Physics 53
Summer 2009
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 carries a value of 4 points.
1.
The total external force on a system is zero. It follows necessarily that:
Total mechanical energy is conserved.
√
Total linear momentum is conserved.
Total angular momentum is conserved.
All of these quantities are conserved.
2.
In the situation shown, M > m , and the pulley has mass but no
friction in its bearings. Which of the following is NOT true as
the system moves?
√
The tension in the left hand string is less than mg.
[ T ! mg = ma > 0 .]
m
M
The tensions in the two strings are different.
The tension in the right hand string is less than Mg.
Mechanical energy is conserved.
3.
A platform diver leaves the platform rotating slowly forward, and then pulls her
arms and legs tightly against her body while in the air. This reduces her moment
of inertia about the axis of her rotation. Her rotation speed increases because:
Energy is conserved.
External torques produce an angular acceleration.
Linear momentum is conserved.
√
Angular momentum is conserved.
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Physics 53
4.
Summer 2009
Three objects with the same mass
roll without slipping from the
same height down an incline as
shown: a hoop of radius R, a
solid sphere of radius R, and a
solid sphere of radius 2R. When
they reach the bottom:
The spheres have the same CM speed, which is less than that of the hoop.
√
The spheres have the same CM speed, which is more than that of the
hoop. [ K rot /Ktrans is greater for the hoop, and the same for both spheres.]
The small sphere has the largest CM speed and the hoop has the smallest.
All three have the same CM speed.
5.
Two equal masses are executing SHM with the same frequency, but the
amplitude of #1 is twice that of #2.
The total energy of #1 is twice that of #2. [Four times.]
The maximum acceleration of #1 is four times that of #2. [Twice.]
√
The maximum speed of #1 is twice that of #2.
None of the above is true.
6.
A file card is resting on top of two paper clips on a table as shown. A jet of air
can be directed into the page, either over or under the card.
In either case the card will experience an upward force.
In either case the card will experience a downward force.
If the jet is directed over it the card experiences a downward force, but if
the jet is directed under it the card experiences an upward force.
√
7.
If the jet is directed over it the card experiences an upward force, but if the
jet is directed under it the card experiences a downward force. [Air in jet
has lower pressure than in still air.]
When a metal rod of length L and cross section area A is placed between a hot
and a cold object, the heat flow rate is 10 J/s. If the rod is cut in half and both
pieces are placed between the same two objects, the net heat flow rate will be:
10 J/s.
5 J/s.
√
40 J/s. [Half the length, twice the area.]
20 J/s.
2
Hot
Cold
Physics 53
8.
Summer 2009
In which of the following processes does the internal energy of an ideal gas NOT
change?
Adiabatic expansion. [Decreases.]
Expansion at constant pressure. [Increases.]
√
Isothermal expansion.
Increasing the pressure at constant volume. [Increases.]
Part B: Check True or False. Each question carries a value of 3 points.
1.
The direction of a conservative force on an object is always toward a region
where the potential energy is lower.
√ True
2.
When a stone is dropped from a great height to the earth, gravity does more
work on the stone than on the earth.
√ True
3.
False [Neap tides.]
Heating a house by a heat pump (essentially a refrigerator with the inside of the
house as the hot reservoir) costs less when the difference between indoor and
outdoor temperatures is larger.
True
6.
√ False [Equal magnitude by 3rd law.]
Tidal effects from the sun and the moon tend to cancel when the lines from the
earth to the two bodies make a right angle.
√ True
5.
False [Equal forces, greater distance moved.]
In a collision between a large SUV and a compact car, the force exerted on the car
by the SUV is less than that exerted on the SUV by the car.
True
4.
False
√ False [Costs more.]
The lowest temperature near the ground during a cold winter night will be lower
if the sky is clear rather than covered with clouds.
√ True
False [Radiation into empty space is more effective.]
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Physics 53
Summer 2009
Part C: Problems. Work problems in the space provided, indicating your method clearly. A correct final
answer supported by no argument, or a fallacious one, will receive little or no credit. The problems carry
the point values shown.
1.
A fireworks rocket is launched into the air with initial vertical velocity
component 40 m/s and initial horizontal velocity component 30 m/s.
a.
