Unit 1 Review, pages 150–157
Knowledge
1. (d)
2. (a)
3. (c)
4. (b)
5. (b)
6. (c)
7. (a)
8. (b)
9. (c)
10. (b)
11. (a)
12. False. Acceleration always occurs in the same direction as the net force on the object.
13. False. Use the equation vf 2 = vi 2 + 2a!d when you know the change in position and the
initial and final speeds, and acceleration is constant.
14. True
15. False. To cross a river as quickly as possible, the velocity of the boat relative to the water
should be directed across the river,
!
16. False. A boat travels with velocity vBR with respect to a river. The river moves with velocity
!
vRS with respect to the shore. The equation that describes the boat’s velocity with respect to the
!
!
!
shore is therefore the sum vBR = vBR + vRS .
17. True
18. True
19. False. Static friction resists the motion of an object as long as the applied force is smaller
than the upper limit of the force of static friction.
20. False. For an object kept in a circular path, the centripetal acceleration decreases as the
square of the period of revolution.
21. True
22. (a) (vi)
(b) (viii)
(c) (vii)
(d) (x)
(e) (ix)
(f) (i)
(g) (iii)
(h) (ii)
(i) (v)
(j) (iv)
23. If the position–time graph of an object is a straight line with negative slope, the average
velocity of the object is constant and negative. The object is moving backward at constant speed.
24. If the position–time graph of an object is a straight line with positive slope, the acceleration
of the object is zero. The object is moving forward at constant speed.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-10
25. The position of a freely falling object decreases more and more quickly with time. If up is the
positive direction, the position–time graph of the object is a section of a downward opening
parabola.
26. Given: vi = 0 m/s; vf = 39.6 m/s; Δt = 1.2 s
Required: a
v −v
Analysis: Use the definition a = f i to calculate the acceleration a.
Δt
v −v
Solution: a = f i
Δt
39.6 m/s − 0 m/s
=
1.2 s
a = 33 m/s 2
Statement: The acceleration of the drag racer is 33 m/s2.
27. The kinematics equations apply only to motion in one dimension at constant acceleration.
28. When solving vector addition problems, the sine and cosine law method involves fewer
calculations than the component method if there are only two vectors to sum and if neither is
aligned with a canonical direction. If there are three or more vectors to add, the component
method is less complicated and has fewer calculations. If there are two vectors, with one aligned
in a canonical direction, the two methods are essentially of the same level of complexity.
29. The magnitude of vector displacement in terms of its components is
!
! dT = (! d x )2 + (! d y )2 .
30. Average velocity is calculated from displacement, which can be small if the motion involves
changes in direction. Average speed is calculated from the total distance travelled ignoring any
change in direction. Since the total distance is always at least as great as the magnitude of
displacement, the average speed is always as least as great as the magnitude of the average
velocity.
31. The projectile motion equation, assuming no horizontal motion, is Δd y = vi Δt − 12 g Δt 2 where
vi is the initial velocity. If there is also horizontal motion, the equation becomes
Δd y = v1y Δt − 12 g Δt 2 where viy = vi sin θ and θ is the launch angle as measured from the
horizontal.
32. The ball returns to the level from which it was kicked. The horizontal range equation is
v2
!d x = i sin 2" where vi is the initial velocity, θ is the launch angle, g is the acceleration due to
g
gravity and Δd x is the horizontal displacement or range.
33. (a) When the crate lands the plane will be directly over the landing site.
(b) From the hatch of the plane, an observer sees the crate falling straight down.
34. The ground velocity of the plane is the vector sum of the air velocity of the plane and the
r
r
r
wind velocity: vPG = vPA + vAG .
35. The velocity of the train relative to the platform and the velocity of the platform relative to
the train are opposite. So the velocity of the train is 72 km/h north.
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Unit 1: Dynamics
U1-11
36. (a) The tension in the cable is FT = mg sin θ .
(b) The normal force acting on the car is FN = mg cos θ .
(c) As the steepness of the incline is decreased, the angle θ decreases. With the incline more
nearly horizontal, the component of gravity down the incline decreases and the tension in the
cable decreases. Also, the normal force is more nearly vertical. The incline supports more of the
weight of the car and the normal force increases.
37. Inertia is a measure of an object’s resistance to change in its motion.
38. When you push against a heavy object, you feel the object pushing back on you. If you push
lightly, you feel a slight pressure back on your hands. The harder you push the more the object
pushes back into your palms. This is an action-reaction pair of forces.
39. For the forces on an object to be in a state of equilibrium, the object must not be accelerating.
The object can be moving at a constant velocity or be stationary.
r
r
40. Given: Fg = 1.0 N [down]; Fair = 4.0 N [right]
r
Required: FT
r
Analysis: The vector sum of the three forces is zero, ΣF = 0 N , because the string holds the ball
r
in place. Use ΣF = 0 N in components to calculate the components of the tension. Then use
r
Pythagorean theorem to calculate the magnitude of FT . Use right and up as positive.
Solution:
r
Determine the components of FT :
ΣFy = 0 N
ΣFx = 0 N
Fgx + Fairx + FTx = 0 N
FTx = − Fairx
FTx = −4.0 N
r
Calculate the magnitude of FT :
!
FT = ( FTx ) 2 + ( FTy ) 2
Fgy + Fairy + FTy = 0 N
FTy = − Fgy
FTy = −1.0 N
= (1.0 N) 2 + (4.0 N) 2
!
FT = 4.1 N
Statement: The magnitude of the tension is 4.1 N.
41. A linear actuator is any device that converts energy into a constant force. When you use a car
jack, your stored chemical energy is converted into the upward applied force used to raise or
hold the car. So a car jack is an example of a linear actuator.
42. Answers may vary. Sample answer: A skier reduces sliding friction of the skis on the snow
by the profile of the underside of the ski and the material used on this surface. The skier reduces
air resistance with compact posture and slippery clothing. The reduction of these opposing forces
allows the skier to accelerate faster and maintain high speeds more easily. So the skier can
complete the race more quickly.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-12
43. Answers may vary. Students Venn diagrams should include the distinction that inertial
frames of reference have a constant velocity (no acceleration) while non-inertial frames of
reference are accelerating and therefore have changing velocity. The laws of inertia hold only in
inertial frames of reference. Non-inertial frames of reference experience fictional forces. The
intersection on the Venn diagram may contain a definition of a frame of reference, such as, a
choice of coordinate system for making measurements.
44. You can observe a fictitious force by watching the movement of an object in a non-inertial
frame of reference as it reacts to the acceleration of the frame of reference. Examples include the
centrifugal force experienced by an object in a rotating frame of reference.
v2
45. By the equation for centripetal acceleration ac = , the centripetal acceleration varies
r
inversely with distance from the centre of the circular path, r.
46. Answers may vary. Sample answers: A car may be held along a curve by the static friction
between the tires and the road, by its own normal force if the curve is banked at an angle tilting
the car into the curve, or a combination of the two forces.
47. The speed doubles from 7.5 m/s to 15 m/s. Substitute vnew = 2v in the equation for centripetal
acceleration:
2
vnew
r
( 2v )2
=
r
v2
ac = 4
r
The centripetal acceleration is four times its original value.
ac =
48. To simulate Earth’s gravity, the spacecraft must rotate at a speed such that
v2
= 9.8 m/s2.
r
Understanding
49. Answers may vary. Sample answers:
(a) A car driving backward and slowing down has negative velocity and positive acceleration.
