AP Physics “B”
San Dieguito Academy
2012 – 2013
Book 1
Collisions in Two Dimensions
Center of Mass
Springs
Density - Pressure
Fluid Dynamics
Flight
1
Book 1
Table of contents
Chapter
Topic
Contents
Pages
1
Introduction and Review
1 – 12
2
Momentum
 in one dimension
 in two dimensions
Center of mass
14 – 23
24 - 28
29 - 34
3
Springs
Hook’s Law
Energy in a spring
35 - 37
38 - 46
4
Density
Volume – Density
47 - 56
5
Fluid dynamics
Pressure and Force
Pressure and Energy
Bernoulli’ Principle
Buoyancy
57
66
77
88
-
65
76
87
92
Chapter 1 – General Understandings from first year physics
Basic terms, concepts, and problem solving techniques that you should have learned include:
Inertial mass (m) is an object’s resistance to change in motion, or conversely, its desire to
continue in its current motion. Inertia is a scalar. It has magnitude but not direction. In the
MKS system it is measured in Kilograms
The basic law of inertia as stated by Newton: Objects in uniform motion want to stay in
uniform motion forever; unless acted upon by a net unbalanced force.
Gravitational mass (m) an object’s ability to attract other matter and its ability to curve
space. The gravitational attraction created by mass is experienced as gravitational force. This
force is equal on both masses but opposite in direction. G is a constant 6.67 E-11 in the MKS
system
Fg = G m1 m2 / r2
Note: It has never been proved that inertial mass and gravitational mass are equivalent. On the
positive side, there has never been a case where the two were not equivalent.
Force (F) The amount of push or pull exerted on matter. It is important to note that force is a
vector. It has both magnitude and direction.
The basic law of force as stated by Newton: Force causes objects to accelerate.
Fnet = ma
Ffrict = Fnor  = m g (cos  
Fslope = m g sin 
Fg = m g
Fg = G m1 m2 / r2
Kinematics: the study of how things move.
2
Acceleration (a) is a measure of how fast an object speeds up, slows down, or changes
direction. It is measured in meters/sec2. In general it is given by
a=v/t
For circular motion it is called acceleration centripetal and is given by the equations
ac = v 2 / r
ac = 4 2 R / T2
ac = 4 2 R f2
Basic Kinematics equations (note that both Vickie and Dickie require constant acceleration)
Vickie
Dickie
Arthur
vf2 = vo2 + 2 a (d)
d f = ½ a t 2 + v o t + do
a=v/t
Newtonian reference frame: For Newton motion in the universe is all relative. There is no
prime location in the universe. Each frame of reference works as well as all others. The only
thing that really mattered is that for Newton’s equations to work out you needed to be in a nonaccelerated frame of reference.
dobserved = din reference frame + dof reference frame
vobserved = vin reference frame + vof reference frame
Notice: if you are in an inertial (non-accelerated) frame of reference, all of the kinematics and
dynamics equations will work and give working answers in that frame and in any other
observing inertial frame of reference.
3
Momentum and Force: First Universal method of studying the world.
Momentum () which is an inertial measure of an object’s current motion  = mv is the basic
equation for the determination of momentum. Unlike inertia, momentum is a vector. When
working with momentum you must include direction. In a closed system momentum is always
conserved. Total momentum in a closed system remains constant.
 Momentum before =  Momentum after
1o + 2o + . . . = 1f + 2f + . . .
 = mv so
m1v1o + m2v2o + . . . = m1v1f + m2v2f + . . .
Impulse (J) is the measure of the strength of a mechanical interaction between objects. It is
seen in the momentum change in objects that the interaction produces J =  (mv) It is also
expressed in terms of the force of the interaction J = F t
Impulse = Impulse
 (mv) = F  t
It is important to remember that momentum and impulse are vector quantities. You have to
keep track of positive an negative directions.
4
Energy and Force: the second universal way of studying the world
Energy (E) is an alternative way of measuring the mechanical world. Energy comes in many
forms and it is measured by its ability to do work with the assumption being that in a closed
system the total energy will be constant. The forms of energy that we used were
Kinetic energy
KE = ½ m v2
Gravitational Potential energy
PEg = mgh
Conservation of Energy
Energy Before = Energy After
Work (W) is a measure of how much energy has changed form or has entered or left a system.
Mechanically, Work is found in terms of force W = F d. Most commonly when energy leaves
a system it is called Wout and it is usually in terms of heat generated by friction.
W=Fd
Power (P) is the rate at which energy changes form, enters, or leaves a system
It is measured in Watts.
P=E/t
In situations where velocity is constant we can use
P=Fv
The English units for power are Horsepower. One horsepower is equal to 745 Watts.
5
Review problems. Show your work
1. A 240 kg block initially stationary is acted upon by a 400 Newton force find the
following after 6 seconds has passed:
A. acceleration of the block.
B. displacement of the block.
C. velocity of the block.
D. KE of the block.
E. momentum of the block.
F. impulse on the block.
6
2. A 24 kg block moving to the right with a speed of 12 m/s collides with and sticks to
a 16 kg block moving to the left with a speed of 30 m/sec.
A. What is the total kinetic energy of the objects in the system before the collision?
B. What is the net momentum of the blocks in the system before the collision?
C. What is the final speed of the 24 kg after the collision?
D. What is the total kinetic energy of the objects in the system after the collision?
E. What is the net momentum of the objects in the system after the collision?
F. What is the impulse on the 24 kg object due to the collision?
G. What is the impulse on the 16 kg object due to the collision?
7
3. A 1200 kg car is driving up a 4 degree slope at a speed of 30 meters/second.
Answer the following:
A. What is the minimum coefficient of friction needed to maintain this speed?
B. How much power must the engine produce to maintain this speed?
C. What average force must the engine be producing to maintain this speed?
4. A 2 kg rock is thrown straight upwards from the top of a 50 meter cliff with an initial
speed of 20 m/sec. The rock ends up at the ground at the bottom of the cliff, answer
the following:
A. The rock’s initial KE is.
B. The rock’s maximum elevation above the bottom of the cliff is.
C. The rock’s impact velocity at the bottom of the cliff.
8
D. The rock was in the air for how many seconds.
E. When the rock hits the ground, find the magnitude of the impulse it experiences.
F. If the final collision lasts 0.05 seconds, what is the magnitude of the average force
of impact with the ground?