How long after it is launched does it reach its maximum height?
b.
If it did not explode in flight, how far would it go horizontally before
returning to the same level from which it was launched?
c.
Now suppose that at the top of its flight it explodes into two fragments of
equal mass, A and B, such that B is momentarily at rest. What is the ratio
of the kinetic energy of A immediately after the explosion to the kinetic
energy of the rocket immediately before the explosion?
d.
How far horizontally from the initial launch point does A land (at the
original level)?
[Use g = 10 m/s2 .]
[20 points]
a.
Use v y = v0 ! gt and set v y = 0 to find t = v0 / g = 40/10 = 4 s.
b.
It would return to the original level in 8 s, so it would travel
x = v0 ! 8 = 240 m.
y
y
x
c.
Momentum is conserved, and there is no vertical motion so we have
mv0 = (m/ 2)v , or v = 2v0 . The ratio of kinetic energies is
x
Kf
Ki
d.
=
x
1 (m/ 2)v 2
2
1 mv 2
0x
2
= 2.
It still takes 4 s to return to the original level, so A travels 60 ! 4 = 240 m
after the explosion, plus 30 ! 4 = 120 m before the explosion, a total of 360
m.
[Alternative: The CM of the system continues to follow the original
trajectory, landing at 240 m from the launch point. Mass B drops to earth
at 120 m, so A lands at 360 m.]
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Physics 53
2.
Summer 2009
A wheel of mass m, radius R, and I = 12 mR 2 about its symmetry axis, has an axle
attached as shown to an ideal spring of stiffness k, which is attached to a wall.
The wheel rolls without slipping on the rough floor. You are to show that the CM
executes SHM by showing that the total force
(including friction) has the form F = !m" 2 x , where
x is the displacement of the CM from its position
when the spring is at equilibrium. Then you are to
evaluate ω in terms of the given quantities.
x
+
a.
For a positive value of x, give the directions of: the CM acceleration, the
angular acceleration about the CM (clockwise or counter-clockwise), and
the static friction force. [What direction is the torque?]
b.
Write Newton’s 2nd law for the acceleration of the CM and for the angular
acceleration. [Take care that the positive direction is to the right.]
c.
Use these equations and the rolling condition to find the friction force as a
function of x, then find the total force as a function of x.
d.
Compare to the standard form given above and evaluate ω.
[20 points]
a.
The CM acceleration is to the left; this requires the angular acceleration to
be counter-clockwise, which requires the friction force (which supplies the
only torque) to be to the right.
b.
We have fs ! kx = !ma and fs R = I! .
c.
Use a = R! , eliminate a and α to find fs =
force is thus F = fs ! kx = !
d.
2
kx
3
I
I + mR
2
! kx = 13 kx . The total
.
Using the standard form F = !m" 2 x we find ! =
5
2k
.
3m
Physics 53
3.
Summer 2009
A wheel with a string wrapped around its axle moves from rest down the incline
shown. The wheel has mass m, axle radius R and rim radius 2R. Its moment of
inertia about its CM is I = 3mR 2 . From the original height h to height h1 the
incline is frictionless; from that point on the kinetic friction coefficient is µ k .
a.
Find the acceleration of the CM of the wheel as it moves on the frictionless
part of the incline, in terms of g and θ. [Draw a free body diagram.]
b.
Find the CM speed when it is at height h1 , in term of g and the heights.
c.
Now µ k is such that it moves at constant CM speed down the rest of the
incline. Find µ k in terms of θ. [Which way must friction act?]
d.
In that case, what is the
tension in the string, in
terms of m, g and θ?
[The axle rolls without
slipping on the string, but the
rim does not roll without
slipping on the incline.]
h
h1
[20 points]
a.
The axle rolls on the string, so the linear and angular accelerations are
related by a = R! . We have for the forces down the incline:
mg sin ! " T = ma , and for the torque about the CM: TR = I! . Eliminating T
and α from these three equations and using the value of I, we find
a=
b.
θ
1
4
g sin ! .
Conservation of energy: mg(h ! h1 ) = 12 mv 2 + 12 I" 2 . Using v = R! and the
g(h ! h1 )
. [Can also use v 2 = 2ad , where d is the
2
distance the CM moves along the incline.]
value of I, we find v =
c.