(b) A ball that is rolling at constant speed to the left has negative displacement and zero
acceleration.
(c) Assuming forward is the positive direction, a person walking backward and slowing down
has negative displacement and positive acceleration.
(d) A ball that has just been thrown up into the air has positive velocity and negative
acceleration.
r
r
50. (a) Given: Δd1 = 1.3 km; Δd2 = −1.3 km; Δt = 40.0 min
r
Required: Δd
Analysis: The total displacement for a trip is the vector sum of the displacements for the
!
!
!
individual legs of the trip: !d = !d1 + !d2
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-13
!
!
!
Solution: !d = !d1 + !d2
= 1.3 km + ("1.3 km)
!
!d = 0 km
Statement: The total displacement is 0 km. This makes sense since you returned to your starting
point.
!
!
(b) Given: !d1 = 1.3 km; !d2 = "1.3 km; !t = 40.0 min
Required: vav
Δd
where Δd is the total distance travelled
Analysis: The average speed is given by vav =
Δt
during the motion. This distance is the sum of the individual distances, Δd = Δd1 + Δd 2 . We will
calculate Δd and then vav .
Solution:
Δd
Δt
= 1.3 km + 1.3 km
2.6 km 60 min
=
×
Δd = 2.6 km
1h
40 min
vav = 3.9 km/h
Statement: The average speed is 3.9 km/h.
!
!
!
(c) Given: !d1 = 1.3 km; !d2 = "1.3 km; !t = 40.0 min; !d = 0 km
r
Required: vav
!
!
!d
!
Analysis: The average velocity is given by vav =
where !d is the total displacement. Since,
!t
as seen in part (a), you return to your starting position the total displacement is zero. As a result,
the average velocity is zero.
Statement: The average velocity is 0 m/s.
51. Answers may vary. Sample answer:
(a) The rock on the Moon will experience a lower acceleration due to gravity. Its downward
acceleration will be less and it will take a longer time to fall a given distance than it would on
Earth.
(b) Other than a lower value for g, there is no difference between projectile motion on Earth and
projectile motion on the Moon. The vertical motion is accelerated motion and the horizontal
motion is uniform motion.
!
52. Answers may vary. Sample answer: To determine the sum of two vectors A and
r
B algebraically, use !the fact
in
! !that vector addition is the addition of arrows
!
! a triangle. Draw the
A
+
B
A
B
triangle that shows C =
.
From
the
given
information
about
and
, determine the angle
!
!
2
2
2
θ between the tip of A and the
! tail of B . Now use the cosine law C = A + B − 2 AB cos θ to
determine the magnitude of!C and sine law to calculate another angle in the triangle. Finally,
determine the direction of C using the known angles.
Δd = Δd1 + Δd 2
vav =
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-14
53. Answers may vary. Sample answer:
!
!d
!
(a) The average velocity is given by vav =
. Use the displacement information to determine
!t
the displacement from the starting time to the finishing!time. Also, note the length of time
!
corresponding to the given data. Use these values of !d and Δt in the formula for vav .
Δd
(b) The average speed is given by vav =
. To use this formula, you need to know the distance,
Δt
Δd , that the train travels. Assuming the train travels in a straight line, the distance is the
magnitude of the displacement.
54. (a) The force of gravity is toward Earth while the velocity is in the direction the moon is
heading: the tangent to the circular orbit. The two vectors are perpendicular.
(b) The force of gravity causes acceleration at a right angle to the moon’s velocity. This
acceleration constantly deflects the moon’s velocity toward the Earth, resulting in a circular path.
1
55. (a) !d y = v1 sin " !t # g!t 2
2
1
(b) !d y = " g!t 2
2
1
(c) !d y = " g!t 2
2
56. Answers may vary. Sample answer: The required vector addition triangle is given by
!
!
!
vBG = vBW + vWG . The velocity of the boat with respect to the water is the side of the triangle that
crosses the river perpendicularly. The velocity of the water with respect to the ground is parallel
to the shore. The velocity of the boat with respect to the ground is the hypotenuse of the right
angle.
!
!
!
57. Answers may vary. Sample answer: Draw the vector addition triangle vPE = vPA + vAE . The
velocity of the plane, with respect to the air, is directed north. The velocity of the air with respect
to the ground points E 28° S. The third side of this scalene triangle is the velocity of the plane
with respect to the ground. It points mostly north and somewhat east. You can solve for the
ground velocity of the plane using the component method or the algebraic method of vector
addition.
58. (a) Given: angle of inclination θ = 15°
Required: FBD
Analysis: There are four forces acting on the box: force of gravity (down), normal force of plane
(perpendicular to the plane), applied force (up the plane), and friction (down the plane). Since the
acceleration is up along the plane, use this as the positive x-direction. Use up and perpendicular
to the plane as the positive y-direction.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-15
Solution:
(b) By Newton’s second law, the net force acts in the direction of the acceleration. The net force
acts up along the plane.
59. Answers may vary. Sample Answer: Make a system diagram of the situation showing all of
the forces acting on the object. Draw a FBD for the object, with the tail of each of the force
vectors starting at the point that represents the object. If you know the magnitude of a force or
that it balances another force, you may try to draw its arrow to scale. Label each force with its
name and, if possible, magnitude. Select an axis system where one direction is parallel to the
motion or, if the net force is zero, to some of the important forces. Calculate the components of
the net force by summing the components of the individual forces: ΣFx = F1x + F2x + F3x + ... and
ΣFy = F1y + F2y + F3y + ... . Finally determine the net force from its components using the
!
⎛ ΣF ⎞
Pythagorean theorem and tangent ratio: !F = (!Fx )2 + (!Fy )2 and θ = tan −1 ⎜ y ⎟ .
⎝ ΣFx ⎠
60. Answers may vary. Sample Answer: The motion of the piano depends on the net force acting
on it. When the piano is at rest, Newton’s first law states that the net force must be zero. By
Newton’s second law, at some time there must be a net upward force if the piano is starting to
move up. There are many possible pulley and cable systems for lifting a piano so we can’t
specify the details of the net force here. At the very least there will be one or several vertical
cables. The sum of their tensions must balance the weight of the piano when it is at rest or
moving at constant speed, and be greater than the weight of the piano when it accelerates up.
There are probably other cables that are directed down from the piano and to the sides. These are
used to prevent the piano from swinging as it is lifted. The horizontal components of these
tensions cancel while the vertical components will cause an increase in the tension of the lifting
cables to more than just the weight of the piano.
61. Kinetic friction occurs between two surfaces when one surface slides over the other surface.
Static friction occurs between two surfaces when the surfaces grip each other and have no
relative motion.
62. When the elevator accelerates upward, the normal force increases, resulting in a greater
apparent weight.
63. (a) To an observer on the bus, the book appears to be moving due to a force that pushes it in
the opposite direction of the direction of the bus. However, it is the acceleration of the frame of
reference (the bus) that causes the motion, so the apparent force on the book is fictitious.
(b) If the book does not move, then from Earth’s reference frame, it accelerated into the right
turn. The cause of this acceleration is the force of static friction preventing the book from sliding
in the direction of the bus before the turn.
64. Centripetal acceleration maintains a constant speed with an ever-changing velocity as an
object travels in a circular path. Acceleration along the direction of motion creates an everchanging speed but maintains a constant direction (assuming positive acceleration).
