Download 2010 Midterm 1 Solutions - Physics@Brock

BROCK UNIVERSITY
Mid-term Test 1: February 2010
Course: PHYS 1P22/1P92
Examination date: 12 February 2010
Time of Examination: 19:00–21:00
Number of students: 126
Number of hours: 2
Instructor: S. D’Agostino
SOLUTIONS
1. [6 marks] An object with mass 200 g is attached to a horizontal spring, and oscillates
with a frequency of 2.0 Hz. At a certain time, the object has position x = 5.0 cm and
has velocity v = −30 cm/s. Calculate each quantity.
(a) The period.
(b) The amplitude.
(d) The total energy.
(c) The maximum speed.
(e) The angular frequency.
(f) The spring constant.
SOLUTION: (See Page 479, CP 20.) We’ll do Parts (a), (e), and (f) first, since
they’re the easiest. Then we’ll tackle Parts (b), (c), and (d), since they’re all related
and can be tackled by thinking about the conservation of energy. This will be easiest
to do once we know the spring constant, which we will have already determined in Part
(f).
(a) The period is the reciprocal of the frequency:
T =
1
1
=
= 0.5 s
f
2.0
(e) The angular frequency is
ω = 2πf = 2π(2.0) = 12.57 radians/s = 13 radians/s
r
(f) The spring constant is determined by rearranging the equation 2πf =
k
as
m
follows:
k
= (2πf )2
m
k = 4π 2 f 2 m
= 4π 2 (2.0)2 (0.2)
= 31.58 N/m
= 32 N/m
(d) Now calculate the total mechanical energy. Because energy is conserved, if you
calculate the total energy at any time, it equals the (constant) total energy at all
times. Thus, using the information that the speed is v = 0.3 m/s when the position is
x = 0.05 m, we get
1 2 1 2
mv + kx
2
2
1
1
=
(0.2)(0.3)2 + (31.58)(0.05)2
2
2
= 0.04848 J
= 0.049 J
E =
(b) The amplitude can be calculated by noting that the maximum displacement occurs
when all the energy is potential energy; then x = ±A, and
1 2
kA = 0.04848
2
2(0.04848)
A2 =
k
2(0.04848)
A2 =
√ 31.58
A =
0.0031
A = 0.0554 m
A = 5.5 cm
Note that it’s not necessary to memorize a special formula in this case; we just took
the expression for mechanical energy and used reasoning.
(c) The maximum speed occurs when all the energy is kinetic energy; thus
1 2
mv
= 0.04848
2 max
2(0.04848)
2
vmax
=
m
2(0.04848)
2
vmax
=
√ 0.2
0.4848
vmax =
vmax = 0.696 m/s
vmax = 70 cm/s
Once again, note that no special formula needed to be memorized here; just use the
expression for mechanical energy, and a bit of reasoning.
2. [6 marks] Standing waves on a string that is 1.0 m long and fixed at both ends are
seen at successive frequencies of 24 Hz and 36 Hz.
(a) Determine the fundamental frequency.
(b) Determine the wave speed.
(c) Draw the standing-wave pattern when the string oscillates at 36 Hz.
SOLUTION: (See Page 541, CP 14.) (a) The standing wave frequencies for a string
are multiples of the fundamental frequency. Since 24 Hz and 36 Hz are successive
frequencies, the fundamental frequency must be 36 − 12 = 12 Hz. This means that the
standing wave with frequency 36 Hz is the third harmonic (i.e., m = 3), and therefore
the standing wave pattern asked for in Part (c) looks like the third figure in the diagram
at the top of Page 520 in the textbook.
v
, where f1 is the fundamental
(b) The wave speed satisfies the equation f1 =
2L
frequency and L is the length of the string. Thus,
v = 2Lf1
= 2(1.0)(12)
= 24 m/s
3. [6 marks] Two protons are 2.0 fm apart.
(a) Calculate the magnitude of the electric force exerted by one proton on the other.
(b) Calculate the magnitude of the gravitational force exerted by one proton on the
other.
(c) Calculate the ratio of the forces that you calculated in Parts (a) and (b).
(d) What is the relevance of the result of Part (c)?
SOLUTION: (See Page 680, CP 40.) (a) By Coulomb’s law, the electric force that
each proton exerts on the other is
Fe =
=
=
=
=
kq1 q2
r2
ke2
r2
(9 × 109 )(1.6 × 10−19 )2
(2 × 10−15 )2
57.6 N
58 N
(b) By Newton’s law of gravity, the gravitational force that each proton exerts on the
other is
Gm1 m2
r2
(6.67 × 10−11 )(1.67 × 10−27 )2
=
(2 × 10−15 )2
= 4.65 × 10−35 N
= 4.7 × 10−35 N
Fg =
(c) The ratio of the electric force between the two protons to the gravitational force
between two protons is
57.6
Fe
=
Fg
4.65 × 10−35
= 1.2 × 1036
(d) The stupendous ratio in Part (c) indicates that the gravitational force between the
protons is insignificant compared to the electric force. This is typical of microscopic
situations, and means that in striving to understand and explain atomic and molecular
phenomena (including chemistry), gravitational forces can be safely ignored.
4. [6 marks] An electron with an initial speed of 500 000 m/s moves from left to right.
The electron is brought (for an instant) to rest by a uniform (i.e., constant) electric
field.
