Mark answers in spaces 27-52 on the answer sheet
No class Friday March 12
No class Friday March 12
PHYSICS 221
Spring 2004
EXAM 2: March 11 2004 8:00pm—9:30pm
Name (printed): ____________________________________________
ID Number: ______________________________________________
Section Number: __________________________________________
INSTRUCTIONS:
Each question is of equal weight, answer all questions. All questions are multiple choice.
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. Note also formula sheets pages 11-13.
"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⇒31. 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 (page 14) 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. There are 26
questions on this exam labeled 27-52.
Mark answers in spaces 27-52 on the answer sheet.
Best of luck, David Atwood and Anatoli Frishman
Physics 221 2004 S Exam 1
Page 1 of 17
Mark answers in spaces 27-52 on the answer sheet
[27] A particle of mass 1kg travels in uniform circular motion at the speed of 4m/s.
How much work is done in joules on the particle by centripetal force acting on it during
one quarter of one revolution
(A) 16π J
(B) 8π J
(C) 4π J
(D) 0 J
(E) None of the above
No work is done by centripetal force since it is perpendicular to motion
[28] A bullet with a mass of 3.00g and a speed of 400m/s penetrates a wooden block
horizontally to a depth of 2.00cm. Assume that a constant resistance force slows the
bullet. What is the magnitude of this resistance force?
(A) 6k N
(B) 12 kN
(C) 24 kN
(D) 48 kN
(E) 96 kN
Initial KE =(1/2)(.003kg)(400m/s)²=240J
Using Work Energy Thm F=240J/.02=12kN
[29] A 100g ball initially at rest is dropped from a height of h=50.0cm above a spring
of negligible mass. The ball compresses the spring to a maximum displacement of 5.0cm.
What is the spring force constant k?
(A) 19.6 N/m
There seems to be two ways to interpret this question. In either case (D) is the
(B) 196 N/m
closest answer so it seems the question is not quite defective. The issue is whether
(C) 294 N/m
you take 50cm to be the distance of the release above the top of the spring or
(D) 392 N/m
above the final position of the ball. In either case if you take y to be the distance
(E) 490 N/m
between the release point and the final point of the ball (either y=h or y=h+5cm)
then the gravitational potential energy liberated is mgy. This will be equal to the
final elastic PE so equating the two mgy=kx²/2 where x=5cm, the compression of
the spring. Solving for k, k=2mgy/x². For future questions of this type I guess I
need a diagram--sorry
[30] A 18g bullet is shot vertically upwards into a 10kg block. The block lifts upwards
9.0mm. If the bullet penetrates the block in a time interval of 1ms the initial kinetic
energy of the bullet is closest to:
Final PE of block =mgh=(10kg)(9.8m/s²)(.009m)=.882J
Momentum of block=sqrt(2mK)=sqrt(2(10kg)(.882J))
(A) 0.0016 J
=4.2Ns
(B) 0.88 J
This should be the momentum of the bullet before
collision.
Physics 221 2004 S Exam 1
Page 2 of 17
The KE of the bullet is thus
K=p²/(2m)=(4.2Ns)²/(2x.018kg)=490J
(C) 250 J
(D) 330 J
(E) 490 J
[31] Particle #1 of mass 1kg is initially located on the –y axis and is moving with a
velocity of + 2 ˆj m / s while particle #2 of mass 2kg is initially located on the –x axis and
is moving with a velocity of + 1iˆ m / s . They meet near the origin and collide, after the
collision, the speed of the particle #1 is 1m/s in a direction 45º counterclockwise from the
x-axis. Which of the following statements is true concerning this collision. The system of
particle #1 and particle #2 is not subject to any external forces:
(A) The collision is elastic
(B) The collision is inelastic, but not totally inelastic
(C) The collision is totally inelastic
(D) The kinetic energy of the two objects increases after the collision
(E) This collision is not physically possible
Before the collision the KE of particle 1 is 2J and the KE of particle 2 is 1J giving a total initial KE of 3J.
Using the conservation of Momentum, the final momentum of particle 1 is 0.707(i+j) and for particle 2 the
momentum is 1.292(i+j). The final KE of particle 1 is 0.5J while the final KE of particle 2 is .835J giving a total
final KE of 1.34J. KE was clearly lost in the collision so it is inelastic but the two particles do not have the same
vinal velocity so it is not totally inelastic
[32] A cannon ball is fired at a speed of 80m/s from a cannon located 50m from the
base of a cliff which is 100m high. It follows a trajectory and strikes a spot on the level
ground above the cliff. What is the speed of the cannon ball on impact. Neglect air
resistance.
