STUDENT ID: ______________________________NAME: __________________________________
1010, Fall 2006, Exam 2: GREEN
Write your name and ID on both your exam and your scantron answer sheet.
Write the color of your exam paper in the large blank area of your scantron sheet. DO NOT WRITE ON
THE MARGIN OF THE SCANTRON SHEET.
Be sure to turn your exam in to the proper box and to return both the answer sheet and the exam.
This test is closed book, but you may use a single 3x5 card with your own notes written on it. You
may also use a calculator.
The exam is worth 40 pts total. All multiple choice questions are worth the same number of points,
but not all of them are of the same difficulty. If you have difficulty answering a question, move on
to the next one and come back to the difficult ones later. The last (long hand) question is worth 10
points. Make sure you leave enough time for it.
Answer the last (long hand) question on the exam sheet, not the scantron sheet.
For all of these problems, assume that air resistance is not important unless the problem states
otherwise.
Conversions you may or may not need: 1 pound = 4.45 N,
1 slug = 14.594 kg,
1 mph = 0.447 m/s.
The velocity of sound in air is 330 m/s. The density of water is 1000 kg/m3.
1. Two speakers are located side by side. One generates a sound of 2m wavelength, the other a sound of
4m wavelength. Compared to the time it takes the 2m wave to reach my ear, the 4m wave will reach my
ear in
a. half the time
b. one quarter the time
c. twice the time
d. four times the time
e. the same time
Answer: E
The speed of sound is independent of frequency.
2-5 You hook a speaker up to an amplifier which generates a tone of frequency 50Hz:
2. What is the period of that tone
a. 0.02 s
b. 0.2 s
c. 2.0 s
Answer: A
T = 1/f = 1/50 s = 0.02s.
d. 0.05 s
e. 0.5 s
STUDENT ID: ______________________________NAME: __________________________________
3. What is the wavelength of that tone in air?
a. 6.6 m
b. 66 m
c. 1.65 m
d. 16.5 m
e. some other value
Answer: A
v=λ/T so λ = vT = 330 m/s * 0.02 s = 6.6m.
4. You put the speaker under water, where the speed of sound is 1650 m/s. What is the wavelength of
the tone in water?
a. 330 m
b. 660 m
c. 33 m
d. 165 m
e. some other value
Answer: C
v=λ/T so λ = vT = 1650 m/s * 0.02 s = 33m.
5. What is the period of the tone in water?
a. 3.3 s
b. 0.33 s
c. 0.2 s
d. 0.02 s
e. some other value
Answer: D
T = 1/f = 1/50 s = 0.02s.
The period is independent of the speed of the wave.
6. If the pendulum in the Gamow tower is 30 m long, what is its period?
a. 1 s
b. 2.2 s
c. 11 s
d. 30 s
e. 94.2 s
Answer: C
T = 2π sqrt(l/g) = 2π sqrt(30m/(9.8m/s2)) = 11.0 s
STUDENT ID: ______________________________NAME: __________________________________
7. If we hang a pendulum four times as long as the Gamow pendulum from the top of the Empire State
Building, its period would be
a. twice as long as the period of the Gamow pendulum
b. four times as long as the period of the Gamow pendulum
c. half as long as the period of the Gamow pendulum
d. one quarter as long as the period of the Gamow pendulum
e. the same as the period of the Gamow pendulum
Answer: A
T = 2π sqrt(l/g) so four times the length means twice the period.
8. If we move the Gamow pendulum to planet Zorg, where the surface gravity is four times that of Earth,
the period of the pendulum would be
a. twice as long as the period of the Gamow pendulum on Earth
b. four times as long as the period of the Gamow pendulum on Earth
c. half as long as the period of the Gamow pendulum on Earth
d. one quarter as long as the period of the Gamow pendulum on Earth
e. the same as the period of the Gamow pendulum on Earth
Answer: C
T = 2π sqrt(l/g) so four times the acceleration of gravity means half the
period.
STUDENT ID: ______________________________NAME: __________________________________
9. A water tank, hose, and valve are shown below with points A and B as marked. The faucet is closed.
Water tank
B
A
Valve on faucet is closed
B
A
12 m
20 m
meters
16 meters
10 m
meters
The pressure at the faucet (point B) is:
a. less than the pressure at point A
b. greater than the pressure at point A
c. equal to the pressure at point A
d. it cannot be determined with the information provided.
Answer: A
P + ρgh = const. so the greater the hight (h) the lower the pressure (P).
