Physics 53
Summer 2008
Final Exam
Solutions
In questions or problems not requiring numerical answers, express the answers in terms of the
symbols given, and standard constants such as g. If numbers are required, use g = 10 m/s2.
Part A: Multiple choice questions. Check the best answer. Each question has a value of 4 points.
1.
Which is a correct statement of a fundamental law of mechanics?
If all the external forces on a system are conservative, total mechanical
energy is conserved.
If the total external force on a system is zero the total angular momentum
about any point is conserved.
If only conservative forces do work, the total linear momentum of the
system is conserved.
√
2.
If any component of the total external force is zero, that component of the
linear momentum is conserved.
In simple harmonic motion in one dimension:
√
The acceleration and the displacement are always in opposite directions.
The acceleration and the velocity are always in opposite directions.
The velocity and the acceleration have their greatest magnitudes at the
same time.
None of the above is true.
3.
A billiard ball (a solid sphere of radius R and mass m) moves to the right across a
rough table. It started without rotating, so it is still skidding, although eventually
it will roll without slipping. Which of the following is NOT true?
The torque about its CM is clockwise.
√
Both its CM speed and its angular speed of rotation are decreasing.
[Angular speed is increasing]
It angular momentum about a point on the table
is constant.
Its total kinetic energy is decreasing.
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Physics 53
4.
Summer 2008
Three objects (1,2,3) start from rest at the same time from the same height on an
incline, and roll to the bottom without slipping. Their moments of inertia obey
I1 > I2 > I3 . The order of speeds with which they will reach the bottom is:
1,2,3
√
3,2,1
1,3,2
2,3,1
5.
Shown is an auto with the normal forces on the front and rear wheels indicated.
When the auto is at rest N1 is somewhat larger than N 2 because the CM of the
auto is closer to the front wheels. When the auto accelerates forward:
Both normal forces remain the same.
Both normal forces increase.
N1 increases while N 2 decreases.
√
N 2 increases while N1 decreases.
N1
N2
[Consider torques about the CM.]
6.
Concerning Kepler’s three laws, which of the following is NOT true?
One law can be used to determine the mass of a planet if it has a moon.
√
One law says that the quantity T 2 / a3 (T is the period, a the semi-major
axis) is the same for the earth’s orbit around the sun and moon’s orbit
around the earth. [Must orbit the same object]
One law says the planetary orbits are ellipses with the sun at a focus.
One law is equivalent to conservation of angular momentum of the orbit.
7.
The change in internal energy of an ideal gas is given by !U = ncV !T :
Only for processes at constant volume.
Only for adiabatic processes.
Only for isothermal processes.
√
For any process.
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Physics 53
8.
Summer 2008
A heat engine operates between reservoirs at 300 K and 250 K. In each cycle it
takes heat 300 J from the hot reservoir.
The net entropy change of the engine itself is 1 J/K per cycle. [Zero]
The engine’s efficiency is 1/5. [No greater than 1/6]
√
At least 250 J of heat must be expelled to the cold reservoir.
None of the above is true.
Part B: True-false questions. Check the correct answer. Each question has a value of 3 points.
1.
A particle acted upon only by a conservative force and released from rest will
move toward lower potential energy.
√ True
2.
When you unscrew the top of a bottle, your thumb and fingers exert forces in
opposite directions but torques in the same direction.
√ True
3.
√ False [Twice, also when they are on opposite sides]
Nights in the desert can be quite cold, even though the days are very hot,
because the clear sky allows rapid energy loss by radiation at night.
√ True
6.
√ False [Always exactly true]
Spring tides, when the sun and moon combine to make especially high tides,
occur only once per lunar month, when the sun and moon are on the same side
of the earth.
True
5.
False
Galileo’s finding, that an object dropped from rest in a gravitational field has an
acceleration independent of its mass, is only valid as an approximation for
objects near the earth’s surface.
True
4.
False
False
Any kind of heat flow from higher to lower temperature is an irreversible
process.
√ True
False
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Physics 53
Summer 2008
Part C: Problems. Work problems in the space provided, indicating your method clearly. Problems have
the point values shown.
1.
A hunter with a blow-gun intends to shoot a monkey hanging from a tree limb at
height h above the ground. The hunter is at distance 3h from the base of the tree.
Having studied physics, the hunter knows that if he aims his blow gun directly
at the monkey, and shoots at the instant the monkey releases its grip on the limb
to drop to the ground, then the dart will strike the falling monkey. But the dart
must travel fast enough to hit the monkey before it reaches the ground.
a.
How long does it take the monkey to reach the ground?
b.
What must be the minimum horizontal component of the dart’s initial
velocity?
c.
What must be the minimum initial speed of the dart?
Give all answers in terms of h and g. [Draw a picture.]
[15 points]
a.
