Physics 221 2006 S Exam 1 Solutions
Answer Key
Solutions
Questions
1
2
11
Answer Key
1
2
3
4
5
6
7
8
9
10
B
E
E
C
D
D
B
A
C
D
11
12
13
14
15
16
17
18
19
20
E
E
A
D
A
C
A
E
C
B
Page 1 of 27
21
22
23
24
25
26
27
28
29
30
C
A
C
B
B
A
C
C
D
B
Physics 221 2006 S Exam 1 Solutions
[1](B)
The weight is W=mg. First calculate the mass from the data given:
2
2
g   1kg 
 8.5 in ×11 in   2.54 cm   1 m  
m = (500 sheets) 


 
  75 2  
sheet
m   1000 g 

  1 in   100 cm  
=2.26kg
The weight is therefore
W = mg = (2.26 kg)(9.81 sm2 ) = 22.2 N
[2](E)
The total mass of the system is 6kg. The acceleration of all the blocks is thus
equal to (6N)/(6kg)=1m/s². This is therefore the acceleration of block P.
[3](E)
The net force accelerating P at the above acceleration is T1. The net force
accelerating the PQ system is T2. All blocks are undergoing the same acceleration so
T1:T2=mP a:(mP+mQ)a=mP:(mP+mQ)=1:3.
[4](C)
By the work energy theorem, the net work done on the block is the change in
kinetic energy. Since the speed does not change, the kinetic energy does not change and
so the net work is 0.
[5](D)
Both the circle and the rectangle are moving to the right relative to the
background. Around t = 2s, the spacing between the circles is smaller than the spacing
between the rectangles, which means that the velocities of the circle and the rectangle
relative to the background are something like this:
vC,background
vR,background
Therefore, the velocity of the circle relative to the rectangle, vC,R = vC,bg − vR,bg , points to
the left. The same is true at three seconds (even more so). Thus at both times the relative
velocity is to the left.
Also, we can look at the position of the circle relative to the rectangle at different times.
At t = 1 s, the circle is quite ahead of the rectangle. At t = 2 s, still ahead but not so much.
At t = 3, they are leveled. At t = 4 s, the circle is behind the rectangle. So from the point
of view of the rectangle, the circle is moving left.
Page 2 of 27
Physics 221 2006 S Exam 1 Solutions
[6](D)
The angle between the two vectors is (15º+90º+15º)=120º. The dot product is
Ai B = AB cos θ = 4 × 3 × (− 12 ) = −6
[7](B)
The magnitude of D is:
D = 152 + 162 + 122 = 25
Thus
D
Dˆ = =
D
1
25
( −15iˆ + 16 ˆj − 12kˆ ) = −0.60iˆ + 0.64 ˆj − 0.48kˆ
[8](A)
Since Sally is moving at a constant speed and not accelerating, the net force on
her is 0.
[9](C)
We will apply the work-kinetic energy theorem twice, in the two situations where
a force is doing work.
The kinetic energy of the block after it leaves the spring is just the work that the
spring does on the block:
K = Wspring = 12 k ∆x 2
The (negative) work done on the block by gravity must be equal to that kinetic energy if
the block is brought momentarily to rest. If h is the height of the ramp then, on the one
hand:
Wgravity = W i∆r = (− mgjˆ)i(∆xramp iˆ + hjˆ) = − mgh
while on the other hand
Wgravity = − K
∴ mgh = 12 k ∆x 2
k=
2mgh
∆x 2
Plugging in the numbers:
k=
2mgh 2(0.1kg )(9.81 sm2 )(1m)
N
=
= 196
2
2
∆x
m
(0.1m)
Page 3 of 27
Physics 221 2006 S Exam 1 Solutions
[10](D)
The change in velocity is the integral of acceleration as a function of time.
Graphically, this means the area under the a(t) curve. From the graph we can integrate it
by counting the squares from which we find ∆v=17.5m/s. Since the initial velocity is
10m/s, the final velocity is therefore 27.5m/s.
