PHYSICS 221
Summer 2005
EXAM 1: June 27, 2005
Questions 1-10 are worth one point each, 11-15 are worth two points, as marked.
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. 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 7) 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 the front of the room.
No cell phone calls are 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.
Throughout this exam, use 10 m/s
2
for the acceleration due to gravity.
GOOD LUCK!
Physics 221 2004 S Exam 1 Page 1 of 9

[1] (1 point) The components of vectors A and B are given as follows:
A x
= +
2 B x
= −
3
A y
= −
1 B y
= −
1
A z
=
4 B z
=
5
The components of the vector C
=
2 B
+
A are:
(A) C x
= −
4 C y
= −
3 C z
=
14
(B) C x
(C) C x
(D) C x
= −
1 C y
=
8 C y
= −
2 C z
= −
2 C z
=
9
=
9
= −
4 C y
= −
2 C z
=
14
(E) None of the above
The correct answer is A, as you can confirm by calculating 2B x
+A x
, etc.
[2] (1 point)
(A)
(B)
83
97
(C) 60 o o
The magnitude of vector product is –4m
2 o
A is 7m and that of vector
. What is the angle between the two vectors?
B is 5m. Their dot
(D) There is too little information to tell
The correct answer is B.
Since the dot product is defined as A B A B cos(
θ
AB
) we can solve this for the angle cos
θ
AB
=
−
4 m
2
A B
=
7 m
×
5 m
= −
4 / 35
= −
0.11, or
θ
AB
=
97 o
[3] (1 point) A weight lifter lifts a 100 kg weight a distance of 2.2 m to hold it above his head. He then holds it for 3 seconds. How much work does he do while holding it during the 3 seconds?
(A) 2200 J
(B) None
(C) 220 J
(D) 733 J
(E) None of the above
The correct answer is B, none. Since the force is not exerted over any distance, there is no work done. Just try telling that to the weight lifter!
Physics 221 2004 S Exam 1 Page 2 of 9

[4] (1 point) A truck has a speed of 20 m/s when it starts to brake. Assuming there is no skidding, and that it slows at a constant rate of -2m/s
2
, how far does it go before it stops?
(A) –100m
(B) 200m
(C) 100m
(D) -200m
(E) None of the above
The correct answer is C. Use the equation v
2 − v
0
2 =
2 ax , with v
= m s v
0
= m s a
= − m s
2
2 / .
[5] (1 point) A car is speeding up such that its velocity is given by the function v t a bt
2
. What is the function for its acceleration?
(A) a
+
2 bt
(B) 2 bt
(C) a bt
(D) bt
(E) None of the above
The correct answer is B. Take the derivative of the velocity with respect to time to get the acceleration.
[6] (1 point) A ball on a string is spinning in a circular path at a constant speed. The length of the string is 0.5m and it makes 2 complete orbits every second. What is its centripetal acceleration?
(A) 32
π
(B) 16
(C) 32
(D) 16
π
π
2
Rm/s
2
2
Rm/s
2
Rm/s
π Rm/s
2
2
(E) None of the above
The correct answer is B. The centripetal acceleration is either v
2
/R or
ω 2
R; the two are equivalent. In terms of velocity, we know the orbit is 2
π
R m and there are two orbits per second, so the ball travels with a speed of v=4
π
R m/s. This leads to the correct answer when plugged into the above equation. In terms of radial velocity,
ω
, we observe that the ball goes through 4
π
radians per second, which also leads to answer B.
[7] (1 point) Which of the following is NOT a conservative force?
Physics 221 2004 S Exam 1 Page 3 of 9

(A) Gravity
(B) Friction
(C) Elastic force from an ideal spring
(D) None of the above
The correct answer is B, friction.
[8] (1 point) A 100 g ball is spun in a circle at the end of an 80 cm string. The string will break when the tension exceeds 2 N. What is the maximum linear speed the ball can have without breaking the string?
(A) 2.8 m/s
(B) 4.0 m/s
(C) 5.6 m/s
(D) 16 m/s
(F) None of the above
The correct answer is B. Just plug the numbers into F=mv
2
/R and solve for v.
[9] (1 point) A 50 gram ball is placed in a spring-loaded cannon which has a spring constant of 400N/m. The canon is armed by compressing the spring 10 cm. The cannon fires the ball straight up, how high will it go?
(A) 2.0 m
(B) 400 m
(C) 4.0 m
(D) 80.0 m
(F) None of the above
The correct answer is C. Use conservation of energy. The energy stored in the spring is
W spring
=
1
2 kx
2
, which equals (after converting the units to SI) 2.0 J. The gravitational potential energy at the top of the orbit must be this same amount since there are no losses and there is no kinetic energy at the top point. So 2.0 J = mgh. Plug in the numbers and solve for h.
[10] (1 point) Two balls are released from a tower 125m high. Ball A is simply dropped straight down from rest, ball B is released with a sideways velocity of 2m/s, but no initial vertical velocity. Which ball will hit the ground first?
(A) Ball A
(B) Ball B
(C) They hit the ground together
(D) There is too little information to tell.
Physics 221 2004 S Exam 1 Page 4 of 9

