Download Instructions Grading Scheme

Physics 2210 NAME: _____________________________________________ Fall Semester 2015 UID NUMBER: _______________________________________ Final Exam, Version A TA NAME: __________________________________________ Tuesday, December 15, 2015 Discussion Day (circle one): M/W T/TH Academic Integrity: Giving assistance to or receiving assistance from another student or using unauthorized
materials during a University Examination can be grounds for disciplinary action, up to and including dismissal
from the University.
Honor pledge: On my honor as a University of Utah student, I have neither given nor received unauthorized
assistance on any part of this exam.
______________________________
Name (please print)
______________________________
Signature
____________
Date
Instructions
1) 2) 3) 4) 5) Use a pencil to fill in the circles corresponding to your name and UID on the answer sheet.
In the space above, write your name, UID#, TA name, and circle your discussion days.
Sign and date the honor pledge above.
Use a pencil to fill in the circles on the answer sheet that match your answer choices.
Each problem has only one correct answer, but you can select more than one answer for partial credit on
questions that have 5 choices (MC5): see the grading scheme below.
6) Circle your answers in the exam booklet ALSO as a backup to the answer sheet.
7) Use the space provided in the exam booklet to work out your problems. If you need more space, we can
provide scratch paper.
8) Bring your completed exam booklet and answer sheet to the front of the classroom along with your UID
card. Your UID card must match your face and the name on your exam to receive a grade.
Grading Scheme
MC5: multiple-choice-five-answer questions, each worth 6 points.
Partial credit will be granted as follows:
(a) If you mark only one answer and it is the correct answer, you earn 6 points.
(b) If you mark two answers, one of which is the correct answer, you earn 3 points.
(c) If you mark no answers, or more than two, you earn 0 points.
MC3 and MC4: multiple-choice-three-answer (3 points) or four-answer (4 points) questions.
No partial credit. Choose only one answer!
T/F: true or false questions, each worth 2 points.
No partial credit. Choose only one answer!
2
Scenario 1:
A stick which can rotate around a stationary vertical rod can be used to
measure the speed of a bullet. The stick is mounted onto the rod by drilling a
hole near the end of the stick of the same diameter as the rod so that when the
stick rotates, the two surfaces slide against each other. If the coeffecient of
kinetic friction between these two surfaces is assumed to be constant, then
there is a constant dissipative torque acting on the stick while it rotates.
During the collision, the bullet embeds completely into the end of the stick.
The ultimate goal is to find the speed of the bullet given information about the rotation of the stick after the
collision. (The stick can rotate on the rod, but it cannot slide down the rod.)
Understand the Physics (select only one answer):
1. [2 points] To calculate the angular speed of the stick+bullet system just after the collision, we consider the
situation just before and just after the bullet embeds into the stick. At these two moments in time, the
angular momentum of the system is approximately conserved:
a. True
b. False
2. [3 points] What is the justification for your answer?
a. The collision happens over a very short time, during which friction from the rod does very little work on
the stick+bullet system.
b. The angular momentum is zero just before the collision since nothing is rotating, but is not zero just
afterward since the system rotates.
c. The angular momentum is zero just before and just after the collision.
_________________________________________________________________________________________
3. [2 points] Just after the collision, the stick+bullet system rotates with a particular angular speed. Afterward,
the angular speed of the stick+bullet system is constant for some time and then decreases rapidly:
a. True
b. False
4. [3 points] What is the justification for your answer?
a. The rod exerts an external torque on the system, and this torque increases as time progresses.
b. The rod exerts an external torque on the system, and this torque decreases as time progresses.
c. The rod exerts a constant external torque on the system.
__________________________________________________________________________________________
Plan Your Solution (select only one answer):
On the following page, 𝒎𝒃 is the mass of the bullet; 𝒗𝒃 is the velocity of the bullet just before the collision; 𝑴𝒔
is the mass of the stick; 𝑳 is the length of the stick; 𝝎𝒊 is the angular speed of the stick+bullet system just after
the collision; 𝑰𝒕𝒐𝒕 is the moment of inertia of the stick + bullet system; 𝝉𝒓𝒐𝒅 is the torque on the stick due to
friction with the rod; and 𝒈 is the acceleration due to gravity.