5. A 70 kg coasting bicycle enters a loop the loop with an initial speed of 14 m/sec.
The loop the loop has a radius of 3 meters. Find the following:
A. The initial kinetic energy of the bicycle
B. The initial centripetal acceleration of the bicyclist.
C. The true weight of the bicycle
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D. The apparent weight of the bicycle at the bottom of the loop
E. The speed of the bicycle at the top of the loop
F. The centripetal acceleration of the bicycle at the top of the loop
G. The apparent weight of the bicycle at the top of the loop.
6. A 5 kg block is resting on a horizontal surface. The coefficient of friction between
the block and the surface is 0.7
A. Find the maximum force of friction the surface can exert on the block
B. If you push down on the block with a force of 35 Newtons, now what is the
maximum force the surface can exert?
10
Simple concept problems
1. You are fired upwards with an initial speed of 10 m/sec, what will your acceleration be at the
highest point of your projectile motion?
2. You are fired at an angle of 60 degrees with respect to the horizontal with an initial speed of 10
m/sec. What will your velocity be at the highest point of your projectile motion?
3. Which exerts a greater force on the other. The Earth’s gravity on the Moon or the Moon’s gravity
on the Earth?
4. A bee flies between two bicyclists as shown below. How many Kilometers did the bee travel
before the two bicyclists collide?
5. In which case above, will block A move with a greater acceleration?
6. In the picture above, the blocks have a weight of 100 Newtons each. What is the tension in the
rope connecting them together?
7. In the problem above, Block A is initially moving downward and Block B is initially moving
upwards with a speed of 5 m/sec. How does this effect the tension in the rope?
11
Define in layman’s terms the following:
1. Difference between displacement and distance
2. Difference between velocity and speed
3. Difference between work and power
4. Difference between KE and PEg
5. Describe two ways in which a person can have negative acceleration
6. Give two instances where an object can have zero velocity and
still have acceleration.
7. Fg is
8. Mass can be defined in terms of “inertia” explain
9. Mass can also be defined in terms of “gravity” explain
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Commitment of Understanding
Do not let anybody fool you. AP Physics is tough. It requires commitment, hard work, and
teamwork. Let me make this perfectly clear, AP Physics is a team sport. Look around you
for a moment. Your obligation here is not just to learn, but to teach physics to each and
every individual in this classroom. Holding yourself back holds the whole class back. For
the sake of the rest of us, if you are not willing to do the work and go the distance, drop
this class now!
This is taken from another challenge but it applies here.
I am here to work hard, to work with the best, to teach the best, and to take the AP Exam
with the best of intentions.
Signed: _________________________________________________
13
Chapter 2 Momentum
Ch 7 in Giancoli
One of Newton’s greatest contributions to physics is the concept of momentum
and the argument that in closed systems momentum is always conserved. For
him an object’s mass and velocity share equal importance when discussing the
impact that objects will have on other objects when interactions occur.
Momentum p = m v
Momentum is a vector and we have to include direction.
Lets look at some simple review problems where objects collide and stick
together.
1. A 10 kg cat runs with a speed of 8 m/sec into a stationary 20 kg bale
of hay. Assume that the cat and hay are on a frictionless surface.
Find the resultant speed of the two.
Conservation of momentum
mcat vcat = _______________
vfinal = _______________
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2. A 10 kg cat comes up from behind with a speed of 8 m/sec and runs
into a 20 kg bale of hay. The hay was already going 2 meters per
second in the same direction Assume that the cat and hay are on a
frictionless surface. And they stick upon hitting. Find the resultant
speed of the two.
3. A 10 kg cat runs with a speed of 8 m/sec into a 20 kg bale of hay.
The hay was going in the opposite direction with a speed of 8 meters
per second. Assume that the cat and hay are on a frictionless surface
and they stick when they collide. Find the resultant speed of the
two.
15
From a global view (closed system) momentum is always conserved. From an
individual particle’s point of view that particle either gains or loses momentum
depending on the magnitude of the interaction. He called this change in
momentum Impulse.
Impulse on an object =  (m v)
He also discovered that impulse on an object was equal to the combination of
force and time of interaction.
Impulse on an object = F  t
Lets look at examples of impulse.
4. Return to problem 1 in this section and find the magnitude of the
impulse on the cat.
5. Return to problem 1 in this section and find the magnitude of the
impulse on the bale of hay.
16
6. If the collision in problem 1 lasted one tenth of a second, how much
force did the cat experience?
7. Return to problem 3 in this section and find the impulse on the cat.
8. If the collision lasted one tenth of a second how much force did the
cat experience.
17
Momentum and Impulse apply to all sorts of interactions between objects. An
explosion is another sort of interaction.
9. Assume an explosion deep in space (no mentionable gravity) blows
apart two masses. Mass A is 6 kg and goes upwards with a speed of
10 m/sec. How fast and what direction is the 2 kg Mass B going?
10. What is the impulse on mass B? (Show your work)
11. Which would you rather be hit by: the fastest pitched baseball (40
m/sec) or the fastest served tennis ball? To answer this question,
determine the following quantities …the momentum of each ball the
impulse delivered if the baseball stops dead on impact but the
tennis ball rebounds at nearly the same speed.
Write your answers in the table below.
Quantity
Mass
Velocity
Momentum
Impulse
Baseball
145 grams
Tennis Ball
57 grams
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12. A problem for Americans — because Americans still think non-metric units are fine.
Analyze this …Now, the bears I live with average, the males, eight to twelve hundred
pounds. They're the largest bears in the world …. They've been clocked at 41 [mph]
and they've run a hundred meter dash in 5.85 seconds, which a human on steroids
doesn't even approach. Timothy Treadwell author of Among Grizzlies. The Late Show with David
Letterman. NBC. 20 February 2001.
A. Mr. Treadwell states two different top speeds for the grizzly bear — one
directly and one indirectly. Are they effectively the same? Explain.
B. Pick whichever speed you wish and compute the momentum of a Grizzly bear
(in SI units) using the average mass quoted by Mr. Treadwell.
C. How fast would a 200 lb man have to run to have the same momentum you
calculated in part b? (Do not use a calculator to compute your answer.)
D. How fast would a 2000 lb car have to drive to have the same momentum you
calculated in part b? (Do not use a calculator to compute your answer.)
19
Read this passage about one of the difficulties of interstellar travel.,
the faster we go, the more difficult it is to avoid collisions with small objects and
the more damage such a collision will wreak. Even if we are fortunate enough to
miss all sizable objects, we can scarcely expect to miss the dust and individual
atoms that are scattered throughout space. At two-tenths of the speed of light, dust
and atoms might not do significant damage even in a voyage of 40 years, but the
faster you go, the worse it is — space begins to become abrasive. When you begin
to approach the speed of light, each hydrogen atom becomes a cosmic ray particle
and will fry the crew. (A hydrogen atom or its nucleus striking the ship at nearly
the speed of light is a cosmic ray particle, and there is no difference if the ship
strikes the hydrogen atom or nucleus at nearly the speed of light. As Sancho Panza:
"Whether the stone strikes the pitcher, or the pitcher strikes the stone, it is bad for
the pitcher.") So 60,000 kilometers per second may be the practical speed limit for
space travel.