Use conservation of energy. Initial KE is (1/2)mv²=(1/2)m(80m/s)²
=(3200m²/s²)m
By conservation of energy, at the top of the cliff the KE is reduced
by the PE due to the height difference so the final KE is
(3200m²/s²)m-ghm=(2220 m²/s²)m. The velocity is therefore
sqrt(2K/m)=66.6m/s
(A) 44.4 m/s
Cannon
(B) 50.0 m/s
(C) 66.6 m/s
(D) 73.6 m/s
(E) Cannot be determined with the given information.
Physics 221 2004 S Exam 1
Impact
point
100m
50m
Page 3 of 17
[33] A force of 4.0N acts on a 12kg body initially at rest. Compute the instantaneous
power due to the force at t=3.0s
(A) 3W
(B) 4W
(C) 6W
(D) 8W
(E) 12W
The momentum of the object is p=Ft. Power is given by
P=Fv=Fp/m=F²t/m=(4N)²(3)/(12kg)=4W.
[34] Four blocks of mass m1=1kg, m2=2kg, m3=3kg, m4=4kg are on a frictionless
horizontal surface as shown on the figure below. The blocks are pressing against each
other. A force FL=30N is applied to the left block and is directed to the right. Another
force FR=50N is applied to the right block, and is directed to the left. Initially, the blocks
are moving left at a rate of 20m/s. When the 2kg block has slid 2m to the left, how much
work has the 3kg block done on it?
v0=20m/s
FL=30N
1kg
Displacement = 2m
Work done on 2kg block
by 3kg block =?
(A) –72J
(B) +72J
(C) –40J
(D) +40J
(E) –100J
2kg
3kg
4kg
FR=50N
Frictionless surface
Let us choose an axis where positive is to the right. The total force on the system is
–20N. The total mass is 10kg so the acceleration is –2m/s m²/s². The two blocks on
the left have a mass of 3kg and so the net force acting on them is –6N. The applied
force is +30N so the force of the 3kg block on the 2kg block is –36N. This force acts
through a displacement of –2m so the work done by this force on the 2kg block is (36N)(-2m)=+72J
Physics 221 2004 S Exam 1
Page 4 of 17
(F)
[35] The figure below shows potential energy versus position for a particle that is
confined to move in 1 dimension. Which of the labeled points on the graph correspond to
points of unstable equilibrium?
(A) P and T
(B) R only
(C) R and P
(D) Q and S
(E) T only
Just point
R because
it is a the
only local
maximum
Physics 221 2004 S Exam 1
Page 5 of 17
[36] The graph below shows the potential energy of a 1kg particle confined to move
along the x-axis as a function of position. The particle is released from point O with
initial kinetic energy K=20J and it is moving to the right. Which of the labeled points is a
turning point for the motion of the particle:
The total mechanical energy at
(A) Point A
release is U+K=15J+20J=35J.
(B) Point B
The turning point is where U=E
(C) Point C
which is only true at point E. The
(D) Point D
particle makes it over hump B
(E) Point E
since U(B)<E
[37] A 20g stone has kinetic energy of 100J and is not rotating in any way. What is the
magnitude of its momentum?
(A) 1.0 Ns
p=sqrt(2mK)=sqrt(2(.02g)(100J))=2Ns
(B) 2.0 Ns
(C) 3.0 Ns
(D) 4.0 Ns
(E) 5.0 Ns
[38] If a collision is inelastic then which of the following must be true:
(A) The total kinetic energy is conserved
KE is lost during an inelastic
(B) The total linear momentum is not conserved
collision so total KE is not
(C) The total kinetic energy is not conserved
conserved. The other statements
(D) The velocity of the center of mass of the system changes.
are false or not always true in an
(E) The two particles must stick together after the collision.
inelastic collision.
[39] On a 1 dimensional air track, a particle of mass 100g is moving with a velocity of
3m/s. The particle collides with another particle of mass 200g. After the collision, both
particles have the same velocity of 1m/s. What was the initial velocity of the 200g
particle?
(A) 0 m/s
The final momentum is (0.3kg)(1m/s)=0.3Ns. The initial momentum of 100g
(B) +1 m/s
paraticle is (0.1kg)(3m/s)=0.3Ns so no momentum can be contributed by the
(C) -1m/s
200g block so it must have been stationary. Incidentally, this collision is totally
(D) +2m/s
inelastic
(E) -2m/s
Physics 221 2004 S Exam 1
Page 6 of 17
[40] What is the y coordinate of the center of mass of the 3mX3m metal slab with a
1m × 1m square cut out of it depicted below. The slab is of uniform thickness and density.