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-16
65. Answers may vary. Sample answers: Centripetal forces fix satellites in their orbits (gravity),
move cars around corners (static friction), and keep metal balls in the hammer throw from flying
away from Olympic athletes (tension).
66. (c), (a), (d), (b)
The distance from the centre of the planet to the spacecraft, r, must increase because the
acceleration caused the velocity, v, to increase. Since the centripetal force due gravity is constant,
mv 2
these two values vary with each other according to the equation for centripetal force: Fc =
r
67. (a) The rotating spacecraft is a non-inertial frame of reference. Astronauts in the spacecraft
experience a centrifugal force pushing them toward the outer wall. This fictitious force acts as an
artificial gravity.
(b) To use the spacecraft to simulate Earth’s gravity, choose the speed of rotation, v, such that
v2
= 9.8 m/s2, where r is the distance from the centre of rotation.
r
Analysis and Application
68. (a) Given: position–time graph
!
Required: !d
! ! !
Analysis: Read the initial and final positions from the graph. Use !d = d f " di to calculate the
total displacement with west as the positive direction.
Solution:
The initial point on the graph is (0 s, 0 m). The final point is (10 s, 220 m).
The displacement is
! ! !
!d = d f " di
= 220 m " 0 m
!
!d = 220 m
Statement: The total displacement of the car is 220 m [W].
!
(b) Given: position–time graph; !d = 220 m [W]
!
Required: vav
!
!d
!
to calculate the average velocity.
Analysis: Use vav =
!t
!
!d
!
Solution: vav =
!t
220 m [W]
=
10 s
!
vav = 22 m/s [W]
Statement: The average velocity of the car is 22 m/s [W].
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-17
69. (a) Given: velocity–time graph
!
!
Required: time t when vav = vinst
!
!d
!
Analysis: The average velocity is given by vav =
. To determine the displacement we need to
!t
calculate the area under the velocity–time graph. We will then calculate the average velocity and
use the graph to find the time with this instantaneous velocity.
An alternate method to determine the average velocity may be used in this question because the
truck’s motion involves constant acceleration. Only under very special circumstances is the
average velocity the same as the average of the initial and final velocities—constant acceleration
! !
vi + vf
!
is one such situation. So vav =
is valid here. We will verify that both methods yield the
2
same result.
Solution:
Calculate the area under the velocity–time graph:
Calculate
! the average velocity:
!d
base × height
!
vav =
Δd =
!t
2
100 m [W]
(10 s)(20 m/s)
=
=
10 s
2
!
vav = 10 m/s [W]
= 100 m
Calculate the average of the velocities:
! !
vi + vf
!
vav =
2
20 m/s [W] + 0 m/s
=
2
!
vav = 10 m/s [W]
As expected, both methods give the same value for average velocity. Look at the velocity–time
graph and determine when the instantaneous velocity is 10 m/s. This occurs at t = 5 s , which is
the half-time point only because of the constant acceleration.
Statement: The average velocity matches the instantaneous acceleration at 5 s.
(b) Given: velocity–time graph
!
Required: a
Analysis: The velocity–time graph is straight showing constant acceleration. The acceleration is
Δv
the slope of this line, a =
. Use west as the positive direction.
Δt
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-18
Solution:
Two points on the line are (0 s, 20 m/s) and (10 s, 0 m/s).
Calculate the slope,
!v
a=
!t
0 m/s " 20 m/s
=
10 s " 0 s
a = "2 m/s 2
!
a = 2 m/s 2
Statement: The magnitude of the acceleration is 2 m/s2.
70. Given: vi = 0 m/s; ∆t = 25 s; ∆d = 7.8 × 102 m
!
Required: a
Analysis: We know the initial speed, the time interval and the distance covered. The appropriate
formula for acceleration is Δd = vi Δt + 12 aΔt 2 . Since vi is 0, the equation can be rearranged to
2!d
.
!t 2
Solution:
2!d
a= 2
!t
2(780 m)
=
(25 s)2
a=
a = 2.5 m/s 2
Statement: The magnitude of the race car’s acceleration is 2.5 m/s2.
71. Given: vi = 0 m/s; a = 4.8 m/s 2 ; Δd = 10.0 m
Required: vf
Analysis: We know the initial speed, the acceleration and the distance. The appropriate formula
for the final speed is vf 2 = vi 2 + 2aΔd .
Solution:
vf 2 = vi 2 + 2aΔd
vf = 2(4.8 m/s 2 )(10.0 m)
vf = 9.8 m/s
Statement: The sprinter’s final speed is 9.8 m/s.
!
!
!
72. Given: vi = 50.0 m/s [down]; a = 12.5 m/s 2 [up]; vf = 10.0 m/s [down]
Required: Δt
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-19
Analysis: We know both the initial and final velocities, and the acceleration. The appropriate
Δv
formula for the time interval is a =
. Use up as the positive direction.
Δt
!v
a=
!t
!v
!t =
a
Solution:
!v
!t =
a
("10.0 m/s) " ("50.0 m/s)
=
12.5 m/s 2
!t = 3.2 s
Statement: The acceleration occurs over 3.2 s.
73. Given: ∆t = 0.5 s; ∆dy = 1.225 m at maximum
Required: vi
Analysis: The ball’s speed is zero when it is at its maximum height. Use Δd y = v1y Δt − 12 g Δt 2 to
determine the ball’s initial speed.
!d y = v1y !t " 12 g!t 2
v1y =
!d y + 12 g!t 2
!t
Solution:
!d y + 12 g!t 2
v1y =
!t
1.225 m + (4.9 m/s 2 )(0.5 s)2
=
0.5 s
v1y = 4.9 m/s
Statement: The student threw the ball vertically with initial speed of 4.9 m/s.
!
74. Given: !d = 125 m [E 60.0° N]
Required: !d x ; !d y
!
!
!
Analysis: The components of the vector !d are given by !d x = !d cos" and !d y = !d sin " .
Use east and north as positive.
Solution:
!
!d x = !d cos"
= (125 m)cos60.0°
!d x = 62.5 m
!
!d y = !d sin "
= (125 m)sin60.0°
!d y = 108 m
Statement: The components of the student’s displacement are 62.5 m [E] and 108 m [N].
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-20
!
!
!
75. (a) Given: !d1 = 125 km [E]; !d2 = 85.0 km [E 45.0° N]; !d3 = 94.0 km [S]; !t = 6.75 h
!
Required: !d
Analysis: Calculate the components of the total displacement by adding the components of the
three displacement vectors, Δd x = Δd1x + Δd2 x + Δd3 x and Δd y = Δd1 y + Δd 2 y + Δd3 y . Then
!
$ #d '
!
!
y
combine these components using !d = !d x2 + !d y2 and ! = tan "1 & ! ) to calculate !d . Use
&% #d x )(
east and north as positive.
Solution:
Determine the components of the total displacement.
!d x = !d1x + !d2 x + !d3x
!d y = !d1y + !d2 y + !d3 y
= (125 m) + (85.0 km)cos 45.0° + (0 km)
!d x = 185.10 m (two extra digits carried)
Determine the total displacement.
!
!d = !d x2 + !d y2
= (185.10 km)2 + (33.896 km)2
= 188.18 km (two extra digits carried)
!
