km/h
∆t = 60.0 h
d=?
d = vav ∆t
= (1.74 km/h) (60.0 h)
d = 104 km
The distance travelled is 104 km, or 1.04 × 105 m.
PRACTICE
(Page 10)
Understanding Concepts
7.
Typical examples of different types of vector quantities include a displacement while walking at a constant velocity, an
acceleration while on a school bus, and the force of gravity.
8. (a) It is possible for the total distance travelled to be equal to the magnitude of the displacement if the motion is all in one
direction.
(b) It is possible for the total distance travelled to exceed the magnitude of the displacement if the motion is in one
dimension along a path that turns back on itself, or if the motion is in two or three dimensions. For example, if a car
travels 50 km [N] and then 20 km [S], the distance travelled is 70 km, but the displacement is 30 km [N].
(c) It is not possible for the magnitude of the displacement to exceed the total distance travelled. The total distance is the
sum of the magnitudes of all of the vectors. Thus, the total distance is always equal to or greater than the magnitude of
the final displacement vector.
9. Yes, the average speed can equal the magnitude of the average velocity if the motion is all in one direction.
10. (a) d = 12 km + 12 km = 24 km
 1h 
∆t = 24 min + 24 min = (48 min) 
 = 0.80 h
 60 min 
vav = ?
d
vav =
∆t
24 km
=
0.80 h
vav = 3.0 × 101 km/h
The average speed of the bus for the entire route is 3.0 × 101 km/h.
Copyright © 2003 Nelson
Chapter 1 Kinematics
7
G
(b) ∆d = 12 km [E]
 1h 
∆t = (24 min) 
 = 0.40 h
 60 min 
G
vav = ?
G
∆d
G
vav =
∆t
12 km[E]
=
0.40 h
G
vav = 3.0 × 101 km/h [E]
The average velocity of the bus is 3.0 × 101 km/h [E].
(c) vav = ?
First we must calculate the total displacement:
G
∆d = 12 km [E] + 12 km [W]
= 12 km [E] + ( − 12 km [E])
G
∆d = 0.0 km
Since the total displacement is 0.0 km, the average velocity of the bus for the entire route is 0.00 km/hr.
(d) The answers in (b) and (c) are different because the average velocity is determined by the displacement, which is a
vector. In (b) the bus has reached its maximum displacement, however in (c), the bus returns to its starting position, so
its displacement is zero and its average velocity is zero.
11. ∆t = 0.32 s
G
vav = 27 m/s [fwd]
G
∆d = ?
G G
∆d = vav ∆t
= (27 m/s [fwd])(0.32 s)
G
∆d = 8.6 m [fwd]
The truck is displaced 8.6 m [fwd] during the time it takes for the driver to react.
G
12. ∆d = 1.6 × 104 km
G
vav =21 km/h [S]
∆t = ?
G
∆d
∆t = G
vav
=
1.6 ×10 4 km[S]
21km/h
 1d 
∆t = 7.6 × 102 h 

 24 h 
∆t = 32 d
The tern’s journey takes 7.6 × 102 h or 32 days.
Applying Inquiry Skills
13. (a) The windsock indicates both the approximate speed and the approximate direction of the wind. Speed with a direction
is velocity, a vector quantity.
(b) Calibrating the windsock involves finding how the angle of the sock above the vertical depends on the speed of the
wind. (At a wind speed of zero, the sock hangs vertically downward.) Thus, an experiment must be devised to measure
the angle at various increasing speeds (e.g., at 10 km/h, 20 km/h, etc.). The easiest way to do this would be to hold the
sock securely out of an open window of a car as the car travels at different speeds on a calm day. (Students are not
expected to attempt such an experiment.)
8
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Try This Activity: Graphing Linear Motion
(Page 12)
The required graphs are drawn with the assumption that the zero displacement is the location of the motion sensor, and the
positive displacement direction is away from the sensor.
PRACTICE
(Pages 13–14)
Understanding Concepts
14. (a) The motion, beginning north of the zero displacement position, is southward with constant velocity to a position south
of the zero displacement position.
(b) The motion begins at the zero displacement position and is upward but slowing down (downward acceleration).
(c) The motion begins west of the zero displacement position and is eastward but slowing down (westward acceleration).
15. (a)
Copyright © 2003 Nelson
Chapter 1 Kinematics
9
(b)
(c)
16. The area under the velocity-time graph represents the displacement. Assume two significant digits.
1
Area = (15 m/s [N])(0.20s) + (15 m/s [N])(0.40s − 0.20s)
2
= 3.0 m [N] + 1.5 m [N]
Area = 4.5 m [N]
Thus, the area is 4.5 m [N].
17. In each case, the initial velocity is equal to the slope of the tangent at the time indicated. The diagram with typical
tangents is shown below:
G
∆d
G
Using the equation v = slope = m =
, students should calculate approximate answers of 7 m/s [E], 0 m/s, 7 m/s [W],
∆t
13 m/s [W], and 7 m/s [W].
10
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
PRACTICE
(Page 16)
Understanding Concepts
18. Using a vector scale diagram to solve this problem, the vector sum of the displacements is found to be 5.6 m [24° E of S].
Using components would yield the same result.
19. (a)
(b) Solving Sample Problem 5 (c) using components, with +x eastward and +y northward:
G
G
∆d1 = 22 m
∆d 2 = 11m
∆d1, x = 22 m (cos 33°)
∆d 2, x = 11m (cos 28°)
∆d1, x = 18 m
∆d 2, x = 10 m
∆d1, y = 22 m (sin 33°)
∆d 2, y = 11m (sin 28°)
∆d1, y = 12 m
∆d 2, y = 5 m
∆d x = ∆d1, x + ∆d 2, x
∆d y = ∆d1, y − ∆d 2, y
= 18 m + 10 m
∆d x = 28 m
Copyright © 2003 Nelson
= 12 m − 5.0 m
∆d y = 7.0 m
Chapter 1 Kinematics
11
To calculate the total displacement:
G
∆d = ∆d x 2 + ∆d y 2
= (28 m) 2 + (7.0 m)2
G
∆d = 29 m
tan θ =
∆d y
∆d x
 7.0 m 
θ = tan −1 

 28 m 
θ = 14°
The total displacement is 29 m [14° N of E].
G
20. ∆d1 = 8.5 × 102 m [25° N of E]
G
∆d 2 = 5.6 × 102 m [21° E of N]
 60 s 
∆t = 4.2 min 
 = 252 s
 1 min 
(a) Add the vectors together:
Use the cosine law to find the magnitude of the resultant displacement:
G2
G 2
G 2
∆d = ∆d1 + ∆d 2 − 2 ∆d1 ∆d2 cos φ
G2
∆d = (8.5 × 102 m) 2 + (5.6 × 102 m)2 − 2(8.5 ×102 m)(5.6 × 102 m)(cos136°)
G
∆d = 1.3 × 103 m
Use the sine law to solve for the angle of the displacement’s direction:
sin θ sin φ
G = G
∆d 2
∆d
sin θ
sin136°
=
2
5.6 × 10 m 1.3 × 103 m
 (5.6 × 102 m) sin136° 
θ = sin −1 

1.3 × 103 m


θ = 17°
The angle of the displacement is 17° + 25° = 42°
The skater’s displacement is 1.3 × 103 m [42° N of E].
12
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b) vav = ?
d
∆t
(8.5 × 102 m) + (5.6 ×102 m)
=
252 s
vav =
vav = 5.6 m/s
To calculate the average velocity:
G
vav = ?
G
∆d
G
vav =
∆t
1.3 × 103 m [42° N of E]
=
252 s
G
vav = 5.2 m/s [42° N of E]
The skater’s average speed is 5.6 m/s and average velocity is 5.2 m/s [42° N of E].
Section 1.1 Questions
(Page 17)
Understanding Concepts
1. (a)
(b)
(c)
(d)
(e)
2. (a)
(b)
(c)
(d)
scalar
scalar
scalar
vector
vector (equal to the displacement)
A car travelling at constant speed in one direction is at constant velocity.
A car travelling at a constant speed around a circular track has a velocity that is constantly changing.
A bus travelling from a station to a bus stop and then travelling back along the same route.
In the example in (c) above, if the bus returns to the station its average speed is greater than zero, but its average
velocity is zero.
(e) A car travelling around a circular track has an average speed greater than zero, but when it reaches its starting position,
its average velocity is zero.
3. Measurements with different dimensions can be multiplied; for example, velocity (m/s) × time (s) = distance (m).
Measurements with different dimensions cannot be added; for example, velocity cannot be added to time.
4. (a) vav = 3.00 × 108 m/s
d = 1.49 × 1011 m
∆t = ?
d
∆t =
vav
=
1.49 × 1011 m
3.00 × 108 m/s
∆t = 4.97 × 102 s
Light takes 4.97 × 102 s to travel from the Sun to Earth.
(b) d = 3.84 × 105 km = 3.84 × 108 m
∆t = ?
d
∆t =
vav
=
(
2 3.84 ×108 m
8
)
3.00 × 10 m/s
∆t = 2.56 s
The laser light takes 2.56 s to travel to the Moon and back to Earth.
Copyright © 2003 Nelson
Chapter 1 Kinematics
13
5. (a) ∆t = 4.0 s
∆d = 0 m
vav = ?
∆t = 8.0 s
∆d = 40 m
vav = ?
∆d
∆t
0m
=
4.0s
∆d
∆t
40 m
=
8.0s
vav =
vav =
vav = 0.0 m/s
vav = 5.0 m/s
The average speed between 4.0 s and 8.0 s is 0.0 m/s. The average speed between 0.0 s and 8.0 s is 5.0 m/s.
(b) ∆t = 1.0 s
∆t = 4.0 s
∆t = 16 s
G
G
G
∆d = 2.5 m [E]
∆d =45 m [W]
∆d = 5.0 [E]
G
G
G
vav = ?
vav = ?
vav = ?
G
G
G
∆d
∆d
∆d
G
G
G
vav =
vav =
vav =
∆t
∆t
∆t
2.5 m[E]
45 m[W]
5.0 m[E]
=
=
=
1.0 s
4.0 s
16s
G
G
G
vav = 2.5 m/s[E]
vav = 11m/s[W]
vav = 0.31m/s[E]
The average velocity between 8.0 s and 9.0 s is 2.5 m/s [E]; the average velocity between 12 s and 16 s is 11 m/s [W];
and the average velocity between 0.0 s and 16 s is 0.31 m/s [E].
(c) The instantaneous speed at t = 6.0 s and t = 9.0 s is the slope of the line at that instant. Thus,
∆d
∆d
v=m=
v = slope = m =
∆t
∆t
10 m
0.0 m
=
=
4.0 s
4.0 s
v = 2.5 m/s
v = 0.0 m/s
The instantaneous speed at 6.0 s is 0.0 m/s and at 4.0 s is 2.5 m/s.
(d) The instantaneous velocity at t = 14 s is the slope of the line at that instant. Thus,
G
G
∆d
v = slope = m =
∆t
45 m [W]
=
4.0 s
G
v = 11 m/s [W]
The instantaneous velocity at 14 s is 11 m/s [W].
6. The slope of the line can be calculated from a position-time graph to indicate velocity. If the graph is curved, the slope of
the tangent to the curve indicates the instantaneous velocity.
7. The data to draw the position-time graph are found by calculating the area under the line at several instances and adding
8.0 m [E] to the area in each case. The table shows the results.
t (s)
0.0
1.5
3.0
4.0
5.0
14
G
d (m [E])
The required graph:
8.0
9.5
14
18
22
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
8.
∆t = 2.0 min = 120 s
d1 = 22 m [36° N of E]
d2 = 65 m [25° E of S]
(a) d = ?
d = d1 + d 2
= 22 m + 65 m
d = 87 m
Thus, the total distance travelled is 87 m.
(b) vav = ?
d
vav =
∆t
87 m
=
120 s
vav = 0.73 m/s
The average speed is 0.73 m/s.
(c) Add the vectors together as shown in the illustration:
Use the cosine law to find the magnitude of the resultant displacement:
G2
G 2
G 2
∆d = ∆d1 + ∆d 2 − 2 ∆d1 ∆d 2 cos φ
G2
∆d = (22 m) 2 + (65 m) 2 − 2(22 m)(65 m)(cos 79°)
G
∆d = 65 m
Use the sine law to solve for the angle of the displacement’s direction:
sin θ sin φ
G = G
∆d
∆d 2
G
∆d sin φ
sin θ =
G
∆d 2
 (65 m) sin 79° 
θ = sin −1 

65 m


θ = 79°
The angle of the displacement is 79° − 36° = 43°. Thus, the total displacement is 65 m [43° S of E].
Copyright © 2003 Nelson
Chapter 1 Kinematics
15
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
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
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
G
G
vTW = vTS + vSW
vTW,x = 2.8 m/s
vTW,y = 1.1m/s
G
vTW = vTW,x 2 + vTW,y 2
= (2.8 m/s) 2 + (1.1m/s) 2
G
vTW = 3.0 m/s
 vTW,y
tan θ = 
v
 TW,x
3.



 1.1m/s 
θ = tan −1 

 2.8 m/s 
θ = 21°
When walking toward the starboard, the group’s velocity relative to the water is 3.0 m/s [21° right of fwd].
Use the subscripts S for the ship, W for the water, T for the tourist, and C for the coast. Let the +x direction be north and
the +y direction be east.
G
vTS = 1.1 m/s [E]
G
vSW = 2.8 m/s [N]
G
vWC = 2.4 m/s [N]
G
vTW = ?
G
G
G
G
vTC = vTS + vSW + vWC
vTW,x = 2.8 m/s + 2.4 m/s = 5.2 m/s
vTW,y = 1.1m/s
G
vTW = vTW,x 2 + vTW,y 2
= (5.2 m/s)2 + (1.1m/s) 2
G
vTW = 5.3 m/s
 vTW,y
tan θ = 
v
 TW,x
4.