3
Scenario 1 continued:
5. [4 points] Find an expression for the angular speed of the stick+bullet system just after the collision:
a. 𝜔1 =
𝑚 𝑏 𝑣𝑏 𝐿
𝐼𝑡𝑜𝑡
𝑚 𝑏 𝑣𝑏 𝐿
3𝐼𝑡𝑜𝑡
𝜏
c. 𝜔1 = 𝑟𝑜𝑑
𝐿
b. 𝜔1 =
d. 𝜔1 =
𝑀𝑠 +𝑚𝑏 𝑔𝐿
𝐼𝑡𝑜𝑡 𝜏𝑟𝑜𝑑
2
6. [4 points] Find an expression for the number of rotations of the stick+bullet system after the collision if
its initial angular speed is 𝝎𝒊 :
a. 𝑁 =
𝐼𝑡𝑜𝑡 𝜔2𝑖
2 𝜏𝑟𝑜𝑑
4𝜋2 𝐼𝑡𝑜𝑡 𝜔2𝑖
b. 𝑁 =
𝜏𝑟𝑜𝑑
2𝜋 𝐼𝑡𝑜𝑡 𝜔2𝑖
c. 𝑁 =
𝜏𝑟𝑜𝑑
d. 𝑁 =
Wnet =
1
Itot !i2
2
⌧rod ✓ =
Wnet =
K
K=
!f2
!i2 = 2↵ ✓
✓
◆
⌧rod
=2
(2⇡N )
Itot
Itot !i2
=) N =
4⇡⌧rod
𝐼𝑡𝑜𝑡 𝜔2𝑖
4𝜋 𝜏𝑟𝑜𝑑
⌧rod (2⇡N )
1
Itot !i2
2
Itot !i2
N=
4⇡⌧rod
⌧rod (2⇡N ) =
7. [4 points] Find an expression for the moment of inertia of the stick+bullet system after the collision:
J
a. 𝐼HIH = 𝐿L (𝑀N + 𝑚O )
K
Itot = Istick + Ibullet
1
= Ms L2 + mb L2
3
1
= L2 (Ms + 3mb )
3
J
b. 𝐼HIH = 𝐿L (𝑀N + 𝑚O )
K
c. 𝐼HIH =
J L
𝐿 (𝑀N
K
+ 3𝑚O )
d. 𝐼HIH = 𝐿L 𝑀N + 𝑚O
__________________________________________________________________________________________
Execute the Plan (select up to two answers for partial credit):
In the following question use:
𝑀N = 2 kg;
𝑚O = 0.01 kg;
𝜏TIU = 0.1 N-m;
𝐿 = 1.5 m
8. [6 points: select up to 2 answers] Find the speed of the bullet just before the collision if the stick+bullet
system rotates 40 times after the collision:
a. 330 m/s
b. 580 m/s
c. 645 m/s
d. 780 m/s
e. 1000 m/s
✓
◆2
Itot !i2
Itot
mb vb L
N=
=
4⇡⌧rod
4⇡⌧rod
Itot
2 2 2
mb v b L
=
4⇡⌧rod Itot
3m2b vb2 L2
=
4⇡⌧rod L2 (Ms + 3mb )
3m2b vb2
=
4⇡⌧rod (Ms + 3mb )
s
4⇡N ⌧rod (Ms + 3mb )
=) vb =
3m2b
4
Scenario 2:
He starts
by joining two uniform rods with the same mass and length using a pin, and then
places the joined structure on the (frictionless) ice. A rope of negligible mass is
connected to the centers of the two rods so that the structure does not collapse.
An ice fisherman in Minnesota is erecting a tent on the ice to shield him from the weather.
The ultimate goal is to find the tension in the rope.
45°
45°
Understand the Physics (select only one answer):
9. [3 points] To find all the forces acting on the two rods, we should take advantage of the fact that:
a. the sum of all external forces on the system is zero,
b. the sum of all external torques on the system is zero,
c. the sum of all external forces AND torques on the system is zero,
10. [3 points] because:
a. Newton’s third law requires all forces to come in equal and opposite pairs.
b. the system rests on a frictionless surface.
c. the system is in equilibrium.
__________________________________________________________________________________________
Plan Your Solution (select only one answer):
In the following questions, each rod has mass 𝒎 and length 𝑳. 𝑻 is the tension in the rope.
11. [4 points] Find an expression for the magnitude of the normal force acting on the left-hand rod from the
ice surface:
a. 𝑁 = 𝑚𝑔/2 b. 𝑁 = 𝑚𝑔 c. 𝑁 = 2𝑚𝑔
d. 𝑁 = 𝑚𝑔 − 𝑇
12. [4 points] Find an expression for the magnitude of the horizontal force exerted by the left-hand rod on
the right-hand rod at the top pivot:
a. 𝐹\→T = 𝑚𝑔/2
b. 𝐹\→T = 𝑇
c. 𝐹\→T = 2𝑇
d. 𝐹\→T = 2𝑚𝑔
5
Scenario 2 continued:
Execute the Plan (select up to two answers for partial credit):
13. [6 points: select up to 2 answers] Find the value of the tension if the mass of each rod is 2.0 kg:
a. 5 Newtons
T
b. 10 Newtons
L
L
sin 45 + mg cos 45
2
2
c. 20 Newtons
N L cos 45 = 0
T
d. 30 Newtons
e. 40 Newtons
1
sin 45 = mg cos 45
2
=) T = mg
mg
cos 45
2
14. [6 points: select up to 2 answers] The fisherman hangs his load of fish from the hinge at the top of the tent.