!d = 188 km
= (0 km) + (85.0 km)sin 45.0° + ("94.0 km)
!d y = "33.896 km (two extra digits carried)
!
$ #d '
y
! = tan "1 & ! )
&% #d x )(
$ 33.896 km '
= tan "1 &
)
% 185.10 km (
! = 10.4°
Statement: The total displacement is 188 km [E 10.4° S].
!
(b) Given: !d = 188.18 km [E 10.4° S]; !t = 6.75 h
!
Required: vav
!
!d
!
Analysis: Use the definition vav =
with the answer from part (a).
!t
Solution:
!
!d
!
vav =
!t
188.18 km [E 10.4° S]
=
6.75 h
!
vav = 27.9 km/h [E 10.4° S]
Statement: The average velocity of the sailboat is 27.9 km/h [E 10.4° S].
r
r
r
(c) Given: Δd1 = 125 km [E]; Δd2 = 85.0 km [E 45.0° N]; Δd3 = 94.0 km [S]; Δt = 6.75 h
Required: vav
Δd
Analysis: The average speed is calculated from the total distance travelled using vav =
. The
Δt
total distance is the sum of the distances of the individual legs of the trip, Δd = Δd1 + Δd 2 + Δd3 .
We will calculate the total distance and then the average speed.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-21
Δd
Δt
= 125 km + 85.0 km + 94.0 km
304.0 km
=
Δd = 304.0 km
6.75 h
vav = 45.0 km/h
Statement: The average speed of the sailboat is 45.0 km/h.
!
!
76. Given: vi = 65 m/s [S]; vf = 75 m/s [E]; !t = 15 s
!
Required: aav
Analysis: We know the initial and final velocities and the time interval. The appropriate formula
!
!v
!
to calculate the acceleration is aav =
. We will use vector subtraction by components to
!t
!
determine the change in velocity !v . Use east and north as positive.
Solution:
r
The components of Δv are
Δv y = vfy − viy
Δvx = vfx − vix
Δd = Δd1 + Δd 2 + Δd3
vav =
= 75 m/s − 0 m/s
Δvx = 75 m/s
The change in velocity is
= 0 m/s − (−65 m/s)
Δv y = 65 m/s
!
!v = !vx2 + !v 2y
= (75 m/s)2 + (65 m/s)2
!
!v = 99.25 m/s (two extra digits carried)
The average acceleration is
!
!v
!
aav =
!t
99.25 m/s [E 41° N]
=
15 s
!
2
aav = 6.6 m/s [E 41° N]
$ #v! '
y
! = tan & ! )
&% #vx )(
"1
$ 65 m/s '
= tan "1 &
)
% 75 m/s (
! = 41°
Statement: The average acceleration of the airplane is 6.6 m/s2 [E 41° N].
77. (a) Given: !d y = 98 m [down]; viy = 0 m/s; vx = 2.9 × 102 m/s [E]
Required: Δt
Analysis: We know the displacement and the initial velocity for the vertical motion of a
1
projectile. The appropriate formula for the time taken is !d y = v1y !t " g!t 2 . Use up and east as
2
positive.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-22
1
!d y = v1y !t " g!t 2
2
!t =
"2!d y
g
Solution:
"2!d y
!t =
g
2(98 m)
9.8 m/s 2
= 4.472 s (two extra digits carried)
=
!t = 4.5 s
Statement: It takes the projectile 4.5 s to land.
(b) Given: !d y = 98 m [down]; viy = 0 m/s; vx = 2.9 " 102 m/s [E]; !t = 4.472 s
Required: Δd x
Analysis: The horizontal velocity of the projectile is constant and we calculated the time to fall
in part (a). Use Δd x = vx Δt to find the horizontal distance, or range, travelled by the projectile.
Solution:
!d x = vx !t
"
m%
= $ 290 ' (4.472 s )
#
s&
= 1296 m
!d x = 1.3 km
Statement: The range of the projectile is 1.3 km.
!
78. (a) Given: !d y = 98 m [down]; vi = 2.9 " 102 m/s [S 24° up]
Required: Δt
Analysis: We know the displacement and can calculate the initial vertical velocity for the
1
projectile. The appropriate formula for the time taken is !d y = v1y !t " g!t 2 . We will need to
2
use the quadratic formula Δt =
and south as positive.
−b ± b2 − 4ac
to solve the resulting equation for Δt . Use up
2a
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-23
Solution: The equation for Δt is
1
!d y = v1y !t " g!t 2
2
"98 m = (290 m/s)sin 24°!t " (4.9 m/s 2 )!t 2
(4.9 m/s 2 )!t 2 " (117.95 m/s)!t " 98 m = 0
Solve for Δt .
!t =
"b ± b2 " 4ac
2a
(117.95 m/s) ± (117.95 m/s)2 " 4(4.9 m/s 2 )("98 m)
=
2(4.9 m/s 2 )
= 24.88 s (two extra digits carried)
!t = 25 s
Statement: The projectile lands after 25 s.
!
(b) Given: !d y = 98 m [down]; vi = 2.9 " 102 m/s [S 24° up]; !t = 24.88 s
Required: Δd x
Analysis: The horizontal velocity of the projectile is constant and we calculated the time to fall
in part (a). Use Δd x = vx Δt to find the horizontal distance, or range, travelled by the projectile.
Solution:
The horizontal component of velocity is
The range of the projectile is
2
vx = (2.9 ! 10 m/s)cos 24°
Δd x = vx Δt
= (264.93 m/s)(24.88 s)
vx = 264.93 m/s
= 6591 m
Δd x = 6.6 km
Statement: The range of the projectile is 6.6 km.
r
(c) Given: Δd y = 98 m [down]; vi = 2.9 ×102 m/s [S 45° up]
Required: Δd x
Analysis: We will repeat the calculations of parts (a) and (b) with the revised launch angle. First
we will determine the time for the projectile to land. Then we will calculate its range.
Solution:
The equation for Δt is
Δd y = v1y Δt − 12 g Δt 2
−98 m = (290 m/s)sin 45°Δt − (4.9 m/s 2 ) Δt 2
(4.9 m/s 2 )Δt 2 − (205.06 m/s)Δt − 98 m = 0
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-24
Solve for Δt .
Δt =
−b ± b 2 − 4ac
2a
(205.06 m/s) ± (205.06 m/s) 2 − 4(4.9 m/s 2 )( −98 m)
2(4.9 m/s 2 )
= 42.32 s (two extra digits kept)
Δt = 42 s
The horizontal component of velocity is
The range of the projectile is
!d x = vx !t
vx = (290 m/s) cos 45°
=
vx = 205.06 m/s
= (205.06 m/s)(42.32 s)
= 8678 m
!d x = 8.7 km
Statement: With a launch angle of 45°, the range of the projectile is 8.7 km.
(d) Answers may vary. Sample answer: The range equation for a projectile landing at the same
v 2 sin 2θ
height as it was launched is Δd x = i
. This range has its maximum possible value when
g
the launch angle is 45°. We will use this formula to determine the horizontal distance off the cliff
where the projectile is once again level with the cliff.
v 2 sin 2θ
Δd x = i
g
(290 m/s) 2 sin 90°
9.8 m/s 2
Δd x = 8.6 km
This range is just 100 m short of the full range. Most of the trajectory of the projectile is
described by the simpler range formula. This suggests that the same optimization conditions
should apply, at least approximately, to the real projectile—that is the maximum range occurs at
a launch angle of 45°.
