Download Science Focus 10 Chapter 5 Review KEY

TR 5-60
MHR • Unit 2 Energy Flow in Technological Systems
they weigh about one-sixth as much on the Moon
as they do on Earth.
8. Without realizing it, you estimate an object’s weight
and exert an estimated force on it to lift it. If it
weighs much less than you subconsciously estimated, the force that you exert will lift it higher and
faster than it would the heavier object.
9. When an elevator goes up, the counter weights go
down. Much of the work done on lifting the elevator is done by the counter weights pulling on the
cable. When the elevator gains gravitational potential energy, the counter weights lose gravitational
potential energy and vice versa.
(k)
CHAPTER AT A GLANCE
(l)
(i)
(j)
Student Book Page 217
(a) Distance is a scalar quantity that describes the
length of a path between two points or locations.
Displacement is a vector quantity that describes the
straight-line distance from one point to another;
because displacement is a vector, it includes direction.
(b) Uniform motion is motion in which velocity is
constant.
(c) A position-time graph for uniform motion is a
straight line; a position-time graph for accelerated
motion is a curved line.
(d) To determine speed (velocity) of an object from a
line of best fit, you calculate the slope of the line
using two points on the line that are near opposite
ends of the line.
(e) Accelerated motion involves velocity that is changing (speeding up or slowing down), whereas uniform motion involves velocity that remains constant (unchanging).
(f) For something to have kinetic energy, it must be
moving; thus, speed must have an effect on kinetic
energy. If two objects of different mass are moving
at the same speed, greater force is needed to
change the condition of the larger mass, so mass
must affect the kinetic energy of objects. Mass and
speed affect kinetic energy. The equation for calculating kinetic energy shows that mass and speed
necessarily affect kinetic energy.
(g) A joule is equivalent to a kilogram times a metre
squared divided by a second squared.
(h) Any work that reduces the kinetic energy of an
object (e.g., applying brakes on a bicycle, stopping
a door that is swinging open, stopping a top from
spinning) is negative work. Any work that adds
(m)
(n)
(o)
kinetic energy to an object (e.g., hitting a baseball,
slam-dunking a basketball, pedalling faster on a
bicycle) is positive work.
For example: If you see a soccer ball at rest on the
grass and do work on the ball by exerting a large
force over a short distance, the ball gains kinetic
energy by moving.
If you know that an object is stretched or distorted
from its usual original form, and if you know it will
return to that form when the force causing the distortion is changed, you know the object has elastic
potential energy.
To make chemical compounds from reactants with
a lower chemical potential energy than the products, you have to add energy.
When a nucleus splits (fissions), nuclear potential
energy in the form of mass is transformed into
kinetic energy.
People weigh less on the Moon than on Earth
because the gravitational force of attraction
between the Moon and objects on it is less than the
gravitational force of attraction between Earth and
objects on it.
Mass, the height of an object with respect to lower
reference point, and the acceleration due to gravity
determine the gravitational potential energy of
something.
The work done on an object when it is lifted is the
gravitational potential energy of the object.
Prepare Your Own Summary
• The ideas for summaries offered to students show
the broad range of possibilities students with varied
interests could pursue to demonstrate their understanding of the concepts presented in this chapter.
The first idea might appeal more to math-oriented
students. The second and third ideas might appeal
more to students with stronger visual-spatial or
kinesthetic learning styles. The fourth idea most certainly will appeal to creative writers in the class. Of
course, any student with any learning style and any
personal interests could select any one of these ideas,
or pursue other ideas that are more personally meaningful. Given the interrelated nature of the concepts
in this chapter, it would be valuable for students to
share their summaries with one another — especially
those students who choose a mathematical means to
summarize the chapter (as in the first idea).
Chapter 5 Energy and Motion • MHR
CHAPTER 5 REVIEW
Student Book Pages 218–219
Key Terms
See the glossary in the student book.
Understanding Key Concepts
1. Position describes a specific point in relation to a
reference point; it is a vector quantity.
Displacement is the straight-line distance from one
position to another. Displacement is a vector.
Mathematically, the relationship between displacement and position is ∆dY = dY2 – dY1.
2. The concept of a time interval, the difference
between two points in time, is needed in order to
calculate speed or velocity, because both are defined
in terms of a time interval.
3. The first formula for velocity is written in terms of
displacement and time interval. The second formula
substitutes the definition for displacement in terms
of initial and final positions. It also uses the definition of time interval in terms of initial and final
times. Your choice of formula to use depends on the
form of your given data.
4. Students’ graphs of zero velocity should look like
Figure 5.7A in the student text, and graphs of uniform motion should look like either Figure 5.7B or
C. Their graphs of increasing velocity will be
curved, looking like Figure 5.12A or B.
5. Students should describe how to use the slope of
the graph to calculate the velocity of the object
Y
(slope = }∆∆}dt ). A strong answer will include reference
to using the slope of a line of best fit.
6. Students’ graphs should show a straight line going
upward to the right.
7. Kinetic energy is related (proportional) to the
square of the speed, Ekav2. If you double the speed
and square it, you get four times as much kinetic
energy because (2v)2 = 4v2.
