Physics 221 2005S Exam 2
Name (printed): ____________________________________________
ID Number: ______________________________________________
Section Number: __________________________________________
INSTRUCTIONS:
Each question is of equal weight, answer all questions. All questions are multiple choice.
Choose the best answer to each question.
Before turning over this page, put away all materials except for pens, pencils, erasers, rulers, your calculator and “aid sheet”. An “aid sheet” is one two sided 8½ × 11 page of notes prepared by the student. There is also a list of possibly useful equations at the end of the exam.
"In general, any calculator, including calculators that perform graphing numerical analysis functions, is permitted. Electronic devices that can store large amounts of text, data or equations are NOT permitted." If you are unsure whether or not your calculator is allowed for the exam ask your TA.
Examples of allowed calculators: Texas Instruments TI-30XII/83/83+/89, 92+
Casio FX115/250HCS/260/7400G/FX7400GPlus/FX9750 Sharp EL9900C.
Examples of electronic devices that are not permitted: Any laptop, palmtop, pocket computer, PDA or e-book reader.
In marking the multiple choice bubble sheet use a number 2 pencil. Do NOT use ink. If you did not bring a pencil, ask for one. Fill in your last name, middle initial, and first name. Your ID is the middle 9 digits on your ISU card. Special codes K to L are your recitation section, for the Honors section please encode your section number as follows:
H1 ⇒ 02; H2 ⇒ 13 and H3 ⇒ 25 .
If you need to change any entry, you must completely erase your previous entry. Also, circle your answers on this exam. Before handing in your exam, be sure that your answers on your bubble sheet are what you intend them to be.
It is strongly suggested that you circle your choices on the question sheet.
You may also copy down your answers on the record sheet at the end and take this page with you for comparison with the answer key to be posted later.
When you are finished with the exam, place all exam materials, including the bubble sheet, and the exam itself, in your folder and return the folder to your recitation instructor. No cell phone calls allowed. Either turn off your cell phone or leave it at home.
Anyone answering a cell phone must hand in their work; their exam is over.
Best of luck, David Atwood and Paula Herrera-Siklody
Page 1 of 18

Physics 221 2005S Exam 2
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[28] A kid of mass m = 40 kg standing on a cart that is moving to the right at v
0
= 1.0 m/s suddenly jumps out of the cart to the left at 0.50 m/s relative to the ground. The mass of the cart is M = 20 kg. What is the final speed of the cart the instant after the kid jumps?
(A)1.0m/s (B)2.0m/s (C)3.0m/s (D)4.0m/s (E)5.0m/s
Solution:
[D] Use conservation of momentum. Taking right to be the +x direction, the initial momentum is
0
= ( m + M )
G
0
= ( 40 kg + 20 kg )( 1 m / s ) i ˆ = 60 Ns i ˆ . This will be the total momentum after the jump. To find the final momentum of the cart, we need to subtract off the final momentum of the kid: cart
=
0
− kid
=
0 of the cart is therefore
− v v kid cart m
=
= p i ˆ cart
[
60
/
Ns
M =
− ( −
( 80
0 .
5 m
Nm
/ i ˆ ) s
/
)( 40
20 kg kg =
)
]
4
= m
80
/ s
Ns i ˆ i ˆ . The final velocity
and therefore the speed is v cart
=4m/s.
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[29] Two vectors
G G
A and B angle between them.
are such that
G G
A B 4 and
G G
A B 2 i ˆ . Determine the
(A)13° (B)27° (C)42° (D)63° (E)71°
Solution:
G
A ⋅
G
B
Thus
= AB
AB = cos θ
(
G
A ⋅
G
B
|
)
A
2
×
+ |
B |
G
A
=
×
AB
G
B sin
| 2
θ
= 1 .
From the cross product then
θ = arccos
( G
A ⋅
G
B /( AB )
)
= arccos( 4 / 5 ) = 26 .
6 0 .
Note: working it out this way avoids ambiguities of inverse trig functions. You could also take the ratio of cross product/dot product and get tan θ. In this case you would get the right answer but if A·B were negative you would have to add 180º to be on the correct
Page 2 of 18