!
!
79. (a) Given: vPA = 800.0 km/h [E]; vAG = 75.0 km/h [W 30.0° S]
!
Required: vPG
Analysis: The ground velocity of the plane is the vector sum of its air velocity and the wine
!
!
!
r
velocity, vPG = vPA + vAG . Use vector addition by components to determine vPG . Use east and
north as positive.
Solution: The components of the ground velocity are
vPGx = vPAx + vAGx
vPGy = vPAy + vAGy
=
vPGx
= 800.0 km/h + (−75.0 km/h) cos 30.0°
= 0 km/h + (−75.0 km/h)sin 30.0°
= 735.1 km/h
vPGy = −37.5 km/h
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-25
The ground velocity is
#v
PGy
! = tan %
%$ vPGx
r
vPG = (vPG x ) 2 + (vPG y ) 2
"1
= (735.1 m/s) 2 + (37.5 m/s) 2
&
(
('
# 37.5 m/s &
= tan "1 %
(
$ 735.1 m/s '
r
vPG = 736 km/h
! = 2.92°
Statement: The ground velocity of the plane is 736 km/h [E 2.92° S].
!
(b) Given: vPG = 736 km / h [E 2.92° S]; vPGy = !37.5 km/h; "t = 3.0 h
Required: Δd y
Analysis: The plane is flying mostly east with a small component of its velocity to the south.
Over the 3.0 h, the plane will drift south with at this velocity. The distance is given by
Δd y = vPGy Δt . Continue to use east and north as positive.
Solution:
!d y = vPGy !t
#
km &
= % "37.5
( (3.0 h)
$
h '
!d y = "112 km
Statement: The plane is displaced 112 km south of its due east course.
!
80. Given: vCW = 5.0 m/s; vWE = 3.0 m/s [S]; !d1 = 4.0 km [N]; !d2 = 5.0 km [S]
Required: Δt
!
!
!
Analysis: The canoe trip has two segments. For each segment, use vCE = vCW + vWE to determine
the speed of the canoe with respect to the Earth. Then use the constant speed relation Δd = vΔt
to determine the time taken for each segment of the trip. Finally use Δt = Δt1 + Δt2 to calculate
the total time taken. Use south as the positive direction.
Solution:
Upstream segment of the trip:
!d1 = vCE !t1
vCE = vCW + vWE
vCE
= (−5.0 m/s) + 3.0 m/s
= −2.0 m/s
!t1 =
=
!d1
vCE
"4000 m
"2.0 m /s
!t1 = 2000 s
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-26
Downstream segment of the trip:
vCE = vCW + vWE
vCE
= 5.0 m/s + 3.0 m/s
= 8.0 m/s
!d2 = vCE !t2
!t2 =
=
Calculate the total time of the trip:
!t = !t1 + !t2
!d2
vCE
5000 m
8.0 m /s
!t2 = 625 s
= 2000 s + 625 s
= 2625 s "
1 min
1h
"
60 s
60 min
!t = 0.73 h
Statement: It takes the student
2.6 × 103 s to complete the trip, or 0.73 h.
!
81. (a) Given: ! = 25°; Fg = 6.8 × 103 N
Required: FBD for the car
Analysis: There are three forces to draw on the FBD of the car: the force of gravity (down), the
normal force (perpendicular to the plane) and the applied force (up the plane). Use up the plane
as the positive x-direction and perpendicular to the plane as the positive y-direction.
Solution:
!
(b) Given: ! = 25°; Fg = 6.8 × 103 N
!
Required: minimum magnitude of Fa to pull the car up the plane
Analysis: For the car to move up the plane, the net force on the car must act up the plane or be
zero. Use !Fx " 0 N to determine the minimum applied force.
Solution:
"Fx # 0 N
Fa $ Fg sin ! # 0 N
Fa # (6.8 %103 N) sin 25°
Fa # 2.9 %103 N
Statement: The applied force must be 2.9 × 103 N or greater to pull the car up the plane.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-27
82. Given: Fa space = 1.2 × 103 N; a = 2.4 m/s2
Required: Fa Earth
Analysis: From the given applied force and acceleration, we can determine the mass of the space
probe using ΣF = ma . With the mass and the acceleration, we can find the applied force needed
on Earth, again using ΣF = ma .
Solution:
In space
On Earth
ΣF = ma
ΣF = ma
Fa space = ma
Fa Earth − mg = ma
Fa Earth = m(a + g )
1.2 ×103 N
2.4 m/s 2
m = 500 kg
m=
= (500 kg)(2.4 m/s 2 + 9.8 m/s 2 )
Fa Earth = 6.1×103 N
Statement: The force needed to accelerate
the probe on Earth is 6.1 × 103 N.
!
83. Given: mA = 57 kg; mB = 75 kg; FA on B = 135 N [forward]
Required: aA ; aB
Analysis: By Newton’s third law, the skaters push each other with an action-reaction pair of
forces of magnitude 135 N. When skater A pushes forward on!B, skater B pushes back on A.
!
Each skater accelerates according to Newton’s second law, !F = ma . Use forward as the
positive direction.
Solution:
Skater B
Skater A
FA on B = mB aB
FB on A = mA aA
aB =
FA on B
mB
aA =
=
135 N
75 kg
=
FB on A
mA
−135 N
57 kg
aB = 1.8 m/s 2
aB = −2.4 m/s 2
Statement: The less massive skater has an acceleration of 2.4 m/s2, while the more massive has
an acceleration of 1.8 m/s2.
!
84. (a) Given: m1 = 4.26 × 102 kg; !1 = 18.0°; ! 2 = 33.0°; "F = 0 N
Required: FT
Analysis: Draw a FBD for crate 1 showing the force of gravity (down), the normal force
(perpendicular to the plane) and the force of tension (up the plane). Align the axes down along
and perpendicular to the plane. The system is in equilibrium so we can determine the force of
tension using ΣFx = 0 N .
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-28
Solution: Equilibrium for crate 1
!Fx = 0 N
m1g sin "1 # FT = 0 N
FT = m1g sin "1
= (426 kg)(9.8 m/s 2 )sin18.0°
= 1290 N (two extra digits carried)
FT = 1.29 $ 103 N
Statement: The tension in the cable is 1.29 × 103 N.
!
(b) Given: m1 = 4.26 × 102 kg; !1 = 18.0°; ! 2 = 33.0°; "F = 0 N; FT = 1290 N
Required: m2
Analysis: Draw a FBD for crate 2 showing the force of gravity (down), the normal force
(perpendicular to the plane) and the force of tension (up the plane). Align the axes up along and
perpendicular to the plane. The system is in equilibrium so we can determine the mass for crate 2
using ΣFx = 0 N .
Solution: Equilibrium for crate 2
!Fx = 0 N
FT " m2 g sin # 2 = 0 N
m2 =
=
FT
g sin # 2
1290 N
(9.8 m/s 2 )sin33.0°
m2 = 2.42 $ 102 kg
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-29
Statement: To keep the system in equilibrium the mass of the second crate must be
2.42 × 102 kg.
(c) Given: m1 = 4.26 × 102 kg; !1 = 18.0°; m2 = 1.95 " 102 kg; ! 2 = 33.0°
Required: a
Analysis: The mass of crate 2 is less than that required for equilibrium. So the system will
accelerate to the left and the tension in the cable will change from that found in part (a). Use
ΣFx = max for each crate to find two equations relating the acceleration and the tension. Then
solve these equations for the acceleration of the system.