 1.1m/s 
θ = tan −1 

 5.2 m/s 
θ = 12°
The velocity of the tourist group relative to the coast is 5.3 m/s [12° E of N].
Use the subscripts P for the plane, A for the wind (air), and G for the ground. Let the +x direction be west and the +y
direction be south.
G
vPA = 320 km/h [28° S of W]
G
vAG = 72 km/h [S]
∆t = 2.0 h
G
d PG = ?
G
G
G
vPG = vPA + vAG
vPG,x = (320 km/h)cos28°
vPG,y = (320 km/h)sin28° + 72 km/h
vPG,x = 283 km/h
vPG,y = 222 km/h
Copyright © 2003 Nelson
Chapter 1 Kinematics
59
G
vPG = vPG,x 2 + vPG,y 2
= (283 km/h)2 + (222 km/h)2
G
vPG = 360 km/h
 vPG,y
tan θ = 
v
 PG,x



 222 km/h 
θ = tan −1 

 283 km/h 
θ = 38°
G
Therefore, vPG = 360 km/h [38°S of W] .
G
G
d PG = vPG ∆t
= (360 km/h [38°S of W] )(2 h)
G
d PG =7.2 × 102 km [38°S of W]
The plane’s displacement from Winnipeg is 7.2 × 102 km [38° S of W].
Making Connections
5.
The jet streams are high-speed winds in the high regions of the troposphere. Like other prevailing winds, they generally
move from west to east and influence the motion of most of the major high- and low-pressure weather systems. Pilots can
take advantage of a jet stream only when travelling from west to east (e.g., from Tokyo to Toronto). In travelling
westward, pilots try to avoid the jet stream by flying below it, or north or south of it.
Section 1.5 Questions
(Page 57)
Understanding Concepts
1.
G
Let vKW be the velocity of either kayaker relative to the water and ∆d the magnitude of the displacement across the river,
perpendicular to the shores. For the kayaker that aims straight across the river:
vKW =
∆t1 =
∆d
∆t1
∆d
vKW
For the kayaker who aims upstream at an angle θ to the near shoreline:
vKW sin θ =
∆t 2 =
2.
60
∆d
∆t 2
∆d
vKW sin θ
Since ∆t2 > ∆t1, the kayaker who aims straight across the river reaches the far side first.
Use the subscripts H for the helicopter, A for the wind (air), and G for the ground. Let the +x direction be west and the
+y direction be north.
G
(a) vHA = 55 m/s [35° N of W]
G
vAG = 21 m/s [E]
G
vHG = ?
G
G
G
vHG = vHA + vAG
vHG,x = (55 m/s)cos 35° − 21 m/s
vHG,y = (55 m/s)sin 35°
vHG,x = 24 m/s
vHG,y = 32 m/s
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
G
vHG = vHG,x 2 + vHG,y 2
= (24 m/s) 2 + (32 m/s) 2
G
vHG = 40 m/s
 vHG,y
tan θ = 
v
 HG,x



 32 m/s 
θ = tan −1 

 24 m/s 
θ = 53°
The velocity of the helicopter relative to the ground is 4.0 × 101 m/s [53° N of W].
G
(b) vHA = 55 m/s [35° N of W]
G
vAG = 21 m/s [22° W of N]
G
vHG = ?
G
G
G
vHG = vHA + vAG
vHG,x = (55 m/s)cos 35° + (21 m/s)sin 22°
vHG,y = (55 m/s)sin 35° + (21 m/s)cos 22°
vHG,x = 53m/s
vHG,y = 51m/s
G
vHG = vHG,x 2 + vHG,y 2
= (53m/s)2 + (51m/s)2
G
vHG = 74 m/s
 vHG,y
tan θ = 
v
 HG,x



 51m/s 
θ = tan −1 

 53m/s 
θ = 44°
The velocity of the helicopter relative to the ground is 74 m/s [44° N of W].
3. 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 = 0.75 m/s [across the river]
∆y = 72 m
∆x = 54 m
G
(a) vWG = ?
∆y
∆t =
vSW,y
=
72 m
0.75 m/s
∆t = 96s
Copyright © 2003 Nelson
Chapter 1 Kinematics
61
G
∆x
vWG =
∆t
54 m
=
96s
G
vWG = 0.56 m/s
The speed of the river current is 0.56 m/s.
G
G
G
(b) vSG = vSW + vWG
vSG,x = vWG = 0.56 m/s
vSG,y = vSW = 0.75 m/s
G
vSG = ?
G
vSG = vHG,x 2 + vHG,y 2
= (0.56 m/s)2 + (0.75 m/s) 2
G
vSG = 0.94 m/s
 vSG,y
tanθ = 
v
 SG,x



 0.75 m/s 
θ = tan −1 

 0.56 m/s 
θ = 53°
The swimmer’s velocity relative to the shore is 0.94 m/s [downstream, 53° from the initial shore].
G
(c) vSG = ?
G
G
G
vSG = vSW + vWG
G 2 G 2 G 2
vSG = vSW + vWG
G 2 G 2 G 2
vSG = vSW − vWG
G
vSG = (0.75 m/s)2 − (0.56 m/s)2
G
vSG = 0.50 m/s
G
 vSG
tan θ =  G
v
 WG



4.
 0.50 m/s 
θ = tan −1 

 0.56 m/s 
θ = 42°
The direction in which she should have aimed to land directly across from her starting position is upstream, 42° from
the shore.
Use the subscripts P for the plane, A for the wind (air), and G for the ground. Let the +y direction be south and the +x
direction be east.
G
∆ d = 1.4 × 103 km [43° E of S]
∆t = 3.5 h
G
vAG = 75 km/h [E]
G
vPA = ?
62
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
G
d
G
vPG = PG
∆t
1.4 × 103 km [43° E of S]
=
3.5 h
G
2
vPG = 4.0 ×10 km/h [43° E of S]
G
G
G
vPG = vPA + vAG
G
G
G
vPA = vPG − vAG
To determine the x- and y-components:
vPA,x = (4.0 ×10 2 km/h) sin 43° − 75 km/h
vPA,y = (4.0 × 102 km/h) cos 43°
vPA,x = 2.0 × 102 km/h
vPA,x = 2.9 × 102 km/h
To calculate the velocity of the plane:
G
vPA = vPA,x 2 + vPA,y 2
= (2.0 × 102 km/h)2 + (2.9 × 102 km/h)2
G
vPA = 3.5 × 102 km/h
 vPA,x
tan θ = 
 vPA,y





 2.0 × 102 km/h 
θ = tan −1 

2
 2.9 × 10 km/h 
θ = 34°
The velocity of the plane relative to the air is 3.5 × 102 km/h [34° E of S]
Applying Inquiry Skills
5. (a) vtrain = 64 km/h
G
vrain = ?
Referring to Figure 9 on page 57 of the text, we see that the drops appear to be at an angle θ below the horizontal such
that:
2 hand spans
tan θ =
1 hand span
tan θ =
Copyright © 2003 Nelson
2
1
Chapter 1 Kinematics
63
As shown in the diagram:
G
vtrain
tan θ = G
vrain
G
vtrain
G
vrain =
tan θ
=
64 km/h
2
 