8. The kinetic energy of A will be double that of B.
9. Kinetic energy most likely transforms into thermal
energy.
10. When you are going faster it takes much more distance to stop than when you are going slower. For
example, if you double your speed, the distance that
you need to stop will be four times greater.
11. Students likely will suggest an example of gravitational potential energy (such as water behind a dam)
for energy that results from position, and an example of chemical or nuclear energy for energy that
results from an object’s condition.
12. Energy should be written on the left side of the
equation, because the chemical potential energy of
the products will be higher than the chemical
potential energy of the reactants.
TR 5-61
13. Students’ answers should trace both energy sources
back to the Sun. Fossil fuels were formed from
plant and animal matter; the plants used the energy
of the Sun in the process of photosynthesis to make
food for their life processes, and the animals ate
photosynthetic plants. Hydro-electric energy
depends on water that results from the Sun-driven
water cycle.
14. A large nucleus, as a result of absorption of a neutron, becomes unstable and splits (fissions) into two
smaller nuclei; this process also releases several neutrons. The total mass of the products is smaller
than the total mass of the reactant particles. The
extra mass is converted into kinetic (thermal) energy. The chain reaction that results from the original
fission reaction releases large amounts of thermal
energy that may be used to convert water into
steam for an electric generator.
15. Kilograms are units of mass, not weight. The label
should say either that the potatoes have a mass of 5
kg or that the potatoes weigh 49 N.
16. The gravitational potential energy of the book
depends on its position in relation to a reference
point. The book could be on a shelf. It could have
12 J of gravitational potential energy relative to a
desk top and still have 19 J of gravitational potential
energy relative to the floor.
Developing Skills
17. You could make several objects that have the same
mass but different shapes. For example, you could
take several objects that have the same mass and
attach pieces of cardboard that make very different
shapes. You could then measure the time that it
takes the different objects to fall a certain distance.
Any difference in the time they take to reach the
ground would be due to the difference in air friction on the different shapes.
18. The technology referred to in this question is called
a drop tube, which is used to simulate microgravity
conditions to study the solidification characteristics
of molten solids or fluids of different kinds. Air is
removed from the tube to prevent friction.
Depending on the length of the tube (the longest,
in Japan, is 490 m), microgravity conditions for the
experiment that experiences free fall may last from
a few seconds up to 11.7 seconds for the facility in
Japan. Larger experimental samples are tested in
facilities called drop towers. Various NASA web
sites are good, but not the only, sources of information about these simulation technologies.
19. The graph will all be different. The important
result of the work is to analyze the graphs. If the
position versus time graphs are straight lines, the
motion is uniform motion. If the velocity versus
time graphs are straight lines, the acceleration is
constant.
TR 5-62
MHR • Unit 2 Energy Flow in Technological Systems
20. Students will find that Aristotle believed objects
with different weights (masses) fall at different
rates, with heavier objects falling faster than lighter
ones. Galileo showed that objects fall at the same
rate, regardless of their weight (mass). Students
could provide details of Galileo’s reasoning and the
experimental evidence he provided to support his
reasoning.
Problem Solving/Applying
21. Given: Distance, ∆d = 1.8 km, time interval,
∆t = 2 min + 18 s
Required: Speed, v
Analysis: Convert to SI units. Then, since the distance and time interval are known, you can use the
formula, v = }∆∆}dt
1000 m
}) = 1800 m
Solution: (1.8 km)(}
km
60 s
}) + 18 s = 120 s + 18 s = 138 s
∆t = (2 min)(}
min
∆d
v = }∆}t
1800 m
}
v=}
138 s
}
v = 13.043 }m
s
m
v < 13 }s}
Paraphrase: The horse’s speed was 13 m/s.
22. Given: Distance, ∆d = 25 km - 5.0 km = 20 km,
time interval, ∆t = 15 min.
(from 1:16 p.m. to 1:31 p.m.)
Required: Speed, v
Analysis: Convert to SI units then use v = }∆∆}dt
1000 m
}) = 20 000 m
Solution: (20 km)(}
km
60 s
}) = 900 s
(15 min)(}
min
v = }∆∆}dt
20 000 m
}
v=}
900 s
Solution: Ek = }12}mv2
})2
Ek = }12}(0.002 kg)(3.8}m
s
2
kg•m
}
Ek = 0.01444}
s2
Ek < 0.014 J
Paraphrase: The kinetic energy of the PingPong™ is 0.014 J.
}, kinetic energy,
25. Given: Speed, v = 40.0}m
s
Ek = 4.8 × 108 J
Required: Mass, m
Analysis: You can use the formula Ek = }12}mv2 but
you will have to rearrange it to solve for mass.
Solution: Ek = }12}mv2
2Ek = 2(}12}mv2)
2Ek mv2
}}
= }v}
2
v2
2Ek
m = }v}
2
2(4.8 × 108 J)
}}
m=
})2
(40 }m
s
m = 6.0 ×
kg•}
m2
}
2
s
105 }
m}2
}
s2
m = 6.0 × 105 kg
Paraphrase: The mass of the freight train is
6.0 × 105 kg.
