PAP Physics/Physics
Measurement and Problem Solving
Supplemental Worksheet
I.
Mathematics Review
1. Convert each of the following measurements to meters.
a. 42.3 cm
d. 0.023 mm
b. 6.2 pm
e. 214 m
c. 21 km
f. 570 nm
2. Rank the following mass measurements from smallest to largest: 11.6 mg, 1021 g,
0.000 006 kg, 0.31 mg.
3. Express the following numbers in scientific notation:
a. 5 000 000 000 000 m
c. 2 003 000 000 m
b. 0.000 000 000 166 m
d. 0.000 000 103 0 m
4. Add or subtract as indicated. Give your answers using the correct number of
significant figures.
a. 5.80 x 109 s + 3.20 x 108 s
c. 3.14 x 10-5 kg + 9.36 x 10-5 kg
-6
-6
b. 4.87 x 10 m – 1.93 x 10 m
d. 8.12 x 107 g – 6.20 x 106 g
5. State the number of significant figures in each of the following measurements.
a. 0.000 03 m
c. 80.001 m
b. 64.01 fm
d. 0.720 g
6. State the number of significant figures in each of the following measurements.
a. 2.40 x 106 kg
c. 4.07 x 1016 m
6
b. 6 x 10 kg
d. 0.0037 m
7. Add or subtract as indicated. Give your answers using the correct number of
significant figures.
a. 16.2 m + 5.008 m + 13.48 m
c. 78.05 cm2 – 32.046 cm2
b. 5.006 m + 12.0077 m + 8.0084 m
d. 15.07 kg – 12.0 kg
8. Multiply or divide as indicated. Give your answers using the correct number of
significant figures.
a. (6.2 x 1018 m)(4.7 x 10-10 m)
c. (8.1 x 10-4 km)(1.6 x 10-3 km)
-7
-12
b. (5.6 x 10 m)/(2.8 x 10 s)
d. (6.5 x 105 kg)/(3.4 x 103 m3)
9. Using a calculator, Chris obtained the following results. Give the answer to each
operation using the correct number of significant figures.
a. 5.32 mm + 2.1 mm = 7.4200000 mm
b. 13.597 m x 3.65 m = 49.62905 m2
c. 83.2 kg – 12.804 kg = 70.3960000 kg
d. 196.15 m  39.23 s = 5 m/s
10. Solve the following equations for v.
a. d  vt
d
b. t 
v
c.
d.
v2
2d
v b

a c
a
Measurement and Problem Solving, Supplemental Worksheet, page 2
11. Solve each of these equations for E.
E
a. f 
s
2E
b. m  2
v
12. Solve each of these equations for a.
a. v  vo  at
b.
y  vo t  1 / 2at 2
c.
E
m
c2
c.
v 2  vo2  2ay
d.
v  2as
II. Mathematical Relationships
13. Given the equation E 
a. E and m?
b. E and v?
14. In the equation, a 
1
mv 2 , what relationship exists between
2
c. If m is doubled, what happens to E?
d. If v is halved, what happens to E?
F
, a is the dependent variable and F and m are the
m
independent variables.
a. If F is doubled, what happens to a?
b. If m is halved, what happens to a?
c. If F is quadrupled, what happens to m?
15. During a class demonstration, an instructor placed a 1.0-kg mass on a horizontal
table that was nearly frictionless. The instructor then applied various horizontal
forces to the mass and measured the rate at which the mass gained speed (was
accelerated) for each force applied. The results of the experiment are shown below.
Force (N)
5.0
10.0
15.0
20.0
25.0
30.0
Acceleration (m/s2)
4.9
9.8
15.2
20.1
25.0
29.9
a. Plot the values given in the table and draw the curve that best fits all points.
b. Describe, in words, the relationship between force and acceleration according to
the graph.
c. Write the equation relating the force and the acceleration that results from the
graph.
d. Find the units of the slope of the graph.
Measurement and Problem Solving, Supplemental Worksheet, page 3
16. The teacher who performed the experiment in the previous problem then changed
the procedure. The mass was varied while the force was kept constant. The
acceleration of each mass was then recorded. The results are shown in the table
below.
Mass (kg)
1.0
2.0
3.0
4.0
5.0
6.0
Acceleration (m/s2)
12.0
5.9
4.1
3.0
2.5
2.0
a. Plot the values given in the table and draw the curve that best fits all points.
b. Describe the resulting curve.
c. According to the graph, what is the relationship between mass and the
acceleration produced by a constant force?
d. Write the equation relating acceleration to mass given by the data in the graph.
e. Find the units of the constant in the equation.
17. The relationship between the circumference and the diameter of a circle is shown
below. Could a different straight line describe a different circle? What is the meaning
of the slope?
Circumference
(cm)
Circumference vs. Diameter of a
Circle
20
15
10
5
0
0
2
4
6
Diameter (cm)
III. Dimensional Analysis
18. How many seconds are in one year?
19. The space shuttle orbits the earth at an altitude of 30.0 km. What is this distance in
miles?
20. The speed of light is 3.00 x 108 m/s. Convert this to km/h.
Measurement and Problem Solving, Supplemental Worksheet, page 4
IV. Problem Solving
21. Wave Formula: v  f  
v: wave speed, in feet per second
f: frequency, in cycles per second
: wavelength, in feet/cycle
The shrill cries of help are heard in Gotham City. (This looks like a job for Batman.)
The Boy Wonder, wanting some help, locates the sound analyzer. Robin must know
the frequency of the sound and its wavelength. With an oscilloscope, he has been
able to fix the frequency at 4400 cycles per second. He cannot find the wavelength.
If the speed of sound is 1100 feet per second, can you help the Boy Wonder so he
can help the person in distress?
22. Volume Expansion of a liquid: V  Vo T
V : the change in volume
 : the coefficient of volume
expansion = 9 x 10-4/C
Vo : the original volume
T : change in temperature
An Edsel fills up its 20.-gallon tank with gas at 20C. If the car sits in the sun until
the temperature of the gas is 37C, how much gas would overflow?
23. Focal Length of Mirrors and Lenses:
1
1
1