1
G
vrain = 32 km/h
The speed of the raindrops is 32 km/h.
(b) Estimating an angle by using hand spans is a major source of error. Another source of error, not shown in the text, is
that different drops move at different angles. (This corresponds to the fact that lighter water droplets have a lower
terminal speed than heavier ones.) Another source of error is knowing or estimating the speed of the train.
Making Connections
6.
Refer to the solution to question 5(a) above. From the video of the weather report, estimate the angle θ, and then apply the
equation:
G
vwind
tan θ = G
vrain
G
G
vwind = vrain tan θ
G
vwind = (32 km/h)( tan θ )
CHAPTER 1 LAB ACTIVITIES
Investigation 1.3.1: Comparing Terminal Speeds
(Page 58)
This investigation is one of the few in this text in which the students are required to display all of the Inquiry Skills listed at
the top of the page. Thus, it provides a good opportunity for assessing students’ abilities to design, carry out, and report on a
relatively easy experiment.
Question
(a) A typical question is “How does the terminal speed of a falling coffee filter depend on the mass of the filter?”
Hypothesis/Prediction
(b) Since the surface area of the flat part of the falling filter in contact with the air is constant for all three trials, it is
reasonable to say that the terminal speed increases as the number of filters added increases. As the mass increases the
downward force of gravity increases, but the upward force of air resistance may not increase in proportion to the increase
in gravity.
(c) A typical graph of the predicted motions:
64
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Experimental Design
(d) The main safety concern is to be sure that electric wires that may be along the floor are concealed so nobody trips over
them.
(i) Position the motion sensor on the floor so it faces upward. Connect the sensor to the computer and set up the program
so it can record the motion (speed-time data and graph) of the filter dropped from a preset position. Hold the first filter
directly above the sensor, and then let it drop flat side down toward the sensor. Repeat for accuracy.
(ii) Add a second filter inside the first one, and then repeat step (i).
(iii) Add a third filter inside the first one, and then repeat step (i).
(e) Refer to the text.
Materials
(f) The materials and apparatus needed per group are:
• 3 flat-bottom coffee filters
• motion sensor
• computer with software to operate the motion sensor
Analysis
(g) Students can generate their own graph based on their observations, or they can have the computer generate the graph.
The graph above in part (c) shows the shapes of the curves on a typical graph of the results.
Evaluation
(h) Each student’s evaluation depends on his or her hypothesis and prediction.
(i) As students soon discover, one of the biggest sources of error in this investigation is trying to get the filters to drop
straight downward. The air resistance on a filter is maximum when the bottom surface remains horizontal during the fall.
Furthermore, if the filter is not horizontal, it moves away from the vertical and the motion might not even be picked up by
the sensor. It takes practice to be able to allow the filter to fall horizontally. Repeating trials helps to minimize the effects
of this type of error.
Another source of error is moving the filter vertically while holding it as the sensor is activated. This error can be
minimized by activating the sensor just prior to letting the filter(s) drop.
(j) One part of the experiment would involve spheres of equal diameter but different masses all falling through the same
liquid, say water at room temperature. Rather than using a motion sensor, students could use a metre stick and a stopwatch
to collect data to determine the approximate terminal speed.
The second part of the experiment could involve a single sphere falling through liquids of different density but the
same temperature, or falling through a single liquid at different, controlled temperatures.
Investigation 1.4.1: Investigating Projectile Motion
(Pages 58–60)
The air table shown on page 59, Figure 1 of the text is highly recommended for studying projectile motion. With good
organization and student cooperation, an entire class can perform the observations in a single 70-min period, even if there is
only one air table. The instructions have been written so that after students gather data for one motion, they can begin
analyzing the data as other groups use the air table, and so on.
The table should be elevated a small amount at the back of the table, say between 5 cm and 8 cm higher than the front. If
the table is set at too high an angle, the puck will accelerate too quickly and may crash into the lower edge. Such collisions
must be avoided.
Question
See page 59.
Hypothesis
(a) The “correct” answers to the questions are given here, although students may not be expected to know the answers before
the investigation is complete.
(i) The direction of the acceleration of the puck is down the inclined plane. This is true as long as the horizontal
component of the motion is at a constant speed (or zero), which in turn is true in the absence of any friction.
(ii) One way to show the independence is to find the magnitude and direction of the acceleration of the puck on the
inclined plane. If that acceleration is straight down the inclined plane and is constant in magnitude for all motions on the
Copyright © 2003 Nelson
Chapter 1 Kinematics
65
plane, then the horizontal motion must not be affected by the vertical motion. Another way is to compare the individual
horizontal displacements covered in equal time intervals, and then compare the observations with the individual vertical
components of the displacements covered in equal time intervals.
Materials
The instructions suggest three sheets of construction paper per student, but if supplies are limited, only one sheet per student
would be needed if the first two motions were placed on one side of the sheet and the third motion were placed on the reverse
side.
Procedure
1.
A typical calculation of the angle of the table is given below.
height of front of table = 3.7 cm
height of rear of table = 9.8 cm
change in elevation, ∆y = 9.8 cm − 3.7 cm = 6.1 cm
length of table (front to rear), L = 61 cm
Let the angle of the table above the horizontal be θ.
∆y
sin θ =
L
∆y
θ = sin −1
L
6.1
cm
= sin −1
61 cm
θ = 5.7°
Steps 2–6: The three different motions and mathematical manipulations of the vectors are shown in the text. When
students perform vector subtractions by completing vector scale diagrams, it is important to move the vectors exactly
parallel to themselves. In this investigation, that is fairly easy to do because the vectors are close together. However,
students may benefit from the Learning Tip about transferring vectors on pages 264–265 of the text (in Investigation 5.3.1
where the vectors are relatively far apart).
Analysis
(b) Within experimental error, the magnitude and direction of the accelerations should be the same for all three motions.
(c) A sample calculation is shown here using the sample shown in Procedure 1.
a y = g sin θ
= (9.8 m/s 2 )(sin 5.7°)
2
a y = 0.97m/s
In this example, the acceleration down the table is 0.97 m/s2.
(d) A sample calculation is shown here for any one of the motions. Assume that the average of the accelerations for motion 3
is 1.05 m/s2 [down the inclined plane].
difference in values
% difference =
× 100%
average of values
=
1.05 m/s 2 − 0.97 m/s 2
1
(
1.05 m/s 2 + 0.97 m/s 2
)
× 100%
2
% difference = 8%
(e) (i) The direction of the acceleration of the projectile is straight down the inclined plane.
(ii) Since the acceleration in all three cases is the same (i.e., down the inclined plane and equal in magnitude), the
projectile has acceleration in the vertical plane that is independent of whether the initial velocity was zero, in the +x
direction, or directed at an angle above the horizontal.
Another way of showing the independence of the vertical and horizontal components is to compare the horizontal
displacements between the dots for motions 2 and 3 with the vertical displacements between the dots. (For motion 2, this
analysis is similar to what is shown in the text, page 42, Figure 4.) The horizontal displacements are constant, while the
vertical displacements are constantly changing.
66
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Evaluation
(f) Answers will depend on the initial hypothesis.
(g) The main random sources of error and how to minimize them are:
• levelling the table horizontally (Use a level to be sure the left-to-right orientation of the table is level.)
• twisting effects of the cord (Be sure that the cord connected to the air puck is not twisted.)
• measuring the angle of the air table (Be sure the ruler is vertical and reduce parallax error in measuring the elevation of
the front and rear edges of the table above the bench top. )
• drawing and measuring the vectors (Use a sharp pencil to draw the vectors, exercise care in drawing the tangents to the
curve, and measure the vectors as accurately as possible without parallax error by viewing the ruler straight on.)
• transferring the vectors parallel to themselves to perform vector subtractions (Use one or two rules as accurately as
possible.)
• rounding errors (Interim answers should be kept in the calculator and answers should be rounded off at the end of the
calculations.)
The main source of systematic error is in using dots on the construction paper created when the puck was still being
pushed by hand. The puck is a projectile only after that force has been removed. If the dots are included in the vector
analysis, both the magnitude and direction of the acceleration will be wrong. To minimize this error, do not use the first
3 or 4 dots of the motion.
Synthesis
(h) Theoretically, it is better to use smaller values of ∆t because we are trying to find the instantaneous acceleration of the
puck at various instants. In practice, the advantage of using relatively small ∆t values is that vector subtractions can be
carried out more accurately. However, if the ∆t values become too small, the percent uncertainty in measuring the lengths
G
of the vectors, especially the ∆v vectors, becomes quite high. So the best value of ∆t is small but not too small.
(i) The question asks for percent difference rather than percent error because both values result from experimental
measurements.
(j) The diagram below shows the situation. The acceleration due to gravity has a magnitude of g, which has two components,
ay perpendicular to the inclined plane, and ax parallel to the inclined plane.
From the diagram:
θ1 + β1 = 90°
θ2 + β2 = 90°
However, β1 = β2, therefore θ1 = θ2 = θ.
Now the component of g parallel to the inclined plane is g sinθ.
Lab Exercise 1.4.1: Hang Time in Football
(Pages 60–61)
The following references are highly recommended for teachers and motivated students.
“The Physics of Kicking a Football” by Peter J. Brancazio. The Physics Teacher, October 1985, pp. 403–407. Although this
article is relatively old, it still acts as a standard for this topic.
“Physics buys a ticket to the Super Bowl” by Steven Strauss. The Globe and Mail, Saturday, January 28, 1995, Science and
Physics section.
(a) Answers will vary greatly and are not expected to be accurate. Here is a potential response: The range of angles will go
from about 45° to 60° in order to maximize both the hang time and the horizontal range.
Copyright © 2003 Nelson
Chapter 1 Kinematics
67
Analysis
(b)
(c) The angles between 45° and 60° yield horizontal ranges from 49 m to 60 m and hang times from 3.8 s to 4.6 s.
The horizontal range drops substantially at launch angles higher than 60°, while hang time continues to rise.
(d) For an “ideal” projectile, the maximum horizontal range occurs for a launch angle of 45°, and the maximum hang time
occurs as the launch angle approaches 90°. (Refer to the results of the Try This Activity on page 49 of the text.) Between
45° and 90° there is a compromise between these two variables. A good value of both range and hang time occurs at
launch angles around 60°.
For a football punt, the maximum horizontal range is reached at launch angles just below and up to 45°, and hang
time increases as the launch angle increases. Thus, the best launch angle is likely close to that of an “ideal” projectile, in
the region of 55° to 60°. (In actual football games, hang time is more important for punts than kickoffs because in
kickoffs all the players on the kicking team are running at nearly top speed when the ball is kicked, whereas with punts
the players are blocked by the opposing team and need more time to get downfield. Thus, the best launch angles for
kickoffs are around 45° and the best angles for punts are around 60°.
Evaluation
(e) The answer depends on the prediction. Accept any reasonable explanation.
(f) The main assumptions are:
• the data were analyzed for games in which the wind was negligible
• the TV camera was located so its line of sight was along a field line so the launch angle could be estimated from a
straight-line view
• the height above the field that the ball was kicked and caught was approximately constant
• the ball was kicked straight downfield, parallel to the sidelines
(g) Sources of random error are:
• estimating the horizontal range of the ball in each trial
• estimating the angle of each launch
• measuring the time the ball was in the air in each trial
• any angle between the camera’s line of sight and the kick
• any lateral motion of the ball relative to the sidelines
The main source of systematic error occurs when estimating the angle of launch based on the view of a camera that is at a
level well above the playing field.
(h) Some ways of obtaining accurate data are:
• use an actual activity that can be analyzed using a combination of direct measurements and measurements from a video
or a digital camera
• have several students aid in gathering data (e.g., some students can be responsible for determining the horizontal range
while other students are responsible for obtaining a video to determine the launch angle and measuring hang time)
Synthesis
(i) The closed stadium would prevent the source due to error of any wind blowing outdoors.
(j) Analysis helps to determine how to maximize certain variables, and then the athletes can practise until they approach the
ideal. For example, punters can practise punting so that the launch angle of the football is between 55° and 60°.
68
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
CHAPTER 1 SUMMARY
(Page 62)
Make a Summary
This illustration shows the start of a projectile motion diagram with some of the labels suggested. Many of the concepts and
equations in Chapter 1 are included in the diagram.
CHAPTER 1 SELF QUIZ
(Page 63)
True/False
1.
2.
3.
4.
5.
6.
7.
8.
F This motion provides an example of one-dimensional motion (not two-dimensional motion) because the ball has
travelled along a single line even though it has reversed directions
F The magnitude of the velocity just before landing is equal to (not greater than) the magnitude of the initial velocity as
the ball leaves your hand.
F The acceleration at the top of the flight is equal to the acceleration due to gravity and is constant throughout the ball’s
motion.
T
F A jogger running four laps around a circular track at 4.5 m/s is not an example of motion with constant velocity because
the direction is always changing even though the magnitude is constant.
T
T
F At sea level, the magnitude of the acceleration due to gravity at Miami, Florida, is less than that at St. John’s,
Newfoundland, because Miami is closer to the equator where the distance to Earth’s centre is greater.
Copyright © 2003 Nelson
Chapter 1 Kinematics
69
CHAPTER 1 SUMMARY
(Page 62)
Make a Summary
This illustration shows the start of a projectile motion diagram with some of the labels suggested. Many of the concepts and
equations in Chapter 1 are included in the diagram.
CHAPTER 1 SELF QUIZ
(Page 63)
True/False
1.
2.
3.
4.
5.
6.
7.
8.
F This motion provides an example of one-dimensional motion (not two-dimensional motion) because the ball has
travelled along a single line even though it has reversed directions
F The magnitude of the velocity just before landing is equal to (not greater than) the magnitude of the initial velocity as
the ball leaves your hand.
F The acceleration at the top of the flight is equal to the acceleration due to gravity and is constant throughout the ball’s
motion.
T
F A jogger running four laps around a circular track at 4.5 m/s is not an example of motion with constant velocity because
the direction is always changing even though the magnitude is constant.
T
T
F At sea level, the magnitude of the acceleration due to gravity at Miami, Florida, is less than that at St. John’s,
Newfoundland, because Miami is closer to the equator where the distance to Earth’s centre is greater.
Copyright © 2003 Nelson
Chapter 1 Kinematics
69
9.
F The quadratic formula must be used to solve problems involving the quadratic equation of the form a ∆t 2 + b∆t + c = 0 ,
which has the solution ∆t =
−b ± b 2 − 4ac
.
2a
10. T
G
G
11. F If vAB = 8.5 m/s [E], then vBA = 8.5 m/s [W] .
Multiple Choice
12. (a) A ball is tossed vertically upward from your hand: the initial position is your hand; +y is upward.
13. (d) A rubber stopper is dropped from your raised hand: the initial position is your hand; +y is upward.
14. (b) A ball is tossed vertically upward from your hand: the initial position is your hand; +y is downward.
15. (d) A cart is released from rest at the top of a ramp: the initial position is at the top of the ramp; +y is up the ramp.
G
16. (c) vi =25 m/s [E]
G
a = 2.5 m/s2 [W] = –2.5 m/s2 [E]
∆t = 2.0 × 101 s
G
vf = ?
G G
vf − vi
G
aav =
∆t
G G G
vf = vi + aav ∆t
= 25 m/s[E] + (−2.5 m/s 2 [E])(2.0 × 101 s)
G
vf = −25 m/s [E]or 25 m/s [W]
17. (e) ax = 2.5 m/s2
ay = 6.2 m/s2
θ=?
tan θ =
ay
ax
 6.2 m/s 2 
θ = tan −1 
2 
 2.5 m/s 
θ = 68°
The direction of the acceleration is 68° N of E.
18. (e) We can use components to solve the problem.
G
vi = 7.2 m/s [N]
G
vf = 7.2 m/s [W]
∆t = 2.0 s
Let +x be west and +y be north.
∆v y = vfy − viy
∆v x = vfx − vix
∆v x = 7.2 m/s
∆v y = −7.2 m/s
G
∆v = ∆v x 2 + ∆v y 2
= (7.2 m/s)2 + (−7.2 m/s)2
G
∆v = 10.2 m/s
tan θ =
∆v y
∆vx
 7.2 m/s 
θ = tan −1 