There are several ways to do this. I will decompose the system
into two pieces as shown. Let U be the mass of a 1x1 square.
Piece A has ycm=2m, M=6U. Piece B (which is disconnected)
has ycm=0.5m, M=2U. The total
ycm=((6U)(2m)+(2U)(0.5m))/(8U)=1.625m
(A) 1.375m
(B) 1.500m
(C) 1.625m
(D) 1.750m
(E) 2.000m
[41] The angular position of a turntable in radians is given as a function of time by
θ = Ae − Bt where A = 5 and B=2 s-1. What is the angular acceleration, α, of the turntable
at t=1s?
(A) +10.0 s−2
Angular acceleration is the second derivative of angular position
(B) +2.71 s−2
so α = d 2θ / dt 2 = AB 2e − Bt Plugging in the given numbers the
−2
(C) –2.71 s
answer is 2.71s-2.
−2
(D) +1.35 s
(E) –1.35 s−2
[42] Two blocks are free to slide along a frictionless track along the x-axis. At t=0s
block A is located at x=0m and block B is located at x=10m. Block A has a mass of 1kg
and an initial velocity of +5m/s. Block B has a mass of 4kg and an initial velocity of
+1m/s. The two blocks collide with a perfect elastic collision. What is the final velocity
of block A?
(A) +1.8 m/s
Vcm=((1kg)(+5m/s)+(4kg)(+1m/s))/(5kg)=1.8m/s. The final velocity of
(B) +2.6 m/s
block 1 is 2Vcm-Vi=2(1.8m/s)-5m/s=-1.4m/s
(C) +1.4 m/s
(D) –2.6 m/s
(E) –1.4 m/s
Physics 221 2004 S Exam 1
Page 7 of 17
[43] A wheel rotating with a constant angular acceleration turns through 12 revolutions
during a 3 second time interval. Its angular velocity at the end of the time interval is
13rad/s. What is the angular acceleration?
(A) -4.1 rad/s²
Average angular velocity is (24π rad)/(3s)=25.13rad/s. The
(B) -5.1 rad/s²
initial angular velocity is thus 2(25.13 rad/s)(C) -6.1 rad/s²
13rad/s=37.26rad/s. The angular acceleration is thus
(D) -7.1 rad/s²
[13rad/s-37.26rad/s]/(3s)=-8.2rad/s2.
(E) -8.1 rad/s²
[44] A particle moves on a circle of radius 2m. The angular position of the particle is
given by θ = (3 s −4 )t 4 − (1 s −1 )t . What is the magnitude of the radial component of the
acceleration of the particle (towards the center of the circle) at t=1s?
(A) 3.0 m/s²
(B) 4.5 m/s²
(C) 22 m/s²
(D) 121 m/s²
(E) 242 m/s²
The angular velocity is the derivative of angular position ω=(12s-4)t3-1. The
centripetal acceleration is a=rω2=r[(12s-4)t3-1]2 . Plugging in the numbers
a=242 m/s²
[45]
What is the cross product
(A) + 15iˆ − 10 ˆj − 12kˆ
(B) + 15iˆ + 10 ˆj − 12kˆ
(C) − 15iˆ + 10 ˆj + 12kˆ
(D) − 15iˆ − 10 ˆj + 12kˆ
(E) + 15iˆ − 10 ˆj + 12kˆ
(2iˆ + 3 ˆj ) × (4iˆ + 5kˆ) ?
Denote this vector C. Using the cross product formula:
Cx=3 x 5=15
Cy=-2 x 5=-10
Cz=-3 x 4 =-12
Answer is therefore (A)
[46] Consider the configuration below. The circles represent particles of mass 1kg.
The empty squares do not represent any mass but are proposed centers of mass of the
configuration. Which of the squares is the correct center of mass?
(A) Square A
(B) Square B
(C) Square C
(D) Square D
(E) Square E
Physics 221 2004 S Exam 1
The C of G is the average of the radius vectors
for the seven circles since they are of the same
mass. The answer is thus:
(1/7)[ (i+4j)+ (2i+5j)+ (5i+4j)+ (5i)+ (4i+j)+
(3i)+ (i)]=3i+2j i.e. point C (all in meters)
Page 8 of 17
[47] A 2kg projectile is fired from a cannon resting on level ground as shown in the
diagram below. When the projectile reaches the highest point of its trajectory where the
vertical component of its velocity is 0, it
has move a distance of 100m
Impact of second piece
horizontally. At this point an explosive
charge breaks it into two 1kg pieces.
One of the pieces lands at the same
location as the cannon. At the same
moment the other piece of the projectile
hits the ground. How far from the
cannon does the other piece land?