Isaac Asimov. "Sail On! Sail On!" The Magazine of Fantasy & Science Fiction, Vol. 72
(February 1987):
4. The density of the interstellar medium is about one hydrogen atom per cubic
centimeter. Imagine a 1000 tonne (1 tonne is 1000 kg), 4 by 6 meter, classroomsized interstellar spacecraft traveling at 60,000 km/s on its way to Proxima Centauri
(the nearest solar system to our own (4.0 E13 km away)).
A. How long would it take our hypothetical spacecraft to complete its
hypothetical journey?
impulse–momentum
B. Determine the momentum of our spacecraft.
C. What mass of interstellar medium is swept up during the journey?
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D. What impulse does the interstellar medium deliver to the spacecraft?
E. How does this impulse compare to the momentum of the spacecraft?
work–energy
F. Determine the kinetic energy of our spacecraft.
G. What is the effective drag force of the interstellar medium during the
journey?
H. How much work does the interstellar medium do on the spacecraft?
I. How does this work compare to the kinetic energy of the spacecraft?
21
conceptual
1.In older passenger cars, body panels were attached to a single frame around the
perimeter, making them very rigid. This is known as body-over-frame construction.
In newer cars, different body parts have stress-bearing elements within them and
these parts are then welded to each other. This is known as unitized body
construction. Repairing "unibody" cars after collision is comparatively difficult as
stress (and thus damage) are distributed throughout the different parts. Why then
are cars now built this way
2.To escape from a horrible fire, two people are forced to jump from the third story of a
burning building on to solid concrete. Which person is more likely to sustain serious
injuries: the jumper who comes to an abrupt halt when he lands or the jumper who
bounces after impact?
numerical
1.When hit, the velocity of a 0.145 kg baseball changes from +20 m/s to −20 m/s. What is
the magnitude of the impulse delivered by the bat to the ball?
22
2. A rubber ball of mass 0.025 kg traveling at 4.0 m/s down strikes the floor and bounces
straight up at 2.0 m/s. Find the magnitude of the impulse that the floor gave to the
ball.
3. A model rocket has mass of 1.5 kg. The engine exerts an effective upward thrust of
120 N for 3.2 seconds. (Assume a negligible amount of air resistance while the
rocket is ascending.) Determine …
A. the net force on the rocket
B. the net impulse on the rocket
C. the speed of the rocket when the engine stopped
D. the height of the rocket above the ground when the engine stopped
E. After the engine shuts down, the rocket is still moving upward. What
maximum height above the ground did the rocket reach?
23
Momentum in Two Dimensions
Working with collisions that happen in two dimensions is similar to what we
have done in the past. The major difference is that we must keep track of both
horizontal and vertical aspects of momentum.
1. Imagine that two masses A (10 kg) and B (4 kg) are moving
horizontally with a speed of 2 meters per second. Then, while
they are moving a small explosion pushes then apart. Object A
ends up going upwards with a speed of 2 m/sec. Find the
downward speed of B. (assume positive is upwards and forwards
Initially the vertical momentum of the combined blob is? ___________
In the vertical direction conservation of momentum looks like
Conservation statement
Before
=
After
General equation
0
=
ma va +
Substitutions
0
= _____ _____ + _____ vb
m b vb
vb in the vertical direction is __________________
24
2. In the previous problem the final speeds of the two blobs are actually
greater than 2 m/sec and 5 m/sec. This is because the true speeds are
combinations of horizontal and vertical components.
Magnitude
(Vel A)2 = 22 + 22
Vel A = _________
Direction
Tan  = opp / adj
Tan  = 2/2
 = _________
(Vel B)2 = 22 + 52
Vel A = _________
Direction
Tan  = opp / adj
Tan 
 = _________
25
3. Two masses collide and stick as shown below. Initial conditions
mass A (20 kg) is moving to the right with a speed of
12 m/sec
mass B (8 kg)
is moving upward with a speed of
20 m/sec
Find the final magnitude and direction of the combined mass.
Horizontal
Before = After
Vertical
Before = After
MaVa + MbVb = ( Ma+ Mb)Vbf
MaVa + MbVb = ( Ma+ Mb)Vbf
____ ____ + ____ ____ = ( _____ + _____) Vbf
_____ _____ + _____ _____ = ( _____ + _____) Vbf
Vbf giruz = __________
Vbf vert = __________
Recombine to create the resultant vector
V2vert
+
V2hor
= V2resultant
_________ + _________ = V2resultan
Vresultant = ____________
Tan  = Vver / Vhor = _____ / _____
 = ______
26
3. Two masses collide and stick as shown below.
mass A (20 kg)
mass B (8 kg)
is moving with a speed of
is moving with a speed of
12 m/sec
20 m/sec
They hit and stick, find the magnitude and direction of the velocity of
the combined mass.
Va horiz = __________
Va vert. = ___________
Vb horiz = __________
Vb vert. = ___________
Momentum in the Vertical
Momentum in the Horizontal
27
4. Imagine you are playing billiards. The white ball starts in motion and hits
a stationary black ball. Each ball has a mass of 0.4 kg Find the magnitude
and direction of the velocity of the white ball.
28
Center of Mass
Of you look at a the generic conservation of momentum law it looks something like this
Before
After
ma va + mb vb + mc vc + . . . = ma va + mb vb + mc vc + .
Momentum is (m v) has units of kg m /sec as expressed below
kga m/sa + kgb m/sb + kgc m/sc + . . . = kga m/sa + kgb m/sb + kgc m/sc
If both sides of the equation are multiplied by time, the equation reduces to
kga ma + kgb mb + kgc mc + . . = kga ma + kgb mb + kgc mc + . . .
Let us return to the generic form of the above equation where distance is given by the letter r
ma ra + mb rb + mc rc + . . . = ma ra + mb rb + mc rc + .
If you combined all of the term on the before side the equation reduces to
Mtotal rCofM = ma ra + mb rb + mc rc + . .
The above equation is the statement of center of mass for a system.
29
1. Let us find the location of the center of mass of a barbell composed of two equal
masses (3 kg each) separated by a distance of 2 meters.