 7.2 m/s 
θ = 45°
G
Thus, ∆v = 10.2 m/s [45° W of N].
70
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.
Copyright © 2003 Nelson
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 = ?
74
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=?
76
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
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].
78
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
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.
80
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(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 =
82
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
84
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
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
86
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 = ?
88
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
CHAPTER 2 DYNAMICS
Reflect on Your Learning
(Page 68)
G
G
G
G
1. (a) A is an applied force or tension force, B is a normal force, C is the force of friction, and D is the force of gravity on
the loaded toboggan.
(b)
2. (a)
(b) The vector sum of all the forces must be zero because the box is initially at rest and remains at rest.
3.
4.
The applied force on the two books would have to be two times as great as that on one book in order to achieve a constant
velocity. If the mass is two times as great, then the normal force pushing upward on the books is two times as great, so the
kinetic friction acting against the motion of the two books is two times as great. Thus, the force needed to overcome
friction and maintain constant velocity must be two times as great.
Copyright © 2003 Nelson
Chapter 2 Dynamics
93
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.”
94
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
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
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
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
CHAPTER 2 LAB ACTIVITIES
Activity 2.2.1: Static Equilibrium of Forces
(Page 112)
As stated in the text, page 79, an object in static equilibrium is at rest and has a net force of zero acting on it. In this activity,
students use vertical and horizontal components of vectors to test the condition of static equilibrium of a common point,
referred to as the origin.
Students will likely recall from Grade 11 physics that they can calculate the force of gravity on a mass by using the
gravitational field strength. Thus, for example, the magnitude of the force of gravity on a 100-g mass, to two significant digits,
is:
Fg = mg
= (0.10 kg)(9.8 N/kg)
Fg = 0.98 N
Materials
See page 112.
Procedure
1.
After setting up the apparatus according to the instructions and measuring angles a and d, the students can perform the
calculations and compare the components. Following is a start to the calculations.
G
G
G
Let the tensions in the strings attached to masses m1, m2, and m3, be FT1 , FT2 , and FT3 , respectively. In the vertical plane
with +y up:
?
ΣFy = ma y = 0
?
FT1 sin a + FT2 sin d − F3 = 0
?
FT1 sin a + FT2 sin d = F3
G
G
G
In other words, does the sum of the upward components of FT1 and FT2 equal the downward component of FT3 ? Within
experimental error, the answer is yes.
In the horizontal plane, with +x to the right:
?
ΣFx = max = 0
?
− FT1 cos a + FT2 cos d + 0 = 0
?
FT2 cos d = FT1 cos a
2.
3.
G
G
In other words, are the horizontal components of FT1 and FT2 equal? Within experimental error, the answer is yes.
The data collected are analyzed as described in #1 above.
After the students follow the instructions and measure angles a, d, and f, they can perform the calculations as shown
below.
Considering the vertical components of the forces, with +y up:
?
ΣFy = ma y = 0
?
FT1 sin a + FT2 sin d − F3 cos f = 0
136 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Considering the horizontal components of the forces, with +y to the right:
?
ΣFx = max = 0
?
4.
FT2 cos d + FT3 sin f − FT1 cos a = 0
Students should find that in each case the components add to zero.
The data collected are analyzed as described in question 3 above.
Analysis
(a) Static equilibrium occurs for a stationary object when the net force acting on that object is zero.
(b) Friction between the strings and the pulleys can affect the tension in the strings, and thus affect the calculations. In step 1,
for example, if there is friction involving the strings holding m1 and m2, then m3 would have to be larger than is predicted
by theory to achieve static equilibrium.
Evaluation
(c) The following ways would help to improve accuracy:
• use very smooth strings and pulleys to reduce friction to a minimum
• check carefully to be sure the circular protractor is properly aligned horizontally
• eliminate parallax error as much as possible when checking to be sure the origin of the protractor coincides with the
common point of the three strings
Investigation 2.4.1: Measuring Coefficients of Friction
(Page 113)
The intent of this investigation is to have students design and carry out their own controlled experiment to determine the
coefficients of static and kinetic friction. (Notice that all of the Inquiry Skills are highlighted.)
Other parts of the text that relate directly to this investigation are Sample Problem 2 on page 100, question 8 on page 101,
and question 8 on page 107.
Question
(a) A typical question is: How do the coefficients of static and kinetic friction for a running shoe on wood compare to each
other and to other objects on the same wood?
Hypothesis/Prediction
(b) For each set of surfaces in contact, the coefficient of static friction should be greater than the coefficient of kinetic friction
because objects tend to move more easily once their motion has begun. Most running shoes on wood likely have higher
coefficients of friction than other objects, especially wooden or plastic blocks or books, because rubbery substances tend
to stick to other surfaces more. For example the coefficient of static friction of rubber on wood may approach 1.0.
Experimental Design
(c) Refer to the answer to Practice question 8 earlier in this manual (and text page 101) with sight changes. In that question
only kinetic friction was investigated, whereas here both static and kinetic friction are investigated. Thus, students must
work together carefully to obtain two sets of data for each pair of surfaces tested, one in which one end of the wooden
board is raised until the object in it begins to slide down the board (to determine the coefficient of static friction), and the
other to lower the end of the board slightly until the object appears to move down the slope at a constant velocity (to
determine the coefficient of kinetic friction).
The safety concerns for this investigation are rather simple. Students should be sure that any breakable object tested
does not fall off the raised board or the lab bench. Also, students should ensure that the raised board does not drop onto
any other object or somebody’s fingers.
(d) The student report should include the Investigation title, Question, Hypothesis/Prediction, Experimental Design (approved
by the teacher), calculations (samples of which are shown below), a summary table of data and calculations (a sample of
which is shown below), Analysis, Evaluation, and Synthesis.
Copyright © 2003 Nelson
Chapter 2 Dynamics 137
Sample calculations for a running shoe on a wooden board:
For the coefficient of static friction, the rise of the board = 1.63 m and run of the board = 1.81 m.
rise
µS =
run
1.93 m
=
1.61 m
µS = 0.901
For the coefficient of kinetic friction, the rise of the board = 1.47 m and run of the board = 1.95 m.
rise
µS =
run
1.47 m
=
1.95 m
µS = 0.754
The following is a sample data table for objects on a wooden board:
Object Tested
running shoe
math text
varnished wood
Type of Friction
static
kinetic
static
kinetic
static
kinetic
Rise (m)
1.63
1.47
0.95
0.84
0.81
0.67
Run (m)
1.81
1.95
2.24
2.29
2.30
2.34
Coefficient of Friction
µS = 0.901
µK = 0.754
µS = 0.42
µK = 0.37
µS = 0.35
µK = 0.29
Materials
See page 113 of the text.
Analysis
(e) Questions and answers may vary. Three examples are given here.
(i) For each object tested, how does the coefficient of kinetic friction compare to the coefficient of static friction? Answer:
In all cases, the coefficient of kinetic friction was less than the coefficient of static friction. The percentage drop in the
coefficient for the running shoe is:
µ − µK
× 100%
% drop = S
µS
0.901 − 0.754
× 100%
0.901
% drop = 16.3%
For the other two examples given, the corresponding drops are 12% and 17%.
(ii) How do the coefficients of static and kinetic friction relate to the materials in contact? Answer: Rubber-soled shoes
have much higher coefficients of static and kinetic friction than wood, plastic, or most other materials on the same surface
(wood in this investigation).
(iii) What procedure could be used to check the validity of the coefficients of friction found in this investigation? Answer:
The coefficients of static and kinetic friction for the same sets of surfaces can be found by pulling horizontally on the
moveable object with a force sensor or a force meter attached and the board set up horizontally. The force required to just
start the stationary object moving can be used to calculate the coefficient of static friction, and the force needed to keep
the object moving at a constant velocity can be used to determine the coefficient of kinetic friction. The equations needed
are:
F
F
µS = S,max
and
µK = K
mg
mg
=
138 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Evaluation
A major source of random error is that the running shoe 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 velocity. The effects of these errors can be minimized by taking an average of several
readings. 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. This error can be minimized by looking straight at the metre stick and checking to be sure the metre stick is
as vertical as possible when measuring the rise. A minor systematic error occurs if the metre stick has a worn end. If this is the
case, the error can be added to the measured quantities.
Synthesis
(g) Students can use a mathematical derivation to verify that the statement is true. They should begin with a set of diagrams
like those in the text, Figure 6 on page 100, but with the force of kinetic friction rather than static friction. Then they
should analyze the forces by considering the components perpendicular and parallel to the inclined plane, as shown in
Sample Problem 2, page 100. As shown there, mg cancels out.
Students can verify the statement experimentally by carrying out the same steps they used in this investigation, but
adding more mass to the object’s tested. (For example, they could secure two books together and repeat the
measurements.
CHAPTER 2 SUMMARY
Make a Summary
(Page 114)
Individual summaries may vary somewhat, but the FBDs and basic facts should resemble what is depicted below. Urge the
students to add enough detail to help them understand and remember the contents of the chapter.
Copyright © 2003 Nelson
Chapter 2 Dynamics 139
CHAPTER 2 SELF QUIZ
True/False
1.
2.
3.
T
F The magnitude of the tension in the string is 16 N.
F The magnitude of the normal force is now greater than 155 N because the y-component of the applied force would add
to the force due to gravity. With +y up:
ΣFy = 0
FN + (− Fapp sin θ ) + (−mg ) = 0
FN = Fapp sin θ + mg
4.
F The coefficient of kinetic friction still equals 0.18 because it is independent of the force of gravity acting against the
skier’s motion.
5. F It is possible for an object to be travelling eastward while experiencing a net force that is westward. The object would
be decelerating.
6. T
7. F Maximum static friction (i.e., starting friction) tends to be greater than kinetic friction.
8. F One possible SI unit of weight is the newton.
9. T
10. T
11. F Fictitious forces must be invented to explain observations in an accelerating frame of reference.
12. (a) According to Newton’s first law of motion, the net force on the person must be zero because the person is moving with
a constant velocity.
13. (d) A force exerted by the air and the downward force of gravity are acting on the football.
14. (b) A skier of mass m is sliding down a snowy slope that is inclined at an angle φ above the horizontal. With +y defined as
perpendicular to and upward from the hillside, the magnitude of the normal force on the skier is:
ΣFy = 0
FN + (−mg cos φ ) = 0
FN = mg cos φ
15. (d) The reaction force is the upward force gravity on Earth exerted by the can.
16. (b) The magnitude of the minimum horizontal force that must be applied is 61 N using the following equations, with +y up
and +x the direction of the applied force:
ΣFy = 0
FN − mg = 0
FN = mg
ΣFx = 0
Fapp + (− Ff ) = 0
Fapp = µS FN
= µS mg
= (0.65)(9.5 kg)(9.8 m/s 2 )
Fapp = 61N
17. (a) During the time interval that the spring acts on the double cart, the double cart exerts an equal but opposite force of
2.0 N [W] on the single cart.
18. (b) During the spring interaction, the net force acting on the single cart is 2.0 N [W].
140 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
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
CHAPTER 3 CIRCULAR MOTION
Reflect on Your Learning
(Page 120)
1. (a) The FBD is shown in below. The choice of positive directions is explained in (b) and (c) below. (At this stage, it is
acceptable to draw the diagram with +x down the inclined plane and +y perpendicular to that axis. However, that
choice would not be convenient for analysis of the truck’s acceleration.)
(b) This is not an easy question to analyze the first time it is seen. The truck is not sliding down a hill; rather it is moving
in a horizontal arc, experiencing acceleration toward the centre of radius of that arc. Thus, we will let the direction of
the horizontal acceleration be called +x, which means that the +y direction is up. Now:
ΣFy = ma y = 0
FN cos φ − mg = 0
FN =
mg
cos φ
mg
.
cos φ
(c) The net force acting on the truck is in the same direction as the acceleration, as described in (b) above. It is caused by
the horizontal component of the normal.
2. (a) The system diagram in the noninertial frame of reference
The normal force acting on the truck is
The FBD of the middle bead in the noninertial frame of reference
Copyright © 2003 Nelson
Chapter 3 Circular Motion 155
(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
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
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
3.4 SATELLITES AND SPACE STATIONS
PRACTICE
(Pages 147–148)
Understanding Concepts
GmE
1
, so as the radius increases the speed
, the speed of the satellite is proportional to
r
r
decreases in a nonlinear fashion.
(b) rM = 3.84 × 105 km = 3.84 × 108 m
mE = 5.98 × 1024 kg
GmE
vM =
rM
1. (a) From the equation v =
=
(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(5.98 ×1024 kg)
3.84 × 108 m
vM = 1.02 ×103 m/s
The speed of the Moon is 1.02 × 103 m/s. The speed of the HST is 7.46 × 103 m/s (from Sample Problem 1, page 146).
Thus, as the distance of a satellite from Earth’s centre increases, the speed of the satellite decreases.
2. r = 4.50 × 105 m + 6.38 × 106 m = 6.83 × 106 m
(a) mE = 5.98 × 1024 kg
v=
=
GmE
r
(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(5.98 ×10 24 kg)
6.83 ×106 m
v = 7.64 × 103 m/s
The speed of the ISS is 7.64 × 103 m/s.
(b) Using the data from (a) above:
2π r
T=
v
2π (6.83 ×106 m)
=
7.64 × 103 m/s
= 5.62 × 103 s
3.
T = 1.56 h
It takes the ISS 1.56 h to make one revolution around Earth.
The force of gravity causes the centripetal acceleration.
GmS mE mSv 2
=
r
r2
GmE
r= 2
v
180 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Since v =
2π r
, we can substitute for v:
T
GmE
r=
2
 2π r 


