PHYSICS 221(SOLUTIONS)
Spring 2004
EXAM 1: Feb 12 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.
"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 12) 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.
Total number of questions is 26.
Best of luck, David Atwood and Anatoli Frishman
Physics 221 2004 S Exam 1
Page 1 of 27
G
G
[1] The components of vector A and B are given as follows:
Ax = +3.1 B x = −5.6
Ay = −7.0 B y = −3.1
G G
The magnitude of the vector difference B − A , is closest to:
(A) 1.3
(B) 9.5
(C) 4.6
(D) 91.0
(E) 100.0
G G
The components of B − A = (−8.7,+3.9,0) taking the magnitude:
G G
| B − A |= (−8.7) 2 + (+3.9) 2 = 9.5
Physics 221 2004 S Exam 1
Page 2 of 27
G
G
[2] Consider the vectors A and B shown in the diagram below lie in the xy plane. Both
G G
vectors have the same magnitude A=B=3. The vector product A ⋅ B is:
y
3
G
A
x
30º
30º
(A) 7.8
(B) 4.5
(C) 3.0
(D) 9.0
(E) 15.6
3
G
B
G G
The angle between the vectors is 60º. Thus A ⋅ B = AB cos(θ ) = (3)(3)(0.5) = 4.5.
Physics 221 2004 S Exam 1
Page 3 of 27
G
G
[3] Consider the vector C = 2.00iˆ − 5.00 ˆj − 1.00kˆ , the unit vector in the direction of C
is:
(A) 0.54iˆ − 0.80 ˆj − 0.27kˆ
(B) 0.37iˆ − 0.91 ˆj − 0.18kˆ
(C) 0.50iˆ − 1.30 ˆj − 0.25kˆ
(D) 0.43iˆ − 1.10 ˆj − 0.21kˆ
(E) 0.21iˆ − 0.97 ˆj − 0.11kˆ
Finding the unit vector parallel to C:
G
C
2.00iˆ − 5.00 ˆj − 1.00kˆ
ˆ
C= =
= 0.37iˆ − 0.91j − 0.18kˆ
2
2
2
C
(2.00) + (5.00) + (1.00)
Physics 221 2004 S Exam 1
Page 4 of 27
G G
G
G G G
[4] If the vectors P , Q and R satisfy the relations P + Q = R and P 2 + Q 2 = R 2 , what
G
G
is the angle between P and Q ?
(A) 0º
(B) 15º
(C) 30º
(D) 90º
(E) 120º
G G G
The relation P + Q = R implies that the three vectors form a triangle while P 2 + Q 2 = R 2
implies that it is a right triangle with R as the hypotenuse. The angle between P and Q is
thus 90º
Q
P
R
Physics 221 2004 S Exam 1
Page 5 of 27
[5] A child stand on a bridge throws a rock straight down with a downwards velocity of
5m/s. The rock leaves the child’s hand at t=0. In a coordinate system where up is positive
which of the graphs shown here best represents the velocity of the stone as a function of
time?
(B)
(A)
(D)
(C)
(E)
The velocity is negative and thereafter declines linearly due to the acceleration of gravity therefore the answer
is E, the only graph which has these features
Physics 221 2004 S Exam 1
Page 6 of 27
[6] John runs 1 km north at 6 m/s and then runs an additional 1 km in the same direction
at 4 m/s. What is the magnitude of John’s average velocity for his 2km run?
(A) 3.0 m/s
(B) 4.0 m/s
(C) 4.8 m/s
(D) 5.0 m/s
(E) 5.4 m/s
The time for the first part of the trip is 1000m/(6m/s)=167s. The time for the second part
of the trip is 1000m/(4m/s)=250s. The total time is thus 417s and since the trip is 2000m
the average velocity is (2000m)/(417s)=4.8m/s.
Physics 221 2004 S Exam 1
Page 7 of 27
[7] The position of a particle along the x-axis as a function of time is given by the
expression: x(t)=at3+bt+c where a=4m/s3, b=2m/s and c=5m. What is the acceleration of
the particle at the time t = 4 s ?