The time is given by h = 12 gt 2 , so t = 2h/ g .
b.
It must go 3h in this time or less, so 3h = v0 t , or
x
v0 = 3h ! g / 2h = 3 gh/ 2 .
x
c.
To shoot at the monkey, we must have v0 / v0 = h/3h = 1/3 , so
y
v0 =
y
1
v
3 0x
v 2 = v0
2
x
=
+ v0
x
gh/ 2 . The speed is given by
2
y
= 9gh/ 2 + gh/ 2 = 5gh , or v = 5gh .
3h
4
h
Physics 53
2.
Summer 2008
A ball of mass m attached to a massless string of length R is moving as shown in
a vertical circle. We are interested in the situation at the three points indicated.
a.
The ball has the minimum speed necessary
to stay in the circle at point A. What is that
speed? Explain.
b.
What is its speed at point B?
c.
Suppose when it reaches point C the string
breaks. To what height h above C will it rise
before it falls again?
A
v
C
Give answers in terms of R and g.
[15 points]
a.
B
The minimum downward acceleration is g (when the tension in the string
is zero) so the minimum speed is given by g = vA 2 /R , or vA = Rg .
b.
By conservation of energy,
1
mvA 2
2
+ mg(2R) = 12 mvB 2 . This gives (using
the previous answer) vB = 5Rg .
c.
The speed at C is given by
1
mvA 2
2
+ mg(2R) = 12 mvC 2 + mgR , so
vC = 3Rg . It is moving vertically so it is at rest at height h.. Using
conservaton of energy again we find mgh = 12 mvC 2 , or h = 32 R .
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Physics 53
3.
Summer 2008
A crate is resting on the bed of a truck as shown. The truck is moving to the left
with speed v0 when the driver applies the brakes to avoid hitting a stopped car
in front of him. He wants to stop quickly, but not to have the crate either slide
forward on the bed or tip over. Assume the braking acceleration is constant. The
line from the CM of the crate to the front corner makes the angle θ shown. The
coefficient of static friction between the crate and the truck bed is µs .
a.
If the crate does not tip over before it slides, what is the smallest distance x
in which the truck can stop without making the crate slide?
b.
If the crate tips over before it slides, what is the smallest distance x in
which the truck can stop without making the crate tip over? [Consider
torques about the CM in an inertial frame, or else use the frame of the
truck and g eff .]
c.
For what value of µs are these distances the
same?
θ
[15 points]
a.
v0
The only horizontal force on the crate is friction, so fs = ma . Using the
maximum friciton force we find amax = µs mg /m = µs g . The stopping
distance is given by v0 2 = 2ax , so xmin = v0 2 / 2amax = v0 2 / 2 µs g .
b.
Torques about CM: Friction acts to the right along the bed surface, with
magnitude ma. Its torque is counter-clockwise, with moment arm h, the
height of the CM. The normal force has magnitude mg and is upward. It
can act effectively at any point where the crate contacts the bed. Its
maximum clockwise torque occurs when it acts at the left corner, when its
moment arm is w / 2 , half the width of the crate. Just before the crate tips
over the torques are mg ! w / 2 = ma ! h , or a = g ! w / 2h = g tan " .
Frame of the truck: Take torques about the left lower corner. Friction has
no moment arm, and if the normal force acts at that corner it has no
moment arm. The only torque would come from gravity, given by g eff . If
this vector is along the line from the CM to the corner, there is no moment
arm, so the total torque is zero. This happens when a = g tan ! . If a is
larger, there is a counter-clockwise torque about the corner and the crate
rotates that way.
In either case, the minimum stopping distance is
xmin = v0 2 / 2amax = v0 2 / g tan ! .
c.
The two distances are the same if µs = tan ! .
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Physics 53
4.
Summer 2008
Shown is a bicycle plus rider of total mass M accelerating to the right at rate a,
both wheels rolling without slipping. The wheels have the same radius R and
moment of inertia I about their axles. A clockwise torque ! 0 is applied by the
rider to the rear wheel through the pedal mechanism. You are to show that the
increase in the kinetic energy of the system comes from the power supplied by
this torque, according to the formula P = !" , where ω is the angular velocity of
the object the torque τ is applied to (in this case, the rear wheel).
a.
Call the friction forces by the road on the front and rear wheels fF and fR .
Show their directions on the drawing and write the equation relating these
forces to a. [What causes the angular acceleration of the front wheel?]
b.
Write the two equations relating the total torque on each wheel to its
angular acceleration α, in terms of fF , fR and given quantities.
c.
Eliminate fF and fR to find ! 0 in
terms of a, α and given quantities.
d.
Finally, write the expression for the
total kinetic energy K of the system
and show that dK /dt = ! 0" . [Use
the fact that the wheels are rolling.]
a
[20 points]
a.