 T 
r=
r3 =
r=
( )
GmE T 2
2 2
4π r
T GmE
2
4π 2
3
T 2GmE
4π 2
4. (a) Since a digital TV satellite follows a geosynchronous orbit, it has a period of 24 hours or 8.64 × 104 s.
(b) mE = 5.98 × 1024 kg
rE = 6.38 × 106 m
Using the equation derived in question 3(b):
r=
3
=
3
T 2 GmE
4π 2
(8.64 × 104 s) 2 (6.67 ×10 −11 N ⋅ m 2 /kg 2 )(5.98 × 10 24 kg)
4π 2
r = 4.22 × 107 m
h = r − rE
= 4.22 × 107 m − 6.38 × 106 m
h = 3.59 × 10 4 m
The altitude of the orbit above the surface of Earth is 3.59 × 104 km.
Applying Inquiry Skills
5. (a) Speed versus mass: v ∝ m
(b) Speed versus radius: v ∝
Copyright © 2003 Nelson
1
r
Chapter 3 Circular Motion 181
Making Connections
6. (a) r = 5.7 × 1017 m
v = 7.5 × 105 m/s
v=
Gm
r
v2r
G
(7.5 ×105 m/s) 2 (5.7 ×1017 m)
=
6.67 ×10 −11 N ⋅ m 2 /kg 2
m=
m = 4.8 × 1039 kg
The mass of the black hole is 4.8 × 1039 kg.
(b) mS = 1.99 × 1030 kg
m
4.8 ×1039 kg
=
mS 1.99 × 1030 kg
m
= 2.4 × 109
mS
Thus, the ratio of the black hole’s mass to the Sun’s mass is 2.4 × 109:1. According to the estimate calculated here, this
black hole has a mass equivalent to 2.4 billions Suns. This incredibly huge mass implies than the black hole grew more
and more massive by sucking in the materials from numerous surrounding stars, so its makeup is likely stellar and
interstellar material.
(c) Any body that is relatively small yet is so massive is also extremely dense. Probably the word “hole” does not describe
an object with such high density.
PRACTICE
(Pages 150–151)
Understanding Concepts
7.
m = 56 kg
Let the +y direction be upward.
ΣFy = ma y
FN − mg = ma y
FN = m( g + a y )
(a) ay = –3.2 m/s
2
FN = 56 kg(9.8m/s 2 − 3.2m/s 2 )
FN = 3.7 × 10 2 N
The magnitude of the apparent weight of the student when the acceleration is downward is 3.7 × 102 N.
(b) ay = 3.2 m/s2
FN = 56 kg(9.8m/s 2 + 3.2m/s 2 )
FN = 7.3 × 10 2 N
8.
The magnitude of the apparent weight of the student when the acceleration is downward is 7.3 × 102 N.
The astronauts are in free fall toward Earth’s centre, just as the ISS and all other objects in it, so they appear to float in
space.
182 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
9.
r = 450 km = 4.5 × 105 m
rE = 6.38 × 106 m
(a) m = 64 kg
FG =
=
GmmE
(r + rE )2
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(64 kg)(5.98 × 1024 kg)
(4.5 × 105 m + 6.38 × 10 6 m) 2
FG = 5.5 × 102 N
The magnitude of the gravitational force is 5.5 × 102 N.
F
5.5 × 10 2 N
(b) G =
mg (64 kg)(9.8 N/kg)
=
5.5 × 10 2 N
6.3 × 10 2 N
FG
= 0.87
mg
Therefore, the force calculated in (a) is 87% of the astronaut’s Earth-bound weight.
3.24 × 103 m
= 1.62 × 103 m
10. (a) r =
2
g = 9.80 m/s2
mv 2
= mg
r
v = rg
= (1.62 × 103 m)(9.80 m/s 2 )
v = 126 m/s
The speed of an astronaut would be 126 m/s.
(b) Using the data from (a):
2π r
v=
T
2π r
T=
v
2π (1.62 × 103 m)
=
126 m/s
T = 80.8s
The period of rotation of the spacecraft is 80.8 s.
Applying Inquiry Skills
11. Any experiment that would be done on Earth’s surface to determine your weight could be done on the interior surface
of the rotating spacecraft. For example, if you had a bathroom scale calibrated in newtons, as it would be on Earth
(i.e., 9.8 N/kg), you could measure your own weight. Alternatively, if you had a spring scale with the same calibration,
you could find the weight of a standard 1.0-kg mass. If the frequency of rotation is too high, the artificial gravitational
field would be greater than 9.8 N/kg, and adjustments to the frequency would be required.
Making Connections
12. In each case, the exercise program must consider the person’s bones, muscles, respiratory system, and circulatory system.
Thus, for example, the program would involve a balanced, health-conscious diet, plenty of stretching and strengthening
exercises, and plenty of cardiovascular exercise. Since the ISS is so small and has no artificial gravity, an exercise
program there must involve machines that provide resistance to forces exerted by the astronaut’s arms, legs, etc. On a
large, rotating spacecraft to Mars, however, the exercises can be done along the interior surface of the craft where the
artificial gravity provides the normal force for many of the exercises. Of course, exercise machines could still be used,
and some of them could resemble ones used here on Earth’s surface.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 183
Section 3.4 Questions
(Page 151)
Understanding Concepts
1.
2.
3.
A space station is a satellite when it revolves around a central body (e.g., the ISS travelling in orbit around Earth).
A space station is not a satellite when it is travelling from one body toward a second, distant body (e.g., when it is on a
mission to Mars).
1
The order of decreasing speed corresponds to increasing distance from Earth’s centre (since v ∝
). Thus, the required
r
order is: a weather-watch satellite; the ISS; a geosynchronous satellite; the Moon.
1.23
mM =
mE
100
Since gravity causes the centripetal acceleration:
GmS mE mSv 2
=
r
r2
GmE
GmM
, where v1 is the speed of the satellite in orbit around Earth and v2 =
, where v2 is the speed of
r
r
the satellite in orbit around the Moon.
Thus, v1 =
The ratio of the speeds is:
v1
=
v2
v1
=
v2
Since
GmE
r
GmM
r
mE
mM
mE 100
=
,
mM 1.23
v1
100
=
v2
1.23
v1 9.02
=
v2 1.00
The ratio of the speed of an artificial satellite in orbit around Earth to the speed of a similar satellite in orbit around the
Moon is 9.02:1.00.
4. T = 1.88 Earth years = 5.93 × 107 s
r = 2.28 × 108 km = 2.28 × 1011 m
2π r
(a) v =
T
2π (2.28 ×1011 m)
=
5.93 ×107 s
v = 2.42 × 104 m/s
The orbital speed of Mars (relative to the Sun) is 2.42 × 104 m/s. (The speed to several significant digits is left in the
calculator for the next part of the question.)
184 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b)
mv 2
r
r2
2
v r
mS =
G
(2.42 × 104 m/s)2 (2.28 × 1011 m)
=
6.67 × 10−11 N ⋅ m 2 /kg 2
GmS m
=
mS = 1.99 × 1030 kg
The mass of the Sun is 1.99 × 1030 kg.
5. (a) v = 1.05 × 104 km/h = 2.92 × 103 m/s
mE = 5.98 × 1024 kg
GmE m mv 2
=
r
r2
GmE
r= 2
v
(6.67 ×10 −11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
=
(2.92 ×103 m/s) 2
r = 4.68 × 107 m, or 4.68 × 104 km
The orbital radius of each satellite is 4.69 × 104 km.
(b) rE = 6.38 × 106 m = 6.38 × 103 km
Let d represent the distance from the surface of Earth of each satellite. Thus,
d = orbital radius of satellite – rE
= 4.69 × 104 km – 6.38 × 103 km
d = 4.05 × 104 km
Each satellite’s distance from the surface of Earth is 4.05 × 104 km.
2.8 km
6. r =
= 1.4 ×103 m
2
mv 2
= mg
(a)
r
2π r
v = rg =
T
2 2
4π r
rg =
T2
T=
=
4π 2 r
g
4π 2 (1.4 × 103 m)
3.8 m/s 2
T = 1.2 ×10 2 s
The period of rotation should be 1.2 × 102 s.
(b) Using the period from (a) above:
1
f =
T
1
=
1.2 × 102 s
f = 8.3 × 10−3 Hz
The frequency of rotation should be 8.3 × 10−3 Hz.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 185
Applying Inquiry Skills
7. (a) Answers will vary. If students have a difficult time getting started, ask them to brainstorm the following categories of
toys:
• games
• colliding objects
• spinning objects
• vibrating objects
• magnetic toys
• toys with wheels
• flying toys
• toys that move through a fluid (air or water, for example)
• jumping or flipping toys
• physics demonstration toys
A highly-recommended resource for this topic is Toys in Space: Exploring Science with the Astronauts, by
Dr. Carolyn Sumners, published by TAB Books, a division of McGraw-Hill, Blue Ridge Summit, PA, 17294-0850
(1994), ISBN # 0-8306-4533-0 (hardcover) and ISBN # 0-8306-4534-9 (paperback).
One neat example of the experiments conducted in space is demonstrating longitudinal pulses on a coiled spring
(such as a Slinky toy). The coil just lies still in free fall, and when a longitudinal pulse is generated, the pulse travels
easily along the axis.
(b) There are many Web sites suitable for this research. The first example listed below is also a source of videos showing
many of the experiments with toys conducted by astronauts in space. The second site listed shows wonderful photos of
many of the toys.
http://spacelink.nasa.gov/Instructional.Materials/Curriculum.Support/.index.html
http://aerospacescholars.org/Cirr/SS/L6/toys.htm
http://www.spintastics.com/HistoryofYoYo.asp
Making Connections
8. (a) Since the speed is slightly greater during our winter, Earth must be closer to the Sun in our winter than in our summer
1
because v ∝
.
r
(b) The answer in (a) does not explain the temperature differences between summer and winter. In fact, if Earth’s surface
temperature depended solely on the distance to the Sun, our winters would be slightly warmer than our summers. The
temperature differences between winter and summer depend mainly on the tilt of Earth’s axis of rotation to its orbital
plane of revolution around the Sun. The Northern Hemisphere faces away from the Sun in winter, resulting in less
direct sunlight and lower temperatures.
CHAPTER 3 LAB ACTIVITIES
Investigation 3.1.1: Analyzing Uniform Circular Motion
(Page 152)
This important investigation is a controlled experiment in which the frequency of an object in circular motion is measured as
the independent variables (the net for on the object, the radius of the circle, and the mass of the object) are varied, one at a
time. It is highly recommended that the instructions be changed from one semester to the next so that the period of revolution
replaces the frequency.
Notice that this investigation involves force, which is not discussed until Section 3.2. The reasoning is that students will
understand centripetal acceleration much better if they experience it in a hands-on activity. But the best way to analyze the
variables in a controlled way is to include force. (The force on the stopper in this investigation is tension in the string, but its
value is easily determined by calculating the force of gravity on the mass attached to the string.) Furthermore, the investigation
prepares the students for Section 3.2.
When introducing the investigation to the students, it is a good idea to demonstrate the apparatus and show how the
motion of the rubber stopper can be controlled and measured. Students will discover soon enough that the procedure requires a
lot of cooperation within the group.
186 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Before performing this investigation, it is important that the students understand thoroughly at least one method of
analyzing data in a controlled experiment involving four variables. For example, if log-log graphing is used, students should
be shown how to plot lines on the graph and just how convenient it is to determine the relationships between the variables
plotted. For details on analyzing data, refer to Appendix A in the text, pages 751 to 755.
Question
Refer to the text.
Hypothesis/Prediction
(a) Accept whatever the students predict, as long as they attempt an explanation. A typical response is
• The frequency must increase as the force increases because a greater force means a greater acceleration, which would
require more revolutions per second.
• The frequency must decrease as the radius of the circle increases because the object will have a greater distance to
cover with each revolution and will need more time to cover it.
• The frequency will decrease with an increase in mass because pulling on a greater mass would require a greater force to
maintain the same frequency, but the force is controlled to be constant, so the frequency should decrease.
• Students who, for whatever reason, apply mathematical relationships may predict the following relationships:
1
1
f ∝ ΣF
f ∝
f ∝
r
m
(b) The two lines required are shown on each of the graphs below. The straight lines represent a simple translation of the
relationships described in words in (a). The curved lines represent the (correct) relationships indicated by the variation
statements in (a).
Materials
The materials listed for each group are straightforward. The materials needed for the analysis depend on which method of
analysis you intend the students to apply.
Procedure
1.
A typical data table, based on the instructions in steps 4 to 7, is shown below. In the table, space is left for three trials for
each controlled test.
Step
Trial
4
1
2
3
average
1
2
3
average
1
2
3
average
5
5
Copyright © 2003 Nelson
Mass of
Stopper
(kg)
Radius of
Circle
(m)
0.75
Weight of
Hanging Mass
(N)
1.96
0.75
1.47
0.75
0.98
Time for 20
Cycles
(s)
Frequency of
Revolution
(Hz)
Chapter 3 Circular Motion 187
6
1
2
3
average
1
2
3
average
1
2
3
average
1
2
3
average
6
7
7
0.60
0.98
0.45
0.98
0.75
1.96
0.75
1.96
2.
The masses of rubber stoppers of the same size tend to differ by unpredictable amounts. Students should not assume that
the total mass of two rubber stoppers is twice that of one rubber stopper.
3. The “count-down” method is recommended for counting a predetermined number of revolutions, such as 20. As the
stopper is twirled at its (attempted) constant speed, the person counting, who is likely the person with the stopwatch,
counts 3, 2, 1, 0, 1, 2, 3, . . . 20, starting the stopwatch at 0 and stopping it at 20.
4–7. It is unrealistic to expect good results with every trial. A good way to ensure a rewarding analysis for the students is to
have them check their data with you or with a computer program you can write to check their data. For example, you
can use the equation stated below to check the time interval for 20 revolutions of each trial without telling then students
how you are evaluating their data. If their value is not within 10% of the calculated value, have the students repeat the
trial. The time for 20 complete revolutions is:

mr 
t = 20T = 20  2π

FT 

where T is the period of revolution, m is the mass of the stopper, r is the radius of the circle, and FT is the tension in the
string, which equals the force of gravity on the suspended mass.
Using this equation, the average results of a typical investigation are given below, using a mass of 14 g (0.014 kg)
for each stopper.
Step
Mass of
Stopper
(kg)
Radius of
Circle
(m)
Weight of
Hanging Mass
(kg)
Time for 20
Cycles
(s)
Frequency of
Revolution
(Hz)
4
5
5
6
6
7
7
0.014
0.014
0.014
0.014
0.014
0.028
0.042
0.75
0.75
0.75
0.60
0.45
0.75
0.75
1.96
1.47
0.98
0.98
0.98
1.96
1.96
9.2
10.6
13.0
11.6
10.1
13.0
15.9
2.2
1.9
1.5
1.7
2.0
1.5
1.3
Analysis
(c) Whichever method the students use to analyze the data, the relationships they discover should be as follows:
1
1
f ∝ FT
f ∝
f ∝
r
m
where f is the frequency of revolution, FT is the magnitude of the tension in the string, r is the radius of the circle, and m is
the mass of the stopper.
If regular graphing techniques are used, students would plot and replot the data until they obtain a straight line, and
then determine the proportionality statements. If log-log graphing techniques are used, the slopes of the lines on the graph
188 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
can be used to write the proportionality statements. (The slope of the graph of f versus FT is +
1
and the slope for the
2
1
other two relationships is − .)
2
(d) Considering the proportionality statements in (c):
FT
f ∝
m r
f ∝
f =k
FT
mr
FT
mr
Evaluating the constant k using data from the first trial:
mr
k= f
FT
= ( 2.2 Hz )
(
= 2.2 s −1
)
(0.014 kg )(0.75 m )
1.96 N
(0.014 kg )(0.75 m )
1.96 kg ⋅ m/s 2
k = 0.16
Notice that the units cancel out, so k is dimensionless.
Students should calculate an average of at least three different trials to verify the constant. Then they can write the
equation relating the variables:
F
f = 0.16 T
mr
Students can check this equation using the data for the remaining trials.
(e) Starting with the equation derived in (d):
F
f = 0.16 T
mr
F
f 2 = 0.0256 T
mr
mrf 2
FT =
0.0256
FT = 39mrf 2
The equation given in the text is:
ΣF = 4π 2 mrf 2
ΣF = 39mrf 2
Assuming the tension in the string is the net force causing the stopper to undergo its centripetal acceleration, the equations
are the same. Of course, experimental results may differ somewhat.
(f) The FBD is shown below. The tension force is close to being horizontal, but it has a small downward component due to
gravity, so it is not truly horizontal.
The following example shows that the force of gravity on the stopper is smaller than the tension in the string.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 189
Fg,stopper = mstopper g
= (0.014 kg )(9.8 N/kg )
Fg,stopper = 0.14 N
The tension is caused by the force of gravity on the hanging mass:
Fg,weight = mweight g
= (0.10 kg )(9.8 N/kg )
Fg,weight = 0.98 N
The ratio of these forces is
0.98 N
= 7.0 :1.0 .
0.14 N
Evaluation
(g) With a low frequency (or a low speed of motion), the effect of gravity pulling down on the stopper is quite noticeable.
At higher frequencies, the motion becomes more horizontal. Thus, the tension is closer to being horizontal at higher
frequencies.
(h) There are several sources of random error in this investigation:
• trying to keep the stopper moving at a constant speed; this takes practice and coordination
• using the incorrect radius of the circle whenever the paperclip touches the bottom of the glass tube; the setup should be
checked constantly
• measuring the radius of the circle; the distance should be measured to the middle of the rubber stopper
• measuring the time for 20 revolutions of the stopper; an average of at least three trials should be taken
• keeping the vertical part of the string holding the metal mass from swinging; fingers wrapped gently around the string
help
Sources of systematic error are:
• keeping the plane of motion as horizontal as possible; this takes practice
• friction between the string and glass tube; a very smooth string helps
Sources of human error are:
• counting the number of revolutions; the count-down method described above helps
• reaction time using the stopwatch; it helps to have the same student counting and using the stopwatch
Synthesis
(i) The three laws of motion are illustrated as follows:
• An object maintains its state of rest or constant velocity only if the net force is zero. When this investigation is
conducted properly, the net force on the hanging mass is zero, so it maintains its state of rest.
• If the net force on an object is not zero, the object accelerates in the direction of the net force. In this case, the net force
on the stopper is toward the centre of the circle, which causes the centripetal acceleration of the stopper.
• For every action force, there is a simultaneous reaction force equal in magnitude but opposite in direction. For
example, if the action force is the string pulling on the stopper toward the centre of the circle, then the reaction force is
the stopper pulling on the string away from the centre of the circle.
Activity 3.4.1: Simulating Artificial Gravity
(Page 154)
Materials
See page 154.
Procedure
1.
2.
3.
The distance measured gives an approximate value of the radius of the circle.
Times will vary. A typical time interval for five revolutions is 6.0 s.
Values will vary, depending on the distance between the student’s shoulder and the bottom of the bucket. A typical time
interval for five revolutions is 10.0 s.
190 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Analysis
(a) For Step 2
(b) Using typical values:
r = 1.0 m
T = 1.2 s
v=
=
2π r
T
2π (1.0 m )
1.2 s
v = 5.2 m/s
ac =
=
v2
r
(5.2 m/s )2
1.0 m
ac = 27 m/s 2
The magnitude of the centripetal acceleration in this example is 27 m/s2.
(c) Answers will vary, depending on the values in (b).
Let FA represent the apparent weight and Fg the Earth-bound weight.
FA mac
=
Fg
mg
=
ac
g
=
27 m/s 2
9.8 m/s 2
FA
= 2.80
Fg
Thus, for the example shown, the ratio of the apparent weight to the Earth-bound weight is 2.8:1.0.
Copyright © 2003 Nelson
Chapter 3 Circular Motion 191
(d) The FBD is shown below.
Using an estimated radius of 1.0 m, the situation in step 3 involves a normal force of zero. A typical value for five
revolutions is 10.0 s, so T = 2.0 s.
2π r
v=
T
2π (1.0 m )
=
2.0 s
v = 3.1 m/s
ac =
=
v2
r
(3.1 m/s )2
1.0 m
ac = 9.9 m/s 2
The magnitude of the centripetal acceleration in this example turns out to be 9.9 m/s2, which is certainly within
experimental error of the theoretical value of 9.8 m/s2. To see why the theoretical value is 9.8 m/s2, consider that FN = 0.
ΣFy = ma y
FN + mg = mac
mg = mac
ac = g
Thus, the ratio of the apparent weight to the Earth-bound weight is 1.0:1.0.
Evaluation
(e) One strength of this model is that it illustrates in a concrete way that artificial gravity created by a normal force on an
object as it undergoes centripetal acceleration increases as the frequency of revolution of the motion increases. Another
strength is that it is a very simple model to demonstrate. The main weakness of the model is that it is done in a
gravitational field, so the situation involving ac = 9.8 m/s2 occurs when the normal force on the object is zero rather than
9.8 N/kg as it would be in outer space where g = 0.
CHAPTER 3 SUMMARY
Make a Summary
(Page 156)
Legend
PIF: passenger on a merry-go-round (inertial frame)
PNF: passenger on a merry-go-round (noninertial frame)
B: bead in a horizontal accelerometer
RT: rubber stopper at top of vertical circle
RB: rubber stopper at bottom of vertical circle
192 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
RS: rubber stopper at side of vertical circle
C: car on a banked curve
G: geosynchronous satellite
A: astronaut on the inside of a rotating space station
VT: vertical accelerometer at inside top of a loop (Earth’s frame)
VB: vertical accelerometer at bottom of a dip (Earth’s frame)
Sample diagrams are shown below.
CHAPTER 3 SELF QUIZ
(Pages 157–158)
True/False
1.
2.
3.
4.
5.
6.
7.
8.
9.
T
F Centripetal acceleration is in a direction that is toward the centre of the circle.
T
F The centrifugal force on an object in uniform circular motion is a fictitious force in the accelerating frame of reference
of the object and is directed away from the centre of the circle.
F Centripetal force is a net force that applies to all objects, both natural and human-made, in circular motion.
T
F Only the centripetal force is real. The centrifugal force is fictitious, and is made up in the accelerating object’s frame of
reference.
T
F The magnitude of your weight calculated using the equation F = mg equals the magnitude of the force of gravity
GmmE
between you and Earth calculated using the equation FG =
.
r2
Copyright © 2003 Nelson
Chapter 3 Circular Motion 193
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
UNIT 1 PERFORMANCE TASK
APPLYING PRINCIPLES OF MOTION AND FORCES
(Pages 162–163)
Like the other performance tasks in the textbook, this task is set up for ease of student assessment. It is recommended that you
discuss the task, the options, and the assessment with the students early in the unit, and then have them work on the task as the
unit progresses.
The three options within the task are quite different in design. Options 1 and 2 involve building devices, but Option 2 is
much more open-ended than Option 1. Option 3 involves research and analysis of existing devices.
Option 1: A Model Roller Coaster
This option involves numerous principles of forces and motion, especially those involving circular motion (Chapter 3). The
task is designed so that it can stand on its own or be used in conjunction with the model roller coaster contest held at
Paramount Canada’s Wonderland each May on Physics/Math Day. Each year of this competition the rules and the dimensions
of the ride are altered slightly to promote student creativity and originality. More information about this competition can be
found at www.canadaswonderland.com.
It is important to decide on the criteria used to evaluate the success of the model before the students begin the task. Some
of the quantitative criteria used to compare coasters of the same outside dimensions are:
• total length of the track
• number of vertical loops
• a minimum number of horizontal loops (e.g., 2)
• a maximum allowable mass
Some of the qualitative criteria are:
• originality/creativity
• thrill level of the ride
• aesthetic appeal
• wise use of materials
Students can refer to Appendix A4, text page 767, for a summary of technological problem solving.
Option 2: Toys That Apply Physics Principles
This option applies the concepts of motion and forces, especially friction and air resistance (discussed in Section 2.4). It is an
open-ended task in which students exhibit creativity as they design, build, test, and revamp two different models of the same
toy. Two models are required to show the effects of different amounts of friction or air resistance.
Students can refer to Appendix A4, text page 767, for a summary of technological problem solving.
Option 3: Centrifuges
This option relates most directly to the principles presented in Chapter 3. Students can find numerous resources, including the
school’s resource centre, Internet sites, and local experts who use centrifuges. For a list of Internet sites, refer to the answer to
question 11 (text, page 138) in the Solutions Manual.
Analysis
The answers to the Analysis questions depend on various factors, especially the choice of option. With any of the three
options, students are urged to create their own questions and answer them, a feature that helps to make the task more openended.
Evaluation
Here the students evaluate their own task. Answers will depend on the option chosen.
Copyright © 2003 Nelson
Unit 1 Performance Task 207
UNIT 1 SELF QUIZ
(Pages 164–166)
True/False
1.
2.
3.
4.
5.
6.
7.
8.
9.
T
F The net force on the ball at the top of the flight is the force of gravity, of magnitude mg, pulling downward on the ball.
T
T
F The horizontal component of the velocity of the ball remains constant during the entire projectile motion.
T
T
G
∆v
approaches the instantaneous acceleration as ∆t approaches zero, so the change of position when this
F The quantity
∆t
occurs is from position 4 to position 5.
The magnitude of the tension at position 1 exceeds the magnitude of the tension at position 4 by 2mg. To compare the
tensions at positions 1 and 4, let +y be toward the centre of the circle in each case. Notice that the speed v is constant.
At position 1:
ΣFy = ma y
mv 2
r
mv 2
FT1 =
+ mg
r
FT1 − mg =
At position 4:
ΣFy = ma y
mv 2
r
mv 2
=
− mg
r
FT4 + mg =
FT4
Now:
 mv 2
  mv 2