Acceleration
(m/s²)
6
4
2
10.5m/s
4m/s
1
2
3
4
3m/s
5
Time (s)
6
7
[11](E)
Since the particle is moving at a constant speed, the acceleration is inversely
proportional to the radius of curvature. From looking at the figure we can see that this
will order them aR < aS < aQ < aP (i.e., the tighter the curve, the larger the radial
acceleration)
[12](E)
(A) is false because the weight (=mg, attraction from the Earth) does not change
even if the motion is different. (B) is true because the rocket increases the force into the
ramp but the block does not actually move into the ramp so the normal force must
increase in response to the additional load. From (B) it follows that (C) must be true since
the kinetic friction will increase. Thus that answer is (E).
[13](A)
If the acceleration is positive the curve must be a “smile” shape ☺. This is only
true at point R.
[14](D)
The net force in these cases must be the centripetal force = mRω² so that
FP:FQ= mRω²: (2m)(2R)(2ω)²=1:16
Page 4 of 27
Physics 221 2006 S Exam 1 Solutions
[15](A)
The (negative) work done by the breaking force on the car must cancel the initial
kinetic energy to stop the car. The initial kinetic energy is:
K = 12 mv 2
so
− K = W = − F ∆x
∴F =
mv 2
= 1250 N
2 ∆x
(You can also use the change in speed to figure out the constant acceleration,
v 2 − v02
a=
= −12.5 m/s 2 and then compute the force with Newton’s second law,
2∆x
F = ma = 1250 N )
[16](C)
Using the v-squared equation for the y component of velocity we find that
v y = 2 gh
To get the total speed we take:
2(9.81 sm2 )(5m)
2 gh
v=
=
=
= 11.4 ms
sin θ
sin θ
sin 60o
vy
[17](A)
As in the example in Lecture 14, the work done by gravity depends only on the
height difference. In this case that work will be mgh=(0.1kg)(9.81m/s²)(80m)=78.5J. Air
resistance does not alter the work done by gravity.
[18](E)
Take the x-axis along the slope. Consider the instant when the block comes to a
stop. First we need to know if the block will then start sliding back down the ramp.
Suppose it does not, then the force of static friction cancels out the x component of the
weight so
f s = mg sin θ
This picture is consistent if this is less than the limit of static friction so we need to check
if
mg sin θ = f s < µ s N = µ s mg cos θ
⇔ tan θ < µ s
In this case tan θ = 0.17 so the above inequality is true so the block sticks to the ramp
and never returns to point P.
Page 5 of 27
Physics 221 2006 S Exam 1 Solutions
[19](C)
The work of the tension is W = F i∆r = (300 N )(2m) cos 30o = 520 J
[20](B)
Let R=5m be the radius of the turntable. Since initially the turntable is at rest there
is initially no centripetal acceleration so the only acceleration is tangential. The
acceleration of the block is thus
a=αR
By Newton’s second law the net force must be
F = f s = ma = mα R
This net force is the force of static friction which is the only force on the block in the
horizontal plane.
The block does not slide as long as this is less than the static friction limit. The critical
case, where this is equal to the static friction limit is thus
f s = f s max = µ s mg
∴ mα R = µ s mg
∴α =
µs g
R
=
(0.2)(9.8 sm2 )
5m
= 0.39
Page 6 of 27
Physics 221 2006 S Exam 1 Solutions
[21](C)
Let m=1kg and M=3kg. First check if the system is static. If this is the case then
the force of static friction must balance the tension in the string which, in this case, is the
weight of the 3kg block. Thus
1kg
f s = gM = (9.81 sm2 )(3kg ) = 29.4 N .
The static friction limit, however,
is f s max = µ s mg = 5.9 N . The
fk
T
µs=0.6
µk=0.3
required static friction exceeds the
limit so indeed the block slides.
T
Now that we know it is sliding we
need to draw the free body
diagrams of each of the blocks. If a
is the acceleration of the blocks,
Newton’s second law gives:
3kg
1kg block T − f k = ma
3kg block Mg − T = Ma
Mg
Adding these two equations
together we get
Mg − f k = (m + M )a
⇒ Mg − µ k mg = (m + M )a
∴a = g
M − µk m
= 6.6 sm2
M +m
[22](A)
The net effect of John’s trip is that he is back where he started so his net
displacement is 0. Since average velocity is net displacement over time, his average
velocity is 0.