Physics 221 2005S Exam 2 branch of the arctan function. The point is that without any ambiguity, the angle between two vectors in three dimensions is defined as the arccos(A·B/AB)
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[30] A block starting at rest at a height of 1m slides down a frictionless ramp at angle 30º to the horizontal. The block then slides along a horizontal track with coefficient of kinetic friction µ k
=0.5. How far along the track does it slide before it comes to rest?
Start from rest
Frictionless Distance=?
Height=1m
30º stop
Friction µ k
=0.5
(A) 0.5m (B)1.0m (C)1.5m (D)2.0m (E)4.0m
Solution:
[D] The initial mechanical energy of the block (taking the bottom of the ramp as the 0 of gravitational potential) is E=mgh. This is converted into kinetic energy when the block is at the bottom of the ramp. Friction must do this amount of work against the block in order to bring it to a stop. Thus if the distance is d and f f
is the force of friction (acting to the left), E=d f f so: d = E /
( )
=
µ mgh k mg
=
µ h k
=
1 m
0 .
5
= 2 m
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[31]
A soldier on a firing range fires an automatic weapon that shoots bullets at a rate of
1000 per minute. If each bullet has a mass of 5g and is fired with a speed of 200m/s, what is the average recoil force exerted on the weapon while it is being fired?
(A)1000N (B)556N (C)139N (D)33.3N (E)16.7N
Solution:
[E] The average recoil force is the average impulse/unit time. Therefore:
(average recoil)
=(momentum of bullets)(fireing rate)
=(.005kg)(200m/s)(1000/(60s))
=16.7N
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Page 3 of 18

Physics 221 2005S Exam 2
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The following applies to the questions [32] and [33]: A particle of mass m = 1.2 kg moves along the x -axis under the influence of a single force whose associated potential energy is depicted in the figure below. At t = 0, the particle is released at rest at x = 8 m.
U (J)
50
0
−30 J
-50
3m x (m)
[32] How close to the origin will the particle get?
(A) 3 m
(B) 4 m
(C) 5 m
(D) 8 m, because the particle will not move since U (8 m) = 0 J.
(E) 8 m, because the particle will move away from the origin.
Solution:
[A] Looking at the graph we see that when x=8m, U=0. The other turning point of the motion will be when x is again 0 which happens at x=3m. The particle will thus oscillate between x=3m and x=8m, x=3m being the closest it will get to the origin.
________________________________________________________________
[33] What is the maximum kinetic energy of the particle?
(A)0 (B)10 J (C)20 J (D)30 J (E)40 J
Page 4 of 18

Physics 221 2005S Exam 2
Solution:
[D] The particle has 0 mechanical energy. The maximum kinetic energy will occur when the potential energy is minimum at x=5m where U=-30J. The kinetic energy there is thus K=30J.
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[34] Which of the following is true about a conservative force?
(A) The force does a positive amount of work when a particle moves around a closed loop
(B) The work done by the force as a particle moves from point A to point B depends on the path of the motion
(C) The work done by the force is related to the difference in a potential energy function.
(D) When a conservative force does work on an object, its temperature will change.
(E) None of the above.
Solution:
[C] (A) is false since a conservative force does 0 work around a closed loop. (B) is false because the work done by a conservative force is path independend and (D) is false because a conservative force does not convert energy to or from a non-mechanical form such as heat. On the other hand (E) is true, for any conservative force, there exists a potential energy function.
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[35] Find the center of mass of the lollipop-shaped object shown below, made of a solid
3-kg sphere of uniform density of radius R = 20 cm glued to the end of a thin 1.0-kg stick of length L = 40 cm. The stick has a uniform density and negligible width. y
CM x
L
R
(A) x
CM
= 30 cm (B) x
CM
= 40 cm (C)
(D) x
CM
= 60 cm (E) x
CM
= 70 cm x
CM
= 50 cm
Solution:
[C] The center of gravity of the stick is x cm
(stick)=L/2. The center of gravity of the sphere is x cm
(sphere)=L+R. The center of gravity of the whole system is:
Page 5 of 18

Physics 221 2005S Exam 2 x cm
=
= x cm
( stick ) m stick m stick
+
+ x cm m
( sphere sphere
) m sphere
( L / 2 ) m stick m stick
+ ( L + R ) m sphere
+ m sphere
=
( 20 cm )( 1 kg ) + ( 60 cm )( 3 kg )
4 kg
= 50 cm
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[36] The same object described in question [35] is to rotate about three different axes 1,
2 and 3. All three axes are in the plane of the page. Compare the moments of inertia where I
1
is the moment of inertia about axis 1, I
2
the moment of inertia about axis 2 and I
3 the moment of inertia about axis 3:
Axis 3
Axis 1 Axis 2
(A) I
1
< I
2
< I
3
(B) I
2
< I
1
< I
3
(C) I
2
< I
3
< I
1
(D) I
3
< I
2
< I
1
(E) I
3
< I
1
< I
2
Solution:
[D]
1. I
2
> I
3 because both receive an equal contribution from the sphere but only case 2 gets an additional contribution from the rod. The order is thus I
3
< I
2
< I
1
.
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Page 6 of 18