The rope is frayed and it will break if the tension exceeds 100 Newtons. What is the largest mass of fish
he can hang (the rods each have mass of 2.0 kg)?
a. 3.1 kg
b. 18.4 kg
c. 7.1 kg
d. 10.2 kg
e. 8.2 kg
T
L
L
sin 45 + mg cos 45
2
2
N L cos 45 = 0
T
1
mg
sin 45 = (mg + W/2) cos 45
cos 45
2
2
T /2 = (mg + W/2 mg/2)
1
T
= (mg + W )
2
2
T = mg + W
W = T mg
6
Scenario 3:
A bicycle racer propels his bike by cranking the pedals, which in turn exerts a
torque on the rear wheel. The bicycle wheels are 70 cm in diameter and each
has a mass of 3 kg. The total mass of the rider + bike is 70 kg.
The goal is to relate the torque generated by the rider to the acceleration of the
bicycle.
Understand the Physics (select only one answer):
15. [2 points] When the bicycle accelerates forward, the friction force from the ground on the rear wheel
points in the backward direction.
a. True
b. False
16. [3 points] What is the justification for your answer?
a. Friction from the ground exerts a counterclockwise torque on the wheel, helping to speed it up.
b. Only the torque generated by the rider acts on the rear wheel.
c. Since the rider exerts a counterclockwise torque on the wheel, friction from the ground exerts a reactive
clockwise torque.
__________________________________________________________________________________________
17. [2 points] When the bicycle accelerates forward, the friction force from the ground on the front wheel
points in the backward direction.
a. True
b. False
18. [3 points] What is the justification for your answer?
a. The front and rear wheels experience equal friction forces from the ground.
b. Friction from the ground initially points forward at low speed, but points backward at high speed.
c. The bicycle frame pushes the front wheel forward, and friction acts to resist this motion.
__________________________________________________________________________________________
Plan Your Solution (select only one answer):
On the following page, assume the total weight of the bicycle + rider is equally distributed between the two
wheels.
The following variables are used on the next page:
𝒂=
𝑰𝒘𝒉𝒆𝒆𝒍 =
𝑹=
𝝁𝒔 =
acceleration of the bicycle
moment of inertia of each wheel
radius of each wheel
coefficient of static friction between each wheel and the ground
𝑴𝒕𝒐𝒕 =
total mass of the bicycle + rider
7
Scenario 3 continued:
19. [4 points] Assume the total weight of the bicycle + rider is equally distributed between the two wheels.
Find an expression for the maximum torque the rider can exert on the rear wheel before it begins to slip:
𝑎
𝑅
1
2
a. 𝜏efg = 2𝐼hijj\ + 𝜇N 𝑀HIH 𝑔𝑅
b. 𝜏efg =
𝑎
𝐼hijj\ + 𝜇N 𝑀HIH 𝑔𝑅
𝑅
𝑎
𝑅
⌧max
1
2
c. 𝜏efg = 𝐼hijj\ + 𝜇N 𝑀HIH 𝑔𝑅
fmax R = Iwheel ↵
a
=) ⌧max = Iwheel + fmax R
R
a
1
= Iwheel + µs Mtot gR
R 2
d. 𝜏efg = 𝜇N 𝑀HIH 𝑔𝑅
20. [4 points] Assume the total weight of the bicycle + rider is equally distributed between the two wheels.
Find an expression for the friction force exerted by the ground on the front wheel:
a. 𝑓N = 0
1
2
𝐼𝑤ℎ𝑒𝑒𝑙 𝑎
b. 𝑓N = 𝜇N 𝑀HIH 𝑔
c. 𝑓N =
d. 𝑓N =
𝑅2
fs R = Iwheel ↵
Iwheel a
=) fs =
R2
𝐼𝑤ℎ𝑒𝑒𝑙 𝑎
2 𝑅2
__________________________________________________________________________________________
Execute the Plan (select up to two answers for partial credit):
For the following question, assume the total weight of the bicycle + rider is equally distributed between the two
wheels. Also, use 𝑅 = 𝟎. 𝟑𝟓 meters; 𝑰𝒘𝒉𝒆𝒆𝒍 = 𝟎. 𝟑𝟓 kg-m2; 𝑴𝒕𝒐𝒕 = 𝟕𝟎 kg; and 𝝁𝒔 = 𝟎. 𝟔.