+ mg  − 
− mg 
FT1 − FT4 = 
 r
  r

FT1 − FT4 = 2mg
10. T
11. If the ball is released at position 1, the stopper’s instantaneous velocity will have an eastward component only.
12. The centripetal acceleration is independent of the mass of the object. The equations for centripetal acceleration are:
v
4π 2 r
ac = 2 = 2 = 4π 2 rf 2
r
T
Multiple Choice
13. (c) The acceleration is down the hillside and its magnitude is g sin θ .
14. (e) With vi = 0:
v 2 = vi 2 + 2ax ∆x
= 0 + 2 g sin θ L
2
v = 2 Lg sin θ
v = 2 Lg sin θ
208 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
1
(v + vi )
2
1
=
2 Lg sin θ
2
Lg sin θ
vav =
2
15. (b) Using the +y and +x directions shown in the text:
ΣFy = ma y = 0
vav =
FN − mg cos θ = 0
FN = mg cos θ
Thus, the magnitude of the force of the hill on the toboggan is mg cos θ , so the magnitude of the force of the toboggan
on the hill is the same, mg cos θ . This is an action-reaction pair of forces.
16. (e) vi = 0
ax = g sin θ
∆x = L
1
∆x = vi ∆t + ax ∆t 2
2
1
L = g sin θ∆t 2
2
2L
∆t 2 =
g sin θ
∆t =
2L
g sin θ
17. (c) From question 14, the magnitude of the average velocity is vav =
Lg sin θ
.
2
At halfway:
v 2 = vi 2 + 2ax ∆x
v 2 = 0 + 2 g sin θ
L
2
v 2 = Lg sin θ
v = Lg sin θ
G
G
Thus, vav < v .
18. (a) Using the +y and +x directions shown in the text:
ΣFy = ma y = 0
FN − mg cos θ = 0
FN = mg cos θ
ΣFx = ma x
Fg sin θ − Ff = ma x
mg sin θ − µ K FN = ma x
mg sin θ − µ K mg cos θ = ma x
g sin θ − µ K g cos θ = a x
a x = g (sin θ − µ K cos θ )
19. (c) The forces are equal in magnitude but opposite in direction.
20. (b) Since there is nor air resistance, the acceleration is simply the acceleration due to gravity, 9.8 m/s2 [down].
Copyright © 2003 Nelson
Unit 1 Self Quiz 209
m
2
r1 = 3rE
r2 = 6rE
21. (a) m2 =
 GmmE 


2
F1  r1 
=
F2  Gm2 mE 


2
 r2

 m  r 2 
=  2   2 
 r1   m2 
 m  r2 
=
 
 m2  r1 
2
m  6rE 
=


 m   3rE 
 
2
2
= ( 2 )( 2 )
2
F1
=8
F2
The ratio is 8:1.
22. (e) The force of gravity toward Earth is the only force acting on the satellite.
23. (d) Let +y be down.
ΣFy = mac
mv 2
r
v2
g=
r
v = gr
mg =
24. (d) m = 9.5 kg
µK = 0.49
µS = 0.65
Let +x be the direction of the applied force. To start the box moving, the minimum force needed is the force needed to
overcome the force of static friction.
FS,max = µS FN
= µS mg
= (0.65 )(9.5 kg )(9.8 N/kg )
FS,max = 61 N
Completion
25. (a) 3
(b) 3
(c) 2
(d) 2
26. (a) 109 km/h = 30.3 m/s
(b) 7.16 × 104 km/min = 1.19 × 106 m/s
(c) 3.4 mm/s2 = 3.4 × 10–3 m/s2
(d) 5.7 cm/(ms)2 = 5.7 × 104 m/s2
(e) 4.62 × 10−3 (km/h)/s = 1.28 × 10–3 m/s2
27. A windsock indicates direction and relative magnitude of the wind.
210 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
28. The three controls a car driver has for producing acceleration are brake pedal, gas pedal, and steering wheel.
29. You are facing southward when suddenly a snowball passes in front of your eyes from left to right. The snowball had
been thrown earlier with an initial horizontal velocity. The direction of the instantaneous velocity is westward and down,
and the direction of the instantaneous acceleration is down.
30. If the direction of an object undergoing uniform circular motion is suddenly reversed, the direction of the centripetal
acceleration is unchanged because it is still toward the centre of the circle.
G
G
31. If vLM is 26 m/s [71° W of S], then vML is 26 m/s [71° E of N].
G
G
G
32. vCE = vCD + vDE
33. The horizontal acceleration of a projectile is zero.
34. The acceleration of an object falling vertically through the air at terminal speed is zero.
35. (a) the slope of a line on a velocity-time graph is LT–2
(b) the area under the line on an acceleration-time graph is LT–1
(c) weight is MLT–2
(d) the universal gravitation constant is L3M–1T–2
(e) gravitational field strength is LT–2
(f) coefficient of static friction is dimensionless
(g) frequency is T–1
(h) the slope of a line on an acceleration-force graph is M–1
36. The law of inertia is also known as Newton’s first law of motion.
37. As the speed of a flowing river increases, the pressure in flowing water decreases.
38. An accelerating frame of reference is also known as noninertial. In such a frame, we must invent fictitious forces to
explain an observed acceleration. If the frame is rotating, the invented force is called centrifugal force.
39. A passenger of mass m is standing on an elevator that has an acceleration of magnitude a. The normal force acting on the
passenger has a magnitude m(a + g) if the acceleration is upward and m(a − g) if the acceleration is downward.
40. On the Moon’s surface, your mass would be the same as on Earth’s surface, but your weight would be reduced by a factor
of 6.0.
Matching
41.
42.
43.
44.
45.
46.
(b)
(g)
(e)
(a)
(a)
(c); (d)
Copyright © 2003 Nelson
Unit 1 Self Quiz 211
UNIT 1 REVIEW
(Pages 167–171)
Understanding Concepts
1. (a)
(b)
(c)
(d)
2. (a)
3.
4.
5.
6.
The object is moving southward and is increasing its speed.
The object is moving southward and is decreasing its speed.
The object is moving northward and is increasing its speed.
The object is moving northward and is decreasing its speed.
The average speed exceeds instantaneous speed during the time when an object is slowing down and moving slower
than its average speed.
(b) Average speed is less than instantaneous speed when an object is speeding up and moving faster than its average speed.
(c) Average speed equals instantaneous speed for any object moving at a constant velocity.
The component of a vector is always less than or equal to the magnitude of the vector because the component is a
projection of the vector along the axis of a rectangular coordinate system.
(a) The object starts with a high velocity in the positive direction, slows down uniformly, coming to a brief stop before
accelerating uniformly in the opposite direction until reaching the starting position.
(b) The object starts with a high velocity in the negative direction, slows down uniformly until it comes to a brief stop, and
then accelerates uniformly in the positive direction until reaching the same position it started from.
(c) The object is moving in one direction with a uniform positive acceleration until it reaches maximum velocity. At that
instant, the object changes direction, starting with a high velocity and then moving with uniform acceleration in the
same direction as the first acceleration. (An example of this motion would be a ball vertically falling from rest, striking
the floor, and bouncing back upward with no loss in energy. In this case, the positive direction would be down.)
(d) The object starts at a high velocity in one direction and undergoes uniform negative acceleration, reaching zero velocity
and then undergoing uniform acceleration in the opposite direction until reaching a velocity equal in magnitude to the
initial velocity.
A parachute would not work on the Moon. A parachute relies on air resistance, but the Moon has no atmosphere and thus
there is no air resistance.
Since the radius of the circle is 12 cm, the circumference of the circle is 2πr and half the circumference is:
C
=πr
2
= π (12 cm )
C
= 37.7 cm
2
To determine the length of each side of the square, consider the right-angled triangle, BCY.
BY 2 = BC 2 + CY 2
BY = BC 2 + CY 2
=
(a) For the first person:
(12 m )
2
+ (12 m )
2
BY = 17 m
vav =
=
d
∆t
2 (17 m )
48 s
vav = 0.71 m/s
212 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
For the second person:
d
∆t
37.7 m
=
48 s
vav = 0.79 m/s
The average speeds are 0.71 m/s for the person walking along the edge of the square, and 0.79 m/s for the person
walking along the circumference.
(b) Both average velocities are the same.
G
∆d
G
vav =
∆t
24 m [45° N of E]
=
48 s
G
vav = 0.50 m/s [45° N of E]
vav =
7.
The average velocity of each person is 0.50 m/s [45° N of E].
Using the equation derived for horizontal range derived in the text, page 49:
 vi 2 sin 2θ 