Neglect air resistance.
Impact of
(A) 100m
First piece
(B) 200m
(C) 300m
(D) 400m
(E) Cannot be determined without
more information.
Both pieces strike the ground at the same time
The center of mass follows the parabolic trajectory while all parts of the projectile are in
free fall. The CofM lands 200m downrange at the same time as the pieces hit the ground. Its
position must be the average of the position of the two equal pieces so since one piece hits
at the origin, the other piece must impact 400m downrange.
[48] A uniform square slab is 4.00m on a side with negligible thickness. It has a mass
of 200kg. If it is originally lying on the flat ground, how much work is needed to stand it
on end?
(A) 7.84 kJ
(B) 5.88 kJ
(C) 3.92 kJ
4m
Slab
(D) 2.94 kJ
Mass=200kg
(E) 1.96 kJ
Work=?
Before
After
The CofM is raised 2m and the mass is 200kg so the work required is
mgh=(200m)(9.8m/s²)(2m)=3.92kJ
Physics 221 2004 S Exam 1
Page 9 of 17
4m
[49] Consider a hollow sphere where the mass M which is uniformly concentrated on
the surface. If the sphere is of radius R, what is the moment of inertia around an axis that
is tangent to the surface of the sphere.
(A) (2/3)MR²
(B) MR²
(C) (4/3)MR²
(D) (5/3)MR²
(E) 2MR²
Use parallel axis theorem and data from
table 9.2. Thus
I = I cm + MR 2
= 23 MR 2 + MR 2
= 53 MR 2
[50] As shown in the picture below, four different 1kg objects with radius 5cm are
mounted on horizontal axels have identical cords of 2m attached to identical 1kg weights
wrapped around them. In all cases the cords unroll without slipping and the axels are
frictionless.
In case P, the object is a solid cylinder with uniform density. In case Q the object is a
hoop with the mass concentrated on the rim. In case R the object is a solid sphere of
uniform density. In case S the object is a hollow spherical shell with the mass
concentrated on the
surface. If the four
weights are released from
rest as shown from a
point 2m above the
ground, in which order do
they strike the ground?
(A) P, Q, R, S
(B) R, P, S, Q
(C) R, S, P, Q
(D) Q, S, P, R
(E) S, P, Q, R
Physics 221 2004 S Exam 1
Page 10 of 17
The lower the moment of inertia of the object the more quickly it will spin up and thus the more
quickly will the block strike the ground. Since all the round things are the same mass and radius we
simply refer to table 9.2 and order them from smallest to largest moment of inertial. The order is
thus RPSQ
[51] Consider the arrangement depicted below. A block of mass 1kg is at rest at a point
4m above the ground on a frictionless ramp at angle 30° from the horizontal. The block is
released and slides down the ramp. It then slides along a rough surface where the
coefficient of kinetic friction µk=0.5. How far along the rough surface does the block
slide from the bottom of the ramp before coming to rest?
In this case it happens that sin(30°)=1/2=µ so that the
force accelerating the block along the ramp is equal in
magnitude to the kinetic friction slowing it down in
the straightaway (but in the opposite direction). By the
work energy theorem, the distance will be the same as
the length of the ramp=8m.
(A) 1.0 m
(B) 2.0 m
(C) 4.0 m
(D) 7.0 m
(E) 8.0 m
[52] The figure below shows three forces in the xy plane acting on a blob. Determine
the total torque on the blob due to these forces about the indicated center of torque. Take
the z-axis up, out of the page.
(A) +2.00 k̂
(B) –2.00 k̂
(C) +2.50 k̂
(D) –2.50 k̂
(E) +3.25 k̂
0.5Nm
Nm
Nm
Nm
Nm
Nm
2Nm
0.75Nm
Physics 221 2004 S Exam 1
The total torque is the some of the
indicated torques shown
=(2+0.5+0.75)Nm=3,75Nm
Page 11 of 17
Formula Sheet for Exam 1
1. Physical Constants
(numerical value used to derive answers in exam):
1.1) Acceleration of gravity on Earth’s Surface: g=9.8m/s²
1.2) Radius of Earth: Rearth=6.38×106m
1.3) Mass of Proton: mp=1.67×10-27kg
3. Vectors
G G
G G
3.1) Dot Product: A ⋅ B = Ax B x + Ay B y + Az B z =| A || B | cosθ
G
G
where θ is the angle between A and B .
