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GSCI 1010
MIDTERM EXAM #3 SOLUTIONS
10 FEBRUARY 2010
Part A: Do all parts of the following question (15 parts @ 4 marks each = 60 marks).
1.
For each of the following, indicate whether the statement is true or false, and briefly explain or
justify your answer.
(a)
TRUE
p mv f  mvi 10000  1000 20  20000
F



 5000 N, so the required
t
t
4
4
force is 5000 Newtons, directed west.
1
3
(b)
1
(c)
1
If two 5 kg objects travel toward each other at the same speed of 2 m/s, their total
linear momentum has magnitude 20 kg-m/s.
FALSE
The speeds may be equal, but since they travel toward each other their velocities are
equal and opposite (or + 2 and – 2) . It follows that the total momentum is zero, i.e.
pnet  m1v1  m2 v2  5 2  5 2  10  10  0 kg-m/s .
3
(d)
3
When you are standing still, the reaction force to your weight is the force of the ground
(or floor) pushing up on your feet.
FALSE
Your weight (the Earth’s downward pull on you) and the floor’s upward push on you are
equal and opposite, but they both act on the same body (you), so they can’t be an actionreaction pair. Rather the reaction force to the Earth’s downward pull on you is your
upward pull on the Earth.
3
1
According to Newton’s Second Law, a force of 5000 Newtons (west) is required to
stop a 1000 kg car initially travelling 20 m/s (east) in a time of 4 seconds.
If the Earth had four times its current mass and double its current radius, the acceleration
of gravity g at the Earth’s surface would still have magnitude 9.8 m/s2.
TRUE
GM
 9.8 m/s2. If we
r2
quadruple the mass and double the radius, the new acceleration of gravity (call it g’) will
G 4M  4GM GM

 2  9.8 m/s2.
be g ' 
2
2
4
r
r
2r 
For the current mass M and radius r, we know that g 
2
(e)
According to the “Satellite Equation”, the larger the radius of a circular orbit, the faster
a satellite must travel.
1
FALSE
3
Since the Satellite Equation is v 
GM
, it is obvious that a larger value of r will
r
produce a smaller value of v.
(f)
Here on Earth, an object which displaces 10 kg of any liquid when it is completely
submerged is subject to an upward buoyant force of 98 Newtons.
TRUE
According to Archimedes’ Principle, an object submerged in any liquid must experience
a buoyant force equal to the weight of the liquid displaced. Here, we are told that the
mass of the liquid displaced is m = 10 kg. Therefore the buoyant force B, i.e. the weight
of liquid displaced, is simply
B  W  mg  109.8  98 Newtons (upward).
1
3
(g)
A moving object will quadruple its kinetic energy if it doubles its speed.
TRUE
1
The formula for the kinetic energy of any moving object is K 
speed to 2v produces a new kinetic energy (call it K’) of
1
1