10. A brick of volume 0.002 m3 and density 2500kg/m3 is completely immersed in water. What is the
buoyancy force on the brick? (pick the choice closest to the value you calculated).
a. 20 N
b. 50 N
c. 50 kg
d. 5 kg
e. 2 N
Answer: A
Fb = ρVg = 1000 kg/m3 * 0.002 m3 * 9.8 m/s2 = 19.6 N.
11. A block of wood of volume 0.002 m3 and density 500kg/m3 is floating on water. What is the
buoyancy force on the block? (pick the choice closest to the value you calculated).
a. 20 N
b. 10 N
c. 10 kg
d. 2 kg
e. 2 N
Answer: B
Since the wood is floating, the buoyancy force is equal to its weight:
Fb = Fw = ρVg = 500 kg/m3 * 0.002 m3 * 9.8 m/s2 = 9.8 N.
STUDENT ID: ______________________________NAME: __________________________________
12. The same block of wood as in the previous question is completely immersed in water. What is the
buoyancy force on the block? (pick the choice closest to the value you calculated).
a. 20 N
b. 10 N
c. 10 kg
d. 2 kg
e. 2 N
Answer: A
Fb = ρVg = 1000 kg/m3 * 0.002 m3 * 9.8 m/s2 = 19.6 N.
13. You are putting all of your textbooks on the top shelf of your bookshelf. The books start out on the
ground and the shelf is 3 m up. You have 8 books, each with a mass of 1 kg. How much work does it
take to put the books on the shelf? (Choose the answer that is closest to the value you calculated.)
a. 80 J;
b. 240 J;
c. 8 J;
d. 80 N;
e. 20 N
Answer: B
W = F L = mgL = 1kg * 9.8m/s2 * 3m = 29.4 J per book. So for eight books,
W = 8*29.4 J = 235J
14. A weight oscillating up and down on a spring has the most potential energy when it is:
a. at the top;
b. in the middle;
c. at the bottom;
d. a&c;
e. some other position
Answer: D
Both at the top and at the bottom, all the energy is potential.
15-18. You come home to find that a large crate containing a new piano has been delivered to your
house.
15. The crate says that it has a mass of 200 kg. Once you get the crate moving at a constant speed, how
much force is required to lift it straight up to your third floor apartment?
a. 200 lb
b. 1960 N
c. 200 kg
d. 1960 lb
e. 200 N
Answer: B
To move it straight up at constant speed you need a force equal to its weight,
so F = mg = 200kg * 9.8m/s2 = 1960N
STUDENT ID: ______________________________NAME: __________________________________
16. How much does the piano weigh in pounds?
a. 200 lb
b. 1960 lb
c. 440 lb
d. 890 lb
e. 400 lb
Answer: C
From the conversion factor given at the top: F = 1960 N / (4.45 N/lb) = 440lb.
17. You put the piano on a light cart with very good wheels (i.e. no friction) and push it up a 50 m long
ramp to raise it 8 meters vertically. How much force is required to push the piano up the ramp? (you can
neglect the mass of the cart).
a. 32 lb
b. 32 N
c. 314 lb
d. 314 N
e. 200 kg
Answer: D
W = mgh = F L so F = mgh/L = 200kg * 9.8m/s2 * 8m / 50m = 314N.
18. Just as you get it to the top of the ramp, you stumble and it starts rolling back down the ramp. After
it has gone down 2 m of height (12.5 meters along the ramp) how fast is it moving?
a. 4.0 m/s
b. 3.12 m/s
c. 6.26 m/s
d. 39.2 m/s
e. 28.3 m/s
Answer: C
Conservation of energy: mgh = (1/2)mv2 so v = sqrt(2gh) =
sqrt(2*(9.8m/s2)*2m) = 6.26 m/s.
STUDENT ID: ______________________________NAME: __________________________________
19. A 1300 kg Honda left a skid mark of 82 meters. How fast was the car going when it slammed on its
brakes? The coefficient of friction between the car tires and the pavement was measured to be 0.7.
a. 12 m/s
b. 105 m/s
c. 1122m/s
d. 33.5 m/s
e. 603 m/s
Answer: D
Conservation of energy; all the car’s kinetic energy is used up by the work
done on the car by friction:
(1/2)mv2 = Ff L = µmg L so v = sqrt(2µgL) = sqrt(2*0.7*(9.8m/s2)*82m) =
33.5 m/s.
20. You throw a ball straight up. Right after it leaves your hand, the ball has speed 2.5m/s. Take up to be
the positive direction. What is the graph of velocity versus time and acceleration versus time of the ball.
Answer: A
Since we took up to be the positive direction, acceleration is constant and
negative.