∆xM  g M 
=
∆xE  vi 2 sin 2θ 


gE


gE
=
gM
=
8.
9.
9.8 m/s 2
1.6 m/s 2
∆xM
= 6.1
∆xE
The horizontal range on the Moon is 6.1 times the horizontal range on Earth.
The graph is shown in below, with downward defined as the positive direction. The area under each line on the graph
represents the distance fallen, so all three areas should be the same.
m = 1.2 × 103 kg
vi = 42 km/h
vf = 105 km/h
∆t = 21 s = 5.8 × 10–2 h
1
(a) ∆d = (vi + vf ) ∆t
2
1
= ( 42 km/h + 105 km/h ) (5.8 × 10−2 h)
2
∆d = 0.43 km
The car travels 0.43 km in this time interval.
Copyright © 2003 Nelson
Unit 1 Review 213
vf − vi
∆t
105 km/h − 42 km/h
=
21s
a = 3.0 (km/h)/s
The magnitude of the car’s average acceleration is 3.0 (km/h)/s.
(c) a = 3.0 km/h/s = 0.83 m/s2
ΣF = ma
(b) a =
= (0.83m/s 2 )(1.2 × 103 kg)
ΣF = 1.0 × 103 N
The magnitude of the average force needed to cause this acceleration is 1.0 × 103 N.
G
10. ∆ d1 = 0.44 m [S]
G
∆ d 2 = 0.88 m [N]
G
∆ d3 = 0.12 m [S]
∆t = 2.4 s
(a) d total = d1 + d 2 + d3
= 0.44 m + 0.88 m + 0.12 m
d total = 1.44 m
d total = vav ∆t
d total
∆t
1.44 m
=
2.4s
vav =
vav = 0.60 m/s
The average speed of the ball is 0.60 m/s.
G
G
G
G
(b) ∆d = ∆d1 + ∆d 2 + ∆d 3
= 0.44 m[S] + 0.88 m [N] + 0.12 m [S]
= 0.44 m[S] − 0.88 m [S] + 0.12 m [S]
= −0.32 m [S]
G
∆d = 0.32 m [N]
The final position of the ball is 0.32 m [N] of its original position.
G
G
∆d
(c) vav =
∆t
0.32 m [N]
=
2.4s
G
vav = 0.13 m/s [N]
The ball’s average velocity is 0.13 m/s [N].
11. (a) r = 14 cm
∆t = 60.0 s
d
vav =
∆t
2π r
=
∆t
2π (14 cm )
=
60.0 s
vav = 1.5 cm/s
The average speed of the tip is 1.5 cm/s.
214 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b) The instantaneous velocity has a constant magnitude of 1.5 cm/s and a direction tangent to the circle at that instant.
360°
= 30° .
As the tip sweeps past each hour on the clock, the angle subtended is
12
G
At the 6:00 o’clock position: v = 1.5 cm/s [horizontally left]
G
At the 10:00 o’clock position: v = 1.5 cm/s [30° right from up]
(c) From the 1:00 o’clock position to the 5:00 o’clock position, the second hand moves through 4 h × 30°/h = 120° in 20 s.
Its displacement during this time interval is straight downward, with a magnitude of c:
c 2 = a 2 + b 2 − 2ab cos θ
c = a 2 + b 2 − 2ab cos θ
=
(14 cm ) + (14 cm ) − 2 (14 cm )(14 cm )(cos120°)
c = 24 cm
G
∆d
G
vav =
∆t
24 cm [down]
=
20 s
G
vav = 1.2 cm/s [down]
The average velocity of the tip is 1.2 cm/s [down].
12. Let +x be east and +y be south.
vix = 23 m/s
viy = 0
ax = 0
ay = 0.15 m/s2
∆t = 95 s
G G
G v −v
a= f i
∆t
vfx = vix + ax ∆t
= 23m/s + (0 m/s 2 )(95 s)
vfx = 23m/s
vfy = viy + a y ∆t
= 0 m/s + (0.15 m/s 2 )(95 s)
vfy = 14.25 m/s
G
vf = vfx 2 + vfy 2
= (23 m/s) 2 + (14.25 m/s) 2
G
vf = 27 m/s
tan θ =
vfy
vfx
 14.25 m/s 
θ = tan −1 

 23 m/s 
θ = 32°
The train’s velocity after this acceleration is 27 m/s [32° S of E].
13. The speed of the projectile is greatest just after launch and just before landing. In these cases, the velocity has a (constant)
horizontal component and a vertical component of the same magnitude. The speed of the projectile is least at the top of
the flight when the vertical component of the velocity is zero.
14. The motion of the balloon results from an action-reaction pair of forces. The action force is the force of the inner walls of
the balloon pushing westward on the expelled air, and the reaction force is the expelled air pushing eastward on the
balloon.
Copyright © 2003 Nelson
Unit 1 Review 215
15. m = 63 kg
vi = 0 m/s
vf = 8.0 m/s
∆t = 2.0 s
vf − vi
∆t
8.0 m/s − 0 m/s
=
2.0s
a=
a = 4.0 m/s 2
ΣF = ma
= (63 kg)(4.0 m/s 2 )
ΣF = 2.5 × 102 N
The magnitude of the average horizontal force on the runner is 2.5 × 102 N. This force is exerted by the ground on the
runner’s feet, which is a reaction to the action force of the runner’s feet on the ground.
16. Let F1 be the gravitational force at Earth’s surface and F2 be the gravitational force at distance r above Earth’s surface.
F
Thus, 2 = 0.18.
F1
 GmmE 