Solution:
Equation for crate 1:
Equation for crate 2:
!Fx = max
!Fx = max
m1g sin "1 # FT = m1a
FT " m2 g sin # 2 = m2 a
(426 kg)(9.8 m/s 2 )sin18.0° # FT = m1a
FT " (195 kg)(9.8 m/s 2 )sin33.0° = m2 a
1290 N # FT = m1a
FT " 1041 N = m2 a
Add the two equations to eliminate the cable tension. Solve for the acceleration.
m1a + m2 a = 1290 N ! 1041 N
a=
249 N
621 kg
a = 0.401 m/s 2
Statement: The magnitude of the acceleration of the system is 0.401 m/s2.
85. (a) Given: m = 2.65 ! 102 kg; angle of ramp " = 30.0°; µs = 0.45;
angle of applied force ! = 39.0°
Required: minimum Fa to overcome static friction
Analysis: Draw a FBD of the box showing the following forces: the force of gravity (down), the
normal force of ramp (up perpendicular to ramp), the force of static friction (down ramp) and the
applied force (up φ = 39.0° from ramp). Use axes aligned with and perpendicular to the ramp.
For the minimum possible applied force, the force will! just overcome the force of static friction.
So we know that all the force balance. We will use !F = 0 N in the x- and the y-directions to
relate the four forces and solve for the magnitude of the applied force.
Solution:
Equation for y-direction
!Fy = 0 N
Equation for x-direction
!Fx = 0 N
Fay + FN " mg cos# = 0 N
Fax " mg sin # " Fs = 0 N
FN = mg cos# " Fay
Copyright © 2012 Nelson Education Ltd.
Fax = mg sin # + µs FN
Unit 1: Dynamics
U1-30
Substitute the first equation into the second equation and solve for the magnitude of the applied
force.
Fax = mg sin ! + µs FN
Fax = mg sin ! + µs (mg cos! " Fay )
Fa cos# = mg sin ! + µs (mg cos! " Fa sin # )
Fa =
mg(sin ! + µs cos! )
cos# + µs sin #
(2.65 $ 102 kg)(9.8 m/s 2 )(sin30.0° + 0.45cos30.0°)
cos39.0° + 0.45sin39.0°
= 2109 N (two extra digits carried)
=
Fa = 2.1$ 103 N
Statement: The minimum applied force to slide the box is 2.1 × 103 N.
r
(b) Given: m = 2.65 ×102 kg; angle of ramp θ = 30.0°; Fa = 0 N
Required: minimum µS for system to be in static equilibrium
Analysis: The breaking value for static friction is given by FS = µS FN . The static friction acts up
r
along the ramp to hold the box in place. Use ΣF = 0 N in the x- and the y-directions to relate the
three forces and solve for the coefficient of static friction.
Solution:
Equation for y-direction
Equation for x-direction
!Fx = 0 N
ΣFy = 0 N
FN − mg cos θ = 0 N
FN = mg cos θ
FS " mg sin # = 0 N
µS FN = mg sin #
µS =
mg sin #
mg cos#
= tan30.0°
µS = 0.577
Statement: The minimum value for the coefficient of static friction is 0.577.
86. Given: ∆t = 3.4 s; vf = 7.4 m/s; m = 56 kg
Required: FN
Analysis: Determine the upward acceleration using vf = a∆t. Use up as positive and solve for the
normal force when +FN + (–mg) = ma.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-31
Solution: Determine the upward acceleration:
vf = a !t
v
a= f
!t
7.4 m/s
=
3.4 s
= 2.176 m/s 2 (two extra digits carried)
a = 2.2 m/s 2
Determine the apparent weight:
+ FN + !mg = ma
FN = ma + mg
(
)
(
)(
) (
)(
= 56 kg 2.176 m/s 2 + 56 kg 9.8 m/s 2
2
FN = 6.7 " 10 N
)
Statement: The passenger’s apparent weight is 6.7 × 102 N.
87. Given: r = 6.378 × 106 m; ac = g
Required: T
4! 2 r
Analysis: ac = 2
T
4! 2 r
g= 2
T
4! 2 r
T=
g
Solution: T =
4! 2 r
g
4! 2 ( 6.378 " 106 m )
#
m&
%$ 9.8 2 ('
s
1 min
1h
= 5068 s "
"
60 s
60 min
T = 1.408 h
Statement: The period of rotation would have to become 1.408 h.
88. Given: r = 1.4 m; ac = g
Required: minimum v
v2
Analysis: ac = ; In order to complete the loop, the centripetal force must be at least equal to
r
the gravitational acceleration.
=
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-32
v2
r
v2
g=
r
ac =
v = gr
Solution: v = gr
= ( 9.8 m/s 2 )( 6.53 m )
v = 8.0 m/s
Statement: The skateboarder’s speed must be at least 8.0 m/s.
89. (a) The centripetal acceleration is equal to 90.0 % of gravity on Earth’s surface:
ac = 0.900g
= 0.900 ( 9.8 m/s 2 )
ac = 8.8 m/s 2
The centripetal acceleration is 8.8 m/s2.
(b) Given: r = 57 m; ac = 0.900g
Required: T
4! 2 r
Analysis: ac = 2
T
T=
4! 2 r
ac
Solution: T =
4! 2 r
ac
=
4! 2 (57 m )
0.900 ( 9.8 m/s 2 )
T = 16 s
Statement: The period of rotation is 16 s.
90. (a) Given: f = 22.5 Hz; r = 51.5 cm or 0.515 m; m = 0.656 kg
Required: v
Analysis: v = 2πrf
Solution: v = 2! rf
= 2! ( 0.515 m )( 22.5 Hz )
v = 72.8 m/s
Statement: The speed of the mass is 72.8 m/s.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-33
(b) Given: f = 22.5 Hz; r = 51.5 cm or 0.515 m; m = 0.656 kg
Required: Fc
Analysis: Fc = 4! 2 mrf 2
Solution: FT = 4! 2 mrf 2
= 4! 2 ( 0.656 kg )( 0.515 m )( 22.5 Hz )
FT = 6.75 " 103 N
2
Statement: The magnitude of the centripetal force is 6.75 × 103 N.
91. Given: r = 15.2 m; ac = 40g
Required: v
Analysis: Fc = 40Fg
mv 2
= 40mg
r
v2
= 40g
r
v = 40gr
Solution: v = 40gr
= 40 ( 9.8 m/s 2 )(15.2 m )
v = 77 m/s
Statement: The speed of the centrifuge is 77 m/s.
92. (a) Given: r = 10.0 m
Required: v
v2
Analysis: ac = ; The minimum speed is when the centripetal acceleration equals g.
r
ac = g
v2
=g
r
v = gr
Solution: v = gr
= ( 9.8 m/s 2 )(10.0 m )
v = 9.90 m/s
Statement: The minimum speed of the roller coaster on a clothoid loop is 9.90 m/s.
(b) Given: d = 40.0 m or r = 20.0 m
Required: v
Analysis: The minimum speed is when the centripetal acceleration equals g.
ac = g
v2
=g
r
v = gr
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-34
Solution: v = gr
= ( 9.8 m/s 2 )( 20.0 m )
v = 14.0 m/s
Statement: The minimum speed of the roller coaster on a circular loop is 14.0 m/s.