21. [6 points: select up to 2 answers] What is the acceleration of the bicycle if the rider exerts the maximum
torque on the rear wheel?
1
Iwheel a
a. 𝑎efg = 2.8 m/s2
µs Mtot g
= Mtot a
2 ✓
R2 ◆
2
b. 𝑎efg = 3.4 m/s
Iwheel
1
a Mtot +
= µs Mtot g
2
2
R
2
c. 𝑎efg = 4.1 m/s
µs Mtot g
=) a =
d. 𝑎efg = 4.7 m/s2
2 Mtot + Iwheel
R2
2
(0.6)(70)(9.81)
e. 𝑎efg = 5.6 m/s
2
=
= 2.8 m/s
0.35
2 70 + 0.35
2
Check Your Understanding (select all that apply):
22. [4 points] What would allow the bicycle to accelerate faster? Select all the apply:
a. If the rider exerted a larger torque on the rear wheel, while maintaining equal weight on both wheels.
b. If the rider shifted more of his weight onto the rear wheel, keeping both wheels on the ground.
c. If the rider lifted the front wheel off the ground slightly.
d. If the bicycle tires made more contact with the ground (i.e., if they were wider but not heavier).
8
Scenario 4:
A comet passes near earth along a highly hyperbolic orbit. At its closest
distance from earth, 𝑅e1{ , the comet has maximum speed. At its furthest
distance from earth, 𝑅efg , the comet has minimum speed. In your
calculations, it may help to use 𝐺𝑀j = 4.0 × 10~ km3/s2.
The ultimate goal is to relate the speed of the comet to its position along
the orbital trajectory.
Understand the Physics (select only one answer):
23. [2 points] The comet’s speed depends on its position along the orbital trajectory, but not on its mass.
a. True
b. False
24. [3 points] What is the justification for your answer?
a. The comet’s kinetic energy depends on its mass, so its orbital speed must also.
b. The comet’s potential energy depends on its mass, so its orbital speed must also.
c. The comet’s total mechanical energy is constant, so its mass cancels out.
__________________________________________________________________________________________
Plan your Solution (select only one answer):
25. [4 points] For a comet of mass 𝒎, find an expression that relates the speed of the comet, 𝒗𝟏 and 𝒗𝟐 , at
two corresponding distances from the earth’s center, 𝒓𝟏 and 𝒓𝟐 :
✓
◆
1
1
1
a. −
=
𝑣 L − 𝑣LL
1
1
𝑟2 𝑟1
𝐺𝑀𝑒 J
Wgrav = GMe m
r2
r1
1
1
1
L
L
1
b. −
=
𝑣 − 𝑣J
K = m(v22 v12 )
𝑟2 𝑟1
2𝐺𝑀𝑒 𝑚 L
✓
◆ 2
1
1
1
1
1
1
c. −
=
𝑣LL − 𝑣JL
=
(v 2 v12 )
𝑟2 𝑟1
𝐺𝑀𝑒
r2
r1
2GMe 2
d. 1
𝑟2
−
1
𝑟1
=
1
2𝐺𝑀𝑒
𝑣LL − 𝑣JL
__________________________________________________________________________________________
Execute the Plan (select up to two answers for partial credit):
26. [6 points: select up to 2 answers] When a comet of mass 𝟓×𝟏𝟎𝟏𝟒 kg is very far from the earth (𝑹 ≈ ∞) its
speed is 2 km/s. At what distance from the earth’s center will the comet’s speed be twice this value?
a. 6.7×10„ km
b. 4.0×10~ km
c. 5.0×10„ km
d. 3.3×10„ km
e. 1.3×10~ km
✓
1
r2
1
r1
◆
1
(v 2 v12 )
2GMe 2
1
1
1
1
=
+
(v22 v12 ) =
(42
r2
r1
2GMe
8.0 ⇥ 105
8.0 ⇥ 105
=) r2 =
= 6.67 ⇥ 104 km
12
=
22 )
Scenario 5:
𝑦
9
One simple way to model the motion of atoms in a crystal is to imagine that the atomic nuclei are compact spheres
connected to each other with tiny springs. The image shows a single “unit cell” of a 2-dimensional crystal with a
small (low mass) atom connected to four large (high mass) nearest neighbor atoms. Since the large atoms have
much more mass, we can assume that they are fixed, so only the small atom in the center moves.
If the
oscillation amplitude of the center atom is small, then its motion along the horizontal
and vertical axes are independent. Also, each set of two springs (those along the same
axis) behave as a single spring with twice the spring constant. An 𝑥-𝑦 coordinate
system is centered on the equilibrium position of the center atom.
