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```KEY
NNHS Introductory Physics: MCAS Review
Packet #2
2. Conservation of Energy and Momentum
Broad Concept: The laws of conservation of energy and momentum provide alternate approaches
to predict and describe the movement of objects.
1.) Which of the following objects has the most 2.) Calculate the kinetic energy of a dog,
kinetic energy?
mass = 10 kg, running at a speed of 2 m/s.
A. Toy car with a mass of 1 kg and a speed of
A. 10 J
1 m/s.
B. 20 J
B. Toy car with a mass of 1 kg and a speed of
C. 40 J
5 m/s.
D. 80 J
C. Real car with a mass of 1000 kg and a speed
of 1 m/s.
D. Real car with a mass of 1000 kg and a speed
of 5 m/s.
D. KE= 1/2 * mass * speed
squared.
Bigger mass and bigger speed
means bigger KE.
B. KE=1/2*mass*speed squared
1/2*10*4=20J
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NNHS Introductory Physics: MCAS Review
Packet #2
Broad Concept: The laws of conservation of energy and momentum provide alternate approaches
to predict and describe the movement of objects.
3.) A person is sitting at rest at the top of the
4.) You hold a 0.5 kg mass 1 meter above the
biggest hill of a rollercoaster. If the person has ground. Its gravitational potential energy is
a weight of 600 N, and the hill is 30 m high,
approximately:
what is the person’s gravitational potential
A. .5 J
energy?
B. 5 J
A. 18000 J
C. 10 J
B. 20 J
D. 50 J
C. 180000 J
D. 9000 J
A. GPE=Weight*Height
GPE = 600 N * 30 m = 18000 J
B. GPE=mass * 10m/s2 * height
GPE =.5 kg*10m/s2*1 m
GPE = 5 J
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NNHS Introductory Physics: MCAS Review
Packet #2
Broad Concept: The laws of conservation of energy and momentum provide alternate approaches
to predict and describe the movement of objects.
6.) Which one of the following objects has
5.) Three different boxes are lifted to
mechanical energy (KE +GPE) that remains
different heights.
• Box X weighs 115 N and is lifted to 15 m. constant?
• Box Y weighs 210 N and is lifted to 10 m. A. A crate being lifted vertically upwards at a
constant velocity.
• Box Z weighs 305 N and is lifted to 5 m.
B. An apple in free-fall.
Which of the following statements best
C. A car accelerating on a level (flat) highway.
describes the boxes’ change in
D. A sky-diver falling to Earth with his
mechanical energy?
parachute open.
A. Box X had the greatest change in
mechanical energy.
B. Box Z had the smallest change in
mechanical energy.
C. Boxes X and Y had the same change in
mechanical energy.
D. Boxes Y and Z had the same change in
mechanical energy.
B.
Box X: Change in GPE
= weight * height
= 115 N * 15 m
= 1725 J
Box Y: Change in GPE
= 210 N * 10 m
= 2100 J
Box Z: Change in GPE
= 305 N * 5 m
= 1525 J
B. An apple in free-fall does
not have air resistance acting
on it. Therefore, as it falls
the GPE is converted to KE and
the total remains constant.
KEY
NNHS Introductory Physics: MCAS Review
Packet #2
Broad Concept: The laws of conservation of energy and momentum provide alternate approaches
to predict and describe the movement of objects.
7.) An astronaut drops a 1.0 kg object and a 5.0 8.) The figure below shows a wagon that
kg object on the Moon. Both objects fall a total moves from point X to point Y.
distance of 2.0 m vertically. Which of the
following best describes the objects after they
have fallen a distance of 1.0 m?
A. They have each lost kinetic energy.
B. They have each gained the same amount of
potential energy.
C. They have each lost the same amount of
Which of the following best describes the
potential energy.
wagon’s change in energy as it coasts from
D. They have each gained one-half of their
point X to point Y?
maximum kinetic energy.
A. The wagon has the same kinetic energy at
point Y and at point X.
B. The wagon has more kinetic energy at point
Y than at point X.
C. The wagon has the same gravitational
potential energy at point Y and at point X.
D. The wagon has more gravitational potential
energy at point Y than at point X.
D. As the objects fall, their
PE is converted to KE. When
they have fallen 1.0m they are
halfway to the ground. So half
of their GPE has been converted
to KE. So they have gained half
of their KE.
D. At point Y the wagon is
higher above the ground. Thus
it has more GPE than it did at
point X.
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NNHS Introductory Physics: MCAS Review
Packet #2
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NNHS Introductory Physics: MCAS Review
Packet #2
Broad Concept: The laws of conservation of energy and momentum provide alternate approaches
to predict and describe the movement of objects.
9.) At a weightlifting competition, two
10.) An archer pulls back the bowstring to
competitors lifted the same weight to the same prepare to shoot an arrow as shown below.
height. The second competitor accomplished
the lift 2 seconds faster than the first
competitor. This demonstrated that the second
A. energy than the first.
B. inertia than the first.
C. power than the first.
D. work than the first.
She uses an average force of 40 N, moving the
bowstring 0.2 m. How much energy is stored
in the bow?