A
B
C
2. The center of mass of a system is the point at which the system will balance. Find the balance
point for the following:
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3. A mallet is composed of a cylindrical wooden handle and a wooden head. The
handle is 22 cm long and has a mass of 0.5 kg. The wooden head has a diameter of 8
cm and a mass of 2 kg. Find the center of mass of the mallet.
4. Objects orbit around their center of mass. This explains why there is a high tide on
the side of the earth that is facing away from the moon. Find the center of mass for
the Earth and Moon. Mass moon is 7.26 E22 kg Mass of Earth is 5.97 E 24 kg
31
Find the center of mass in the following situations:
5.
6.
A girl stands on the end of a barge. The bottom of the barge is on a
frictionless surface and the barge is free to move. She starts running toward
32
the dock at the other end of the barge. It takes her 2 seconds to reach the
other end of the barge. Answer the following:
8. What is the location of the original center of mass of the system?
9. How fast is she moving relative to the barge?
33
10. How fast is she moving relative to the dock that was at the end of the
barge?
11. How far away from the dock will she be when she reaches the end of the
barge?
12. When she stops at the end of the barge, what happens to the barge?
34
Chapter 3 Springs
Giancoli Ch 6
In this section we will be investigating the nature of springs. Pick up a sample spring
from the front of the classroom and hang it from a fixed horizontal rod. Place a hanger at
the end of the spring and measure the distance to the ground (or reference point)
Fg
Do
Df
Disp (X)
1 Newton
2 Newtons
3 Newtons
4 Newtons
5 Newtons
6 Newtons
Graph the line and determine its slope.
35
The relationship between spring restorative force and its stretch is given by Hook’s Law.
Fsp = - k x
“k” is called the spring constant and it is a measure of the stiffness of the spring. Notice
that Hook’s law is stated as a negative. This is because the direction a spring acts in
always in opposition to the displacement of the spring. Pull on a spring and it pulls back.
Push on a spring and it springs back.
Lets look at several places where springs are
used.
1. Explain how you would find the spring constant of a short length of surgical tubing.
2. Find the spring constant of a piece of surgical tubing.
Data:
Eq:
Ans:
3. Find the spring constant of twice the length of the same tubing.
Data:
Eq:
Ans:
36
4. A retractable ball point pen has a spring inside of it. Find the spring constant of this
spring.
Data:
Eq:
Ans:
5. A car has four springs, one for each wheel. They are used to cushion the car’s ride.
When a 100 kg person sits down on inside a car the car settles down 1 cm.
37
Energy stored in a spring
Energy can be stored in springs. The stored energy in a spring is given by the following
equation.
PEsp = ½ k x2
Imagine you have a spring that has a spring constant of 500 n/meter is being stretched
one cm at a time. Graph the energy stored in the spring.
Notice that most of the energy stored in a spring is stored in the last part of its stretch.
38
Springs and energy problems.
1. A 10 gram pellet is fired from a pellet gun with a speed of 50 m/sec. The gun’s energy
comes from a spring that is stretched 10 cm. What is the spring constant of the
spring?
2. A 2 kg block is pushed back against a spring. The spring compresses 20 cm. If the
spring constant is 300 n/m, how fast will the block be going when it is released?
39
Given the picture below, find the following for a 4 kg block at the top of a 10 meter high
hill:
3. The speed of the block at the bottom of the hill.
Equation
Substitutions
Answer
4. The magnitude of the spring constant needed to stop the block in a distance of 1
meter.
Equation
Substitutions
Answer
5. The acceleration of the block when the spring has compressed 0.25 meters.
Equation
Substitutions
Answer
6. The acceleration of the block when the spring has compressed 0.5 meters.
Equation
Substitutions
Answer
40
Continued from previous page
7. The acceleration of the block when the spring has compressed 0.75 meters.
Equation
Substitutions
Answer
8. The acceleration of the block at the moment the block has stopped moving into the
spring.
Equation
Substitutions
Answer
9. How fast is the block going when the spring is compressed by 0.5 meters?
Equation
Substitutions
Answer
10. How much work has been done on the block by the time it has compressed the spring
by 0.5 meters?
Equation
Substitutions
Answer
41
11. A spring with a spring coefficient of 100 is compressed by a distance of 0.5 meters and placed
between two blocks. Each block has a mass of 1 kg. When the blocks are released the spring
decompresses. How fast are the blocks going afterwards?
Conservation of Energy
Conservation of Momentum
12. A spring with a spring coefficient of 100 is compressed by a distance of 0.5 meters and placed
between two blocks. One of the blocks (A) has a mass of 120 grams, and the other block (B) has a
mass of 200 grams. When the blocks are released the spring decompresses. How fast are the blocks
going afterwards?
Conservation of Energy
Conservation of Momentum
42
13. A 10 kg block is pushed with a force of 50 Newtons into a spring that has a
spring constant of 62.5 and then released. The block slides away and runs
into another block which has a mass of 8 kg. The two blocks stick together.
Find the following:
A. The distance the spring compresses.
B. The initial acceleration of the spring when the block is released
C. The total energy stored in the compressed spring system
D. The velocity of the block upon leaving the spring
E. The velocity of the combined blocks after the collision
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Problem 13 continued
F. The impulse experienced by the 10 kg block during the collision
G. If the collision lasts 0.2 seconds what average force did the 10 kg block
experience during the collision?
44
14. A 10 kg block is moving around a loop the loop. At the top it has a speed of
8 m/sec. The loop has a radius of 5 meters.
Find the following:
A. The centripetal acceleration at
the top of the loop.
B. The apparent weight of the block at the top (Froad)
C. The velocity at the bottom of the loop.
D. The coefficient of friction needed to stop the block in 5 meters distance.
E. The maximum acceleration of the block while in contact with the spring if
the spring has a “k” of 250.
45
15. Two masses (mass A is 12 kg, mass B is 3 kg) are stationary on a frictionless
flat surface. Between them is a spring that has been compressed 10
centimeters. The spring has a spring constant of 200. The spring expands
sending the two blocks in opposite directions. Find the velocity of each.
46
Chapter 4 Density
Terms we will be using
Terms
Sub-units
M-K-S units
Mass
m = kg
Surface area
A = m2
Volume
V = m3
Surface density
(mass/area)
 = kg/m2
Volumetric density
(mass/volume)

Pressure
(Force/Area)
Pascal = N/m2
Pressure
(Energy/volume)
Joules/meter3
3
Volume
One of the most important concepts to understand in if we are going to be living in and
studying in a three dimensional world is volume. You have been introduced to this
concept numerous times in math classes and introductory science classes. We will quickly
review the basics. In the M-K-S (Meter, Kilogram, Second) version of the metric system
the unit of distance is the meter. Volume starts with this basic unit cubed. This is fairly
large so we often break it down into smaller units. The cubic decimeter – often called the
liter – is a common one.