[23](C)
Let us denote a=4s-4 and b=2s-2 so θ=at4-bt2. The tangential component of
acceleration is given by:
atan = Rα = R
d 2θ
= R 12at 2 − 2b
2
dt
(
)
at t=0.5s then
atan = (2m)(12(4 s −4 )(0.5) 2 − 2(2 s −2 ))
= 16 sm2
Page 7 of 27
Physics 221 2006 S Exam 1 Solutions
[24](B)
The final kinetic energy is mv²/2 =36J. The work done on the object as it moves
from x=0m to x=10m is the area under the curve. By adding up the area of the various
triangular parts of the graph I get W=12J. By the work-energy theorem this must be the
increase in kinetic energy so the initial kinetic energy is 36J-12J=24J.
Force (N)
4
2
10J
8
0
2
4
6
−2
Page 8 of 27
-2J
Position (m)
4J
10
Physics 221 2006 S Exam 1 Solutions
[25](B)
Take the origin at the cannon with the y-axis vertical and x axis horizontal.
Following the logic in the solution of written assignment 2, if there was no gravity the
cannon ball would pass at y=300m when it is over the castle, i.e. x=300m. This is 300m124m=176m above the impact point. Thus, the time the cannon ball is in the air is the
2(176m)
same as the time it takes an object to fall 176m. This time is t =
= 6.0 s
9.8m / s 2
[26](A)
The acceleration points to the inside of the curve (which is not the same as the
inside of the shape). Therefore only PR and T are correct.
[27](C)
The angle between the two ropes is
90°. Let us take a coordinate system where
the x axis is parallel to the right rope and
the y axis is parallel to the left rope. In this
coordinate system the tension in the left
rope must cancel the y component of the
weight. This tension is therefore
(10kg )(9.8 sm2 ) cos 30o =84.9N
x axis
y axis
60°
TL?
Wx
30°
60°
10kg
Wy
W
[28](C)
In both cases the right block is in equilibrium so the tension in the right hand rope
is 40N. Since the scales are in equilibrium it must also be the case that the left hand ropes
have a tension of 40N. Since both scales have a rope with 40N tension on each end, they
must read the same, i.e. 40N.
Page 9 of 27
Physics 221 2006 S Exam 1 Solutions
[29](D)
Following the logic of written assignment 3, let us consider a system of axes
where the x-axis is parallel to the slope and the y-axis is perpendicular to it. Since we are
interested in how long it takes for the projectile to hit the slope, that is return to y=0, we
need only consider the equation for the time evolution of y.
The acceleration vector is
ˆ sin θ − ˆjg cos θ
a = ig
where θ = 30° is the incline of the slope. The time evolution y is thus:
y = v y t − 12 ( g cos θ )t 2
so y=0 at t=0 and t =
2v y
. Clearly it is the second solution that we want to maximize.
g cos θ
Let φ be the angle between the velocity vector and the slope (x-axis). Then,
vy = v sinφ. In order to maximize the time, we need to maximize vy. At a fixed speed this
will happen when the velocity is parallel to the y-axis or φ=90°.
[30](B)
Let us denote the final angle with respect to the horizontal as φ=−60º.
Note that
v0 x
v0
As the projectile moves, the x-component of the velocity does not change, so
v0 x = v fx = v f cos φ . Thus
cos θ =
cos θ =
v f cos φ
v0
 v f cos φ 
∴θ = arccos 

 v0 
 49.0 ms cos(−60o ) 
= arccos 

29.4 ms


o
= 33.6
Note: I take the positive angle since that was required by the wording of the question;
otherwise –33.6° is also a valid solution.
Page 10 of 27
Physics 221 2006 S Exam 1 Solutions
PHYSICS 221
Spring 2006
EXAM 1: Feb 16 2006 8:00pm—10:00pm
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 a piece of paper to take with you and compare
with the posted answers. You may use the table at the end of the exam for this.
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.
