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\begin{center}
{\bf Supplimentary Notes II for Newtonian Mechanics}
\end{center}

In the first four weeks of the course we discussed Newton's
laws of motion, in which the approach taken is one of forces and 
motion.  We discussed three experiments that lead to 
the relationship between force and motion: $\vec{F}_{Net} = m 
\vec{a}$, and the "symmetry" of interactions: $\vec{F}_{12} =
- \vec{F}_{21}$. The general method of analyzing systems is
to first determine all the forces on each object in the system,
then use $\vec{a} = \vec{F}_{Net}/m$ on each object to find the motion of 
that particular
object.  The "physics" of the interactions between particles is
described by the {\it force} that one particle exerts on another,
which can usually be expressed in a simple way.
As objects move, the forces change in time, and the resulting
motion can be complicated.  The benefit of using
Newton's approach is that the equations for the accelerations
$\vec{a} \equiv d^2\vec{r}/dt^2$ of objects are much simplier than the 
equations for the position functions $\vec{r}(t)$.  We are lucky
that nature turned out to be this simple (at least for the realm of classical
mechanics).

The "force-acceleration" approach is helpful in understanding interactions
in "classical" mechanics and electromagnetism.  However, there are other
approaches for analyzing interacting particles.  In the next 3 weeks, we
consider another approach which deals with the energy and momentum of
objects.  We will discuss how energy and momentum are related
to an objects mass and velocity, and at the same time how the laws
of physics can be expressed in terms of energy and momentum.  The
"force-acceleration" and the "energy-momentum" formalisms contain
the same physics.  Sometimes one approach is better than the other
in analyzing a particular system.  

\bigskip

\noindent{\bf An Example to motivate Mechanical Energy}

\bigskip

Consider dropping an object of mass $m$ from rest from a height $h$
above the earth's surface.  Let $h$ be small compared to the
earth's radius so we can approximate the gravitational
force on the object as constant, $m\vec{g}$.  The object will
fall straight down to the ground.  Let $y$ be the objects height
above the surface (+ being up) and $v$ be the objects speed.  As the object
falls, it's speed increases and it's height decreases.  That is,
$y$ decreases while $v$ increases.  Question: is there any combination
of the quantities $y$ and $v$ that remain constant?

We can answer this question, since we know how speed $v$ will changes 
with height $y$.  If the gravitational force doesn't change, the 
objects acceleration is constant, and is equal to $-g$.  The minus
sign is because we are choosing the up direction and positive.
Let $y_i$ and $v_i$ be the height and speed at time $t_i$, and let
$y_f$ and $v_f$ be the height and speed at time $t_f$.  From the
formula for constant acceleration:

\begin{equation}
  v_f^2 = v_i^2 + 2 (-g)(y_f - y_i)
\end{equation}

\noindent This equation can be written as:

\begin{equation}
  {{v_f^2} \over 2} + gy_f = {{v_i^2} \over 2} + gy_i  
\end{equation}

Wow!  This is a very nice result.  Since $t_i$ and $t_f$ could
be any two times, the quantity $v^2/2 + gy$ does not change
as the object falls to the earth!  As the object falls, 
$y$ decreases and $v$ increases, but the combination
$v^2/2 + gy$ does not change.  As we shall see, it is convenient
to multiply this expression by the mass of the object $m$.  Since
$m$ also does not change during the fall, we have

\begin{equation}
   m {{v^2} \over 2} + mgy = constant
\end{equation}

\noindent while an object falls straight down to earth.  In physics
when a quantity remains constant in time, we say that the quantity
is {\bf conserved}.  The quantity that is conserved in this case is
the mechanical {\bf energy}.  The first term, $mv^2/2$, depends
on the particle's motion and is called the kinetic energy.  The
second term, $mgy$, depends on the particle's position and is
called the potential energy.  In these terms, we can say that
the sum of the object's kinetic energy plus its potential
energy remains constant during the fall.  