Constant speed requires constant angular speed, because the axle is still
rolling on the string. Thus the total torque about the CM must be zero, so
the friction force must be down the incline. (The point of contact on the rim
is slipping up the incline.) We have mg sin ! + f k " T = 0 , N ! mg cos" = 0 ,
RT ! (2R) f k = 0 , and f k = µ k N . Eliminating T, N and f k , we find
µ k = tan ! .
d.
Since (from the torque equation) f k = 12 T , the equation for force down the
incline gives T = 2mg sin ! . [There are other ways to get this answer.]
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Physics 53
4.
Summer 2009
A uniform rod of length d and mass 6m is lying at rest on frictionless ice, as
shown from above. At one end is attached a small ball of mass m. A second small
ball of mass m moves with velocity v 0 as shown, perpendicular to the rod. It
strikes the end of the rod and sticks to it. The moment of inertia of the rod alone
about its center is I rod = 12 md 2 .
a.
What quantities are conserved in the collision?
b.
What is the velocity (magnitude and direction) of
the CM of the system after the collision?
c.
What is the angular speed of rotation of the system
about its CM after the collision?
[The rod is not pivoted, and is free to move.]
v0
m
6m
d
m
[15 points]
a.
The only external forces are gravity and the normal forces, which cancel
and give no torque about the CM of the system. Thus linear momentum
and angular momentum about the CM are conserved.
b.
The total momentum is that of the ball initially, and we have
Ptot = Mtot vCM , so mv0 = (8m)vCM , or vCM = v0 /8 . It moves to the right.
c.
At the instant of collision the CM of the system is at the center of the rod.
Conservation of angular momentum about that point gives
mv0 ! (d / 2) = Itot" , where Itot = 12 md 2 + 2 ! m ! (d / 2)2 = md 2 . This gives
!=
v0
.
2d
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Physics 53
5.
Summer 2009
Two identical speakers emit sound of wavelength λ in phase with each other.
They are separated by distance 5λ/2 as shown. Points A and B are on part of a
circle centered at the midpoint between the speakers. At that distance the
intensity from one speaker alone is I0 .
a.
What is the intensity at A?
b.
What is the intensity at B?
c.
At how many other places
on the quarter circle
between A and B is the
intensity the same as at A?
[What is the phase
difference between the
waves as they arrive at B?]
A
[15 points]
B
5λ/2
a.
The phase differences are entirely due to path differences: ! = k"x . Since
point A is equidistant from both speakers, !x = 0 and ! = 0 . This gives
constructive interference, so I = 4I0 .
b.
For B, !x = 5" / 2 so ! =
so I = 0 .
c.
2" 5#
$
= 5" . This gives destructive interference,
# 2
Constructive interference occurs when ! = 0, 2" , 4" ,... Between A and B
the phase difference goes from 0 to 5π, which includes 2π and 4π. Thus
there are two other places where the intensity is also 4I0 .
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Physics 53
6.
Summer 2009
A spacecraft of mass 103 kg is moving in a circular orbit of radius R = 106 m
around a small dense planet of mass M, for which (in SI units) GM = 4 ! 1014 .
[All answers are numerical, in SI units.]
a.
What are the spacecraft’s speed v0 and the total energy of the orbit?
b.
The spacecraft’s engines are fired, quickly reducing the speed to v0 / 2
without changing the spacecraft’s location appreciably. What is the total
energy of the new orbit?
c.
What is the distance of closest approach to the planet in the new orbit?
d.
Sketch the new orbit carefully on the figure.
[20 points]
R
a.
From the general formula for energy in a bound orbit we have
GMm
4 " 1014 #103
E=!
=!
= !2 " 1011 J. The potential energy is
6
2a
2 " 10
GMm
4 " 1014 #103
U=!
=!
= !4 " 1011 J. The kinetic energy is thus
6
R
10
K = E ! U = 2 " 1011 J. Thus we have
m/s.
1
(103 )v02
2
= 2 ! 1011 , or v0 = 2 ! 10 4
[Can also use F = mv 2 /r .]
b.
