Problem 1: Chapter 1, Case 6a-d (Pg. 73)
Two cars are driving at 88 km/h (55 mph) toward an intersection. A 1500 kg sedan is heading east and a 700 kg subcompact is heading north. The subcompact has a green light and drives into the intersection. The tired driver of the sedan runs the red light and crashes directly into the driver's side of the subcompact. The subcompact experiences a sudden, strong force to its right (toward the east).
a. Which car experiences the largest force due to the collision?
Answer: The forces on the two cars have equal amounts (equal magnitudes).
Why: In accordance with Newton's third law, the two cars exert equal but
oppositely directed forces on one another.
b. Which car experiences the largest acceleration due to the collision?
Answer: The subcompact accelerates more than than the sedan.
Why: Since the cars experience forces of equal magnitudes and the subcompact
has the smaller mass, the subcompact experiences the larger acceleration.
c. Which car is most likely to be thrown off the roadway because of
the collision?
Answer: The subcompact.
Why: The subcompact accelerates more than the sedan and experiences the
greater change in its velocity. It is thus more likely to go in a direction
it hadn't intended and drive off the roadway.
d. If the force on the subcompact is directly toward that car's right
(directly toward the east), what happens to the north/south component of
the subcompact's velocity?
Answer: The north/south component of the subcompact's velocity doesn't
change.
Why: The eastward force causes only an eastward acceleration. Since there
is no north/south component of acceleration, that portion of the car's
velocity doesn't change. It continues to coast northward at its original
rate after the collision, but it acquires an eastward component of velocity
as well.
Problem 2: Chapter 1, Case 11a-f (Pg. 74)
An escalator is essentially a moving staircase. The individual steps are supported by metal tracks that run on either side of the escalator. These steps follow one another in a complete loop, driven by an electric motor. When you step onto an escalator at the ground floor, it soon begins to carry you upward and forward at a constant velocity toward the second floor.
a. While you're moving toward the second floor at a constant velocity,
what is the net force exerted on you by all outside forces (specify the
amount and the direction of the net force)?
Answer: Zero.
Why: If you are at constant velocity, and thus not accelerating, the net
force on you must be zero.
b. You know that gravity gives you a weight in the downward direction.
What force does the escalator exert on you as you move toward the second
floor at a constant velocity (specify the amount and the direction of the
force)?
Answer: A force equal in amount (magnitude) to your weight, but in the
upward direction.
Why: Since the net force on you is zero and the only two forces you're
experiencing are (1) your weight downward and (2) a support force from
the escalator upward, those two forces must exactly cancel. Thus the support
force must have exactly the same magnitude as your weight, but they must
point in opposite directions.
c. Is the escalator doing work on you as you move toward the second
floor?
Answer: Yes.
Why: The escalator is pushing upward on you and you are moving (at least
partially) upward. Whenever something exerts a force on you and you move
at least partially in the direction of that force, that something is doing
work on you.
d. As you first step onto the escalator, you begin to accelerate toward
the second floor. Is the net force exerted on you by all outside forces
the same as in a?
Answer: No.
Why: Since you are accelerating, the net force on you must not be zero.
e. If the rapidly moving escalator suddenly stopped moving, you would
be thrown forward and might even fall over. What causes you to be thrown
forward?
Answer: Inertia (or your momentum).
Why: If the escalator were to stop suddenly, it would be accelerating backward
at an enormous rate. Without some external force to stop you, you'll continue
forward at constant velocity. As a result, you will appear to be "thrown"
forward, when in fact you are simply obeying the laws of motion and the
concept of inertia. No force "throws" you forward--just inertia.
f. You have more energy when you reach the second floor than you had
on the first floor. Why aren't you moving faster as a result?
Answer: Your new energy takes the form of gravitational potential energy.
Why: The escalator does work on you as you rise upward and this energy
becomes stored in the forces between you and the earth as gravitational
potential energy. This energy isn't visible, but it can be released by
descending back down to the first floor.