f do di
f:
focal length
do: object distance
di: image distance
A substitute teacher, to entertain the class, is going to show an educational film on
light. The film is put into a motion picture projector that uses a convex lens with a
50.0 cm focal length to project the image on a screen 4.0 m away. How far from the
lens must the film be in order for it to be in focus? (Once it was in focus, it appears
that someone has switched the film on the “sub.” Well, the class would rather see
the “Bugs Bunny Cartoon.”)
Measurement and Problem Solving, Supplemental Worksheet, page 5
24. Work: W  F  d
W: work, in Joules (1 J = 1 Nm)
F: force, in Newtons (1 N = 1 kgm/s2)
d: distance
Newton’s Second Law of Motion: F  ma
F: force, in Newtons
a: acceleration = 9.8 m/s2
Big Bird has six eggs in his nest. However, the nest is too close to the door where a
draft blows and chills the eggs. (The eggs will never hatch this way. Big Bird is left
with only one decision. Move the nest!) The mass of the nest is 500. kg, and each
egg has a mass of 200. kg. If Big Bird moves the nest to the second floor (5.0 m
above), how much work is done? (In case you’re wondering about Big Bird’s wife,
she’s out working on a construction gang so she can feed the family!)
1
25. Energy: E  mv 2
E: energy, in foot-pounds
2
m: mass, in slugs
v: velocity, in feet per second
Newton’s Second Law of Motion: F  ma
F: force (weight), in pounds
(1 pound = 1 slugfoot/second2
m: mass, in slugs
a: acceleration = 32 feet/second 2
As an ancient torture to unfaithfuls, the Romans used to tie a man’s hands to the reins
of two horses (each has a weight of 1100 pounds). At some time, unknown to the man,
the horses were whipped so that they would run away. However, one time the man
was able to run with the horses, and after 2.0 minutes he was able to stop them by
exerting his great strength. (Egads! It’s...It’s...HERCULES!) The horses were running
at 30.0 miles per hour. How much energy was used up to stop the horses?
Measurement and Problem Solving, Supplemental Worksheet, page 6
Answers to Some of the Problems:
1.
a. 0.423 m
b. 6.2 x 10-12 m
c. 2.1 x 104 m
d. 2.3 x 10-5 m
e. 2.14 x 10-4 m
f. 5.7 x 10-7 m
2. 0.31 mg, 1021 g, 0.000 006 kg, 11.6
mg
3. a. 5 x 1012 m
b. 1.66 x 10-10 m
c. 2.003 x 109 m
d. 1.030 x 10-7 m
4. a. 6.12 x 109 s
b. 2.94 x 10-6 m
c. 1.25 x 10-4 kg
d. 7.50 x 107 g
5. a. 1
b. 4
c. 5
d. 3
6. a. 3
b. 1
c. 3
d. 2
7. a. 34.7 m
b. 25.022 m
c. 46.00 cm2
d. 3.1 kg
8. a. 2.9 x 109 m2
b. 2.0 x 105 m/s
c. 1.3 x 10-6 km2
d. 1.9 x 102 kg/m3
9. a. 7.4 mm
b. 49.6 m2
c. 70.4 kg
d. 5.000 m/s
d
10. a. v 
t
d
b. v 
t
c. v   2ad
ab
d. v 
c
11. a. E  f  s
b.
c.
12. a.
b.
c.
mv 2
E
2
E  mc 2
v  vo
a
t
2( y  v o t )
a
t2
v 2  v o2
a
2y
v2
2s
a. E  m
b. E  v 2
b. doubled
c. quartered
a. doubles
b. doubles
c. no change
a. student graph
b. a is directly proportional to F
c. a  kF
m 2
m 2
1
s
s


d.
N
kg  m 2 kg
s
There is only one straight line. The
slope is .
18.6 mi
0.25 feet
0.57 m
67 000 foot-pounds
d. a 
13.
14.
15.
17.
19.
21.
23.
25.