Physics 221 2005S Exam 2
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[37] A solid cylinder of mass m =20kg, length L =0.5m and radius R =0.1m and uniform density is lying on its side with the axis parallel to the ground. How much net work must be done to tip it to the vertical so that the axis is perpendicular to the ground?
After
Before
Work=? L
(A) 29J (B)35J (C)39J (D)49J (E)0J
Solution:
[A] The work done on the cylinder is the increase in the gravitational energy which is W=mg ∆ y cm
where m is the mass of the cylindar and ∆ y cm
is the change in the height of the center of mass. Initially the height of the center of mass is R. After the height of the center of mass is L/2 so ∆ y cm
W=(20kg)(9.8m/s²)(0.15m)=29.4J.
=½L − R=0.15m. The work is thus
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[38] A hollow cylinder of radius R and mass m rolls without slipping up an incline of angle θ with an initial velocity v
0 the net torque, τ
G as shown below. Which of the following is true about
,acting on the cylinder with respect to the center of the cylinder?
(A) |
(B) |
(C) |
(D) |
τ
G
τ
G
τ
G
τ
G
|
|
|
| >
<
>
< mgR mgR sin sin
θ
θ mgR mgR sin sin
θ
θ
, direction out of the page
, direction out of the page
, direction into the page
, direction into the page x y
(E) none of the above
Hollow Cylinder
Mass=m
Radius=R
Solution:
[D] The net acceleration of the cylinder is down the ramp at a rate less than gsin θ thus, using the coordinate system shown: 0 > a x
> g sin θ . Using Newton’s second
θ
Page 7 of 18

Physics 221 2005S Exam 2 law, we therefore conclude that 0 > F net x
> − mg sin θ . The two forces which have components in the x direction are gravity and static friction. The x component of the weight is w x
= − mg sin θ and since F net x
= w x
+ f static , x we find that
0 < f static , x
< + mg sin θ , i.e. in a direction up the ramp. Since static friction is the only force which produced a torque, the magnitude of the torque is | τ
G
| < + mgR sin θ because the length of the lever arm is R. Using the right hand rule the direction of this torque is into the page. You can also see the direction by noting that initially the angular velocity is out of the page but as the object slows down the magnitude decreases so the angular acceleration, hence the torque, is into the page.
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[39] A horizontal disk of radius R and mass M rotates with an angular speed ω . Another disk with the same axis, same radius, mass M /2 and no initial speed is dropped on top of the first one. What is the final angular velocity, ω f
, of the two disk system when the disks stop rubbing and rotate as a single unit. Assume that there is no friction on the axel or other external torques.
(A) ω f
=ω/2 (B) ω
(D) ω f f
=2ω/3 (C) ω
=3ω/2 (E) ω f
=2ω f
=ω
ω
Solution:
[B]Angular momentum is conserved in this case. The initial angular momentum is only due to disk 1 so L= ω I
Equating these two:
1
. After, both disks are rotating together so that L= ω f
(I
1
+I
2
).
ω I
1
= ω f
( I
1
+ I
2
)
ω f
= ω
1 +
1
I
2
/ I
1
For the disks, I
1
=MR²/2 and I
2
=MR²/4 thus ω f
=2ω/3.
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The following applies to questions [40], [41], and [42]: A 2.0-kg block is attached to one end of an ideal, massless spring with k = 400 N/m. The other end of the spring is fixed to a wall. The block can slide on a frictionless, horizontal table. m x x = 0
Disk 2
Disk 1
Page 8 of 18