(c) A clothoid loop have a smaller radius of curvature than a circular loop of the same height.
The smaller r-value means that the roller coaster can travel at slower speeds and still pass
through the loops.
Evaluation
93. Answers may vary. Sample answers:
I would write a computer program to perform vector addition of N vectors. The underlying
method is that of vector addition by components.
Step 1: The user specifies whether the vectors are given in components (if so, proceed to step 2).
If not, for each vector, transform the direction angle to standard mathematical form as measured
counter-clockwise from the x-axis. Then compute the x- and y-components of the vector using
vx = v cos !
v y = v sin !
With the angles in standard form, the signs of the components are correct automatically.
Step 2: Sum all of the x-components to determine the x-component of the final vector. Sum all of
the y-components to determine the y-component of the final vector.
Step 3: Convert the components of the final vector to magnitude and an angle using
# vy $
!
&
v = vx 2 + v y 2 and ! = tan "1 %
% vx &
'
(
Step 4: Interpret the resulting angle using the CAST rules from mathematics to give the direction
of the final vector.
94. (a) Given: aav = 6.37 m/s 2 ; v = 35.0 m/s; "! = 45°
Required: minimum !t without slipping
Analysis: The car is travelling on an arc of a circle at constant speed. To relate the car’s
maximum acceleration to a time interval we need to determine the change in velocity vector. The
r r
r
triangle for vf = vi + Δv has two sides of length 35.0 m/s separated by 45°. Use the cosine law
r
C 2 = A2 + B 2 − 2 AB cos θ to find the magnitude of Δv . Then use the definition of average
r
Δv
r
acceleration aav =
to calculate the required time interval.
Δt
!v
a=
!t
!v
!t =
a
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-35
Solution:
!
! !
!
Solve the triangle of vf = vi + !v for !v = !v :
C 2 = A2 + B 2 ! 2 ABcos"
#v 2 = v 2 + v 2 ! 2v 2 cos"
#v = 2(35.0 m/s)2 (1! cos 45°)
#v = 26.79 m/s (two extra digits carried)
The maximum acceleration on the curve corresponds to the minimum time interval. Find the
time interval.
!v
!t =
a
26.79 m/s
=
6.37 m/s 2
= 4.206 s (two extra digits carried)
!t = 4.2 s
Statement: The minimum time for the driver to follow the curve without slipping is 4.2 s.
(b) Given: Δt = 4.206 s for 45°
Required: time to complete whole circle, T
Analysis: The arc of road from part (a) corresponds to one eighth of a complete circle. Multiply
the time interval Δt to determine the time required to complete the whole circle.
Solution:
T = 8!t
= 8(4.206 s)
= 33.65 s (two extra digits carried)
T = 34 s
Statement: It would take the car 34 s to complete the circle without slipping.
(c) Given: v = 35.0 m/s; T = 33.65 s
Required: radius of circle, R
Analysis: The car moves at constant speed in a circle. Use Δd = vΔt to calculate the
circumference of the circle. Then use C = 2π R to determine the radius.
Solution:
!d = v!t
C = 2π R
m
C
R=
C = (35.0 )(33.65 s )
2π
s
1178 m
C = 1178 m (two extra digits carried)
=
2π
R = 190 m
Statement: The radius of the circular track is 190 m.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-36
95. Answers may vary. Sample answer: A building as a whole is subject to a variety of forces.
The walls of the building are subject to the varying forces of the wind. The structure of the
building is subject to the weight of the objects and people in the building. The foundation of the
building is subject to the upward normal force of the ground beneath holding it up; so, the
designers need to know if they are building on bedrock or bog. In addition, individual parts of
the building are subject to forces exerted by other parts of the building. For example, the roof
structure needs to support the elevator mechanism. The following FBD represents the building as
a whole.
96. (a) Given: mtot = 2.0 × 106 kg; Fa = 2.5 × 107 N
Required: a
Analysis: Two forces act on the space shuttle, the force of gravity and the applied force of the
rockets. Use ΣF = ma to determine the acceleration. Use up as positive.
!F = ma
Fa " mg = ma
a=
Fa
"g
m
Solution:
F
a= a !g
m
2.5 " 107 N
=
! 9.8 m/s 2
6
2.0 " 10 kg
a = 2.7 m/s 2
Statement: The shuttle accelerates at 2.7 m/s2.
(b) Given: vi = 0 m/s; a = 2.7 m/s 2 ; !t = 5.0 s
Required: vf
Analysis: The appropriate formula is vf = vi + aΔt . Continue to use up as positive.
Solution:
vf = vi + a!t
"
m%
= $ 2.7 2 ' (5.0 s)
#
s &
"
m%
3600 s
1 km
(
= $ 13.5
(
'
1h
s & 1000 m
#
vf = 49 km/h
Statement: The speed of the shuttle after 5.0 s is 49 km/h.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-37
(c) Given: vi = 0 m/s; a = 2.7 m/s 2 ; !t = 15 s
Required: vf
Analysis: The appropriate formula is vf = vi + aΔt . Continue to use up as positive.
Solution:
vf = vi + a!t
"
m%
= $ 2.7 2 ' (15 s)
#
s &
"
m%
3600 s
1 km
(
= $ 40.5
(
'
1h
s & 1000 m
#
vf = 150 km/h
Statement: The speed of the shuttle after 15 s is 150 km/h.
97. Answers may vary. Sample answer:
(a) I will discuss the fundamentals of driving using Newton’s laws. Newton’s first law says that
the car will stay at rest or continue moving at a constant speed when all of the forces balance
each other. I see this when I have to maintain a constant pressure on the accelerator (forward
force of static friction) to counter air resistance in order to maintain a constant speed. When I
park on a hill, I need to use the parking brake and leave the car in gear to provide enough friction
to balance the force of gravity down hill. Newton’s second law states that my acceleration is
determined by the net force on the car. I certainly see this as I accelerate from rest or when I
brake. I also see this as I drive up or down a hill. If I don’t adjust the amount of gas, I either slow
down or speed up respectively. A similar situation happens if there is a strong headwind or
tailwind. Newton’s third law discusses action–reaction pairs of forces. I am able to drive forward
because my tires push back on the road and the road pushes the car forward. This is really
obvious if I accelerate from rest on a gravel surface—sometimes gravel kicks out behind the car.
Braking is another situation where there is an action–reaction pair of forces. Slowing my tires
increases the force of static friction in a forward direction, causing the vehicle to accelerate
backward.
(b) Now I have a much better understanding of how my gasoline consumption is governed by my
driving habits and by the design of my car. The amount of gasoline needed to produce the
applied force that pushes the car depends on the acceleration I demand or the highway speed I
choose. I have driven in a hybrid car watching the engine switch from battery to gas going uphill
and from gas to battery going downhill. I also saw how wind conditions affect the gas/battery
balance. This made me realize the importance of air or wind resistance when designing a vehicle.
98. Solutions may vary. Sample solution:
(a) Given: Fg = 2.5 × 103 N; ∆d = 22 m
Required: FW
Analysis: Looking at the lower pulley, there are four ropes supporting the weight of crate. This
is the same factor of four that we are given between the length of rope the worker has to pull
(88 m) and the height increase of crate (22 m). Assume the crate is lifted at constant velocity.