Can the energy considerations for this special case of an object 
falling straight down be generalized to other types of motion?  
Yes it can.  The situation is a little more complicated in
two and three dimensions since {\bf the direction of the net force is
not necessarily in the same direction as the objects motion ($\vec{v}$)}. 
The analysis is facilitated by using a mathematical operation with vectors:
the "dot" (or "scalar") product.  Let's summarize this vector operation, then
see how it helps describe the physics.

\bigskip

\noindent{\bf Vector Scalar Product}

\vspace{4mm}
The vector scalar product is an operation between two vectors that
produces a scalar.  Let $\vec{A}$ and $\vec{B}$ be two vectors.  We
denote the scalar product as $\vec{A} \cdot \vec{B}$.
There are a number of ways to derive a scalar quantity from two vectors.
One can use $|\vec{A}|$ and $|\vec{B}|$.  However, if we define the
scalar product as the product of the magnitudes, then the angle does not
play a role.  If we call $\theta$ the angle between the vectors, then
we could use cos($\theta$) or sin($\theta$) in our definition.  If we
want $\vec{A} \cdot \vec{B}$ to equal $\vec{B} \cdot \vec{A}$ then we
need a trig function that is symmetric in $\theta$.  Since cos(-$\theta$)
equals cos($\theta$) it is the best choice.  The scalar product is 
defined as

\begin{equation}
  \vec{A} \cdot \vec{B} \equiv |\vec{A}| |\vec{B}| cos(\theta)
\end{equation}

\noindent where $\theta$ is the angle between the two vectors.

The scalar product can be negative, positive, or zero.  It is the 
magnitude of $\vec{A}$ times the component of $\vec{B}$ in the 
direction of $\vec{A}$.  If the vectors are perpendicular, then
the scalar product is zero.  The scalar products between the unit
vectors are:

\begin{equation}
 \hat{i} \cdot \hat{j} = \hat{i} \cdot \hat{k} = \hat{j} \cdot \hat{k} = 0
\end{equation}

\noindent and

\begin{equation}
 \hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1
\end{equation}

If the vectors are expressed in terms of the unit vectors (i.e. their components),
$\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and 
$\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$, the distributive property 
of the scalar product gives

\begin{equation}
 \vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z
\end{equation}

\bigskip

\noindent {\bf Work-Energy Theorum}

\vspace{4mm}
Newton's second law is a vector equation, it relates the acceleration of
an object to the net force it experiences, $\vec{F}_{net} = m \vec{a}$.
It is also interesting to consider how a scalar dynamical quantity changes
in time and/or position.  A useful quantity to consider is the speed
of an object squared, $v^2 = \vec{v} \cdot \vec{v}$.  The time rate of
change of $v^2$ is:

\begin{eqnarray*}
   {{d v^2} \over {dt}} & = & {{d(\vec{v} \cdot \vec{v})} \over {dt}}\\
                      & = & {{d \vec{v}} \over {dt}} \cdot \vec{v} +
                               \vec{v} \cdot {{d \vec{v}} \over {dt}}\\
                     & = & \vec{a} \cdot \vec{v} + \vec{v} \cdot \vec{a}\\
  {{d v^2} \over {dt}} & = & 2 \vec{v} \cdot \vec{a}
\end{eqnarray*}

The result above is strictly a mathematical formula.  The last line states that
the increase in the speed squared is proportional to the component of
the acceleration in the direction of the velocity.  This makes sense.  If 
there is no component of $\vec{a}$ in the direction of $\vec{v}$ then the objects
speed does not increase.  For example in uniform circular motion, $\vec{v}$ and
$\vec{a}$ are perpendicular and the speed does not change.