Physics 221 2005S Exam 2
Let x = 0 be the position of the block when the spring is relaxed, and take the positive x axis in the direction away from the wall. We define the potential energy of the spring to be 0 when x=0, i.e. U(x=0)=0. At t = 0, the block is at x = 0 and is moving at 2.1 m/s in the positive x direction.
[40] What is the mechanical energy of the system?
(A) 2.0 J (B)4.4 J (C)6.1 J (D)8.7 J (E)11 J
Solution:
[B]If the block is at x=0 then U=0 so E=K+U=K=mv²/2=(2kg)(2.1m/s) ²/2=4.4J.
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[41] If the speed of the block at x = 0 was twice as large, the period of the motion would change by a factor of:
(A) 4 (B) 2 (C)1 (would not change) (D)1/2 (E)1/4
Solution: [C] for a harmonic oscillator the period of oscillation does not depend on the amplitude of the motion.
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[42] The position of the block as a function of time can be written as x t = A cos
( )
, with A > 0.
Determine the phase angle ϕ in radians.
(A) −
π
2
(B)
π
2
(C) +
π
3
+ π (E)
Soluton: [A] since x(t=0)=0 cos( φ )=0 so φ = −π/2 or +π/2 modulo 2π . The velocity is v ( t ) = − A ω sin( ω t + ϕ ) so that v ( 0 ) = − A ω sin( ϕ ) . Since the initial velocity is positive, we must take φ=−π/2 .
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Page 9 of 18

Physics 221 2005S Exam 2
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[43] You are baby-sitting little Nicole who plays with a toy of mass m = 0.100 kg that hangs at the end of a vertical string of length 20cm.
L = 20cm
m = 0.100 kg
The toy is swinging freely 10º to the left and right of the equilibrium position in simple harmonic motion. What is the period of the swinging toy?
(A) 0.13s (B) 0.36s (C) 0.90s (D) 3.58s (E) 8.98s
Solution:
[C]Using the formalism for a simple pendulum,
T = 2 π L / g = 2 π 0 .
20 m / 9 .
8 m / s 2 = 0 .
90 s
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[44] The horizontal beam in the figure at right is massless and is hinged to a vertical wall as shown.
If the box has a weight of 200N, what is the tension in the 5m wire?
(A)100N
(B)167N
(C)200N
(D)333N
2.00m
2.00m
(E)400N 200N
Solution:
[B]First, let us define θ to be the angle between the wire and the beam. Looking at the figure, it is clear that sin( θ )=3m/5m=0.60. Now let use the torque condition where we take the net torque around the hinge.Thus:
0 = τ net
= TL sin θ − ( L / 2 ) w where w is the weight (200N), T=the tension in the wire and L is the length of the beam.
Solving for T we obtain:
T = w
θ
= 167 N
2 sin
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Page 10 of 18

Physics 221 2005S Exam 2
[45] A wooden rod of uniform density, length 1m and mass 3kg is suspended with a frictionless hinge from one end and hangs motionless. A bullet of mass 2g traveling at
200m/s horizontally hits the center of the rod and becomes embedded in the rod. What is the angular velocity of the rod about the hinge the instant after the bullet strikes?
(A) 0.033 rad/s (B) 0.10 rad/s (C) 0.20 rad/s (D) 0.30 rad/s (E) 0.40 rad/s
Hinge
50 cm
Rod
Mass=3kg
Length=1m
Bullet
Mass=2g
Velocity=200m/s
Solution:
[C]Use conservation of angular momentum about the hinge. The bullet initially has angular momentum L=m b v
0
R/2 where m b
is the mass of the bullet and v
0
is its initial velocity and R is the length of the rod. After the collision L=I ω where I is the moment of inertia of the rod with the bullet embeded in it. The rod is much more massive than the bullet so we can safely neglect the bullet when calculating the final moment of inertial which is given from the equation sheet by I=M r
R²/3 where M r
is the mass of the rod.
Setting up the conservation of angular momentum equation: m b v
0
R / 2 = ω I
∴ ω = m b v
0
R
2 I
=
3 m b v
0
2 M r
R
= 0 .
2 rad / s
Page 11 of 18