47
Conversions
Cubic meters
( m3 )
Cubic Decimeters
( liters )
Cubic Centimeters
(CCs or milliliters)
Cubic Millimeters
1 meter cubed
____________
____________
____________
____________
1 liter
____________
____________
____________
____________
1 cc
____________
____________
____________
____________
1 cubic millimeter
Lets do some quick measurements of volume. How would you find the volume of each of
the following and what equipment would you need?
This classroom
Explain the method
Its approximate volume is _________________ meters3 ± __________m3
48
Finding the volume of your lungs
Explain the method
Its approximate volume is _________________ meters3 ± __________m3
An irregular granite rock
Explain the method
Its approximate volume is _________________ meters3 ± __________m3
49
Real Life Measurements
Length is a measure of distance. We measure length in meters. Your ability to
approximate length is important.
Guess
Actual length
The length of your pencil
______________
________________
Your height
______________
________________
East West
_______________
________________
North-South
_______________
________________
Ground-Ceiling
_______________
________________
The dimensions of this room
Volume is the amount of space that is occupied or enclosed. We measure volume in cubic
meters. The liter is 1/1000 of a cubic meter. It is equal to a cube that is 10 cm on a side.
1. How many gallons are in one cubic meter?
Rough guess ____________________
We will use ratios
50
2. What is the volume of a basketball?
Rough guess ____________________
A standard basketball has a radius of 12 cm
The volume of a sphere is 4/3  r3
= ____________________ meters 3
3. What is the volume of the cubic zirconium sphere that your teacher is holding?
Rough guess ____________________
A standard basketball has a radius of 12 cm
3
= ____________________ meters 3
4. What is the approximate volume of this room?
Rough Guess ____________________
(East West) x ( North South) x (height)
__________ x _____________ x ____________ = __________________ m3
51
Density
Density is a measure of how much material is found in a given space. We will deal with
two forms of density; surface density and volume density.
2
Below are pictures of 1 kg disks placed evenly on table tops. The first table is 16 meters
by 16 meters. Find the density of the material in each figure.
Figure 1
Density
______________ Kg/m2
Figure 2
______________ Kg/m2
Figure 3
______________ Kg/m2
52
Volumetric density
3
Below are pictures of 64 one kg spheres placed evenly inside of cubes. The first cube is 16
meters on a side. The second is 8 meters on a side, The third is 4 meters on a side. Find
the density of the material in each figure.
Figure 1
Figure 2
Figure 3
Density ____________ Kg/m3
____________ Kg/m3
___________ Kg/m3
Density ____________ Kg/liter
____________ Kg/liter
____________ Kg/liter
___________ Kg/cc
__________ Kg/cc
Density ____________ Kg/cc
_
53
Below are the densities of several common materials. If you were to have a cube of each.
How big would it have to be to have a mass of 1 kg? (remember that on the Earth one
kilogram will have a weight of approximately 2.2 lbs)
Density
Volume of 1 kg
Meters cubed
Volume of 1 kg
Liters
Volume of 1 kg
CCs
Air
1.2 kg/m3
______________
_______________
_________________
Water
1000 kg/m3
______________
_______________
_________________
Aluminum 2.7 gr/cm3
______________
_______________
_________________
Granite
2.75 gr/cm3
______________
_______________
_________________
8 gr/cm3
______________
_______________
_________________
Mercury
13.5 gr/cm3
______________
_______________
_________________
Uranium
19.1 gr/cm3
______________
_______________
_________________
Gold
19.3 gr/cm3
______________
_______________
_________________
Steel
Granite
Mercury
Gold
54
Volumetric Density
 = Mass / Volume
Units are Kg/m3
1. What is the approximate mass of the air in this room?
Rough guess ______________________
Equation
Substitutions
Answer _________________
2. What is the mass of a block of gold
that is the size of a Xerox paper box?
Rough guess ______________________
Equation
Substitutions
Answer _________________
55
3. What is the density of the block of foam
that your teacher is holding?
Rough guess ______________________
Equation
Substitutions
Answer __________________
4. What is the density of the cubic zirconium
ball that your teacher is holding?
Rough guess ______________________
Equation
Substitutions
Answer __________________
5. Find the following information about the rock sitting on Mr. Stimson’s desk
Volume
_______________________
Mass
_______________________
Density
_______________________
56
Chapter 5: Pressure
Pressure is a measure of how much force is acting at a representative surface.
Pressure
 =
Pascal
=
PSI = 1 lb/in2
Force / Area
1 Newton / meter2
1 pascal = 0.000145037738 PSI
We are going to do a lot of interesting stuff with pressure. However, before we
do, we need to get some practice dealing with pressure. Lets look at some
situations.
1. At sea level the pressure of the atmosphere acting downward on a one square
meter section of pavement is 101325 Pascals ( Newtons / square meter )
a. How much total force is this?
_______________________
b. How many kilograms of air are there above the 1 square meter
section of pavement?
______________________
2. Your sister is of average size, 65 kg. Normally she wears tennis shoes. Each shoe
has a bottom surface area of 116 cm2. What average pressure does she exert on
the ground when she stands on one foot?
Your sister’s weight _______________ Newtons
The pressure is she exerting on the ground while
standing on one foot
________________ Pascals
________________ PSI
57
Your sister buys a cool pair of Stiletto high heels. The stiletto heels
get very narrow where they hit the ground. The bottom surface is
only 0.3 cm2. Your sister is of average size, 65 kg. When she walks
she places all of her weight on the heel. Answer the following:
Your sister’s weight _______________ Newtons
The pressure is she exerting on the ground while
standing on one foot
________________ Pascals
________________ PSI
BTW, when Disneyland first opened to the public in 1955. Stiletto heels were in fashion. The weather
was hot. And, the asphalt was fresh. Women were leaving deep marks in the asphalt. Some had their
shoes get stuck in the asphalt.
58
3. The neat thing about liquids is that if there is not a substantial elevation
difference. The pressure at one location is the same as the pressure at all
other locations. This is how a hydraulic lift works. Below is a picture of a
system with a 10 kg mass on top of a small diameter piston. Pressure
caused by this mass translates into a substantial lifting force acting on the
larger piston.
a. What is the weight of
the 10 kg mass?
___________________
b. What is the pressure
created by the mass
___________________
c. What is the upwards
force acting on the large
piston?
___________________
d. How many kg of material
must be on the larger
piston to keep it from
moving upwards?