The kinetic energy is reduced by a factor of 2, so the new energy is
E! = 1 " 1011 # 4 " 1011 = #3 " 1011 J.
c.
4 " 1014 #103
, so
2a
2a = 43 ! 106 m. Since the present location is the point of farthest recession
From the general formula we have !3 " 1011 = !
(distance 106 m) the point of closest approach is at distance
from the center of the planet.
d.
Possible orbit shown.
9
1
3
! 106 m
Physics 53
7a.
Summer 2009
A cork ball of volume V and mass density !c is fastened by a short string to the
bottom of a bucket nearly full of water, so the cork is held submerged in water.
Take the mass density of water to be !w , where !w > !c .
a.
Find the tension in the string, in terms of the given quantities and g.
b.
Now suppose the bucket is accelerating to the right at rate a. On the
second drawing, sketch the surface of the water and the configuration of
the string and cork.
–a
[10 points]
g eff
a
g
a.
The ball is at rest, so the total force is zero. We have FB ! mg ! T = 0 , where
FB = V !w g and mg = V !c g . This gives T = FB ! mg = Vg( "w ! "c ) .
b.
The surface of the water is perpendicular to g eff and the buoyant force is
opposite to g eff . The directions are as shown.
7b.
A refrigerator freezes a quantity of water to ice at 273 K by extracting 2730 J of
heat from it. To do this it has input 520 J of work, and the expelled heat goes into
a room at 300 K.
a.
Find the change in entropy of the universe in this process.
b.
If the same quantity of water were frozen by a Carnot refrigerator, using
the same two temperatures, how much work input would be required?
[10 points]
2730
= "10 J/K. For the room
273
2730 + 520 65
J/K. The total change in entropy is
!Sroom =
=
300
6
65
5
!Stot =
" 10 = J/K.
6
6
a.
For the ice !Sice = "
b.
Use QH = QC
TH
300
= 2730 !
= 3000 J. The work required is
TC
273
W = QH ! QC = 3000 ! 2730 = 270 J.
10
Physics 53
8.
Summer 2009
A heat engine uses one mole of an ideal monatomic gas ( cV = 3R / 2 ), operating
on the cycle shown in the P-V diagram. Call the temperature at the lower left
corner T0 , and give all answers in terms of R and T0 .
a.
Find the temperatures at the other two corners and mark them on the
diagram.
b.
For each step (1,2,3) find: the change in internal energy, the work done by
the gas, and the heat input to the gas.
c.
Find the total heat taken in and the total work done by the gas.
d.
Find the efficiency of the engine,
and compare it to the Carnot
efficiency of an engine operating
between the temperatures at the top
corner and the lower left corner.
2P0
2
1
P0
3
[20 points]
V0
2V0
4V0
a.
We have P0V0 = RT0 . At the top corner the pressure and volume have both
been doubled, so the temperature is 4T0 . At the right corner we have
PV = P0 ! 4V0 = 4P0V0 , so the temperature is again 4T0 .
b.
Step 1: !Eint = cV !T = 32 R(4T0 " T0 ) = 92 RT0 ;
W(1) = P0V0 + 12 P0V0 = 32 P0V0 = 32 RT0 (area under curve);
Qin (1) = !Eint (1) + W(1) = 6RT0 .
Step 2: !Eint (2) = 0 (temperature does not change);
W(2) = 2P0V0 + P0V0 = 3P0V0 = 3RT0 (area under curve);
Qin (2) = W(2) = 3RT0 .
Step 3: !Eint (3) = 32 R(T0 " 4T0 ) = " 92 RT0 ; W(3) = !3P0V0 = !3RT0 (area
RT0 . (This is negative, so heat it
under curve); Qin (3) = ! 92 RT0 ! 3RT0 = ! 15
2
taken out).
c.
Heat input (steps 1 and 2): Qin = 9RT0 . Total work: Wtot = 32 RT0 .
d.
Efficiency: ! =
Wtot 1
T
3
= . Carnot efficiency: ! C = 1 " 0 = .
Qin 6
4T0 4
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Physics 53
Summer 2009
Median = 74.5
SD = 14.3
Median = 73.5
SD = 11.1
12