The rear wheel will tend to slip backwards on the road because of the
applied torque ! 0 , so friction is forward. The wheels have a clockwise
angular acceleration ! = a/R (rolling). For the front wheel, the torque to
produce ! can only be caused by friction, which must therefore be
backward. Friction is the only external horizontal force, so fR ! fF = ma .
b.
For the rear wheel ! 0 " fR R = I# . For the front wheel fF R = I! .
c.
Substituting for the friction forces in the equation in (a) we find
! 0 = maR + 2I" .
d.
We see that P = ! 0" = maR" + 2I# w = mav + 2I#" (using the rolling
condition v = R! ). Now K = 12 mv 2 + 2 ! 12 I" 2 , so dK /dt = mav + 2I!" ,
which is equal to P as claimed.
[The momentum of the system is provided by the friction forces, but the
energy comes from the work done by the torque ! 0 , which of course comes
indirectly from the muscles of the rider.]
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Physics 53
5.
Summer 2008
A child of mass m runs at speed v0 and jumps onto the rim of a playground
turntable at rest, as shown from above. The turntable has radius R and moment
of inertia I = 9mR 2 about its axle. The child’s velocity is tangent to the rim when
she jumps on. Give all answers in terms of m, R and v0 .
a.
What is the angular speed of the rotation when the child comes to rest
relative to the turntable?
b.
What fraction of the child’s initial energy is lost in bringing her to rest
relative to the turntable?
c.
Later the child jumps off, but it such a way that her
velocity relative to the ground is zero as she leaves the
turntable. How much energy do her muscles contribute
in making this jump? [What is the final angular speed
of the turntable?]
[15 points]
a.
Angular momentum about the axle is conserved (nothing else is) so we
have mRv0 = Itot! = (9mR 2 + mR 2 )! = 10mR 2! , or ! = v0 /10R .
b.
The final kinetic energy is K = 12 Itot! 2 = 5mR 2 " (v0 /10R)2 =
is 1/10 of the original kinetic energy
the kinetic energy.
c.
1
mv0 2 .
2
1
mv0 2 ,
20
which
The “collision” lost 9/10 of
Angular momentum is conserved again, and the child’s final angular
momentum is zero, so we have Itot! = I! " , or
10mR 2 ! (v0 /10R) = (9mR 2 )# " , which gives " ! = v0 /9R . The final kinetic
1
mv0 2 . She has added energy
energy is K f = 12 I" ! 2 = 12 # 9mR 2 # (v0 /9R)2 = 18
equal to
1
mv0 2
18
!
1
mv0 2
20
1
= 180
mv0 2 by jumping off.
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Physics 53
6.
Summer 2008
Two questions about fluids, 5 points each.
a.
A bubble forms at the bottom of a glass of champagne and rises toward
the surface. As it rises it occupies more volume, and its upward
acceleration becomes larger. Explain both of these facts.
The volume increases because the pressure from the liquid decreases as
the bubble rises, while the temperature stays constant. Because the
volume increases, so does the buoyant force, and the upward acceleration
also increases.
b.
If the long arm of a siphon exceeds about 10 m in
length the siphon will not work. Explain why.
L
Compare points at the top and bottom of the long arm. The speed of the
fluid is the same at both points because the area of the tube is the same
(continuity). At the bottom the pressure is air pressure. We have from
Bernoulli Ptop + 12 ! v 2 + ! gL = P0 + 12 ! v 2 , or Ptop = P0 ! " gL . If L is more
than about 10 m, Ptop will become negative and the flow will form cavities
and stop.
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Physics 53
7.
Summer 2008
A block of mass m is attached to a spring of stiffness k as shown.
a.
If the surface the block slides on is frictionless and the block is set into
oscillation, what is the angular frequency ω?
b.
Suppose the surface has friction coefficients µ k = 0.25 and µs = 0.6 . The
block is pulled out from the wall, stretching the spring by A = mg / k , and
released from rest. How far will the block slide past the equilibrium point
of the spring before coming momentarily to rest? [Consider the change in
total mechanical energy, and recall that a2 ! b 2 = (a + b)(a ! b) .]
c.
Will it stay at rest at that point? Explain with a free body diagram.
Give answers in terms of m, g and k.
[15 points]
a.
The usual formula, ! = k /m .
b.
Let the distance beyond the equilibrium point be x. The work done by
friction is W f = ! f k (A + x) = ! µ k mg " (A + x) . Use the fact that the change in
total mechanical energy is the work done by non-conservative forces. The
initial and final kinetic energies are zero, so we have
1
kx 2 ! 12 kA 2 = W f = ! µ k mg " (x + A) . Using
2
1
k(x ! A) = ! µ k mg , so x = A ! 2(m/ k)µ k g .
2
the hint this becomes
Using the valus of A and µ k
we find x = 0.5(mg / k) .
c.