Total number of questions is 30. Question 30 is “extra credit”
Best of luck, David Atwood and Paula Herrera-Siklody
Page 11 of 27
Physics 221 2006 S Exam 1 Solutions
Formula sheet for Exam 1 – Phys 221 – Spring 2006
z
Vectors and math
A = A +A +A
2
x
2
y
2
z
k̂
A ⋅ B = AB cos θ = Ax Bx + Ay By + Az Bz
iˆ
A × B = AB sin θ
A × B = ( Ay Bz − Az By ) iˆ + ( Az Bx − Ax Bz ) ˆj + ( Ax By − Ay Bx ) kˆ
x
−b ± b 2 − 4ac
2a
d
d
sin x = cos x
cos x = − sin x
dx
dx
ax 2 + bx + c = 0
⇒
d n
x = nx n −1
dx
Geometry
x=
10−15
10−12
10−9
10−6
10−3
103
106
109
1012
perimeter circle: 2π R
area circle: π R 2
area sphere: 4π R 2
4
volume sphere: π R 3
3
1 revolution = 2π radians = 360
Conversion factors (for barbaric units)
1 yard = 3 foot = 36 inches
1 inch = 2.54 cm
1 mile = 1.609 km
1 gallon = 3.788 liters
1 lb = 4.448 N
y
ĵ
Physical constants
g = 9.81 m/s 2
G = 6.67 ×10−11 Nm 2 /kg 2
General kinematics
vaverage =
∆r
∆t
v=
dr
dt
Constant acceleration
1
r = r0 + v0t + at 2
2
1
x = x0 + v0 x t + ax t 2
2
Circular motion
ω=
dθ
dt
α=
aaverage =
a=
dv
dt
R=
v = v0 + at
v 2 − v02 = 2a ⋅ ∆r
vx = v0 x + axt
In 1D: vx 2 − v02x = 2ax ∆x
dω
dt
s = Rθ
1 2π 2π R
T= =
=
ω
f
v
Constant α:
∆v
∆t
arad =
1
2
θ = θ 0 + ω 0t + α t 2
v = Rω
v2
= Rω 2
R
atan =
ω = ω0 + α t
Page 12 of 27
v02
sin 2θ
g
atan = Rα
dv
dt
a = arad + atan
ω 2 − ω 02 = 2α∆θ
femto- (f)
pico- (p)
nano- (n)
micro- (ì)
milli- (m)
kilo- (k)
mega- (M)
giga- (G)
tera- (T)
Physics 221 2006 S Exam 1 Solutions
Relative motion
rA relative to C = rA relative to B + rB relative to C
vA relative to C = vA relative to B + vB relative to C
aA relative to C = aA relative to B + aB relative to C
Forces
∑ F = ma
Fg (≡ W ) = mg
FHooke = − k ∆x
FNewton = −G
fs ≤ µs N
Mm
rˆ
r2
g =G
fk = µk N
ME
RE2
Work and energy
W = ∫ F ⋅ dl
Wconservative = −∆U
U=
1 2
kx + C
2
E = KE + U
KE =
1 2 p2
mv =
2
2m
Wnet = ∆KE
r
U (r ) − U (r0 ) = − ∫ F ⋅ dl
r0
U = mgy + C
∆E = Wnon-conservative
U = −G
Pave =
W
∆t
F = −∇U
Pinst =
dW
= F ⋅v
dt
( Fx = −
∂U
, etc)
∂x
Mm
+C
r
( When only conservative forces do work: ∆E = 0 )
Page 13 of 27
Physics 221 2006 S Exam 1 Solutions
[1] On the side of a pack of paper you can read: 500 sheets, 75 g/m2, 8 ½ inches × 11
inches. Estimate the weight of the pack (1 inch = 2.54 cm).
(A) 13 N
(B) 22 N
Block P
1kg
(C) 37 N
(D) 62 N
Block R
Block Q
T1
2kg
(E) 112 N
T2
F=6N
3kg
This Figure applies to questions 2 and 3
[2] Block P has mass 1kg, block Q has mass 2kg and block R has mass 3kg. The three
blocks are on a frictionless surface connected with massless strings as shown. A force of
6N is pulling on block R to the right. What is the magnitude of the acceleration of block
P?
(A) 6m/s²
(B) 4m/s²
(C) 3m/s²
(D) 2m/s²
(E) 1m/s²
[3] Continuing with the system in the last problem, if T1 is the tension of the string
connecting blocks P and Q and T2 is the tension of the string connecting blocks Q and R.
What is the ratio: T1:T2 ?