2
K '  m2v   4 mv 2   4 K .
2
2

3
(h)
3
(i)
3
When a 2 kg body is raised 10 m above the ground, it gains 196 Joules of gravitational
potential energy.
TRUE
For “local” gravitational potential energy, the change is given by the formula
mgy  29.8 10  196 Joules, i.e. a gain of 196 Joules.
1
1
1 2
mv . Doubling the
2
There are other, different kinds of energy besides kinetic and potential energy, including
thermal energy, nuclear energy and electrical energy.
FALSE
While the terms “thermal energy”, “nuclear energy” and “electrical energy” are often
used to describe sources of energy, the only distinct kinds of energy are kinetic energy
(energy due to motion) and potential energy (energy due to position/location or shape).
That is, when viewed microscopically or analyzed more closely, these other energies are
really kinetic energy and/or potential energy.
3
(j)
FALSE
Once airborne and in level flight, the engines of fixed-wing planes do not push
downward on the air at all. They simply keep the plane moving horizontally through the
air. The resulting airflow over and under the wings generates “lift” by (according to
Bernoulli’s Law) creating lower pressure above the wing than below the wing, and
therefore maintaining a net upward force on the aircraft.
1
3
(k)
3
(l)
The Sun, like any other star, will eventually run out of fuel and undergo a supernova
explosion.
FALSE
Only stars which are significantly larger than our Sun will undergo supernova
explosions. At the end of their “Red Giant” phases, large stars begin to collapse so
quickly that the sudden build-up of pressure within the core causes them to explode. In
contrast, small stars like the Sun shrink much more slowly after becoming Red Giants,
and eventually end up as White Dwarf stars without ever “going supernova”.
1
3
(m)
3
Despite the enormous release of energy and matter immediately following the Big Bang,
there would have been no accompanying blinding flash of light.
TRUE
It is estimated that it took about 1 million years after the initial Big Bang before the first
and simplest conventional atoms (hydrogen and helium) had formed. Because the
emission of light only occurs when electrons which have jumped to a higher energy
level (i.e. to a higher shell within the atom) return to their original shell and emit visible
radiation in the process, light could not be produced until these atoms were present.
Prior to this, the temperature in the Universe was too high for hydrogen and helium
nuclei to “capture” the necessary electrons and form atoms.
1
1
Fixed-wing airplanes can fly because their engines push downward on the air below,
causing the air to produce an upward reaction force on the plane.
Neutron stars and black holes would generally be expected to spin very rapidly.
TRUE
Virtually all stars have a (relatively slow) rotation. Thus, by conservation of angular
momentum (just like slow-spinning figure skaters who spin faster when they pull their
arms in tight to their bodies), collapsing stars will rotate faster and faster as they
contract. Since neutron stars and black holes are extremely tiny compared to the
original stars that produce them, they can end up spinning amazingly fast—up to
several hundred rotations per second, in fact!
4
(n)
It is because light has no mass, and is therefore not subject to Newton’s Law of
Universal Gravitation, that the existence of black holes is only predicted by Einstein’s
theory of General Relativity.
TRUE
1
GMm
, correctly predicts the existence of
r2
an attractive force between any two masses. But since “light particles” (or photons) are
massless, a star would exert no attractive force on any light which is trying to escape. In
Einstein’s theory of General Relativity, however, gravity is considered not as a force but
as a distortion in the geometry of space-time. Such a distortion would affect all particles
(massive or massless), and this distortion becomes so severe near a black hole that all
light is prevented from escaping the star, thereby making it “invisible”.
Newton’s Law of Universal Gravitation, Fg 
3
(o)
The cloud of gas and dust that eventually came together to form our solar system is no
different than the original clouds that formed the very first stars after the Big Bang.
FALSE
The clouds of gas that formed the earliest (i.e. “first generation”) stars in the Universe
contained only hydrogen and helium. Moreover, the ensuing nuclear fusion within the
cores of these early stars produced no chemical elements heavier than iron. Elements
heavier than iron can therefore only be produced during supernova explosions of large
dying stars, and since we have many such heavier elements here on Earth (lead,
mercury, uranium, gold, etc.), it follows that the cloud of gas that formed our solar
system had previously undergone at least one supernova event.
1
3
Part B: Do any four questions in this section (10 marks each, as indicated).
2.
A golf ball (mass 0.045 kg) at rest on a tee is struck by a driver and leaves the tee at 60 m/s
(east). If the collision lasts 0.0005 s, find:
(a)
the change in momentum of the golf ball (3 marks);
p  p f  pi  mv f  mvi  0.045 60  0.0450  2.7 kg-m/s (east).
(b)
the average force exerted on the ball by the club during the collision (3 marks);
p
 2.7
F

 5400 N (east).
t 0.0005
(c)
the acceleration experienced by the ball during the collision (2 marks); and
F  5400
 120000  1.2  10 5 m/s2 (east).
Since F also equals ma, then a  
m 0.045
(d)
the average force exerted on the club by the ball during the collision (2 marks).
By Newton’s Third Law, the average force on the club by the ball is –5400 N (west).
3
3
2
2
5
3.
Imagine a tiny, super-dense object with mass 2.00  1030 kg (i.e. comparable to the mass of the
Sun) but with a radius of only 10,000 m, i.e. 10 km. (Note that these numbers would not be
unrealistic for a neutron star!). Calculate and express in scientific notation to 3 decimals:
(a)
the volume of the object in m3, using the formula for the volume of a sphere
4
V  r 3 (3 marks);
3
4
4
3
V  r 3   10000  4.189  1012 m3.
3
3
(b)
the density (mass per unit volume) of the object in kg/m3 (3 marks); and
M
2.00  10 30
Density  