The ultimate goal is to solve for the motion of the center “atom”.
Understand the Physics (select only one answer):
27. [3 points] Imagine the atom is initially displaced horizontally from equilibrium along the 𝑥-axis and given
an initial push in the horizontal (𝑥) direction. The amplitude of the atom’s motion:
a. will depend only on its initial displacement,
b. will depend on both its initial displacement and speed,
c. cannot be determined from the information given,
28. [3 points] because:
a. the atom’s initial velocity will cause it to exceed its initial displacement in subsequent oscillations.
b. the atom’s displacement from equilibrium will never be larger than the initial value.
c. the amplitude depends on the direction of the initial velocity, which is not specified.
__________________________________________________________________________________________
29. [3 points] Now imagine the atom is again displaced horizontally from equilibrium along the 𝑥-axis and
given an initial push in the vertical (𝑦) direction. The subsequent motion of the atom:
a. will proceed along a diagonal line at a fixed angle relative to the 𝑥- and 𝑦- axes,
b. will proceed along a horizontal line parallel to the 𝑥-axis,
c. will follow an elliptical path in the 𝑥-𝑦 plane,
30. [3 points] because:
a. since the atom starts on the 𝑥-axis, it will continue to oscillate back and forth on that axis.
b. the atom oscillates in the 𝑥- and 𝑦-directions independently, and the 𝑥- and 𝑦- oscillations are in phase
(have the same phase angle, 𝜙).
c. the atom oscillates in the 𝑥- and 𝑦-directions independently, and the 𝑥- and 𝑦- oscillations are out of
phase (have different phase angles, 𝜙g and 𝜙ˆ ).
𝑥
10
Scenario 5 continued:
Plan Your Solution (select only one answer):
In the following question, all four springs have the same spring constant, 𝒌, and equilibrium length. The mass
of the “atom” is designated by 𝒎.
31. [4 points] Find an expression for the oscillation period, 𝑷, the time it takes for the “atom” to complete
one complete oscillation (hint: this period is independent of the initial displacement and velocity):
a. 𝑃 = 2𝜋
b. 𝑃 = 2𝜋
c. 𝑃 = 2𝜋
d. 𝑃 = 2𝜋
𝑚
𝑘
2⇡
!
r
r
kef f
2k
!=
=
m
m
r
2⇡
m
=) P = p
= 2⇡
2k
2k/m
P =
𝑚
4𝑘
𝑚
2𝑘
𝑚
2𝑘
__________________________________________________________________________________________
Execute the Plan (select up to two answers for partial credit):
In the following 2 questions, the oscillation period is 𝑷 = 𝟒𝝅 time units (this is a very short time for a real
atom, but the unit doesn’t matter here). The initial position of the atom is 𝒙𝟎 , 𝒚𝟎 = 𝟏, 𝟎 distance units
(again the unit doesn’t matter). The initial velocity of the atom is 𝒗𝟎𝒙 , 𝒗𝟎𝒚 = 𝟎, 𝟎. 𝟓 velocity units.
32. [6 points: select up to 2 answers] What is the position of the atom at 𝒕 = 𝝅 time units (1/4 oscillation)?
a. 𝑥, 𝑦 = 0, 1 distance units
b. 𝑥, 𝑦 =
2, 2 distance units
c. 𝑥, 𝑦 = −1, 0 distance units
d. 𝑥, 𝑦 = 0, .5 distance units
vmax,y
!
vmax,y P
=
2⇡
(0.5)(4⇡)
=
=1
2⇡
vmax,y = !Ay =) Ay =
e. 𝑥, 𝑦 = 0.5, 0.5 distance units
33. [6 points: select up to 2 answers] What is the speed of the atom at 𝒕 = 𝝅 time units (1/4 oscillation)?
a. 𝑣 = 2 velocity units
b. 𝑣 =
1
2
velocity units
c. 𝑣 = 2 2 velocity units
d. 𝑣 = 0.5 velocity units
e. 𝑣 = 1 velocity units
11
Scenario 6:
𝜔
œ⃗
A children’s toy spins rapidly with angular speed 𝜔 along its axis with the bottom point of its axle resting in a
shallow frictionless depression that allows it to spin without resistance. When it is spun at a slight angle from
vertical it will precess: the bottom point of contact will remain in place, but the top of the axle will rotate along a
circular path that lies in a horizontal plane. For clarity, the direction of the angular velocity vector, 𝜔, has been
shown at a particular moment in time when the axle is tilted to the right by an angle 𝜃 relative to the vertical.