The answer is C. Motion in the vertical direction is independent of the horizontal motion.
Remember the monkey falling out of the tree.
[11] (2 points) Referring to problem 10, how far will ball B land from the base of the tower?
(A) 25 m
(B) 20 m
(C) 10 m
(D) 5 m
(E) None of the above
The correct answer is C.
This is a two step problem. You must first solve for how long it takes the ball to fall
125m, then calculate the sideways distance using the sideways velocity. To calculate the fall, use y
=
1
2 at
2
(we have chosen a coordinate system and reference frame such that y
0
=0 and v
0
=0). For x =125m and a =10m/s
2
, we get t= 5 sec. The sideways distance is then x =(2.0m/s)(5s) = 10m.
[12] (2 points) A 2 kg block slides across a rough surface. The only horizontal force on it is friction. In 2m it slows from a speed of 2.0 m/s to 1.0 m/s. What is the coefficient of friction?
(A) 0.005
(B) 0.10
(C) 0.15
(D) 0.075
(E) None of the above
The correct answer is D.
Solve v
2
-v o
2
=2ax to get a=-3/4 m/s
2
. Then
F friction
=(2 kg)(-3/4)m/s
2
=-
µ mg=
µ
(2 kg)(10m/s
2
). Solving for
µ
gives 0.075.
[13] (2 points) Which has the larger gain in mechanical energy: (A) a 5 kg weight raised
10 m, (B) a 2 kg mass accelerated to a speed of 20 m/s, or (C) a spring with a spring constant of 400 N/m which has been compressed 0.8 m?
(A) A
(B) B
(C) C
(D) Both A and C
(E) Both B and C
Physics 221 2004 S Exam 1 Page 5 of 9

The correct answer is A.
We need to calculate the energy in each case.
For A, W=mgh=(2 kg)x(10 m/s
2
For B, KE=mv
2
)x(10 m)=500 J
/2=(2 kg)x(20 m/s)
2/
For C, W=kx
2
/2 =(400 N/m)(0.8 m)
2
2 = 400 J
=128 J.
[14] (2 points) A canoeist crosses a river and can paddle with a speed of 0.5m/s. There is a cross-current with a speed of 0.3m/s, so the canoeist compensates by heading partly into the current so that her net motion goes straight across. What is the canoeist’s speed in the direction straight across the river?
(A) 0.2m/s
(B) 0.4m/s
(C) 0.5m/s
(D) There is too little information to tell.
The correct answer is B. To head directly across, the canoeist has to paddle somewhat into the current to compensate. The vectors are shown below.
Bank
Canoeist,
0.5 m/s
Net motion,
0.4 m/s
Current,
0.3 m/s
Bank
[15] (2 points) A 2 kg block slides on a rough surface with a coefficient of friction of 0.2 and stops in 0.5 m. What was its initial speed?
(A) 1 m/s
(B) 2 m/s
(C) 1.4 m/s
(D) 2.8 m/s
(E) None of the above
The correct answer is C.
We first need to solve for the work done by friction:
W frictio n
= Fx =
µ mg . Putting in numbers we get W friction
=(0.2)(2 kg)(10 m/s
2
)(0.5 m)=2J.
We get the velocity from the initial kinetic energy, which has to equal the work done by friction: KE =(2 kg) v
2
/2=2 J. Solving for v , we get 1.4 m/s.
Physics 221 2004 S Exam 1 Page 6 of 9

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: R earth
=6.38
×
10
6 m
1.3) Mass of Proton: mp=1.67
×
10
-27 kg
3. Vectors
3.1) Dot Product: A ⋅ B = A x
B x
+ A y
B y
+ A z
B z
= | A || B | cos
θ where
θ
is the angle between
3.2) Components: A
=
A x i
ˆ +
A y
ˆ j
+
A z
A k
ˆ
and B .
3.3) Magnitude: | V |
=
V
=
V x
2 +
V y
2 +
V z
2 =
V
⋅
V
5. One Dimensional Motion
5.1) Average Velocity: v
=
5.2) Instantaneous Velocity:
∆ v x /
=
∆ t dx / dt v
5.3) For Constant Acceleration only : v x
2 x x
=
−
− t x x x
0 v
0
2
0 x
=
+
=
= v
0 v
0
1
2 x x
2 a t
( v
+ x x
+
( a x
+
1
2 x t a x t
− v
0 x x
0
)
2
)
2. Calculus
2.1) d dx x n = nx n − 1 d dx sin x = cos x
d dx x n + 1 x n dx
= cos x n
+
1
= − sin x
4. Algebra
4.1) The solutions to are x =
2
1 a
(
− b ± b
2 − ax
2
4 ac
)
+ bx
+ c
=
0
6. Forces
6.1) Newton’s Second: F
= m a
6.2) Newton’s Third:
6.2) Kinetic Friction:
6.4) Static Friction: f s
F
AB f k
=
= −
F
BA
µ k
N
≤ µ s
N
6.5) Centripetal Force: F
= mv
2
R
7. Three Dimensional Motion
7.1) Position Vector: r
= x i
ˆ + y j
ˆ + z k
ˆ
7.2) Velocity and Acceleration: v
= d dt r a
= d dt v
= d
2 dt
2 r v
= v
0
+ a t r
= r
0
7.3) Constant Acceleration only: v
2 r
−
− v
0 r
2
0
+ v
0 t
=
7.4) Circular Motion: f
7.4a) Angular Velocity:
= 1 / T
ω = d
θ
7.5) Centripetal Acceleration: a rad
/
ω t
= dt
=
R
ω
2
π f
2
=
=
2 a
1
2 v
2
( v
/
+
⋅
( r
1
2
+
R v a t
2
− r
0
) v
0
)
=
= R
ω
( 4
π 2
R ) / T
2
7.6) Changing Reference Frames: v
PA
= v
PB
+ v
BA
Physics 221 2004 S Exam 1 Page 7 of 9

4
5
6
2
3
9
10
7
8
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
1 11
14
15
12
13
Physics 221 2004 S Exam 1 Page 8 of 9

Scratch Paper (intentionally left blank)
Physics 221 2004 S Exam 1 Page 9 of 9