Now comes the physics.  Newton's second law relates the acceleration to the
{\it Net Force}, $\vec{a} = \vec{F}_{net} /m$:

\begin{eqnarray*}
  {{d v^2} \over {dt}} & = & 2 \vec{v} \cdot \vec{a}\\
                       & = & 2 \vec{v} \cdot {{\vec{F}_{net}} \over {m}}
\end{eqnarray*}

\noindent Multiplying both sides by $m/2$ and rearrainging terms gives

\begin{equation}
  \vec{F}_{net} \cdot \vec{v} = {{d(mv^2/2)} \over {dt}}
\end{equation}

\noindent For an infinitesmal change in time, $\Delta t$, we have

\begin{equation}
  \vec{F}_{net} \cdot (\vec{v} \Delta t) = \Delta ({{mv^2} \over 2})
\end{equation}

\noindent However, $\vec{v} \Delta t$ is the displacement of the object in
the time $\Delta t$, which we will call $\Delta \vec{r} = \vec{v} \Delta t$.
Substituting into the equation gives:

\begin{equation}
  \vec{F}_{net} \cdot \Delta \vec{r} = \Delta ({{mv^2} \over 2})
\end{equation}

This is a very nice equation.  It says that the (component of the force
in the direction of the motion) times the displacement equals the change
in the quantity $mv^2/2$.  The quantity on the right side, $mv^2/2$, and
the quantity on the left side, $\vec{F}_{net} \cdot \Delta \vec{r}$, are
special and have special names.  $mv^2/2$ is called the kinetic energy
of the object. $\vec{F}_{net} \cdot \Delta \vec{r}$ is called the net
work.  Both are scalar quantities, and the equation is a result of
Newton's second law of motion.  The above equation is the infinitesmal
version of the work-energy theorum.

One can integrate the above equation along the path that an object moves.
This involves subdividing the path into a large number $N$ of segments.
If $N$ is large enough, the segments are small enough such that
$\Delta \vec{r}$ lies along the path.  Adding up the results of the above
equation for each segment gives:

\begin{equation}
  \int  \vec{F}_{net} \cdot d \vec{r} = \int d ({{mv^2} \over 2})
\end{equation}

\noindent If the initial position is $\vec{r}_i$ and the final position is
$\vec{r}_f$, we have

\begin{equation}
  \int_{\vec{r}_i}^{\vec{r}_f}  \vec{F}_{net} \cdot d \vec{r} = {{mv^2_f} \over 2} -
                                                  {{mv^2_i} \over 2}
\end{equation}

\noindent This equation is called the {\bf work-energy theorum}.  It is true for any path
the particle takes and derives from Newton's second law.  Note that there is no explicit 
time variable in the equation.  The work-energy
theorum does not directly give information regarding the time it takes for
the object to move from $\vec{r}_i$ to $\vec{r}_f$, nor does it give information
about the direction of the object.
The equation relates force and the distance through which
the net force acts on the object to the change in the objects (speed) squared.  It
is a scalar equation which contains the physics of the vector equation 
$\vec{F}_{net} = m \vec{a}$.  Since scalar quantities do not have any direction,
it is often easier to analyze scalar quantities.  From the
work-energy theorum {\it we discover} two important scalar quantities, $mv^2/2$ and
$\vec{F}_{net} \cdot \Delta \vec{r}$, and their relationship to each other.
One is called the kinetic energy and the other the net work.  
Many situations for which this equation is applicable will be discussed in lecture.

\bigskip

\centerline{\bf Potential Energy and Conservation of Mechanical Energy}

\vspace{4mm}
Often one can identify different forces that act on an object as it moves along its
path, and the net force is the sum of these forces: $\vec{F}_{net} = \vec{F}_1
+ \vec{F}_2 + ... $.  For example the forces $\vec{F}_j$ can be the weight
or gravitational force, the force a surface exerts on an object, an electric
or magnetic force, etc.  The net work on an object will be the sum of the
work done by the different forces acting on the object:

\begin{equation}
 \int_{\vec{r}_i}^{\vec{r}_f} \vec{F}_{net} \cdot d\vec{r} = 
 \int_{\vec{r}_i}^{\vec{r}_f} \vec{F}_{1} \cdot d\vec{r} +
 \int_{\vec{r}_i}^{\vec{r}_f} \vec{F}_{2} \cdot d\vec{r} + ... 
\end{equation}

\noindent or letting $W_{j}$ represent the work done by the 
force $j$:

\begin{equation}
 W_{net} = W_1 + W_2 + ...
\end{equation}

\noindent Note this equation motivates us to examine the work done
by {\it one force} as it acts along a particular path from the position
$\vec{r}_i$ to $\vec{r}_f$.  We now consider the work done by particular
forces.