___________________
59
4. Remember when we determined the mass of my car by pushing it and
measuring the car’s acceleration? Well, it turns out that you can
determine the car’s mass by measuring
the contact area between the wheels and
the ground. This can be done if you
know the air pressure in the tires. Lets
try this out. What we will do is to lift one
side of the car and place carbon paper
underneath the wheels. Then we will
lower the car down.
a. What is the contact area of the front wheel?
________________
b. What is the contact area of the back wheel?
________________
c. What is the pressure in the tires?
________________
d. What is the mass of the car?
___________________
60
Energy and Fluids
In the past we studied energy as it related to particles. There was kinetic
energy (KE) which is a measure of the energy in a particle’s motion and
gravitational potential energy PEg which is a measure of the energy stored in
a particle’s elevation. The equations for these two forms of energy are
PEg = m g h
KE = 1/2 m v2
With fluids it is often impossible to measure the total energy in a fluids
system because one does not know exactly how many particles there are in
the system. For example, how do we measure the total energy in the
Mississippi River system? Fortunately, we can do a lot of energy stuff , even if
we don’t know about every drop in the stream. What we do need to know is
the energy density of the fluid.
Energy of
particles
Energy density
of fluids
PEg = m g h
PEg/vol =  g h
KE = 1/2 m v2
KE/vol = 1/2
2
Let us start by looking at PEg/vol (Pressure) Below are three glass tubes.
Each has a certain amount of water in it. The pressure of the water at the
h
Pressure due to water = ____________,
_____________,
_____________
Absolute pressure = ____________,
_____________,
_____________
61
5. Lets combine what we have just learned with some older material.
Imagine that your neighbor has a tall water tank (shown below) She is
bored and decides to shoot it up with her assault rifle. She fires five holes
into the tank at different heights. Find the following for each hole: the
pressure (Energy/vol), the velocity at which the water leaves (KE/vol),
and the distance away from the tank at which each stream of water lands.
We do not need to find absolute pressure – can you explain why?
Hole height
H20Pressure
Velocity
Distance
6m
_________________
__________________
_________________
5m
_________________
__________________
_________________
4m
_________________
__________________
_________________
3m
_________________
__________________
_________________
2m
_________________
__________________
_________________
1m
_________________
__________________
_________________
62
6. At the surface of the ocean you are experiencing a pressure of 1
atmosphere. As you go deeper into the ocean, the weight of the water
above you increases. Consequently you feel greater pressure. Water has
a density of 1000 kg/m3. How far down do you have to go for the total
pressure to be equal to 2 atmospheres?
Hint: one atmosphere is 101325 Pascals
_________________ meters
_________________ feet
7. The Marianna Trench is the deepest
point in the pacific ocean. It is 10,924
meters deep (35,840 feet). What is the
pressure at the bottom of the trench?
a. __________________ Pascals
b. __________________ PSI
.
8.
63
8. The street water pressure to the average house is 280,000 pascals (40
psi.) above air pressure The pressure gets less as you go from the first to
the second floor. If it is 3 meters from the ground to the shower on the
second floor. Find the following
a. The pressure of the water at
the ground level in Pascals
___________________
b. The pressure of the water at
the shower head in Pascals
___________________
c. The maximum height that
water can be pumped to
___________________
64
Above is a sphere with particles bouncing around inside. Pressure on the sphere is related
to the outwards force of the collisions of the myriads of molecules that are inside of the
sphere. But, moving molecules have KE. This implies that there must be a relationship
between pressure and energy. Below we will be answering these questions in a not so
scientific but very effective way.
Unit matching – An effective tool in science
In college it was common practice among students, when you were up against a wall and
could not solve a problem, to reverse engineer through units. The procedure went like
this. Look at the units of the answer that you are trying to find. Next, look at the units of
what you are given. Then manipulate the units so that they end up in the form of those
found in the answer. This approach bothered me. For, while it consistently worked, it
demonstrated no real knowledge of the underlying physics. This being said, I have found
unit matching a valuable problem solving method. The following is a good example. We
have been studying pressure. So, is pressure more like energy or more like force? Lets
look at the units and how they came about.
Pressure
Pascals
Constituent Units
Kg / ( m sec2 )
We will derive them on the next page using Newton’s second law and then see how they
are related to energy.
65
Pressure in terms of Force
We know that
Force = m
a
Newtons = kg (m/sec2)
We can view pressure in terms of force
Pressure =
Force / Area
Pascals = (Newtons) / ( m2 )
Pascals =
Pascals =
(Kg m/sec2) / m2
Kg / ( m sec2 )
From this point of view Pressure acts as a force if we are looking at its effects on a
surface..
66
Pressure in terms of Energy
We know that
PEg
= m
g
h
So,
Joules are kg m/sec2 m
Or
Joules
are
Kg m2 /sec2
Below we shall see how Force/Area is the unit equivalent of Energy/Vol
Force / Area
Energy / Volume
Newtons / meter2
Kg (m/ sec2) / ( m2 )
(Joules) / ( m3 )
=
(Kg m2/sec2) / ( m3 )
Kg / ( m sec2 ) =
Kg / ( m sec2 )
Basically,
Pressure is Force if we are looking at its effect on surfaces.
But,
Pressure is Energy if we are looking at its effect on volumes.
67
Lets see how we can use the pressure – energy relationship in problem solving.
To the left is a picture of a high pressure gas cylinder. We decide to fill it up
with air. How much energy is stored in the cylinder?
Volume
English
Metric
2640 ci
0.0433 m3
Max Pressure
3600 psi
2.48 E 7 J/m3
Remember that the pressure is a measure of Energy per unit volume.
Pressure = Energy / volume
Energy
or
= (Pressure) x ( volume)
Energy = ____________ J/m3 _______________ m3
Energy = ______________ Joules
This sounds like a lot of energy but it compares to only 1/8th of a cup of gasoline.
Inside closed systems, Energy is conserved. If we slowly empty the cylinder into a large
mylar balloon how much volume would it fill when the pressure is finally reduced to one
atmosphere.
Before (in cylinder) = After (in balloon)
___________ = ( Pressure ) ( Volume )
___________ = ___________ J/m3 ( Volume )
Volume = ________________ m3
this would fill a spherical balloon with a radius of ______________ meters.
68
Lets talk about water
An Incompressible fluid
In 1738 Daniel Bernoulli published a conservation of energy law for fluids. It came to be
known as Bernoulli’s principle. Basically he figured that fluids held energy in three
different ways: Gravitational, Kinetic, and Pressure.