26. Given: Mass, m = 55 kg, kinetic energy,
Ek = 25 000 J
Required: Speed, v
Analysis: You can use the formula Ek = }12}mv2 but
you will have to rearrange it to solve for speed.
Solution: Ek = }12}mv2
1
2Ek = 2(}2}mv2)
2Ek = mv2
2E
mv2
}}k = }m}
m
}
v = 22.222 }m
s
}
v < 22 }m
s
2(2.5 × 10 J)
}§
}
v = !§
§
55 kg
4
Paraphrase: The driver’s speed in 22 m/s which is
equivalent to 80 km/h.
}
23. Given: Mass, m = 0.046 kg, speed, v = 5.1}m
s
Required: Kinetic energy, Ek
Analysis: Since the mass and speed are known, you
can use the formula, Ek = }12}mv2
Solution: Ek = }12}mv2
} )2
Ek = }12}(0.046 kg)(5.1 }m
s
2
kg•m
}
Ek = 0.59823}
s2
Ek < 0.60 J
Paraphrase: The kinetic energy of the golf ball is
0.60 J.
}
24. Given: Mass, m = 2.0 g, speed, v = 3.8}m
s
Required: Kinetic energy, Ek
Analysis: Since the mass and speed are known, you
can use the formula, Ek = }12}mv2
v=
!§§§§
9.0909 ×
kg•m2
}}
s2
102 }
kg
!§§§
2
m
}
v = 9.0909 × 102 }
s2
m
v = 30.151 }s}
}
v < 3.0 × 101 }m
s
Paraphrase: The antelope was running at 30 m/s.
27. Given: Mass, m = 75 kg, relative height, ∆h = 25 m,
}
acceleration due to gravity, g = 9.81 }m
s2
Required: Gravitational potential energy, Eg
Analysis: Since the mass and relative height are
known, you can use the formula, Eg = mg∆h.
Solution: Eg = mg∆h
})(25 m)
Eg = (75 kg)(9.81 }m
s2
2
kg•m
}
Eg = 1.8394 × 104 }
s2
4
Eg < 1.8 × 10 J
Chapter 5 Energy and Motion • MHR
Paraphrase: The gravitational potential energy of
the climber is 1.8 × 104 J.
28. Given: Mass, m = 550 kg, relative height,
}
∆h = 24 m, acceleration due to gravity, g = 9.81 }m
s2
Required: Gravitational potential energy, Eg
Analysis: Since the mass and relative height are
known, you can use the formula, Eg = mg∆h.
Solution: Eg = mg∆h
})(24 m)
Eg = (550 kg)(9.81}m
s2
2
kg•m
}
Eg = 1.2949 × 105}
s2
Eg < 1.3 × 105 J
Paraphrase: The gravitational potential energy of
the container is 1.3 × 105 J.
29. Given: Mass, m = 314 kg, gravitational potential
energy, Eg = 4000.0 J, acceleration due to gravity,
}?
g = 9.81 }m
s2
Required: Relative height, ∆h
Analysis: Since the mass and gravitational potential
energy are known, you can use the formula,
Eg = mg∆h. However, you will have to solve the formula for relative height.
Solution: Eg = mg∆h
Eg
mg∆h
}}
}
=}
mg
mg
Eg
}
∆h = }
mg
4000 J
∆h = }}
m
(314 kg)(9.81}s}2 )
∆h = 1.29856
kg•m2
}}
s2
}
m
}}
s2
∆h < 1.30 m
Paraphrase: The piano movers lifted the piano
1.30 m.
30. Given: Relative height, ∆h = 0.22 m, gravitational
potential energy, Eg = 3800 J, acceleration due to
}
gravity, g = 9.81 }m
s2
Required: Mass, m
Analysis: Since the relative height and gravitational
potential energy are known, you can use the formula, Eg = mg∆h. However, you will have to solve the
formula for mass.
Solution: Eg = mg∆h
mg∆h
Eg
}}
= }
g∆h
g∆h
Eg
}
m=}
g∆h
3800 J
m = }}
})(0.22 m)
(9.81}m
s2
m=
kg•}
m2
}
s2
1760.7265 }
m}2
}
s2
m < 1.8 × 103 kg
TR 5-63
Paraphrase: The mass of the car was 1.8 × 103 kg.
Critical Thinking
31. The amount of mass that would be equivalent to
the amount of energy released from a chemical
reaction is much too small to be measurable.
Therefore, you could not test a hypothesis that the
source of thermal energy released in a chemical
reaction is mass. However, there is no other
hypothesis that is as reasonable as the hypothesis
that the thermal energy comes from mass.
32. The following are some possible answers.
• Set designers need to know how much energy will
be required to move scenery, lighting, and other
objects so they know how to plan for quick set
changes between scenes of a play.
• In an action movie, the director and especially the
stunt person needs to know how much energy is
needed to get an object or a person up to a certain speed.
• A dance choreographer needs to know how much
energy will be required to generate certain speeds
and jumps so that the plan is reasonable for a person to achieve.
• A swimming coach, or any coach, needs to understand how much energy is required for many different moves.
Students could discuss any modifications
to their Focussing Questions with a
partner or in small groups.