(A) 3:1
(B) 2:1
(C) 1:1
Page 14 of 27
(D) 1:2
(E) 1:3
Physics 221 2006 S Exam 1 Solutions
[4] A motor uses a massless string to pull a 5kg block at a constant speed of 2m/s across a
level table where the coefficient of kinetic friction between the block and the table is
µk=0.3. When the block has been pulled a distance of 1.5m, how much net work has been
done on it?
(A) 22. 1J
(B) 14.7
(C) 0.0J
(D) –14.7J
(E) –22.1J
[5] These are the snapshots of two objects that travel along parallel straight tracks. The
snapshots are taken at t = 1, 2, 3 and 4 seconds. Assume the motion is smooth (nothing
unexpected happens between the snapshots).
1
1
2
2
3
3
4
4
What is the direction of the velocity of the circle relative to the rectangle, at t = 2 s and at
t = 3 s?
(A)
(B)
(C)
(D)
(E)
t = 2 s:
t = 2 s:
t = 2 s:
t = 2 s:
t = 2 s:
→
→
→
←
←
t = 3 s:
t = 3 s:
t = 3 s:
t = 3 s:
t = 3 s:
→
←
0
←
0
Page 15 of 27
Physics 221 2006 S Exam 1 Solutions
[6] Consider the vectors A and
B shown in the diagram which
lie in the xy plane. If A=3 and
B=4, what is the scalar
product A • B ?
y
B
B=4
15º
(A) +12
(B) +6
(C) 0
(D) −6
(E) −12
x
15º
A
A=3
[7] If the vector D = −15 iˆ + 16 ˆj − 12 kˆ , what is the unit vector D̂ ?
(A) Dˆ = +0.60 iˆ + 0.64 ˆj + 0.48 kˆ
(B) Dˆ = −0.60 iˆ + 0.64 ˆj − 0.48 kˆ
(C) Dˆ = +0.35 iˆ + 0.37 ˆj + 0.28 kˆ
(D) Dˆ = −0.35 iˆ + 0.37 ˆj − 0.28 kˆ
(E) Dˆ = −0.15 iˆ + 0.16 ˆj − 0.12 kˆ
[8] Sally weighs 1000N. She is standing in an elevator that is moving downwards at a
constant speed of 4.9m/s. What is the magnitude of the net force acting on Sally?
(A) 0N
(B) 500N
(C) 1000N
Page 16 of 27
(D) 1500N
(E) 2000N
Physics 221 2006 S Exam 1 Solutions
[9] A 100g block is propelled from rest along a frictionless track by a spring that is
compressed by 10cm from its relaxed length. Further along the track there is a frictionless
upwards incline sloped at 30º above the horizontal. The block slides up the ramp until it
is at a height of 1m above the ground level and then comes momentarily to a stop. What
is the force constant, k, of the spring?
Before
30º
10cm
1m
After
30º
(A) 49J
(B) 98N/m
(C) 196 N/m
(D) 392 N/m
(E) 588 N/m
[10] Consider the following acceleration versus time graph for an object constrained to
move along the x-axis. If at t=0s the object has a velocity of vx = 10 ms , what is the xcomponent of velocity at t=6s?
Acceleration
(m/s²)
6
4
2
1
(A) 15.5m/s
2
(B) 25.5m/s
3
4
(C) 17.5m/s
Page 17 of 27
5
Time (s)
(D) 27.5m/s
6
7
(E) 10.0m/s
Physics 221 2006 S Exam 1 Solutions
[11] If a particle travels at a constant speed along the track shown, what is the correct
ordering of the magnitudes
of the acceleration at the
labeled points? Let aP be
the acceleration at point P,
etc.
(A) a R < a S < a P < aQ
(B) aP < aQ < aR < aS
(C) aP = aQ = aR = aS
(D) a P < aQ < a S < a R
(E) aR < aS < aQ < aP
[12] A block slides up an incline where kinetic friction is present. Half way up the incline
a small rocket is fired that exerts a force on the block perpendicular and down into the
incline. Compare the system the instant before the rocket fires to the system the instant
after the rocket fires. Which of the
Velocity of
following statements is true?
Block
(A) After the rocket fires the weight of the
block increases.