PEg
=
m g h
for a static material
 g h
per unit volume
3
KE
=
1/2 m v2 static material
=
1/2  v2
per unit volume
3
Pressure=
for all fluids
Below is a simplified version of Bernoulli’s principle. Essentially it is the law of
conservation of energy applied to liquids. It works any incompressible fluid, as long as
there is little or no friction or drag (Wout).
Before = After
PEgo + KEo + Po = PEgf + KEf + Pf
69
1. Lets start with a very simple problem. We will be looking at a one-liter chunk of
water. We will see what happens to it as it flows through pipes of varying sizes.
Initially the water is flowing through pipes at a rate of 1 liter per second (1 kg per
second). The nominal diameter of the pipes is 4 cm (0.04 meters) at a pressure of
400,000 N/m2. Answer the following:
What is the cross-sectional area
of the pipe?
____________________
Remember: 1 liter is one thousandth
of a cubic meter.
What is the linear velocity of
the water?
Hint: Vol = Vol
(Area) (distance)
(Area) (x) = (1E-3 m3)
____________________
What is the energy per unit volume
For the water in the system?
____________________
Hint: P + KE
= Energy
70
The one liter slug of fluid encounters a constriction in the pipe. The pipe narrows to a
diameter of 0.02 meters.
What is the cross-sectional area
of the constriction?
____________________
What is the linear velocity of
the water?
Hint: Vol in big pipe = Vol in small pipe
____________________
What is the new pressure at the
constriction point?
____________________
Hint: Po + KEo
= Pf + KEf
71
72
2. On the previous page the water pressure in the pipe determined the height of the
water in the stand pipes. The higher the water level, the greater the pressure. How
high would the water be in each pipe
Po = PEgo
Water height above
the larger pipe
_________________ =

g
h
_________________ = _________________ ______ h
ho = _____________________________
Pf = PEgf
Water height above
the narrower pipe
_________________ =

g
h
_________________ = _________________ ______ h
ho = ______________________________
If a hole were to be drilled into the top portion of the pipe, water would shoot out and
attain a maximum height calculated in this problem.
For an applet showing what tube constriction does to the flow of fluids go to
http://library.thinkquest.org/27948/bernoulli.html
73
74
The following problem was taken from the free response section of the 2008 AP Exam.
Show your work on the back of the previous page.
An underground pipe carries water of density
1000 kg m3 to a fountain at ground level, as
shown above.
At point A, 0.50 m below ground level, the
pipe has a cross-sectional area of 1.0 -4 m2.
At ground level, the pipe has a cross-sectional
area of 0.50 E-4 m2.
The water leaves the pipe at point B at a
speed of 8.2 m/s.
(a)
Calculate the speed of the water in the pipe at point A.
_______________________
(b)
Calculate the absolute water pressure in the pipe at point A.
_______________________
(c)
Calculate the maximum height above the ground that the water reaches. Assume
air resistance is negligible and that the water goes straight up.
_______________________
(d)
Calculate the horizontal distance from the pipe that is reached by water exiting
the pipe at 60 degree from the horizontal. Assume air resistance is negligible.
________________________
75
76
3. (From 2008 AP Exam) A fountain with an opening of radius 0.015 m shoots a
stream of water vertically from ground level at 6.0 m/s. The density of water is 1000
kg m3 Show your work on the back side of the previous page.
(a)
Calculate the volume rate of flow of water.
________________________
(b)
The fountain is fed by a pipe that at one point has a radius of
0.025 m and is 2.5 m below the fountain’s opening. Calculate the
absolute pressure in the pipe at this point.
________________________
(c)
The fountain owner wants to launch the water 4.0 m into the air
with the same volume flow rate. A nozzle can be attached to change
the size of the opening. Calculate the radius needed on this new
nozzle.
________________________
77
78
4. (From 2007 AP Exam) The large container shown in the cross section above is filled
with a liquid of density 1.1E3 kg/m3. A small hole of area 2.5E-6 m2 is opened in the
side of the container a distance h above the liquid surface, which allows a stream of
liquid to flow through the hole and into a beaker placed to the right of the container.
The amount of liquid collected in the beaker in 2.0 minutes is 7.2E-4 m3.
(a)
Calculate the volume rate of flow of liquid from the hole in m3/sec.
____________________
(b)
Calculate the speed of the liquid as it exits the hole.
____________________
(c)
Calculate the height h of liquid needed above the hole to cause the speed
you determined in part (b).
____________________
(d)
Suppose that there is now less liquid in the container so that the height h is
reduced to h/2. In relation to the beaker, where will the liquid hit the
tabletop?
____ Left of the beaker ___ Right of the beaker ___ Into the beaker
Justify you answer.
79
Physics of flight
Bernoulli’’s principle is directly involved in aviation. It is central to the physics of wings
and propellers. The exact calculations are very difficult and require very sophisticated
mathematics. However, a first approximation is relatively easy to make and surprisingly
accurate.
We start by assuming that air that makes up our atmosphere acts like an incompressible
liquid when something slices through it. Imagine our atmosphere sort of like a giant jello
mold. If a knife blade were to be passed horizontally through the jello atmosphere, jello
will be temporarily displaced by the blade. It cannot go downward because the ground is
below and we are assuming that air will not compress. All of the air above gets pushed
upward only to fall back down on top of the same lower block of air after the blade goes
by.
Notice how each vertical section of air moves up as the blade passes by.
The picture above shows what is happening from the point of view of a person who is
stationary with the atmosphere. From the blade’s frame of reference, the blade is
stationary and the blocks of air are moving from the left side of the page to the right side.
80
Below is a picture of a typical airfoil. Notice that the distance from front to back over the
top of the airfoil is greater than the distance below (y > x). . Because both blocks of air
will come back together after the foil has moved past, the top block of air must have a
greater apparent surface speed (Vtop > Vbottom) than the block of air on the bottom.
The fact that Vabove is greater than Vbelow has great consequences on the foil. Remember
that the conservation laws of physics must always be obeyed. This includes Bernoulli’s
principle. Let us look at a representative block of air that will pass underneath the airfoil.
It has both KE and Pressure. Because the speed of the block remains the same below the
foil the air pressure remains the same as well.
Below the airfoil
KEo + Po = KEbottom + Pbottom
If we look at a block of air passing over the top of the foil we see that the KE increases over
the airfoil. There is no free lunch in physics. For energy to be conserved the gain in KE
must be balanced by an equal loss in pressure.
Above the airfoil
KEo + Po = KEabove + Pabove
Notice that there is less downward pressure on the wing from above than there is upward
pressure from below.
81
Air pressure acts on all surfaces of the foil. Because there is less pressure acting down on
the top surface of the foil than acting upward on the bottom of the foil, there is a net
upward push.