Problem 3: Chapter 1, Case 14a-e (Pg. 74)
Revolving doors are popular in northern hotels and office buildings as a way to prevent cold outside air from blowing directly into the lobby. Most revolving doors have 4 panels arranged in a cross, as viewed from above. You step in between two panels of the door and push on the panel in front of you. The revolving door begins to rotate and once you reach the inside of the building, you step out into the lobby.
a. It is much easier to make the revolving door rotate by pushing on
the panel far from the central pivot than it is by pushing near the pivot.
Why?
Answer: The farther from the pivot you push, the more torque you produce
with your force.
Why: It takes a torque to cause the door to begin rotating. The force you
exert on the door produces such a torque, but the amount of that torque
depends on where you push. By pushing far from the pivot, you select a
large lever arm and produce a large torque. As a result, the door undergoes
relatively rapid angular acceleration.
b. As you push on the door, it begins to turn more and more quickly.
What is your pushing doing to the door?
Answer: Your push is causing angular acceleration.
Why: When a door (or any rotating object) changes its rate of rotation,
it is undergoing angular acceleration. That is what you are causing by
pushing on the door.
c. One of the dangers of revolving doors is being hit by the panel behind
you as you step out of the door. The door tends to keep on turning after
you stop pushing, and it can bump you if you're not careful. Why does it
keep on turning after you stop pushing?
Answer: Its angular momentum (or equivalently rotational inertia, or
even simply inertia).
Why: When nothing exerts a torque on the door (or at least the torques
are very small), the door will coast around and around because of rotational
inertia. It has angular momentum and is unable to transfer that angular
momentum elsewhere, so around it goes.
d. What eventually stops the revolving door when no one uses it for
a minute or two?
Answer: Friction and/or air resistance.
Why: The door isn't really free of all torques. As it spins, it rubs against
things and also pushes on the air. Friction and air resistance twist the
door in the direction opposite its rotation and it gradually slows to a
stop.
e. The people who built the revolving door wanted it to be easy to start
and stop. They had to be most careful to minimize the mass of which edge
of each door panel: the upper edge, lower edge, inside edge (nearest the
pivot), or outside edge (farthest from the pivot)?
Answer: They had to minimize the mass at the outside edge.
Why: The ease with which you can start or stop the door from turning depends
on its moment of inertia--the smaller this moment of inertia, the easier
it is to start or stop the door. An object's moment of inertia depends
both on its mass and on the location of that mass. The farther the mass
is from the axis of rotation, the more it contributes to the moment of
inertia. That's why minimizing the mass far from the door's pivot (from
its outer edge) is the best way to minimize its moment of inertia.
Problem 4: Chapter 1, Case 17a-e (Pg. 75)
You're seated in a crowded bus that has stopped at a bus stop. Two people board the bus, one wearing rubber-soled shoes and one wearing inline skates (roller skates - i.e., no friction). They both have to stand in the middle of the aisle and neither one holds onto anything.
a. The bus driver is new and the bus lurches forward from the bus stop.
Which way does the person wearing inline skates move in relationship to
the bus?
Answer: The skater moves backward relative to the bus (but stays in
one place relative to the bus stop).
Why: As the bus accelerates away from the bus stop, the ground pushes it
forward. To remain with the bus, the skater also needs a forward force
but the skates prevent the bus from exerting such a forward force on the
skater. The wheels of the skates just spin. So instead of accelerating
forward, the skater remains in place in front of the bus stop and the bus
drives out from under the skater. To the passengers on the bus, the skater
rolls backward through the isle and hits the back of the bus.
b. The bus is accelerating forward as it pulls away from the bus stop.
Why doesn't the person wearing inline skates accelerate with the bus?
Answer: The bus can't exert the forward force on the skater that the
skater needs to accelerate forward with the bus.
Why: The skate wheels turn as the bus accelerates forward, so that the
skater doesn't experience the forward frictional force that a person in
normal shoes would need to accelerate with the bus.
c. The person wearing rubber-soled shoes remains in place as the bus
starts moving. What provides the force that causes that person to accelerate
with the bus?
Answer: (Static) friction.