𝜃
ℓ𝓁
The central axle of the toy has length ℓ𝓁, and there is a disk of radius 𝑅 located a
distance ℓ𝓁/3 from the bottom. The mass of the disk is much larger than the mass of
the axle, so the axle does not contribute appreciably to the moment of inertia.
The ultimate goal is to solve for the precessive motion of the toy.
Understand the Physics (select only one answer):
34. [3 points] Considering the circular precessive motion of the toy’s center of mass, the direction of the vector
representing the precession velocity (which is an angular velocity) is:
a. changing, horizontal, and radially outward,
b. constant, vertical, and downward,
c. constant, vertical, and upward,
35. [3 points] because:
a. the torque due to gravity points into the page at the moment shown in the figure.
b. the torque due to gravity points out of the page at the moment shown in the figure.
c. the torque due to gravity points downward always.
__________________________________________________________________________________________
36. [3 points] How does the precession frequency change if the mass of the disk is increased (keeping
everything else the same)?
a. The precession frequency increases,
b. The precession frequency stays the same,
c. The precession frequency decreases,
37. [3 points] because?
a. The external torque depends on the disk’s mass, but the disk’s moment of inertia does not.
b. The external torque and moment of inertia of the disk are both proportional to the disk’s mass.
c. The disk’s moment of inertia depends on its mass, but the external torque does not.
__________________________________________________________________________________________
Plan Your Solution (select only one answer):
𝑴 is the mass of the disk and 𝜽 is the angle of the axle relative to vertical. All other quantities defined above.
38. [4 points] Find an expression for the precession frequency:
a. 𝛺•Tj– =
b. 𝛺•Tj– =
c. 𝛺•Tj– =
d. 𝛺•Tj– =
3𝑔ℓ𝓁 cos 𝜃
2𝑅2 𝜔
2𝑔ℓ𝓁 sin 𝜃
𝑀𝑅2 𝜔
2𝑔ℓ𝓁 cos 𝜃
3𝑅2 𝜔
2𝑔ℓ𝓁 sin 𝜃
3𝑅2 𝜔
⌧grav
Idisk !
M g ` sin ✓
= 1 3 2
2MR !
2g` sin ✓
=
3R2 !
⌦prec =
12
Scenario 6 continued:
Execute the Plan (select up to two answers for partial credit):
For the following questions, use these values:
𝓵 = 𝟐. 𝟓 cm
𝑹 = 𝟏. 𝟓 cm
𝜽 = 𝟏𝟓°
𝑴 = 𝟓 grams (𝟓×𝟏𝟎
𝟑
kg)
39. [6 points: select up to 2 answers] How fast should the toy spin on its axle (what is 𝝎) so that it completes
one precession cycle every second?
2g` sin ✓
a. 𝜔 = 10 radians/sec
⌦prec = 2⇡fprec =
3R2 !
b. 𝜔 = 20 radians/sec
2g` sin ✓
=) ! =
c. 𝜔 = 30 radians/sec
6⇡R2 fprec
2(9.81)(0.025) sin 15
d. 𝜔 = 40 radians/sec
=
6⇡(0.015)2 (1)
e. 𝜔 = 50 radians/sec
__________________________________________________________________________________________
40. [6 points: select up to 2 answers] As the toy precesses, its center of mass follows a horizontal circular
path. This implies that there is a centripetal force acting on the toy as it precesses. What is the magnitude
of this force for the same precession motion described above (one precession cycle per second)?
a. 1.6 × 10
K
Newtons
b. 1.3 × 10
K
Newtons
c. 7.7 × 10
„
Newtons
d. 6.0 × 10
„
Newtons
e. 4.3 × 10
„
Newtons
Fcent = M ⌦2prec Rprec
`
= M 4⇡ 2 sin ✓
3
= (5 ⇥ 10
3
)4⇡ 2
0.025
sin 15
3
__________________________________________________________________________________________
Check Your Understanding (select only one answer):
41. [4 points] What is the source of this centripetal force?
a. There is no real force, but rather a fictitious force results from the precessive motion.
b. The force that holds the axle in place at its contact point with the surface.
c. Gravity acting on the disk.
d. The vertical support force from the surface.
13
Scenario 7:
When a tennis ball is hit by a racquet, the racquet strings and ball distort, providing
a restoring force much like a spring. Thus the collision is approximately elastic.
Regulation tennis balls have a mass of approximately 0.06 kg, and modern racquets
have a mass of approximately 0.6 kg and are about 70 cm long.
The ultimate goal is to relate the speed of the ball after being hit to the initial
speeds and masses of the ball and racquet.