Physics 221 2005S Exam 2
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[46] A 3-m long flag pole of mass m = 1.2 kg is hinged to a vertical wall and attached at the other end to a 3.5m long massless string. The pole is perfectly horizontal.
3.5m
Length=3.0m; Mass=1.2kg
Hinge
What is the torque by the pole’s weight about the hinge?
(A) 17.7 Nm, out of the page (B) 17.7 Nm, into the page
(C) 35.3 Nm, out of the page (D) 35.3 Nm, into the page
(E) 0
Solution:
[B]The weight of the pole acts at the center of gravity, 1.5m from the hinge. Since gravity goes down, using the right hand rule, the torque is into the page. The magnitude of the torque is τ = mg ( L / 2 ) = ( 1 .
2 kg )( 9 .
8 m / s 2 )( 3 .
00 m / 2 ) = 17 .
6 Nm
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[47] A 1kg block moving at 5m/s along a frictionless track collides with a stationary 1kg block at the base of a frictionless ramp at 30º to the horizontal and sticks to it. How far up the ramp will the two blocks slide before momentarily coming to a stop?
(A)0.64m (B)1.28m (C)2.55m (D) 5.10m (E)6.38m
Solution:
30º
Distance=?
[A] If v is the initial velocity (5m/s), then after the totally inelastic collision with the second block the velocity v
1 of the two block system is v
1
=v cm
=v/2. The initial kinetic
Page 12 of 18

Physics 221 2005S Exam 2 energy of the two block system is thus K=(1/2)(2m)(v/2)²=mv²/4. If d is the distance it slides up the ramp the height above the bottom of the ramp is h=d sin θ.
At the highest point, the kinetic energy is converted into potential energy thus:
1
4 mv 2 = ( 2 m ) gh = ( 2 m ) gd sin θ
∴ d = = 0 .
64 m
8 g v 2 sin θ
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The following applies to questions [48] and [49]: A steel ball of mass m with initial speed v
0
hits head-on another steel ball of mass 3 m which is initially at rest. The collision can be considered elastic.
[48] What is the final speed of the ball that was at rest?
(A)2 v
0
(B) v
0
(C) v
0
/2 (D) v
0
/3 (E) v
0
/9
Solution:
[C]Using the formalism for elastic collisions in 1 dimension, v f
=2v cm
-v i
. In this case the velocity of the center of mass is vcm=(mv
0
)/(4m)=v
0
/4. The initial velocity of the ball at rest is v i
=0 so v f
=v
0
/2.
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[49] When does the center of mass of the two-ball system move faster, before or after the collision?
(A) Before, because some energy is lost in the collision.
(B) Before, because one of the balls is moving backwards after the collision.
(C) After, because then the two balls are moving.
(D)The velocity of the center of mass is the same before and after the collision, because the forces of a collision are internal forces.
(E)The velocity of the center of mass is the same before and after the collision, because the collision is elastic.
Solution:
[D]Since all the forces are internal, the momentum of the system is conserved and so the velocity of the center of mass is constant. This would also be true for an inelastic collision so (E) is wrong.
Page 13 of 18

Physics 221 2005S Exam 2
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[50] A flywheel has moment of inertia I=8kg m 2 . At t=0 it rotates with a period of
T=1.5s and due to friction, the period lengthens at a rate of dT/dt=1.0x10
−3 rate of kinetic energy loss of the flywheel at t=0?
. What is the
(A)47mW (B)70mW (C)94mW (D)188mW (E)94W
Solution: loss is
[C] The kinetic energy of rotation is
P = − dK dt
= − d dt


2 π 2 I
1
T 2


=
4 π
T 3
2 I
K = 1
2
I ω 2 = 2 π 2 I
1
T 2
. The rate of energy

 dT dt


= .0936W=93.6mW
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[51] A block of mass 1kg slides along a frictionless track at 10m/s. It collides and sticks to a motionless block of mass 3kg attached to the end of a spring with spring constant
100 N/m. What is the amplitude of the oscillations of the system after the collision?
(A)400cm (B)250cm (C)200cm (D)100cm (E)50cm
1kg
3kg k=100N/m
10m/s
________________________________________________________________
Solution: v cm
[E]The velocity of the two block system the instant after the collision is
= (10m/s)(1kg)/4kg=2.5m/s. The kinetic energy of the two block system is then
K=m tot v cm
2 /2=(4kg)(2.5m/s) 2 /2=12.5J. The instant after the collision the potential energy is 0 (taking the conventional U=0 when x=0) so this is also the total mechanical energy
E=12.5J. When the displacement is equal to the amplitude all the energy is potential so
E=kA 2 /2. Thus, A= 2 E / k = 0 .
5 m =50cm.
Page 14 of 18