Why: Static friction between the bus floor and the person's rubber-soled
shoes pushes the person forward and the bus floor backward. As a result,
the person accelerates with the bus and the bus accelerates somewhat more
slowly than it would have had the person not been in the bus.
d. Once the bus has reached a constant velocity (it's traveling along
a straight road at a steady pace), the person wearing inline skates is
able to stand comfortably in the aisle without rolling anywhere. What is
the net force on the person wearing skates?
Answer: Zero.
Why: Since the bus and the skater are both moving at constant velocity,
neither one is accelerating. The skater (like the bus) is experiencing
zero net force.
e. The driver accidentally runs into a curb and the bus stops so abruptly
that everything slides forward, including the person wearing rubber-soled
shoes. In stopping quickly, the bus experiences an enormous backward acceleration.
Why doesn't the person wearing rubber-soled shoes accelerate backward with
the bus?
Answer: The force of static friction can't provide enough backward force
to slow the person with the bus.
Why: When the bus stops gradually, static friction between the shoes and
the bus floor exert enough backward force on the person wearing the shoes
to make that person accelerate backward with the bus. However, when the
bus stops suddenly, the force required to accelerate the person backward
with the bus is greater than the maximum force that static friction can
supply. The shoes begin to slide across the floor and the person skids
forward in the bus.
Problem 5: Chapter 2, Case 5a-e (Pg. 119)
A trampoline has an elastic surface, supported at the edges by springs or elastic bands so that it's normally flat. This surface stores energy during a bounce so that you can jump very high on it.
a. When you get on a particular trampoline and stand in its center,
its surface distorts downward 10 cm. It's behaving like a spring that's
distorted away from its equilibrium shape. The distorted trampoline surface
is exerting a restoring force on you in which direction?
Answer: Upward.
Why: Like any spring that has been distorted downward from its equilibrium
shape, the trampoline exerts a force upward toward its equilibrium shape.
b. If someone with twice your weight climbed into the middle of the
empty trampoline, how far downward would its surface distort?
Answer: 20 cm.
Why: The trampoline's distortion is proportional to its restoring force.
Since a 10 cm distortion produces a restoring force equal in magnitude
to your weight, a 20 cm distortion produces a restoring force equal in
magnitude to twice your weight.
c. You begin bouncing up and down on the trampoline. As you land on
the trampoline during one of the bounces, its surface distorts downward
30 cm. Your weight hasn't changed so how can you make it distort so far
downward?
Answer: You simply push downward extra hard against it (and you accelerate
upward as a result) and it distorts downward until its restoring force
is equal in magnitude to the one you are exerting on it.
Why: The force you exert on the trampoline isn't limited by your weight.
If you jump, you will push downward on the trampoline with a force larger
than your weight. As a result, it distorts downward further and pushes
up harder on you. You accelerate upward as a result.
d. While you are in the air above the trampoline, nothing is pushing
on you except gravity. On both your way upward and your way downward (while
in the air), do you feel weightless, your normal weight, or particularly
heavy?
Answer: You feel weightless, both on the way up and on the way down.
Why: While you are above the trampoline's surface, the only force acting
on you is gravity. Since gravity is exerted throughout your body, the various
parts of your body do not have to support one another and you feel weightless.
It doesn't matter whether you are on your way up or your way down, so the
only force acting on you in either case is gravity and it always acts directly
downward.
e. While you are touching the trampoline during a bounce, its surface
is pushing upward on you. Near the bottom of the bounce, you distort the
trampoline's surface downward more than 10 cm. On both your way downward
and your way upward (while the surface is distorted downward more than
10 cm), do you feel weightless, your normal weight, or particularly heavy?
Answer: You feel particularly heavy, both on your way downward and your
way upward.
Why: When you have dented the trampoline's surface far below 10 cm, it
is exerting an upward force on you that more than balances your downward
weight. You are thus accelerating upward. Since it is pushing you upward
through your feet, you feel your ankles supporting your legs and your legs
supporting your knees and so on. The compression that results from all
of these internal support forces is relatively severe because the forces
are larger than they are when you are just standing on the ground. Thus
you feel extra heavy.