Understand the Physics (select only one answer):
42. [2 points] For a perfectly elastic collision, the approach speed of the ball and racquet immediately before
the collision is the same as the separation speed immediately after the collision.
a. True
b. False
43. [3 points] What is the justification for your answer?
a. Momentum is not conserved in this collision.
b. Since the ball has much less mass than the racquet+player, the separation speed after the collision will
be much larger than the approach speed before the collision.
c. The approach and separation speeds are the same for a perfectly elastic collision in the center-of-mass
frame and any other inertial reference frame.
__________________________________________________________________________________________
Plan Your Solution (select only one answer):
In the following questions, the mass of the ball, 𝒎𝒃 , is much smaller than the mass of the racquet, 𝑴𝒓 . The
initial speeds of the ball and racquet just before the collision are given by 𝒗𝒃,𝒊 and 𝑽𝒓,𝒊 , respectively. The final
speeds of the of the ball and racquet just after the collision are given by 𝒗𝒃,𝒇 and 𝑽𝒓,𝒇 , respectively. All speeds
are zero or positive: direction is given by the sign with the positive direction to the right. The collision is
assumed to be perfectly elastic and one-dimensional.
44. [4 points] Find an expression for the speed of the ball just after the collision if the initial velocity of the
ball is to the left and the initial velocity of the racquet is to the right (all speeds are zero or positive!):
a. 𝑣O,£ = 𝑉T,1 − 𝑣O,1 + 𝑉T,£
b. 𝑣O,£ = 𝑉T,1 + 𝑣O,1 + 𝑉T,£
c. 𝑣O,£ = 𝑉T,1 + 𝑣O,1 − 𝑉T,£
Vr,i
( vb,i ) = vb,f
Vr,f
=) vb,f = Vr,i + vb,i + Vr,f
d. 𝑣O,£ = 𝑉T,1 − 𝑣O,1 − 𝑉T,£
45. [4 points] Find an expression for the average force on the ball during the collision described above if
that collision takes a time 𝜟𝒕 (all speeds are zero or positive numbers!):
a. 𝐹f¦§ =
b. 𝐹f¦§ =
c. 𝐹f¦§ =
d. 𝐹f¦§ =
𝑚𝑏 (𝑣𝑏,𝑓 − 𝑣𝑏,𝑖 )
𝛥𝑡
Favg t =
𝑚𝑏 (𝑣𝑏,𝑓 + 𝑣𝑏,𝑖 )
Favg
2𝛥𝑡
𝑚𝑏 (𝑣𝑏,𝑓 − 𝑣𝑏,𝑖 )
2𝛥𝑡
𝑚𝑏 (𝑣𝑏,𝑓 + 𝑣𝑏,𝑖 )
𝛥𝑡
pball
pball
=
t
mb (vb,f + vb,i )
=
t
14
Scenario 7 continued:
Execute the Plan (select up to two answers for partial credit):
For the following questions, the ball is thrown straight upward by the player during a serve and the racquet
hits the ball when the ball is at the peak of its trajectory. The ball has a mass of 0.06 kg and the racquet has a
mass of 0.6 kg (and is being held by a player of mass 70 kg).
46. [6 points: select up to 2 answers] How fast must the player swing the racquet to achieve a ball speed of 65
meters per second ? In other words, what is the approximate speed of the racquet head just before it hits
the ball ?
a. 𝑉T,1 ≈ 70 meters/sec
b. 𝑉T,1 ≈ 130 meters/sec
c. 𝑉T,1 ≈ 100 meters/sec
d. 𝑉T,1 ≈ 35 meters/sec
Mr Vr,i = Mr Vr,f + mb vb,f
mb
Vr,f = Vr,i
vb,f
Mr
vb,f = Vr,i + vb,i + Vr,f
= Vr,i + 0 + Vr,i
e. 𝑉T,1 ≈ 15 meters/sec
mb
vb,f
Mr
✓
◆
mb
vb,f
1+
=
2
Mr
mb
vb,f
Mr
= 2Vr,i
=) Vr,i
47. [6 points: select up to 2 answers] For the serve described above (final ball speed of 65 meters per second),
what is the average force exerted on the racquet strings by the ball if they are in contact for 𝟓×𝟏𝟎 𝟑
seconds?
a. 𝐹f¦§ = 390 Newtons
b. 𝐹f¦§ = 490 Newtons
c. 𝐹f¦§ = 585 Newtons
d. 𝐹f¦§ = 650 Newtons
e. 𝐹f¦§ = 780 Newtons
pball
t
mb vb,f
=
t
Favg =
15
𝑣hf¦j
Scenario 8:
A transverse wave pulse on a string is shown traveling to the
left. The pulse was produced by shaking the right end of the
string up and down once while maintaining a constant tension
in the string.
The ultimate goal is to relate different properties of the wave pulse to each other.