A. 8 J
B. 16 J
C. 24 J
D. 36 J
C. The second competitor has
more power since the same
amount of work was done in less
time:
Power = Work/time.
A. The amount of energy stored
is based on the work done.
W = F*d
= 40 N * 0.2 m = 8 J.
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NNHS Introductory Physics: MCAS Review
Packet #2
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NNHS Introductory Physics: MCAS Review
Packet #2
Broad Concept: The laws of conservation of energy and momentum provide alternate approaches
to predict and describe the movement of objects.
11.) The Watt is the unit for which quantity:
12.) One Joule is equal to
A. energy
A. One Watt
B. work
B. One kg•m/s
C. force
C. One Newton-meter
D. power
D. One Newton
D. The Watt is the unit for
power.
C. Work is measured in Newtonmeters. Energy is measured in
Joules. Work is a transfer of
energy. Thus you know that
Joules and Newton-meters are
the same.
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NNHS Introductory Physics: MCAS Review
Packet #2
Broad Concept: The laws of conservation of energy and momentum provide alternate approaches
to predict and describe the movement of objects.
13.) What is the mass of an asteroid with a
14.) A bowling ball with a mass of 8.0 kg rolls
speed of 200 m/s and a momentum of
down a bowling lane at 2.0 m/s. What is the
2,000 kg•m/s?
momentum of the bowling ball?
A. 10 kg
A. 4.0 kg • m/s
B. 1,800 kg
B. 6.0 kg • m/s
C. 2,200 kg
C. 10.0 kg • m/s
D. 400,000 kg
D. 16.0 kg • m/s
A.
m
=
=
p = m v, so
= p / v
(2000 kg m/s) / 200 m/s
10 kg
D. p = m v
p = 8 kg * 2 m/s
p = 16 kg m/s
KEY
NNHS Introductory Physics: MCAS Review
Packet #2
Broad Concept: The laws of conservation of energy and momentum provide alternate approaches
to predict and describe the movement of objects.
15.) A student is standing on a skateboard that 16.) You are at an ice skating rink and are
is not moving. The total mass of the student
gliding towards a friend who is initially at rest.
and the skateboard is 50 kilograms. The
When you reach your friend, you grab your
student throws a ball with a mass of
friend around the waist and the two of you
2 kilograms forward at 5 m/s. Assuming the
continue gliding forward. Which one of the
skateboard wheels are frictionless, how will the following is true:
student and the skateboard move?
A. Your speed after the collision is greater than
A. forward at 0.4 m/s
B. forward at 5 m/s
B. Your speed after the collision is the same as
C. backward at 0.2 m/s
D. backward at 5 m/s
C. Your speed after the collision is smaller
than your speed before the collision.
D. Not enough information has been provided.
C. Here are two ways to answer:
C. When you grab your friend,
Conceptual Reasoning: Due to
conservation of momentum, if
constant but the amount of mass
the ball is thrown forward, the that moves is increasing. Thus
student will go backward. Since your speed must be decreasing.
the boy has much more mass he
will have much less speed in
order to have the same
momentum.
Mathematical Reasoning: The
total momentum in the system
must remain to be zero so the
momentum of the ball forward
must be equal and opposite to
the momentum of the boy:
pball = 2 kg (5 m/s) = 10 kg m/s
therefore pboy = -10 kg m/s =
(50 kg) v therefore v = (-10 kg
m/s)/(50 kg)
= -0.2m/s, or 0.2m/s backward.
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NNHS Introductory Physics: MCAS Review
Packet #2
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NNHS Introductory Physics: MCAS Review
Packet #2
Broad Concept: The laws of conservation of energy and momentum provide alternate approaches
to predict and describe the movement of objects.
17.) An engineering student is gathering
18.) Mike, who has a mass of 75 kg, is
data on the motion of a model car traveling running north at 2.6 m/s. He accidentally
down a ramp. If energy is conserved, the
collides with Tom, who has a mass of 125
potential energy of the car at the top of the kg and is not moving.
ramp should equal the kinetic energy of
Which of the following statements
the car at the bottom of the ramp. After the describes how much momentum each
first trial, the student calculates that the
person has before the collision?
kinetic energy at the bottom of the ramp is A. Mike has a momentum of 130 kg • m/s
less than the potential energy at the top of north, and Tom has no momentum.
the ramp.
B. Mike has a momentum of 195 kg • m/s
Which of the following can best explain
north, and Tom has no momentum.
this difference?
C. Both Mike and Tom have a momentum
A. The car gained a small amount of mass of 130 kg • m/s north.
as it moved down the ramp.
D. Both Mike and Tom have a momentum
B. The student accidentally accelerated
of 195 kg • m/s north.
the car at the top of the ramp.
C. The measured height of the ramp was
less than the actual height.
D. The student did not include the effect of
frictional force in the calculation.
Rule out wrong choices:
Choices A, B, and C would all
result in more KE of the car at
the bottom of the ramp. KE
depends directly on mass and
velocity, and “A” and “B” would
increase those quantities.
Choice “C” would mean the car
lost more GPE, therefore having
more KE at the bottom.