(A) 0 m/s2
(B) 24 m/s2
(C) 48 m/s2
(D) 96 m/s2
(E) 192 m/s2
The acceleration is the second derivative of position with time: a =
the numbers from the question, the acceleration at t=4s,
dx
dt
= 6at. Plugging in
a = 6ax = 6(4m / s 3 )(4 s ) = 96m / s 2
Physics 221 2004 S Exam 1
Page 8 of 27
[8] In the graph of position versus time, at which points is the acceleration is zero:
(A) P, Q, R and S
(B) P, Q, S
(C) R only
(D) P and S
(E) Q only
The acceleration will be 0 when the line is not curved. This is only the case at point R.
Physics 221 2004 S Exam 1
Page 9 of 27
[9] Ball #1 is thrown vertically upwards with a speed of v0 from the top of a building and
hits the ground with speed v1. Ball #2 is thrown vertically downwards from the same
place with the same speed v0 and hits the ground with speed v2. Which one of the
following three statements is true. Neglect the effects of air friction.
(A) v1>v2
(B) v1=v2
(C) v1<v2
(D) Depends on which ball is more massive
(E) None of the above
The final velocities are the same! To see this use the magic v² formula
v x2 − v 02x = 2a x ( x − x0 ) . Note that the difference in velocity squared does not depend on
the whether the initial velocity is positive or negative.
Physics 221 2004 S Exam 1
Page 10 of 27
[10] A person kicks a soccer ball off level ground at an angle of 45° above the
horizontal. In order that the ball reaches a maximum height of 30m above the ground
during its flight, what is the initial speed of the soccer ball?
(A) 25 m/s
(B) 29 m/s
(C) 34 m/s
(D) 37 m/s
(E) 41 m/s
Let v be the initial speed of the ball. The y component of v is v y = v sin(θ ),
θ = 45 o .
At the peak, vy is 0 so using the magic v squared formula, v y2 = 2 gd where d=30, and
Putting these two relations together,
v=
2 gd
= 34.3m .
sin(θ )
Physics 221 2004 S Exam 1
Page 11 of 27
[11] A straight 100m wide river flows at a speed of 2.0m/s with respect to the bank. A
canoe is paddled at a speed of 4.0m/s with respect to the water. The trajectory of the
canoe is directly across the river, perpendicular to the flow of water, as viewed from the
bank. How long does it take for the canoe to cross the river?
(A) 22 s
4m/s
(B) 25 s
(C) 29 s
(D) 32 s
y
(E) 50 s
From the question, vcanoe / river = 2m / s . The velocity of
the canoe with respect to the ground is:
G
G
G
vcanoe / ground = vcanoe / river + v river / ground since we know that
the canoe moves in the x direction wrt the ground and
the water moves in the y direction (using coordinate
system at right):
Path of canoe
x
G
vcanoe / river = iˆvcanoe / ground − ˆjv river / ground
Taking the magnitude of each side we therefore find
(vcanoe / river ) 2 = (vcanoe / ground ) 2 + (v river / ground ) 2
(vcanoe / ground ) = (vcanoe / river ) 2 − (v river / ground ) 2 = (4m / s ) 2 − (2m / s ) 2 = 3.46m / s
The time it takes to cross the river is thus 100m/(3.46m/s)=28.9s
Physics 221 2004 S Exam 1
Page 12 of 27
[12] A cannon ball is fired at an angle of 30º to the horizontal on level ground with an
initial velocity of 490m/s. Assuming air resistance is negligible, how long will the cannon
ball remain aloft?
(A) 25s
(B) 50s
(C) 100s
(D) 200s
(E) The answer depends on the mass of the cannon ball
From the magic v squared formula the cannon ball will return to earth with the y
component of velocity reversed. The initial vertical component of the velocity is
v y 0 = v sin(θ ) so that t = 2v 0 y / g = 2v sin(θ ) / g = 50s.
Physics 221 2004 S Exam 1
Page 13 of 27
[13] A box that weighs 500N sits on the floor of an elevator. The elevator is moving
upwards at a constant velocity of 5m/s. What is the magnitude of the net force acting on
the box?
(A) 1000 N
(B) 750 N
(C) 500 N
(D) 250 N
(E) 0 N
By Newton’s 1st law, since it is moving at a constant velocity, there is no net force acting
on it.