G
3.2) Components: A = Ax iˆ + Ay ˆj + Az kˆ
G
G G
3.3) Magnitude: | V |= V = V x2 + V y2 + V z2 = V ⋅ V
5. One Dimensional Motion
5.1) Average Velocity: v = ∆x / ∆t
5.2) Instantaneous Velocity: v = dx / dt
2. Calculus
2.1)
d
dx
x n = nx n −1
d
dx
sin x = cos x
x n +1
n +1
d
dx cos x = − sin x
n
∫ x dx =
4. Algebra
4.1) The solutions to ax 2 + bx + c = 0
are x =
1
2a
(− b ±
b 2 − 4ac
)
6. Forces
G
G
6.1) Newton’s Second: F = ma
G
G
6.2) Newton’s Third: FAB = − FBA
6.2) Kinetic Friction: f k = µ k N
6.4) Static Friction: f s ≤ µ s N
6.5) Centripetal Force: F =
v x = v0 x + a x t
mv 2
R
x = x0 + v0 x t + 12 a x t 2
5.3) For Constant Acceleration only: v 2 − v 2 = 2a ( x − x )
0x
0
x
x
x − x0 1
= 2 (v x + v 0 x )
t
7. Three Dimensional Motion
G
7.1) Position Vector: r = xiˆ + yˆj + zkˆ
G
G
G
G
2 G
7.2) Velocity and Acceleration: v = dtd r
a = dtd v = dtd 2 r
G G G
v = v0 + at
G G G
G
r = r0 + v 0 t + 12 at 2
7.3) Constant Acceleration only: v 2 − v 2 = 2aG ⋅ (rG − rG )
0
G G0
r − r0 1 G G
= 2 (v + v 0 )
t
ω = 2πf
v = Rω
7.4) Circular Motion: f = 1 / T
7.4a) Angular Velocity: ω = dθ / dt
7.5) Centripetal Acceleration: a rad = Rω 2 = v 2 / R = ( 4π 2 R ) / T 2
G
G
G
v PA = v PB
7.6) Changing
Reference
Frames:
Physics
221 2004
S Exam
1 + v BA
Page 12 of 17
Formula Sheet for Exam 2
8. Kinetic Energy and Work
8.1) Linear Motion: K = 12 mv 2
8.2) Rotational Motion: K rot = 12 Iω 2
G G
8.3) Work by a constant force W = F ⋅ s = Fs cosθ
8.4) Work done by a variable force in 1 dim:
x2
9. Potential Energy
9.1) Gravitational: Ugrav=mgy
9.2) Spring: Uspring=kx²/2
dU
dx
9.4) Conservative force from potential in 3d: a
9.3) Force from potential in 1D: Fx ( x) = −
W = ∫ Fx dx
x1
G G P2
8.5) Work in 3D: W = ∫ F ⋅ dl = ∫ F cos φ dl
P2
P1
8.6) Power: P=dW/dt
P1
G G
P = F ⋅v
10. Momentum and Impulse
G
G
G G
10.1) Momentum: p = mv F = ddtp
G t2 G
G
G
10.2) Impulse: J = ∫ Fnet dt = p 2 − p1
t1
G
G
10.3) Center of mass position: M tot rcm = ∑ mi ri
G
G
10.4) Center of mass velocity: M tot vcm = ∑ pi
G
G
G
10.5) Center of mass acceleration: M tot a cm = ∑ Fi = Fexternal
11. Collisions
11.1) 1-dimensional totally inelastic collision: v1 f = v 2 f = v cm
11.2) 1-dimensional elastic collision:
v1 f = 2vcm − v1i
v 2 f = 2vcm − v 2i
G
G
G
11.3) 3-dimensional totally inelastic collision: v1 f = v 2 f = v cm
Physics 221 2004 S Exam 1
12. Rotation
12.1) Angular velocity ω = ddtθ
12.2) Angular Acceleration α =
dω
dt
2
12.3) Circular motion: a rad = Rω ; a tan = rα .
12.4) Moment of Inertia: I = ∑ mi Ri2
12.5) Parallel Axis Thm.: I P = I cm + Md 2
G G G
12.6) Torque: τ = r × F
Page 13 of 17
Physics 221 2004 S Exam 1
Page 14 of 17
Record Sheet
You may fill in this sheet with your choices, detach it and take it with you after the exam
for comparison with the posted answers
21
31
41
51
22
32
42
52
23
33
43
53
24
34
44
54
25
35
45
55
26
36
46
56
27
37
47
57
28
38
48
58
29
39
49
59
30
40
50
60
Physics 221 2004 S Exam 1
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Scratch Paper (intentionally left blank)
Physics 221 2004 S Exam 1
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Scratch Paper (intentionally left blank)
Physics 221 2004 S Exam 1
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