Understand the Physics (select only one answer):
Two pieces of string have the same length and same mass. Each string has a wave pulse like the one shown
above. The two pulses have the same amplitude and wavelength, but one travels twice as fast as the other.
48. [3 points] The fast pulse:
a. travels on a string with twice the tension as the other string,
b. travels on a string with half the tension as the other string,
vwave
c. travels on a string with 4 times the tension of the other string,
49. [3 points] and was produced by shaking one end of the string:
!
= = f=
k
v1 = f1 =
a. twice as fast as for the slower pulse.
b. at the same rate as for the slower pulse.
v2 = f2 =
c. half as fast as for the slower pulse.
50. [3 points] The kinetic energy carried by the faster pulse:
a. is the same as the kinetic energy carried by the slower pulse,
s
s
s
s
T
µ
T1
µ
T2
µ
f2 = 2 f1 =) f2 = 2f1
s
T2
T1
=2
=) T2 = 4T1
µ
µ
b. is twice as large as the kinetic energy carried by the slower pulse,
c. is four times as large as the kinetic energy carried by the slower pulse,
51. [3 points] because:
a. the amplitude and frequency of both pulses is the same.
Kwave / A2 ! 2
b. The amplitude of both pulses is the same, but the frequency differs by a factor of 2.
c. The amplitude of both pulses is the same, but the frequency differs by a factor of 4.
__________________________________________________________________________________________
Plan Your Solution (select only one answer):
A wave pulse travels to the left (negative 𝒙 direction) on a string as described at the top of the page. The pulse
is one cycle of a harmonic wave with wavelength 𝝀 and amplitude 𝑨. The string has tension 𝑻, mass 𝑴, and
length 𝑳 ≫ 𝝀.
52. [4 points] Find an equation that describes the vertical displacement (𝒚 direction) of the pulse as a
function of position along the string, 𝒙, and time 𝒕:
y(x, t) = A sin(kx + !t)
𝜆
2𝜋
𝑥 − 𝑇𝐿/𝑀 𝑡
b. 𝑦 𝑥, 𝑡 = 𝐴 sin 𝜆
2𝜋
𝑥 + 𝑇𝐿/𝑀 𝑡
c. 𝑦 𝑥, 𝑡 = 𝐴 sin 2𝜋
𝜆
𝑥 − 𝑇𝐿/𝑀 𝑡
d. 𝑦 𝑥, 𝑡 = 𝐴 sin 2𝜋
𝜆
𝑥 + 𝑇𝐿/𝑀 𝑡
a. 𝑦 𝑥, 𝑡 = 𝐴 sin
h ⇣
! ⌘i
= A sin k x + t
k s
"
!#
T
2⇡
= A sin
x+
t
µ

⌘
p
2⇡ ⇣
x + T L/M t
= A sin
16
Scenario 8 continued:
Execute the Plan (select up to two answers for partial credit):
The right end of the string is shaken up and down such that it takes 0.25 seconds to create the pulse shown on
the previous page. The string is 5 meters long and has a mass of 0.5 kg. The wavelength of the pulse is
measured to be 0.75 meters.
53. [6 points: select up to 2 answers] What is the tension in the string?
a. 𝑇 = 0.9 Newtons
b. 𝑇 = 0.8 Newtons
c. 𝑇 = 0.7 Newtons
d. 𝑇 = 0.6 Newtons
f=
p
T L/M =) T =
2 2M
f
L
= (0.75)2 (4)2
e. 𝑇 = 0.5 Newtons
0.5
5
54. [6 points: select up to 2 answers] Now imagine that instead of a wave pulse, we have a standing wave on a
guitar string with the shape shown. The mass of the string is 5×10 K kg and its length is 0.7 meters. The
string has tension 𝑇 = 75 Newtons. What is the vibration frequency of the string?
a. 𝑓 = 103 Hz
b. 𝑓 = 125 Hz
c. 𝑓 = 146 Hz
d. 𝑓 = 176 Hz
e. 𝑓 = 206 Hz
f =
p
T /µ
r
1 TL
=) f =
M
r
1 TL
=
L M
r
T
=
ML
17
Self Assessment:
55. [4 points] How do you think you did on this exam?
a. I scored higher than 90%
b. I scored between 80% and 90%
c. I scored between 70% and 80%
d. I scored between 60% and 70%
e. I scored below 60%
56. [3 points] I thought the difficulty of this exam was:
a. about the same as the midterm exams.
b. more difficult than the midterms.
c. less difficult than the midterms.
57. [3 points] Choose one:
a. I liked the format of this exam and it was about the right length.
b. I liked the format of this exam, but it was too long.
c. I didn’t like the format of this exam.