82
Sample problem: Imagine you own a private airplane. The physical specifications of the plane are given
below.
Wingspan = 30.0 feet (9.2 m)
Wing area: 160 sq ft (15.14 m2)
Distance X = 1.65 m
Distance Y = 2.145 m
Empty weight: 1201 lb (544 kg)
Loaded weight: 2150 lb (975 kg)
There is a minimum speed that you must have for the plane to lift off from the ground.
On the next pages we are going to do the physics necessary in order to calculate the
minimum takeoff speed of your airplane. Here we go.
We start with that air particles cross the top surface of the wing in exactly the
same amount of time as it takes for air molecules to pass underneath the
bottom surface of the wing.
(A)
Remember that the geometry of the wing determines how fast the air must
move on the top and bottom surfaces of the wing. Where the air moves more
quickly the kinetic energy of the air is greater and the potential energy (air
pressure) is less. Let us start by determining the relative air speeds.
Remember that air traveling over the top of the wing must reach the trailing
(back) edge of the wing at the same time as air that moves under the wing.
Since D = v t
Time above
Dabove / Vabove
=
=
Time below
Dbelow / Vbelow
Solve for Vabove in
Terms of Vbelow
V above = (
) Vbelow
83
Write Bernoulli’s equation for air flowing over the top of the wing vs. air flowing
underneath the wing.
Energy Below Wing = Energy Above Wing
Pbelow + KEx
(B)
Pabove
P (N/m2)
(D)
Pabove
+
KEy
Wings provide lift because the bottom surfaces have a larger air pressure
pushing them upward than the top surfaces feel pushing downwards. We can
find this difference in pressure by bringing all the pressures to the left side of
the equation and the KE’s to the right.
Pbelow -
(C)
=
=
KEy
=
1/2  vabove2 - 1/2  vbelow2
The density of dry air at sea level is
- KEx
3
 P (N/m2)
=
1/2 (1.2) vabove2 - 1/2 (1.2) vbelow2
 P (N/m2)
=
________ ( vabove2 - vbelow2 )
We can now find how much lift you get per square meter of wing in terms of
the plane’s speed. Substituting for Vabove we get
 P (N/m2) =
_____ ((____ vx) 2 - vx2 ))
 P (N/m2) =
_____ (vx) 2 (___ - 1 )
 P (N/m2) =
________ ( vx2 )
84
(E)
 P is the pressure difference between the top and bottom of the wing.
Since
pressure is measured in force per unit area, since we know the total area
(bottom surface) of the wing we can find the amount of lift at any speed.
 P = Force / surface area
________
= Forcelift / __________
Forcelift = ___________
(F)
When fully loaded the plane has a mass of 975 kg. What is its weight in
Newtons?
Fg =
m
g
Fg = _________ _________
Fg = _________
(G)
Calculate the minimum take off velocity for a fully loaded plane.
(should be somewhere near 40 m/sec)
85
Sample car problem:
Below is the Stimson mobile. It is an airfoil of sorts. Because of this it appears to get
lighter at higher speeds (not necessarily a good thing) When the car is moving at 40
m/sec, how much lighter will it be? And, should we worry?
86
1. Another plane problem. The MIT human powered plane the Daedalus weighs only
69 lbs. Fully loaded it is about 117 kg. In 1988 it flew the 199 km from the Iraklion
Air Force Base on Crete in the Mediterranean Sea, to the island of Santorini. What
minimum speed does it need to be able to fly?
The wing has a span of 34 meters and a surface area of 31 m2
The average underside wing width is 1 meter
The average topside width is 1.35 meters
Answer ________________m/sec _________________ mph
Show you work on the back side of this page
87
2. Below is a picture of a spin ball which is moving toward the right side of the page.
Using you knowledge of the Bernoulli principle, draw the direction of the net force
acting on the ball. Explain your answer.
.
88
3. You have probably been into stores during the summer where there is a beach ball
floating up above a fan. This is usually done as an advertisement for fans. Not only
is it an effective advertisement for fans but it is also a great demonstration of
physics. Explain.
Notice that the ball still floats even when
the fan is at an angle
89
We need to pause for a moment. While it is true that airfoils use Bernoulli’s principle
for a substantial portion of their lift, it is also true that the attack angle of a wing has an
enormous impact on lift. Angle of attack works on the concept of Impulse (remember
conservation of momentum?)
.
Diverting the mass of the air flow downward places an upward force on the wing. In
the next section three we shall review the one dimensional studies of momentum and
extend them to two and three dimensions.
For a more complete explanation of the physics of flight go to
http://www.amasci.com/wing/airfoil.html
http://www.av8n.com/how/htm/airfoils.html
90
Buoyant Forces
Have you ever wondered why some objects float and other objects sink? For that matter,
have you ever wondered big ships float when they are made out of steel, a material that
sinks in water? The math for determining is very simple understanding.
Objects sink into water until the mass of the water they displace is equal to the
object’s own mass. It is all a matter of density (sometimes known as specific gravity. In
the metric system the specific gravity of water is measured in one of three ways
1000 kg/m3
or
1 kg /decimeter3
1 kg / liter
or
1 gram per cm3
1 gr/ml
Below are objects that will be placed into water.
1. Balsa wood has a density of 110 kg/m3. A block of balsa (one meter on a side) is
placed in water. How deep will it sink in the water?
2. Oak has a density of 770 kg/m3. A block of oak (one meter on a side) is placed in
water. How deep will it sink in the water?
91
3. Steel has a density of 7800 kg/m3. A block of steel (one meter on a side) is placed in
water. How deep will it sink in the water?
4. You have a sheet of steel that is 0.002 meters thick. The steel is folded into the
shape of a cube with the top side missing. The dimensions of the cube are one
meter by one meter by one meter.
a. What will the mass of the hollow cube be?
b. How deep will the cube sink into the water?
92
Buoyant force = the weight of the displaced fluid
5. Let us revisit the one cubic meter block of balsa wood from problem 1.
a. Calculate the buoyant force on the block.
Picture
Equations and work
b. If the block is pulled completely under water by a rope force the rope must
exert.
Picture
Equations and work
93
6. Let us revisit the one cubic meter block of oak from problem 2
a. Calculate the buoyant force on the block.
Picture
Equations and work
b. If the block is pulled completely under water by a rope calculate the force the
rope must exert.
Picture
Equations and work
94
7. Let us revisit the one cubic meter block of steel from problem 3
a. Calculate the buoyant force on the block.
Picture
Equations and work
b. Calculate the net force on the block.
Picture
Equations and work
95