(B) After the rocket fires, the normal force
of the ramp on the block increases
(C) After the rocket fires, the magnitude of
the friction force which the ramp exerts on
the block increases
(D) Both A and B are true
(E) Both B and C are true
Page 18 of 27
Rocket
Force of
rocket
Block
Physics 221 2006 S Exam 1 Solutions
[13] The figure below shows a position versus time graph for a particle that is constrained
to move along the x-axis. At which of the labeled point(s) is the acceleration positive?
(Note that the segments of the curve which cross the horizontal axis are linear)
Position
Time
(A) R only
(B) P, R and T
(C) S only
(D) Q and U
(E) P and T
[14] In case P a ball of mass m moves in a circle of radius R at a constant angular speed
of ω. In case Q a ball of mass 2m moves in a circle of radius 2R at a constant angular
speed of 2ω. What is the ratio:
(Net force on ball P):(Net force on ball Q)
angular
velocity=2ω
angular
velocity=ω
R
2R
Mass=m
Case P
(A) 1:2
(B) 1:4
(E) None of the above.
Case Q
(C) 1:8
(D) 1:16
Page 19 of 27
Mass=2m
Physics 221 2006 S Exam 1 Solutions
[15] A car of mass 1000kg traveling at a speed of 10m/s brakes to a stop over a distance
of 40m. What is the magnitude of the braking force acting on the car? Assume that the
braking force is constant.
(A) 1250N
(B) 2500N
(C) 125N
(D) 250N
(E) 625J
[16] A football is kicked from ground level at an angle of 60º with respect to the
horizontal. To make a field goal it must reach a height of at least 5m above the ground.
What is the minimum speed that the ball can be kicked at in order to achieve this height?
Neglect air resistance.
(A) 19.8m/s
(B) 14.0m/s
(C) 11.4m/s
(D) 10.0m/s
(E) 8.1m/s
[17] Robinhood shoots a 0.1kg arrow at the Sheriff of Nottingham from the top of a cliff
of height h=80m. The arrow is launched at a speed of 20m/s. The Sheriff is standing at a
distance d=70m
v0=20m/s
from the base of the
Robin
cliff when the arrow
Hood
strikes him. How
much work does
gravity do on the
arrow while it is in
h=80m
flight?
(A) 78.5J
d
(B) 19.6J
(C) 98.4J
(D) 9.8J
(E) The answer depends on the air resistance the arrow experiences in flight.
Page 20 of 27
Sheriff of
Nottingham
Physics 221 2006 S Exam 1 Solutions
[18] At t=0s a 5kg block is at point P on a ramp inclined at 10º with an initial velocity
sliding up the ramp of v0=10m/s. The coefficient of static friction between the block and
the ramp is µs=0.2 and the coefficient of kinetic friction between the block and the ramp
is µk=0.1. How long does it take the block to return to point P?
v0=10m/s
µk=0.1
µs=0.2
5kg
10º
P
(A) 2.0s
(B) 7.5s
(C) 10.9s
(E) The block never returns to point P.
(D) 27.2s
[19] A man exerts a 300-N force to pull an 80-kg crate up an incline with a rope that
makes an angle with the incline as shown below. What is the work done by the man when
he pulls the crate 2.0 m along the incline?
300N
30°
80kg
15°
(A) 400 J
(B) 420 J
(C) 520 J
Page 21 of 27
(D) 580 J
(E) 600 J
Physics 221 2006 S Exam 1 Solutions
[20] A block is placed near the edge of a horizontal turntable of radius 5m that is initially
at rest. It is attached to a motor that gives it an angular acceleration α. The coefficient of
static friction between the block and the turntable is µs=0.2. What is the maximum value
of the angular acceleration α that the motor can deliver so that the block will not slide the
instant after the turntable begins to move?
(A) 0.20 rad/s²
(B) 0.39 rad/s²
(C) 0.63 rad/s²
[21]
In the system depicted at right, a 1kg block
rests on a horizontal table. The coefficient of
static friction between the block and the table
is µs=0.6. The coefficient of kinetic friction
between the block and the table is µk=0.3.
The 1kg block is attached via an ideal
massless string over an ideal massless pulley
to a 3kg block which can move vertically
without resistance. Initially the 1kg block is
held fixed and is released at t=0. What is the
magnitude of the acceleration of the 1kg
block after it is released?