The spring force is kx = 0.5mg , while the maximum static friction force is
fs = µs mg = 0.6mg . The block will stay at rest.
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Physics 53
8.
Summer 2008
Two questions concerning waves.
a.
The frequencies of the 7 notes in the “diatonic” scale in music are such
that harmonics 2-6 of the lowest note in the scale also produce notes in the
scale, in higher octaves. But the 7th harmonic is not a note in the scale; for
that reason the piano is designed to suppress it. How far from one end
should the hammer strike the string (of length L, fixed at both ends) so no
energy goes into the 7th harmonic? Explain with a picture. [The point
where the hammer strikes cannot be a node for any harmonic.]
[5 points]
The 7th harmonic will have a node at distance L/7 from each end. The
hammer should strike at one of those points, forcing an antinode and
eliminating the 7th harmonic from the sound.
b.
Two waves of equal wavelength and frequency and moving in the same
direction interfere. The waves have separate intensities I1 and I 2 .
i.
Show that the maximum resultant intensity is ( I1 + I 2 )2 .
ii.
Show that the minimum resultant intensity is ( I1 ! I 2 )2 .
[Consider the relation between intensity and amplitude.]
[10 points]
For any wave the intensity is related to the amplitude by I = K ! A 2 , where
K is a constant dependent on the type of wave. Thus A1 = I1 /K and
A2 = I 2 /K . The resultant wave will have amplitude A = I /K . For
constructive interference, A = A1 + A2 , and for destructive interference
A = A1 ! A2 . Thus for constructive interference
I /K = I1 /K + I 2 /K ,
so I = ( I1 + I 2 )2 . This is the maximum intensity claimed. A similar
argument gives the minimum intensity claimed.
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Physics 53
9.
Summer 2008
An engine using one mole of an ideal monatomic gas ( cV = 32 R ) operates in the
reversible cycle shown as a circle on the P-V diagram. The cycle starts at a and
proceeds clockwise.
a.
b.
c.
In which steps (denoted as a ! b , etc.) is heat taken into the engine and in
which steps is it expelled? Explain how you know for each step. [Find the
temperatures at the four points shown.]
b
The net work output per cycle (the area
3P0
inside the circle) is ! P0V0 . Find the total
heat taken in per cycle, in terms of P0
a
c
and V0 only. [Each quadrant of the
P0
circle represents ¼ of the total work
d
represented by the circle.]
Find the efficiency of the engine.
[15 points]
a.
V0
3V0
Point a: 2P0 , V0 , so Ta = 2P0V0 /R .
Point b: 3P0 , 2V0 , so Tb = 6P0V0 /R .
Point c: 2P0 , 3V0 , so Tc = 6P0V0 /R .
Point d: P0 , 2V0 , so Td = 2P0V0 /R .
a ! b : W > 0 , !U > 0 (temperature increases) so heat is taken in.
b ! c : W > 0 , !U = 0 , so heat is taken in.
c ! d : W < 0 , !U < 0 (temperature decreases), so heat is expelled.
d ! a : W < 0 , !U = 0 , so heat is expelled.
b.
a ! c : W = 12 ! P0V0 + 4P0V0 , !U = cV !T = 32 R(Tc " Ta ) = 6P0V0 , so
Qin = 10P0V0 + 12 ! P0V0 .
c.
The efficiency is e =
Wtot
! P0V0
2!
=
=
" 0.27 .
1
Qin 10P0V0 + ! P0V0 ! + 20
2
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Physics 53
10.
Summer 2008
Two questions about devices.
a.
The rate at which heat flows by conduction into a house on a hot day is
proportional to the difference between outdoor and indoor temperatures.
Call this amount A(To ! Ti ) , where A is some constant. In order to keep
the indoor temperature constant the air conditioner must remove heat
from the house at the same rate. Assume it runs on a Carnot cycle and
show that the power supplied to the air conditioner (from the electric
power grid) in order to keep the indoor temperature constant is
proportional to (To ! Ti )2 .
[10 points]
The heat removed by the air conditioner is QC , where W = QH ! QC .
Using the Carnot condition we have QH = QC ! To /Ti , so W = QC
and the power required to remove this heat is
To ! Ti
Ti
dW dQC To " Ti
. Setting
=
!
dt
dt
Ti
dQC
equal to the rate at which heat enters the house by conduction, we
dt
(To ! Ti )2
dW
find
as claimed.
=A
dt
Ti
b.
When a refrigerator turns water into ice, the entropy of the freezer and its
contents decreases. Why is this not a violation of the 2nd law of
thermodynamics?
[5 points]
The 2nd law says the total entropy of a closed system cannot decrease. The
freezer compartment is not a closed system, since it interacts with the rest
of the refrigerator and with the external world, e.g., the kitchen.
13