Physics 221 2005S Exam 2
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The following applies to problems [52] and [53]: An ideal massless string unwinds from a drum of mass m and radius R which is free to rotate about a frictionless axel. The drum is a cylinder of uniform density. The string is connected to a block, also of mass m which is free to slide without friction across a horizontal table. The block is in turn connected with an ideal massless string over a massless frictionless pulley to another block, also of mass m which hangs freely. The system is released from rest at t=0s and the masses begin to accelerate. Let T
1
be the tension of the string connecting the drum to the block on the frictionless table. Let T
2
be the tension in the string connecting the two blocks.
Mass=m
Radius=R
Tension
=T
1
T
1
Mass=m
T
2
Tension
=T
2
T
2
Mass=m
[52] Which of the following is true? mg
(A) T
1
= T
2
< mg (B) T
1
= T
2
> mg (C) T
1
= T
2
= mg
(D) T
1
< T
2
= mg (E) T
1
< T
2
< mg
Solution:
[E] The hanging mass is accelerating downwards so mg>T
2
. Likewise the block on the table is acclerating to the right so T
2
>T
1
. Thus T
1
<T
2
<mg.
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[53] How does the tension T
1
depend on the radius of the drum, R ?
(A) proportional to 1/R (B) proportional to R
(C) proportional to R² (D) proportional to 1/R²
(E) none of the above
Solution:
[E] Using dimensional analysis, the only parameters which could enter into the expression for T
1
are g, m, R. The only combination of these parameters which comes to force units is mg. Thus T
1
must be proportional to mg and cannot, in particular, depend on R in any way.
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Page 15 of 18

Physics 221 2005S Exam 2
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[54] A massless, ideal rope is wrapped around a solid, uniform cylinder of mass M and radius R The rope does not slip on the surface of the cylinder. A block of mass m is attached to the free end of the rope. The block is released from rest. Find its speed after falling a distance h .
(A) v =
2
2 mgh
(B) v = mgh
2 +
(C) v =
2 mgh
(D) v =
4 mgh
(E) v =
4 mgh
R
2 + M
Solution: m
[E] The amount of gravitational energy liberated to kinetic energy when the mass falls through a height h is mgh.
The kinetic energy at this point is thus
K=mgh
In terms of the motion of the two parts, this kinetic energy has the form:
K = 1
2 mv 2 + 1
2
I ω 2
The moment of inertia of the drum is given by given by ω = v / mgh
=
=
I = 1
2
MR 2 while the angular velocity is
R . Substituting this into the above we obtain:
1
2
1
2
= K mv
[ m
2
+
=
1
2
+
1
2
1
2
M mv 2 +
1
2
] v 2
MR
1
2
I ω 2
[
2
] [ v / R
]
2
Solving this for v we obtain: v =
4 mgh
2 m + M
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Physics 221 2005S Exam 2
: Consider the contraption shown in the figure. It consists of a rod of uniform density and length L pivoted at point P in initially pulled 5º to the left. An ideal pendulum of the same length L also pivoted at P initially pulled 5º to the right and a brick which is initially at point P. At t=0, the rod, the pendulum and the brick are released simultaniously. The brick falls freely from rest and the rod and pendulum swing from rest without friction.
Eventually each of the elements passes point Q a distance L below point P. In which order do the three elements arrive at point Q?
L
P
5º 5º
(A) Rod; Brick; Pendulum
(B) Brick; Rod; Pendulum
(C) Brick; Pendulum; Rod
(D) Rod; Pendulum; Brick
(E) None of the above
Q
Solution:
[A]Using the formula for uniform acceleration, the amount of time it takes for the brick to fall is given by L = 1
2 gt 2 brick
so t brick
= 2
L g
= 1 .
41
L g
.
For the pendulum, the time it takes to get to Q is ¼ of the period so t pend
= 1
4
T pend
= 1
4
2 π
L g
=
π
2
L g
= 1 .
57
L g
.
Likewise for the rod the time is ¼ of the period but here we need to use the physical pendulum formalism: t rod
= 1
4
T rod thus t rod
=
< t brick
1
4
2 π
< t pend
I
=
π ( 1 / 3 ) mL 2 mg ( L / 2 )
=
π 2 L
3 g
=
π L
= 1 .
283
L mgd 2 2 3 2 g g
which is the order in which the three elements will hit Q.
L
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29 B
30 D
21
24
25
26
22
23
27
28 D
Physics 221 2005S Exam 2
Key
31 E
32 A
33 D
34 C
35 C
36 D
37 A
38 D
41 C
42 A
43 C
44 B
45 C
46 B
47 A
48 C
51 E
52 E
53 E
57
58
54 E
55 A
56
39 B
40 B
49 D
50 C
59
60
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