(A) 0.0 m/s²
(B) 5.9 m/s²
(C) 6.6 m/s²
(D) 23.5 m/s²
(E) 26.5 m/s²
Page 22 of 27
(D) 3.9 rad/s²
(E) 9.8 rad/s²
1kg
µs=0.6
µk=0.3
3kg
Physics 221 2006 S Exam 1 Solutions
[22] John travels 100km north at 50km/hr and then travels 100km south at 30km/hr.
What is the magnitude of his average velocity during the trip?
(A) 0.0 km/hr
(B) 40.0 km/hr
(C) 37.5 km/hr
(D) 35.3 km/hr
(E) 20.0 km/hr
[23] A particle of mass 0.5 kg moves on a circle of radius 2m. The angular position in
radians is given by θ = (4 s −4 )t 4 − (2s −2 )t 2 . What is the magnitude of the tangential
component of the acceleration of the particle at t=0.5s?
(A) 0m/s²
(B) 8m/s²
(C) 16 m/s²
(D) 64 m/s²
(E) 128 m/s²
[24] The graph below shows the position dependent net force acting on a 2kg particle
moving along the x-axis. The object is initially released from x=0m moving to the right.
When it gets to x=10m it has a velocity of 6m/s to the right. What was the object’s kinetic
energy when it was released?
(A) 12J
(B) 24J
(C) 30J
(D) 42J
(E) 48J
Force (N)
4
2
Position (m)
8
0
2
4
6
−2
Page 23 of 27
10
Physics 221 2006 S Exam 1 Solutions
[25] A cannon is fired at a medieval castle perched on a vertical cliff 124m high. The
cannon ball strikes the foot of the castle, just on the edge of the cliff. The cannon ball is
fired from a spot 300m from the base of the cliff and is aimed at an angle 45º above the
horizontal. How long after the cannon ball is fired does it strike the castle? Neglect air
resistance.
Impact
Castle
(A) 4s
Point
(B) 6s
(C) 8s
(D) 10s
t=?
124m
(E) Cannot be determined
without more information
45º
Cannon
300m
[26] Which of the arrows correctly indicates the direction of the acceleration of a particle
that moves clockwise at a constant speed around the path shown below?
(A) P, R and T
(B) P, R, T and U
(C) Q and S
(D) R and U
(E) None of the arrows correctly
indicates the direction of acceleration.
Q
P
R
U
T
S
Page 24 of 27
Physics 221 2006 S Exam 1 Solutions
[27] A 10-kg weight hangs from two ideal massless ropes as shown below. Find the
tension on the left rope.
(A) 33 N
(B) 49 N
(C) 85 N
(D) 98 N
(E) 113 N
60°
TL?
60°
10kg
[28] A 40N block is hung from a rope attached to a scale via a pulley as shown. The scale
is then attached to a wall and reads 40N. What will the scale read when it is instead
attached to another 40N block over another pulley, as shown in the second figure?
(A) 0N
(B) 20N
(C) 40N
(D) 60N
(E) 80N
???N
40N
40N
Page 25 of 27
40N
40N
Physics 221 2006 S Exam 1 Solutions
[29] A cannon that fires a shell at 300m/s is located on the slope of a hill. The slope is
inclined downwards at an angle 30°
E:120
below the horizontal as shown and
D:90°
extends very far in each direction.
C:60°
At which of the indicated angles
between the sloping ground and the
Horizontal
B:30°
initial velocity should the cannon be
Cannon
fired in order to maximize the time
A:15°
that the shell is in the air? Neglect air
resistance.
30°
(A) 15°
(B) 30°
(C) 60°
(D) 90°
(E) 120°
Ground
[30] (Note: Extra credit problem)
A cannon, located on planet X with acceleration of gravity gX , fires a cannon ball from
the top of a cliff of height h at an angle θ>0 above the horizontal. The projectile strikes
the ground with a speed of vf=49m/s at an angle of 60º below the horizontal as shown. If
the cannon ball was fired with a speed of v0=29.4m/s, what angle θ was the cannon ball
fired at? Neglect air resistance.
(A) 20º
(B) 34º
(C) 50º
(D) 56º
(E) The answer cannot be determined without more information.
v0=29.4m/s
Cannon
θ
h
60º
vf=49.0m/s
Page 26 of 27
Physics 221 2006 S Exam 1 Solutions
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