Why D is right:
Friction would do negative
work, which would reduce the
kinetic energy of the car at
the bottom.
B. We can calculate the
momentum each
Person has before the
collision:
Mike: p = m v = 75 kg * 2.6 m/s
= 195 kg m/s
Tom: p = m v = 125 kg * 0 m/s
= zero kg m/s
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NNHS Introductory Physics: MCAS Review
Packet #2
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NNHS Introductory Physics: MCAS Review
Packet #2
Practice: Open-response Question #1
BE SURE TO ANSWER AND LABEL ALL PARTS OF THE QUESTION.
Show all your work (diagrams, tables, or computations)
If you do the work in your head, explain in writing how you did the work.
Start
End
The above diagram shows a simple roller coaster track and one roller coaster car. The car,
when full of passengers, has a mass of 500kg.
A.) If the first hill has a height of 20 meters, calculate the amount of work that must be done
to get the full car to the top of the first hill.
B.) If there were no friction, where on this track would the car be going the fastest? Place an
x on the spot on the diagram. Explain.
C.) In reality, there is friction on a roller coaster track. Explain why this means that the first
hill must be the highest hill on a roller coaster track.
A.)
The work done to get the car to the top of the first
hill is equal to the amount of energy that the car has at
the top of the hill. The amount of energy at the top of
the first hill is GPE=mgh=500*10*20=100,000Joules
B.)
If there were no friction, then the total amount of
energy will remain constant. The car will be going the
fastest when it has the most KE. It has the most KE when it
has the least GPE. It has the least GPE at the lowest
height. It has the lowest height at the end. Thus it is
going the fastest at the end.
C.) Since there is friction, energy will continually be lost to
heat and sound. Therefore the total amount of energy that
the car has will continually decrease. In order to make it
over a hill, a car needs both GPE and KE. If the total is
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NNHS Introductory Physics: MCAS Review
Packet #2
less, the height of a hill will need to be less that the
first hill in order to have any KE.
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NNHS Introductory Physics: MCAS Review
Packet #2
Practice: Open-response Question #2
BE SURE TO ANSWER AND LABEL ALL PARTS OF THE QUESTION.
Show all your work (diagrams, tables, or computations)
If you do the work in your head, explain in writing how you did the work.
The illustrations below show an air track with two carts before and after a collision. The mass
and the initial velocity of each cart are shown below.
Before Collision:
First Cart: mass = 0.20 kg and velocity = 0.10m/s
Second Cart: mass = 0.30 kg and velocity = 0.050m/s
The first cart slides on the air track and collides with the second cart. The two carts stick together
upon impact and move together along the track, as shown below.
a. What is the momentum of the first cart before it collides with the second cart? Show your
b. What is the momentum of the second cart before the collision? Show your calculations and
c. Describe two changes that could be made initially to either one or both carts that would result
in an increase in the momentum of the combined carts after the collision.
p  mv
A.) p  0.2 kg * 0.1m/s
p  0.02 kg m/s
p  mv
B.) p  0.3 kg * 0.050 m/s


p  0.015 kg m/s
C.) You could increase the mass of either one or both of the
carts. You could increase the initial velocity of either one or
both of the carts. Any one of these four changes would increase
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NNHS Introductory Physics: MCAS Review
Packet #2
the total amount of momentum before the collision and thus the
total amount of momentum after the collision.
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NNHS Introductory Physics: MCAS Review
Packet #2
Practice: Open-response Question #3
BE SURE TO ANSWER AND LABEL ALL PARTS OF THE QUESTION.
Show all your work (diagrams, tables, or computations)
If you do the work in your head, explain in writing how you did the work.
In the diagram below, the falling water turns the waterwheel. The turning waterwheel
generates electricity.
The water moves slowly at point A and then falls rapidly past point B.
a. Describe the changes in kinetic and gravitational potential energy of the water as it
travels from point A to point B.
b. Explain why not all of the energy of the moving water available at point A is captured
by the waterwheel to generate electricity.
c. Describe two ways the system can be changed so that more energy from the falling
water is converted into electrical energy.
A.)
As the water travels from point A to point B, its
gravitational potential energy decreases. While falling
from A to B, its kinetic energy increases.
B.)
The waterwheel does not capture all of the energy of
the moving water at point A because some of the energy is
used to turn the wheel itself, and that energy is not
available to generate electricity. Also, it appears that
some of the water is splashing and spilling over the water
wheel, so that water would not transfer its energy to the
wheel. Also, some of the energy goes into turbulence and
heating of the water.
C.)
To get more energy from the falling water converted
into electrical energy, you could build a dam above the
wheel, raising the water level above point A, or move the
wheel lower, so the water would fall from a greater height.
You could build a sluice or channel that would direct all
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NNHS Introductory Physics: MCAS Review
Packet #2
the water smoothly onto the wheel, reducing losses due to
spilling and turbulence. If the wheel is made less massive,
with less friction, then less of the water’s energy would
be used to make the wheel turn. Another possibility is to
make the paddles of the wheel more of a “U-shape” rather
than flat, which lengthens the time the water is in
contact, transferring more momentum to the wheel.
```
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