 4.774  1017 kg/m3.
12
V
4.189  10
(c)
the gravitational acceleration g at the surface of the object (4 marks).
GM 6.67  10 11 2.00  10 30
g 2 
 1.334  1012 m/s2.
r
10000 2
3
3
4
4.

Duplicate the calculations used to determine the mass of the Sun for the first time, as follows:
(a)
convert the time for one Earth orbit (use 365 days) to seconds (2 marks);
Orbit time T  365  24  3600  3.15  10 7 s.
(b)
calculate the circumference 2r of the Earth’s orbit, if the Earth-Sun distance is
r = 1.50  1011 m (2 marks);
Orbital circumference C  2r  2 1.50  1011  9.42  1011 m.
2

2

(c)
find the orbital speed v of the Earth, from the circumference found in (b) and the time
found in (a) (2 marks); and
C 9.42  1011
 2.99  10 4 m/s.
Orbital speed v  
T 3.15  10 7
(d)
use the Satellite Equation v 
2
4

GM
to solve for the mass of the Sun M (4 marks).
r
Solving for M from the Satellite Equation,
M


2
v2r
2.99  10 4 1.50  1011

 2.01 10 30 kg.
11
G
6.67  10
6
5.
3
1
2
2
After its passengers climb aboard a roller coaster, the “cars” are towed up to the highest point of
the ride, at which the coaster is released from rest. From this point on, the ride is “powered”
solely by gravity. State the law of conservation of energy (3 marks), and use this to explain
how numerous exchanges of potential and kinetic energy occur as the coaster descends and
speeds up or ascends and slows down while proceeding around the track before stopping at the
bottom (5 marks). Where did the coaster’s initial potential energy at its original release point
come from (2 marks)?
The general law of conservation of energy states that the total energy of any physical “system”
remains constant. In this particular case, it simply means that the sum of the roller coaster’s
kinetic energy (KE) and gravitational potential energy (PE) remains constant throughout its
“journey” (from the point of initial release to the end of the run). As an equation, this can be
written
1
KE + PE = mv 2  mgy = constant.
2
During the ride, the coaster will undergo a series of descents and ascents (with each “peak”
along the way remaining below the original release point height). During any descent, the
coaster’s height decreases and gravitational potential energy is lost (since the vertical
displacement  y is negative). This decrease in PE must therefore be accompanied by an equal
increase in KE, so that the coaster gains speed (i.e. accelerates). Similarly, during each ascent,
the height, and thus the PE, increases (because  y is positive), so the KE must decrease
accordingly, and the coaster slows down. At the bottom of the run, all of the roller coaster’s
initial PE has been converted to KE, and the coaster is travelling a maximum speed, which is
why brakes must be applied to bring it to a quick stop.
2
The coaster starts from rest at its initial point of release, so its initial energy is entirely
gravitational potential energy. This energy comes from the work done by the motors employed
to tow the coaster from the bottom of the ride to its highest point.
6.
(a)
2
2
2
The densities of lead, mercury and gold are roughly 1.13  104 kg/m3, 1.36  104 kg/m3
and 1.93  104 kg/m3, respectively. At normal Earth temperature ranges, lead and gold
are solids, while mercury is a liquid. Which of the solids sinks in mercury (2 marks)?
Which of the solids floats (2 marks)? What fraction of the floating solid will be
submerged in mercury (2 marks)?
Since gold has higher density than mercury, gold will sink.
Since lead is less dense than mercury, lead will float.
Furthermore, the fraction of lead which will be submerged is
 lead
1.13  10 4 113