Physics 221 2004 S Exam 1
Page 14 of 27
[14] A cannon ball is fired at a speed of 80.0m/s straight up from a cannon located at
the base of a cliff that is 100m high. It proceeds upwards above the cliff and eventually
falls back down. When it passes the edge of the cliff in the downward direction, what is
the speed of the cannon ball? Neglect air resistance.
v=?
v=80.0m/s
100m
Cannon
(A) 44.4 m/s
(B) 50.0 m/s
(C) 66.6 m/s
(D) 73.6 m/s
(E) Cannot be determined with the given information.
Using the v squared formula for constant acceleration:
v 2fy = viy2 − 2 gh = (80m / s ) 2 − 2(9.8m / s 2 )(100m) = 4440m 2 / s 2
The final velocity is therefore v fy = 4440m 2 / s 2 = 66.6m / s
Physics 221 2004 S Exam 1
Page 15 of 27
[15] An elevator with a mass of 2000kg is given an upwards acceleration of 0.98m/s²
by a cable. What is the tension in the cable?
(A) 10 kN
(B) 12 kN
(C) 16 kN
(D) 18 kN
(E) 22 kN
T
The net upwards force is T-mg. Using Newton’s second
law, T-mg=ma so T=m(g+a). Putting in the numbers,
T=2000kg(9.8m/s²+0.98m/s²)=21.6kN
2000kg
mg
Physics 221 2004 S Exam 1
Page 16 of 27
[16] A block of mass m slides down a frictionless ramp that is at an angle θ with
respect to the horizontal, where θ is between 10° and 80°. The magnitude of the
acceleration of the block is
.
(A) 2g cosθ
(B) 2g sinθ
(C) g
(D) g cos θ
(E) g sin θ
The component down the ramp is mg sin θ so the horizontal component of the
acceleration is g sin θ
Physics 221 2004 S Exam 1
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[17]
When Bill drives from home to work he drives through a circular curve in the
road at 60km/hr during which a net force of magnitude F is acting on him. On the trip
back from work to home he rounds the same curve at twice the speed. What is the
magnitude of the net force acting on Bill while rounding the curve on his return trip.
(A) 0
(B) F
(C) 2F
(D) 4F
(E) 8F
Centripetal force is proportional to v² so that doubling the speed multiplies the force
by 4.
Physics 221 2004 S Exam 1
Page 18 of 27
[18] A 130kg block rests on a horizontal frictionless ice surface. Jane pulls the block
north with a force of 120N while Sue pulls the block east with a force of 50N. What is the
magnitude of the acceleration of the block?
(A) 0.00 m/s²
(B) 0.385 m/s²
(C) 0.923 m/s²
(D) 1.00 m/s²
(E) 1.31 m/s²
The total force on the block is F = (120 N )iˆ + (50 N ) ˆj. The acceleration is therefore
a = F / m = (120 N ) 2 + (50 N ) 2 /(130kg ) = 1m / s 2
Physics 221 2004 S Exam 1
Page 19 of 27
[19] At time t=0s a 1kg block is at point P on a frictionless ramp inclined at 30º to the
horizontal. At this time the block is sliding up the ramp with a velocity of 4.9m/s. The
block subsequently slides further up the ramp but due to the action of gravity it
eventually reverses its course and slides back down to point P. At what point in time does
this return of the block to point P take place?
(A) t=0.5 s
(B) t=1.0 s
(C) t=2.0 s
(D) t=4.0 s
(E) t=5.0 s
P
30º
The block is accelerating down the ramp at a rate of a=g sin(θ)=4.9m/s²
Using the magic v squared formula, the velocity is reversed when the block returns to P
therefore t=2v/a=2(4.9m/s)/(4.9m/s²)=2s.
Physics 221 2004 S Exam 1
Page 20 of 27
[20] 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 connected by ideal
massless strings. A force FL=30N is applied to the left block and is directed to the left.
Another force FR=50N is applied to the right block, and is directed to the right. What is
the magnitude of the tension T in the string between m2 and m3.
T=?