(a) State the direction of the electric field.
(b) Did the electron move into a region of higher potential or lower potential?
(c) Calculate the potential difference that stopped the electron.
(d) Calculate the initial kinetic energy of the electron, in electron-volts.
(e) Describe what happens to the electron after it is brought to rest.
SOLUTION: (See Page 717, CP 7.) (a) Since the electron slows down, the force
exerted by the electric field on the electron is to the left. This means that the electric
field is pointing to the right.
(b) The electric field points in the direction of decreasing potential, so the potential is
lower towards the right. Thus, the electron moves into a region of lower potential.
(d) The initial kinetic energy of the electron is
K =
=
=
=
=
=
1 2
mv
2
1
(9.11 × 10−31 )(5 × 105 )2
2
1.13875 × 10−19 J
1.13875 × 10−19
eV
1.60 × 10−19
0.712 eV
0.7 eV
(c) The total mechanical energy is conserved. This means that the magnitude of the
change in kinetic energy is equal to the magnitude of the change in potential energy,
which is in turn equal to q∆V :
0
0
∆U
q∆V
∆V
=
=
=
=
=
∆E
∆K + ∆U
−∆K
−0.7 eV
−0.7 V
Thus, the magnitude of the potential difference that stopped the electron is 0.7 V.
(e) The electron is only stopped for an instant, after which it begins to move back to
the left, accelerated by the electric field.
5. [3 marks]
(a) Draw a rough sketch of the potential of a positive point charge (i.e., V vs. r; no
scales needed on the axes).
(b) Repeat Part (a) for a negative point charge.
(c) Can electric field lines intersect? If so, explain what happens at such an intersection point. If not, explain why not.
SOLUTION: (a) This will be similar to the graph in the margin on Page 694 of the
textbook.
(b) This will be similar to the graph in the margin on Page 695 of the textbook.
(c) No, at least not at a point where the strength of the electric field is non-zero. We
know that the electric field vector points in the direction of the electric field line (i.e.,
tangent to the electric field line), so if two field lines crossed at some non-zero angle,
in which direction would the electric field point?
But wait, maybe it could be possible that two electric field lines intersect as long
as they are tangent at the intersection point??? Hmmm, something to think about.
Maybe there is a loophole in the description at the top of page 670 in the textbook!
Or what about this: Could two field lines intersect at a point where the electric field
strength is zero?
For example, consider the electric field lines for a pair of equal positive point charges.
Isn’t there a field line that points from each charge directly towards the other? And
isn’t there a field line along the perpendicular bisector of the line segment joining the
two point charges? And don’t these field lines intersect at the midpoint of the line
segment joining the two point charges?
The moral here is that there is a distinction between reality and our models of reality.
We do the best we can to create logically consistent models of reality; then we do the
best we can to simplify these models to make them understandable at a first-year level.
However, the model we have for electric field (the mathematical concept of a vector
field), when simplified at the first-year level (in terms of electric field lines) seems to
have a little flaw in it. It’s not a fatal flaw, but it’s a flaw nonetheless, and it should
be (and is) corrected in upper-level physics courses. Recognizing the existence of this
small flaw in the electric field-line model of electric fields drives home the point that
in science we treat our theories as provisional, and we exercise our creative and critical
faculties to continually assess theories, and modify them or completely change them if
necessary. This is the way science works: a continual process of creation, observation,
logical and experimental testing, modifying, more logical and experimental testing,
etc., in an endless cycle. Nothing is ever considered final and complete; we are always
open to new perspectives, new understandings, new ideas.
This is one of the fundamental differences between science and dogmatic ideologies. In
other words, when you think you know everything, you stop learning anything.
6. [6 marks] Write a concise paragraph to explain THREE ONLY of the following
items. Remember, brevity and clarity are courtesy. Use the back of this page
and the following blank page if needed.
(a) Explain what simple harmonic motion is. What are the conditions on the forces
that must be satisfied for simple harmonic motion to occur? Give two examples
of simple harmonic motion from daily life.
(b) Explain what a standing wave is. How does the medium move in a standing wave?
How can a standing wave be produced? Include an example of a standing wave
from daily life.
(c) A wave encountering a boundary between two media is typically partly reflected
and partly transmitted. Give two examples of this in daily life. For each example,
how can you tell that the incident wave is partly reflected and partly transmitted?
(d) Explain the principle of superposition. Give two examples of its use.
(e) Give three every-day examples of charge transfer, explaining how the charge transfer occurs and the noticeable consequences.
(f) Explain the field concept, contrasting it with instantaneous action at a distance.
(g) Use an example to explain how it might be possible for two neutral objects to
attract each other with an electric force.
(h) Explain how the electrostatic air filter on a forced-air gas furnace works.
(i) Explain why the electric field outside a conductor is greatest near sharp points
on the conductor, including an application of this idea.
(j) Describe the properties satisfied by electric field lines and equipotential surfaces.
(k) Explain how electric field and electric potential are related.
(l) Describe and explain the key facts about a conductor placed in a region of space
containing an electric field.
(m) Explain why the electric field inside a conductor is zero. Explain why the electric
field inside a hollow in a conductor is zero.
(n) Explain electric shielding, including a few examples from daily life.
(o) A topic of your choice, taken from the material we’ve studied so far in the course.
(Make it good!)