Use ΣFy = 0 N , with up as positive, to find the force the worker exerts on the rope.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-38
Solution:
!Fy = 0 N
4FW " Fg = 0 N
FW =
Fg
4
2.5 # 103 N
=
4
FW = 620 N
Statement: The worker exerts a force of 620 N on the rope.
(b) The worker pulls on the rope that is wrapped around massless frictionless pulleys. So the
force of the worker and the tension in the rope are the same. The tension in the rope is 620 N.
99. (a) Given: µK = 0.02; m = 122 kg; ! = 30.0°
Required: a
Analysis: There are three forces acting on the box: the force of gravity (down), the normal force
of the plane (up perpendicular to the plane) and the force of kinetic friction (up along the plane).
!
!
Use !F = ma , in components, to relate the forces and solve for the acceleration. Use down the
plane and up perpendicular to the plane as positive.
Solution:
Determine the magnitude of the normal force:
!Fy = 0 N
FN " mg cos# = 0 N
FN = mg cos#
= (122 kg)(9.8 m/s 2 )cos30.0°
FN = 1035 N (two extra digits carried)
Determine the acceleration:
!Fx = ma
mg sin " # FK = ma
mg sin " # µ K FN
m
(122 kg)(9.8 m/s 2 )sin30.0° # 0.02(1035 N)
=
122 kg
a=
a = 4.7 m/s 2
Statement: With the BAM coating, the box slides down the ramp at 4.7 m/s2.
(b) Given: m = 122 kg; ! = 30.0°
Required: a
Analysis: There are two forces acting on the box: the force of gravity (down) and the normal
force of the plane (up perpendicular to the plane). Use ΣFx = ma to solve for the acceleration.
Use down the plane as positive.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-39
Solution:
Equation in the x-direction
!Fx = ma
mg sin " = ma
a = g sin "
a = 4.9 m/s 2
Statement: With no kinetic friction, the acceleration of the box is 4.9 m/s2. The acceleration
with the BAM coating is only about 4 % less than that with a truly frictionless surface.
100. Answers may vary. Sample answer: The best situation for a true inertial frame of reference
is empty space so far from any other object that there are no outside gravitational forces on
objects in the frame of reference. In a true inertial frame of reference, objects would follow
Newton’s laws of motion perfectly and all motion would be explained by real forces acting on
the objects.
101. Answers may vary. Sample answer:
102. Answers may vary. Sample answer: In this unit, I learned that slowing a car is just
acceleration in the direction opposite to the velocity. Also, when an object accelerates in a
specific direction, that does not mean the velocity will be in the same direction. It is possible for
displacement, velocity, and acceleration to all be in different directions. For example, suppose an
object with an initial velocity directed north accelerates southwest until it is directly west of its
starting point. Its acceleration is southwest, its displacement is west, and its final velocity is
directed somewhere between the other two vectors.
103. Answers may vary. Students should discuss what they learned about force and motion. For
most objects, students will recognize that the force of gravity and the equal-and-opposite normal
force balance each other to keep an object from moving. Students should recognize that when an
object is dragged at a constant speed, the force of kinetic friction balances the equal-and-opposite
applied force.
104. Answers may vary. Students should discuss how examining the frame of reference is
important to understanding projectile motion. For example, a ball thrown forward from a moving
vehicle travels farther than a ball thrown backward with the same force because the velocity of
the vehicle contributes to the initial velocity of the ball. In terms of space exploration, relative
motion is important for determining the shortest or most efficient path between two objects,
which reduces travel time. For example, you must not aim a probe at a planet, but where the
planet will be by the time the probe reaches the planet’s orbit. Also, a probe will have a slower
approach to a planet if the two objects are moving in the same direction than if they are heading
toward each other.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
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105. Answers may vary. Students may discuss the applications of a linear actuator (which
converts energy into linear motion), research into reducing friction for skiers, or the circular
motion in a centrifuge.
106. Answers may vary. In the previous grade, students studied kinematics and forces. In this
unit, they expanded on their prior learning by studying acceleration and velocity in two
dimensions and circular motion.
Research
107. Answers may vary. Students’ answers should address the forces (and other physics
concepts) used in a selection of skateboarding maneuvers. For example, during an Ollie, or jump
with the board, the skateboarder accelerates upward by jumping. The rider simultaneously
pushes the rear of the board into the ground causing it to bounce up. Then, using static friction,
the rider guides the board up more while it is in contact with his or her front foot. The deck is
covered with grip tape, which increases the friction over a plain wood board. The frontside 180
and 360 flip are similar, but the force on the board is increased and direction is added to rotate or
twist the board. The upturn of the deck at the rear of the skateboard is a design feature that
allows riders to tilt the board at a greater angle to improve the height of an Ollie or other trick.
108. Answers may vary. Students’ presentations should focus on the importance of acceleration
in drag racing because the displacement is so short. To increase acceleration, a driver or engineer
can increase the total force forward and decrease the mass of the vehicle. The force forward can
be a combination of an applied force, for example, a rocket engine with rear exhaust that applies
an equal-and-opposite force forward, and the static friction between the tires and the ground.
Choosing tires that increase the coefficient of static friction contributes to the increase in the
forward acceleration. Another method of increasing the force of static friction is to increase the
normal force by adding mass, however, this limits the acceleration due to the applied force. In
recent years, aerodynamics has also been applied to reduce air resistance.
109. Answers may vary. Students’ presentation should discuss the physical characteristics of
BAM as well as its advantages. BAM is a super-hard synthetic ceramic that can coat other
materials in a microscopic layer. Its coefficient of kinetic friction is only 0.02. Coating
mechanical parts with BAM reduces the friction that causes wear and tear. This also increases
the efficiency of machinery by reducing the loss of thermal energy due to friction. Students may
include a cost-benefit analysis of coating machinery parts with BAM.
110. Pop bottle rockets can be made by forcing air into a plastic water bottle that contains water
already. Methods of forcing air include using an air compressor or a bicycle pump. The bottles
are then inverted and opened to release the water and air downward. Pop rockets apply Newton’s
third law of motion by creating such a massive downward force pressure that they are propelled
upward.
111. Elizabeth MacGill, born in Vancouver in 1905, was the first female aircraft designer. She
was the Chief Aeronautical Engineer at a Canadian manufacturer and improved aircraft designs
to allow for higher altitude take-offs and flights, and operation during colder conditions (deicing). After World War II, she worked on establishing the Air Worthiness regulations for
commercial aircraft, which dictated the minimum requirements for passenger planes. She also
wrote extensively on women’s rights, discussing her mother, British Columbia’s first woman
judge.
112. (a) Oscar Pistorius holds several gold medals for sprints at the Paralympics Games and has
personal bests for the 100 m, 200 m, and 400 m, all within a couple seconds of the world records
for all athletes.
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Unit 1: Dynamics
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(b) Some people dispute Oscar Pistorius’s records on the grounds that his artificial legs are more
effective for running than a human leg can be.
(c) Answers may vary. Sample answer: As prosthetic technology improves, artificial limbs may
allow for faster, higher, and stronger performances by athletes with the prosthetics than those
without.
(d) Answers may vary. Students should articulate and defend an opinion on the topic of allowing
people with artificial limbs to compete against those without. Students may address the difficulty
of perfectly matching the ability of an artificial limb with the structure of the human leg, and
therefore, the difficulty in attempting to regulate allowable prosthetics.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-42