FL=30N
m1=1kg
m2=2kg
m3=3kg
m4=4kg
FR=50N
(A) T=6N
(B) T=20N
(C) T=30N
(D) T=36N
(E) T=50N
Take right as positive. The acceleration of the whole train is (20N)/(10kg)=2m/s². This is
the acceleration of the first two blocks which has a mass of 3kg so the net force on those
blocks is 6N. The applied force is –30N therefore the force from the center string is
+36N. The tension in the string is therefore 36N
Physics 221 2004 S Exam 1
Page 21 of 27
[21] A 1.00kg wooden block is initially held at rest on a horizontal surface where the
coefficient of static friction with the block is µs=0.46 and the coefficient of kinetic
friction is 0.35. The block is connected to an ideal string that runs over an ideal pulley
and then is connected to a hanging 0.50kg mass, as shown in the figure below. After the
1.00kg block is released, the magnitude of the acceleration of the block is:
1.00kg
0.50kg
(A) 0.0 m/s²
(B) 1.0 m/s²
(C) 3.3 m/s²
(D) 5.6 m/s²
(E) 9.8 m/s²
Physics 221 2004 S Exam 1
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[22] Two boxes of mass m and 4m are pushed together along a frictionless surface by
a constant force of magnitude N as shown in the figure below. Let FL be the magnitude
of the net force on the large box and FS the magnitude of the net force on the small box.
Which of the following relations between these magnitudes is true?
FL
4m
FS
m
(A) FL<N,
(B) FL>N,
(C) FL>N,
(D) FL=N,
(E) FL=N,
N
FS<N
FS<N
FS=N
FS<N
FS=N
Both blocks accelerate at the same rate. By N’s 2nd if acceleration is fixed then net force
is proportional to the mass. Each of the two blocks have mass less than the whole system
of blocks so the force on each force is less than N.
Physics 221 2004 S Exam 1
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[23] A person drives a car over the top of a hill, the cross section of which can be
approximated by a circle of radius 256m, as shown below. What is the greatest speed at
which he can drive without the car leaving the road at the top of the hill?
256 m
(A) 10 m/s
(B) 20 m/s
(C) 25 m/s
(D) 50 m/s
(E) 75 m/s
At the critical velocity, the normal force vanishes and the centripetal force is provided by
gravity. Thus the centripetal acceleration is equal to the acceleration of gravity: g=v²/R so
v = gR = 50 m / s
Physics 221 2004 S Exam 1
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[24] A particle moves at a constant speed along the trajectory shown below. Compare
the magnitude of the acceleration of the particle at points A and B, aA and aB respectively.
(A) aA>aB ≠0
(B) aA>aB =0
(C) aA<aB
(D) aA=aB =0
(E) aA=aB ≠0
The acceleration is inversely related to the radius of curvature. There is nonzero
curvature at each point so neither acceleration is 0. Point A has the greatest
acceleration since it has the smallest radius of curvature hence aA>aB ≠0
Physics 221 2004 S Exam 1
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[25] The figure shows the overhead view of two stones that travel in circles over a
frictionless horizontal surface both with a period of 1s. Each stone is tied to a cord whose
opposite end is anchored at the center of the circle. Which of the following statements
concerning the tension in the two cords is true?
(A) The tension in the long cord is greater than the tension in the short cord.
(B) The tension in the long cord is less than the tension in the short cord.
(C) The tension in the long cord is equal to the tension in the short cord.
(D) The relation between the two tensions cannot be determined from the information
given.
The tension in the long cord is greater than the tension in the short cord because the
tension provides the centripetal force given by T=Fcentripetal=Rω² so at constant
angular velocity F is proportional to R and so the larger circle will have the greater
tension.
Physics 221 2004 S Exam 1
Page 26 of 27
[26] A year (y) is equal to 3.16×107s. A light-year (ly) is a unit of distance used in
astronomy which is equal to 9.47×1015m. What is the correct conversion of the
acceleration of gravity near the earth’s surface (g=9.8 m/s²) into units of (ly)/y²:
(A) g= 3.27×10-8 (ly)/y²
(B) g= 3.10 ×10+8 (ly)/y²
(C) g= 1.03×10-15 (ly)/y²
(D) g= 9.79×10+15 (ly)/y²
(E) g= 1.03 (ly)/y²
g=(9.8m/s²)(1 ly/9.47×1015m)(3.16×107s/1 y)²= 1.03 (ly)/y²
Physics 221 2004 S Exam 1
Page 27 of 27