 0.83 .
 mercury 1.36  10 4 136
7
(b)
Bernoulli’s Law states that at any height (or depth) y in a moving fluid of density ρ,
the fluid pressure P and fluid speed v are related by
1
P  gy  v 2  constant .
2
It follows from this equation that, since the middle term remains constant for any given
value of y, then an increase in v (which makes the last term larger) must be
accompanied by a corresponding decrease in the pressure P.
2
2
7.
State Bernoulli’s Law and explain how it shows that fluid pressure decreases wherever
the speed of a fluid increases (4 marks).
(a)
A star isn't really a star until it "ignites". Briefly describe the process that leads to
ignition (2 marks), and the mechanism that keeps the star in stable equilibrium for
billions of years (2 marks).
2
As the cloud of gas slowly contracts into a dense ball due to mutual gravitational
attraction, the pressure in the core continues to rise, which causes the temperature to
increase. When the core temperature reaches about 10 million degrees, the nuclear
fusion of hydrogen into helium commences, the core “ignites” and begins to “shine”,
and the star is “born”.
2
Moreover, as long as this fusion of hydrogen into helium continues, it produces a steady
outward gas pressure which perfectly balances the inward pull of gravity. Thus, the star
stops contracting and remains in stable equilibrium for anywhere from 1 – 10 billion
years
(b)
What triggers the swelling of a star into a "red giant" (2 marks)? What triggers its final
collapse (2 marks)? What determines whether it ends up as a white dwarf, neutron star
or black hole (2 marks)?
2
When nearly all of the hydrogen in the core has been converted to helium, helium begins
to fuse into carbon. This second fusion reaction produces even more energy, increasing
the outward pressure beyond the inward gravitational pull, and causing the star to begin
slowly expanding outward. As this expansion continues, the star cools slightly and
becomes a “red giant”.
2
In turn, when most of the helium has been converted to carbon, carbon begins to fuse
into iron. However this final fusion reaction consumes more energy than it produces, so
the outward pressure drops dramatically and the star begins to collapse under its own
gravity. This signals that the star is “dying”.
2
A relatively small star (like our Sun) will collapse fairly slowly into a tiny, hot star
called a white dwarf. Significantly larger stars collapse catastrophically, and the sudden
pressure increase in the core causes a supernova explosion, in which most of the star is
blown away. A portion of the core continues to collapse with nothing to stop it, and the
final result is either a neutron star (a solid ball of neutrons) or a black hole (an object in
which even the neutrons are pulverized, and the gravitational distortion is so great that
light can no longer escape, rendering the star invisible).
8
8.
1
1
1
1
1
1
4
(writing)
How can a large, more-or-less spherical cloud of interstellar gas and dust slowly evolve into
a solar system, with a central star and several orbiting planets? Explain in at most 1 page,
including reference to the roles played by gravity, rotation and "clumping" (10 marks, graded
for writing as well as content).
Initially, a large, widely-dispersed cloud of gas and dust begins to contract very slowly due to
the mutual gravitational attraction of all its individual particles. As this contraction continues,
any inherent rotation present in the cloud is magnified, by conservation of angular momentum.
Moreover, as the rate of this rotation continues to increases it causes the cloud to gradually
flatten along the axis of rotation while spreading at the equator. The result is a central, dense
ball of gas surrounded by a wide, relatively flat disc of more distant material. Localized
“clumping” (due to the slightly stronger gravitational attraction between bits of matter which
are in closer proximity) then causes this large disc to begin to organize itself into a number of
distinct rings. Finally, as the dense ball in the center continues its contraction on its way to
becoming a star, continued clumping in the rings leads to larger and larger chunks of material
which eventually come together under gravity, forming individual planets and “sweeping up”
any remaining material as they rotate around the now shining star in their distinct orbits.
9.
Identify something discussed in this quarter of GSCI 1010 (a lecture topic or one of the “class
questions”) (i) which is not specifically covered in questions 2 – 8 above, and (ii) which you
found intriguing, amazing or simply “neat” (4 marks). Briefly describe what you learned about
this topic or question that you didn’t know before, as you might do if you were trying to explain
it to someone who isn’t taking this course (6 marks).
4
Identification of the topic or question.
6
.
Description of what you learned, as if explaining to someone not in the course.
END OF TEST