2 
F2  (r + rE ) 
=
F1
 GmmE 


2
 rE 
 r 
0.18 =  E 
 r + rE 
2
 rE 

 = 0.424
 r + rE 
0.424r = (1 − 0.424)rE
r = 1.4rE
The required distance above Earth’s surface would be 1.4rE.
17. We begin by determining the angle of the rope from the horizontal:
∆y
sin θ =
L
∆y
θ = sin −1
L
2.3
m
= sin −1
7.1 m
θ = 18.9°
Considering the vertical components of the forces on the trapeze artist, with +y up:
ΣFy = ma y = 0
2T sin θ − mg = 0
mg
2sin θ
6.3 × 10 2 N
=
2 (sin18.9° )
T=
T = 9.7 × 10 2 N
The magnitude of the tension in the rope is 9.7 × 102 N.
216 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
18. If the speed of a particle in circular motion is decreasing, 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.
19. ac = 4.49 m/s2
v = 22 m/s
v2
r
v2
r=
ac
ac =
(22 m/s) 2
4.49 m/s 2
r = 1.1×10 2 m
=
The minimum radius of curvature of the curve is 1.1 × 102 m.
20. rM = 2.28 × 1011 m
mM = 6.27 × 1023 kg
FG = 1.63 × 1021 N
TE = 3.16× 107 s
m v2
(a) FG = M
r
FG r
v=
mM
=
(1.63 × 1021 N)( 2.28 ×1011m)
6.27 ×1023 kg
v = 2.42 × 104 m/s
The speed of Mars is 2.42 × 104 m/s.
2π r
(b) vM =
TM
TM =
=
2π r
vM
2π (2.28 ×1011 m)
(2.42 × 104 m/s)
TM = 5.92 × 107 s
Copyright © 2003 Nelson
Unit 1 Review 217
TM 5.92 × 107 s
=
TE 3.16 × 107 s
TM
= 1.88
TE
The period of revolution of Mars around the Sun is 1.88 Earth years.
21. In each case, the magnitude of the instantaneous velocity is the same and the direction is tangent to the circle at that
instant.
r = 6.3 m
(3.0 min)(60 s/min)
T=
= 2.5 s
72 rev
2π r
v=
T
2π (6.3 m)
=
2.5s
v = 16 m/s
(a) The velocity of a passenger at the 3:00 o’clock position would be 16 m/s [down].
(b) The velocity of a passenger at the 6:00 o’clock position would be 16 m/s [left].
(c) The velocity of a passenger at the 7:00 o’clock position would be 16 m/s [30° up from left].
22. (a) T = 80.0 min = 4.80 × 103 s
1.29 × 107 m
r=
= 6.45 × 106 m
2
4.80 × 103 s
∆t =
= 2.40 × 103 s
2
2π r
v=
T
2π (6.45 × 106 m)
=
4.80 × 103 s
v = 8.44 × 103 m/s
G
G
∆v
aav =
∆t
G G
vf − vi
G
aav =
∆t
(8.44 × 103 m/s) − (−8.44 × 103 m/s)
=
2.40 × 103 s
G
aav = 7.04 m/s 2
The magnitude of the average acceleration is 7.04 m/s2.
(b) T = 2.36 × 106 s
v = 1.02 km/s = 1.02 × 103 m/s
2.36 × 106 s
∆t =
= 1.18 × 106 s
2
218 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
G
G
∆v
aav =
∆t
G G
vf − vi
G
aav =
∆t
(1.02 × 103 m/s) − (−1.02 × 103 m/s)
=
1.18 × 106 s
G
aav = 1.73 × 10−3 m/s 2
The magnitude of the average acceleration is 1.73 × 10–3 m/s2.
23. m1 = 55 kg
r = 1.9 m
m2 = 88 kg
f = 0.88 Hz
(a) ∑ F = 4π 2 mrf 2
= 4π 2 (55 kg)(1.9 m)(0.88 Hz) 2
∑ F = 3.2 × 103 N
The magnitude of the force causing the female skater to maintain her circular motion is 3.2 × 103 N.
(b) The force causing the female skater to maintain her circular motion is the action force of 3.2 × 103 N. Thus, the
magnitude of the horizontal force on the male skater is the reaction force, which equals 3.2 × 103 N.
24. (a) In Earth’s frame of reference
(b) In the car’s frame of reference
(c) Considering the vertical components of the forces:
ΣFy = ma y = 0
FT cos θ − Fg = 0
FT =
Copyright © 2003 Nelson
mg
cos θ
Unit 1 Review 219
Considering the horizontal components of the forces:
ΣFx = ma x
FT sin θ = ma x
FT sin θ
m
mg sin θ
=
m cos θ
= g tan θ
ax =
(
= 9.8 m/s 2
) ( tan 5.1°)
2
a x = 0.87 m/s
The acceleration of the car is 0.87 m/s2 [E].
25. m = 27 kg
v = 1.8 m/s
Fapp = 1.12 × 102 N [27° above the horizontal]
Let +y be up and +x be in the direction of the horizontal component of the applied force.
ΣFy = 0
FN + Fapp sin θ − mg = 0
FN = mg − Fapp sin θ
ΣFx = 0
Fapp cos θ − FK = 0
µ K FN = Fapp cos θ
µK =
=
Fapp cos θ
mg − Fapp sin θ
(1.12 ×102 N) (cos 27° )
(27 kg)(9.8 m/s 2 ) − (1.12 × 102 N) (sin 27° )
µK = 0.47
The coefficient of kinetic friction between the garbage can and the sidewalk is 0.47.
26. mA = 2.5 kg
mB = 5.5 kg
µK = 0.54
Let the general direction of the acceleration of the boxes be clockwise around the pulley.
Consider Box B and define the +y direction as down:
ΣFy = mB a
mB g − FT = mB a
FT = mB ( g − a )
Consider Box A and define the +x direction as up the slope:
ΣFy = 0
FN − mA g cos θ = 0
FN = mA g cos θ
ΣFx = mA a
FT − FK − mA g sin θ = mA a
mA a = FT − µ K FN − mA g sin θ
220 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
Substitute for FN and FT:
mA a = FT − µK FN − mA g sin θ
= mB ( g − a ) − µK mA g cos θ − mA g sin θ
(mA + mB )a = (mB − µ K mA cos θ − mA sin θ ) g
a=
=
(mB − µK mA cos θ − mA sin θ ) g
(mA + mB )
(5.5 kg − 0.54(2.5 kg) cos 25.4° − (2.5 kg) sin 25.4° ) (9.8 m/s2 )
(2.5 kg + 5.5 kg)
2
a = 3.9 m/s
The magnitude of the acceleration of the boxes is 3.9 m/s2.
27. m = 62 kg
G
(a) a = 2.5 m/s2 [up]
Let the +y direction be up.
ΣFy = ma
FN − mg = ma
FN = m( g + a )
= 62 kg(9.8 m/s 2 + 2.5 m/s 2 )
FN = 7.6 ×102 N
The magnitude of the apparent weight of the person is 7.6 × 102 N.
G
(b) a = 2.5 m/s2 [down]
Let the +y direction be down.
ΣFy = ma
mg − FN = ma
FN = m( g − a )
= 62 kg(9.8 m/s 2 − 2.5 m/s 2 )
FN = 4.5 ×102 N
The magnitude of the apparent weight of the person is 4.5 × 102 N.
G
(c) a = 0.0 m/s2
G
v = 2.5 m/s [up]
Let the +y direction be up.
ΣFy = 0
FN − mg = 0
FN = mg
= 62 kg(9.8 m/s 2 )
FN = 6.1 × 10 2 N
The magnitude of the apparent weight of the person is 6.1 × 102 N.
Applying Inquiry Skills
28. Answers will vary, depending on the student’s location. For example, for a student living in Ottawa, the closest capital
city of an adjacent province is Quebec City.
(a) Assume the following values:
distance to the capital city: 500 km = 5 × 105 m
diameter of Ferris wheel: 20 m
Copyright © 2003 Nelson
Unit 1 Review 221
Let the number of rotations be N, the distance travelled d, and the circumference of the wheel C.
d
N=
C / rev
d
=
(π D ) / rev
=
5 × 105 m
π ( 20 m ) / rev
N = 8 × 103 rev
The estimated number of revolutions in this example is 8 × 103 revolutions.
(b) The following diagram shows the situation for measuring the diameter (y) of the Ferris wheel.
The distance (d) from the base of the wheel can be estimated using paces or measured using a metric measuring tape.
The angle θ can be determined by using a horizontal accelerometer or a large protractor. The following sample
calculation is made with the assumption that the observer’s eyes are at the same height at the bottom of the wheel.
Observations:
1 pace = 0.65 m
d = 56 paces
θ = 30°
y = x tan θ
= (56 paces )(0.65 m/pace ) tan 30°
y = 21 m
The estimated diameter of the wheel is 21 m.
G
G
29. Since ∆d and vi are known and ∆t can be measured, the equation used to find the average acceleration is
G G
G 1G
1G
∆d = vi ∆t + aav ∆t 2 . With an initial velocity of zero, this equation reduces to ∆d = aav ∆t 2 . Isolating the average
2
2
G
2 ∆d
G
acceleration, we have aav = 2 .
∆t
30. (a) There are many acceptable choices of objects, depending on students’ interests. One example could be different sizes or
shapes of coffee filters. Another example is different surface areas of cloth with string and a small weight attached to
form a parachute.
(b) Answers will vary, depending on the variables tested. For example, the mass of a simple parachute/load combination
could be kept constant while altering the surface area of the parachute and determine the time for the combination to
fall a known distance to the floor.
(c) Safety considerations are fairly obvious. The falling objects should not be breakable and they should not be allowed to
fall onto anything breakable.
222 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
31. The following diagram shows the sine curve. The equation for the horizontal range of a projectile is ∆x =
vo 2 sin 2θ
, so
g
∆x ∝ sin 2θ . According to the graph, there are two values of sin 2θ that are equal on either side of 90° (e.g., for angles of
30° and 60°, sin 60° and sin 120° are both 0.866). However, the exception is 45° because sin 2θ = sin 90° = 1 , which is a
single maximum value.
32. The setup shown in the text can be used to demonstrate Newton’s third law of motion in a concrete way assuming that the
readouts from the force sensors can be displayed simultaneously. That way, any change recorded by one sensor and
balanced by a change in the other sensor is observed immediately. For the most obvious results, it is best to ensure that the
mass of the washers is negligible compared to the mass of the magnet. Furthermore, the action-reaction forces on the
readouts are more obvious if sensor A is zeroed with the magnet suspended from it before the washers are brought near.
For more information on a similar demonstration, refer to the article “An Extraordinary Demonstration of the Third Law”
The Physics Teacher, Volume 39, October 2001, pages 392–393.
(a) The answer depends on how the apparatus is set up. The magnet and the washers attract one another, and the magnetic
force of attraction of the magnet on the washers is equal in magnitude but opposite in direction to the magnetic force of
attraction of the washers on the magnet. Consider the FBDs of the magnet and washers shown below. The force
registered by sensor A in this case is much greater than the force registered by sensor B because the force of gravity on
the magnet must be balanced. However, if the sensors are zeroed before the washers are brought close to the magnet,
then the readings on the two sensors will be virtually the same, with sensor B registering positive values and sensor B
registering negative values.
(b) The following graph illustrates the results if the sensor holding the magnet was zeroed with the magnet suspended from
it and the mass of the washers is negligible compared to the mass of the magnet. The graph illustrates that the force is
1
proportional to 2 , however the computer-generated graphs will show the force as a function of time, so the shape will
r
be similar if the motion suggested is done smoothly. The dual graph nicely illustrates the action-reaction pair of
magnetic forces.
Sensor readout involving the magnetic forces only. The axes depend on the setup, so they are labelled simply x and y.
Copyright © 2003 Nelson
Unit 1 Review 223
33. The acceleration-force graph is shown below.
The magnitude of the slope of the line is
G
∆a
slope =
G
∆ΣF
=
1.95 m/s 2
(
7.0 kg m/s 2
)
slope = 0.28 kg −1
Using unit analysis, it is evident that the mass must be the reciprocal of the slope.
1
m=
slope
1
=
0.28 kg −1
m = 3.6 kg
The mass of the cart is 3.6 kg.
34. The frequency graphs
The period graphs
The relationships are as follows:
f ∝ FT
f ∝
f ∝
1
m
1
r
224 Unit 1 Forces and Motion: Dynamics
T∝
1
FT
T∝ m
T∝ r
Copyright © 2003 Nelson
Making Connections
35. Let +y be down.
(a) viy = 0
∆y = 0.45 m
ay = 9.8 m/s2
1
∆y = viy ∆t + a y ∆t 2
2
1
2
∆y = a y ∆t
2
2 ∆y
∆t =
ay
=
2 ( 0.45 m )
9.8 m/s 2
∆t = 0.30 s
The time interval for the first 45 cm is 0.30 s.
(b) viy = 0
∆y = 0.45 m
ay = 9.8 m/s2
Solve for the final velocity upon reaching the mat using ∆ytotal = 6.0 m.
vfy 2 = viy 2 + 2a y ∆ytotal
vfy = 2a y ∆ytotal
= 2(9.8 m/s 2 )(6.0 m)
vfy = 11m/s
∆y = vfy ∆t −
1
a y ( ∆t ) 2
2
1
a y (∆t )2 − vfy ∆t + ∆y = 0
2
This is a quadratic equation with solution ∆t =
∆t =
=
−b ± b 2 − 4ac
2a
vfy ± vfy 2 − 2a y ∆y
ay
11m/s ± (11m/s) 2 − 2(9.8 m/s 2 )(0.45 m)
9.8 m/s 2
∆t = 0.042s
The time interval for the last 45 cm before reaching the mat is 0.042 s.
(c) The time interval for a short distance at the top of the jump is approximately 7 times longer than the time interval for
the same distance at the bottom of the jump, which explains why the jumper appears to be in slow motion at the top of
the jump.
36. Starting from rest, the swimmer experiences an increasing downward velocity with a uniform acceleration of 9.8 m/s2
[down]. Air resistance is negligible in this short falling distance. The swimmer reaches maximum velocity at the instant
prior to touching the water. The velocity then slows down rapidly with an acceleration that is upward and greater than
9.8 m/s2.
37. Let +y be down and let ∆t1 be the time interval for the stone to drop to the water level and ∆t2 be the time interval for the
sound to travel back to the top of the well.
viy = 0 m/s
vs = 3.40 × 102 m/s
∆ttotal = 4.68 s
ay = 9.80 m/s2
Copyright © 2003 Nelson
Unit 1 Review 225
Considering the stone’s motion:
1
∆y = viy ∆t1 + ay ∆t12
2
1
2
∆y = ay∆t1
2
Considering the sound’s motion:
∆y
vs =
∆t 2
∆y = vs ∆t2
Equating the expressions for ∆y:
1
a y ∆t12 = vs ∆t2
2
1
a y ∆t12 = vs ( ∆ttotal − ∆t1 )
2
( 4.90 m/s ) ∆t = (3.40 ×10
2
(4.90 m/s ) ∆t + (3.40 ×10
2
2
1
2
)
2
1
2
)
m/s ( 4.68 s − ∆t1 )
m/s ∆t1 − 1.59 × 103 m = 0
Applying the quadratic formula:
∆t1 =
=
−b ± b 2 − 4 ac
2a
−3.40 × 102 m/s ±
(3.40 ×10
2
m/s
)
2
(
− 4 4.90 m/s 2
)(−1.59 ×10 m )
3
9.8 m/s 2
∆t1 = 4.40 s
Finally, considering the sound’s motion:
∆y = vs ∆t2
(
= (3.40 × 10
)
m/s ) ( 4.68 s − 4.40 s )
= 3.40 × 103 m/s ( ∆ttotal − ∆t1 )
3
∆y = 95.0 m
The well is 95.0 m deep.
38. (a) Let ∆t1 be the time to reach the maximum height. The total time is ∆t = 2∆t1.
vfy = 0 m/s
∆y = 0.85 m
ay = −9.8 m/s2
1
2
∆y = vfy ∆t1 + a y ( ∆t1 )
2
2 ∆y
∆t1 =
−a y
=
2(0.85 m)
− 9.8 m/s 2
∆t1 = 0.42 s
∆t = 2∆t1
= 2(0.42 s)
∆t = 0.83s
The player is in the air for 0.83 s.
226 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b) Consider only the first half of the motion.
vfy 2 = viy 2 + 2a y ∆y
viy = −2a y ∆y
= −2(−9.8 m/s 2 )(0.85 m)
viy = 4.1m/s
The player’s vertical take-off speed is 4.1 m/s.
39. viy = 0 m/s
ay = 9.8 m/s2
Let +y be down.
vfy 2 = viy 2 + 2a y ∆y
∆y =
vfy 2
2a y
vfy = 5.0 m/s
∆y =
(5.0 m/s) 2
2(9.8 m/s 2 )
vfy = 10.0 m/s
∆y =
(10.0 m/s)2
2(9.8 m/s 2 )
∆y = 1.3m
∆y = 5.1m
The practice platforms would need to have a range of heights from 1.3 m to 5.1 m.
40. The prevailing winds in much of Canada move from west to east, so airplanes travelling westward have a headwind that
slows them down relative to the ground, while airplanes travelling eastward have a tailwind that helps them increase their
speed relative to the ground.
41. viy = 0
∆y = 38 m
ay = 9.8 m/s2
vix = 2.0 m/s
Let +y be down and +x be outward from the top of the falls.
Considering the vertical component of the motion:
1
∆y = viy ∆t + a y ∆t 2
2
1
2
∆y = a y ∆t
2
2 ∆y
∆t =
ay
=
2 (38 m )
9.8 m/s 2
∆t = 2.8 s
Considering the horizontal component of the motion:
∆x = vix ∆t
= ( 2.0 m/s )( 2.78 s )
∆x = 5.6 m
The walkway should be built up to 5.6 m away from the falls.
Copyright © 2003 Nelson
Unit 1 Review 227
G
42. vi = 75.0 km/h [N] = 20.8 m/s [N]
G
vf = 0 m/s
G
a = 4.80 m/s2 [S]
Solve for the distance ∆d needed to stop:
vf 2 = vi 2 + 2a∆d
vi 2
2a
(20.8 m/s)2
=
2(4.80 m/s 2 )
∆d =
∆d = 45.2 m
The reaction distance ∆dr is the distance travelled at constant speed vi before applying the brakes.
∆dr = 49.0 m – ∆d = 2.8 m. The reaction time is
∆d r
∆t =
vi
=
2.8 m
20.8 m/s
∆t = 0.13s
The reaction time is 0.13 s.
43. (a) Earth’s axis of rotation is aligned with the North Star, so as Earth rotates, distant objects in the sky appear to revolve
around the North Star. Since Earth is in motion (i.e., rotating) relative to the stars and planets, the stars and planets are
in motion relative to Earth.
(b) Earth takes 24 h to rotate once on its axis, so the rate of rotation is 360°/24 h, or 15°/h. If we choose an obvious arc and
measure the angle it subtends in the photo, we can estimate the exposure time, ∆t.
 1.0 h 
∆t = (144° ) 

 15° 
∆t = 9.6 h
The exposure time is approximately 9.6 h.
(c) Because of Earth’s curvature, people in Australia and other parts of the Southern Hemisphere are unable to view the
North Star. However, to take a similar photograph people in the Southern Hemisphere would have to aim their cameras
toward the southern sky aligned with Earth’s axis.
44. (a) The principle involved is Bernoulli’s principle. In the region where the air is moving quickly because it is being
dragged downward by the water, the air pressure is reduced. So the air pressure beyond the shower curtain is greater
than the air pressure inside the shower. The pressure difference causes the observed force on the curtain.
(b) Answers may vary. Examples from the text are throwing a curve ball (page 105), blowing air between two empty pop
cans causing them to move toward each other (page 105), demonstrating a venturi flow meter (page 107, question 7),
and demonstrating the way in which lift can be created in an airplane wing (page 118, question 22).
45. The kite must “catch” the wind in order for the force on the surfer to be sufficiently large. Thus, the kite should not be
streamlined; rather it should be designed to offer maximum air resistance for its size and mass.
46. (a) Converting the units:
m   1 km   3600 s 

v =  7.2 ×103



s   1000 m   1 h 

km
v = 2.6 × 104
h
The speed of the satellite is 2.6 × 104 km/h.
228 Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b) mE = 5.98 × 1024 kg
GmE
r
GmE
v2 =
r
Gm
r = 2E
v
v=
(6.67 ×10
=
−11
)(
m/s )
N ⋅ m 2 / kg 2 5.98 × 1024 kg
(7.2 ×10
3
)
2
r = 7.7 × 106 m
The distance above Earth’s surface is 7.7 × 106 m − 6.38 × 106 m = 1.3 × 103 m or 1.3 × 103 km.
(c) Remote sensing is a means of monitoring certain variables from a distance. Some practical examples are checking
weather and climate conditions such as moisture content and water temperatures, tracking pollution, observing forest
fires, checking agriculture production, and researching potential areas for mining.
47. Answers will vary. Students might think of applying the physics principles related to the acceleration due to gravity here
and on other solar system bodies, projectile motion, relative velocity, terminal speeds, Newton’s laws of motion, frictional
forces, Bernoulli’s principle, noninertial frames of reference, banked curves, roller coasters, and artificial satellites. One
specific example could be a realistic speed car race created using images, dimensions, banking angles, and environmental
conditions at a specific track, with a simulation of what might happen if a tire blew out.
48. (a) r = 2.1 m
g = 9.8 m/s2
Let +y be down.
∆Fy = mac
mv 2
r
mv 2
mg =
r
2
v
g=
r
v = gr
Fg =
=
(9.8 m/s ) (2.1 m )
2
v = 4.5 m/s
The speed of the model coaster at the top of the loop is 4.5 m/s.
(b) It is dangerous to have a zero normal force on the passengers in this situation. The passengers would get knocked
around more, and the likelihood of objects falling out of pockets would increase. These objects could become
dangerous to other passengers. The coaster design should be changed so the passengers would feel a normal force, even
at the top of the loop. One way to achieve this is to make the loop lower in maximum height so the coaster does not
lose as much of its kinetic energy to gravitational potential energy.
49. (a) A trip to Mars would take a long time, in the order of a year in each direction. Human muscles, bones, and other body
components deteriorate if they do not have the resistance forces common in a gravitational field. Artificial gravity helps
to overcome the negative effects of space travel in a low gravitational field.
(b) The most common way of creating artificial gravity on a mission to Mars will likely be in a rotating space station.
Astronauts and objects along the inside of the outer circumference of the rotating station would experience a normal
force that would depend on the radius of the station and the rate of rotation.
Copyright © 2003 Nelson
Unit 1 Review 229
Extension
50. ∆x = 6.1 m
vo = 7.8 m/s
Let +y be up and +x be toward the basket.
Considering the horizontal component of the motion, which is constant velocity:
vx = vo cos θ
= 7.8 m/s ( cos 55° )
vx = 4.5 m/s
The time interval for the motion is:
∆x
∆t =
vx
6.1 m
4.5 m/s
∆t = 1.36 s
Considering the vertical component of the motion, with +y up:
ay = −g = −9.8 m/s2
∆t = 1.36 s
1
∆y = viy ∆t + a y ∆t 2
2
1
2
= 7.8 m/s (sin 55° )(1.36 s ) + −9.8 m/s 2 (1.36 s )
2
∆y = −0.4 m
Since the basket is 1.2 m above the starting position and the final position is 0.4 m below, the ball misses the basket by
1.6 m.
=
(
230 Unit 1 Forces and Motion: Dynamics
)
Copyright © 2003 Nelson