\bigskip

\noindent {\it Work done by forces that only change direction}

\vspace{4mm}
Forces that always act perpendicular to the object's velocity will
do no work on the object.  Since there is no component of force
in the direction of the motion, these forces can only change the
direction but not increase the speed (kinetic energy) of the object.
The tension in the rope of a simple pendulum is an example.  If the
rope is fixed at one end and doesn't stretch, the tension is always
perpendicular to the motion of the swinging ball.  If an object
slides along a frictionless surface that does not move, the force that the surface exerts,
(normal force) is always perpendicular to the motion.  If there is
friction, the force the surface exerts on an object is often
"broken up" into a part normal to the surface and one tangential
to the surface.  The part of the force "normal" to the surface will
not do any work on the object.  The tangential part (friction) will
do work on the object.  Note: if the surface moves, the normal
force can do work.

\bigskip

\noindent{\it Work done by a constant force}

\vspace{4mm}
We will consider a particular constant force, the weight of an object
near the earth's surface.  Our results will apply generally to any
force that is constant in space and time.  Also, in our discussion
we will consider "up" as the "+y" direction.  With this notation,
the force of gravity is $\vec{W}_g = -mg \hat{j}$.  Consider any
arbritary path from an initial position $\vec{r}_i = x_i \hat{i}
+ y_i \hat{j}$ to a final position $\vec{r}_f = x_f \hat{i} +
y_f \hat{j}$.  To calculate the work done by $\vec{W}$ from the
initial to the final position, we divide up the path into a large
number $N$ small segments.  We label one segment as $\Delta \vec{r}$.
In terms of the unit vectors, we can write $\Delta \vec{r} =
\Delta x \hat{i} + \Delta y \hat{j}$.  The work, $\Delta W_g$,
done by the force of gravity for this segment is

\begin{equation}
  \Delta W_g = (-mg \hat{j}) \cdot (\Delta x \hat{i} + \Delta y \hat{j})
\end{equation}

\noindent which is

\begin{equation}
  \Delta W_g = -mg \Delta y
\end{equation}

\noindent The work done by the force of gravity for this segment does not
depend on $\Delta x$.  This makes sense, since the force acts only in the "y"
direction.  If all the work done by the gravitational force for all the segments 
of the path are added up, the result will be 

\begin{eqnarray*}
  W_g & = & -mg \sum \Delta y\\
           & = & -mg (y_f - y_i)\\
  W_g & = &  mgy_i - mgy_f
\end{eqnarray*}

\noindent This result is true for any path the object takes.  That is, the work done by 
a constant gravitational force depends only on the initial and final heights of the path.
If the work done by a force depends only on the initial and final positions of the
path, and not the path itself, the force is called a {\bf conservative force}.  The
force $\vec{F}_g = -mg \hat{j}$ is a conservative force.  We see that the work 
done by the gravitational force $\vec{F}_g$ equals the difference in the function
$U_g(\vec{r}) = mgy$ of the initial and final positions:

\begin{equation}
 \int_{\vec{r}_i}^{\vec{r}_f} \vec{F}_g \cdot d\vec{r} = 
                U_g(\vec{r}_i) - U_g(\vec{r}_f) 
\end{equation}

\noindent The function $U_g(\vec{r})$ is called the potential energy
function for the constant gravitational force (i.e. gravity near a
planet's surface).  

In general, whenever the work done by a force depends only on the
initial and final positions, and is the same for any path between
these starting and ending positions, the work can be written as
the difference between the values of some function evaluated at
the initial, $i$, and final, $f$, positions:

\begin{equation}
 \int_{\vec{r}_i}^{\vec{r}_f} \vec{F} \cdot d\vec{r} = 
                U(\vec{r}_i) - U(\vec{r}_f) 
\end{equation}

\noindent This equation is the defining characteristic of conservative forces:
the work done by a conservative force from the position $\vec{r}_i$ to the
position $\vec{r}_f$ equals the difference in a function
$U(\vec{r})$, $U(\vec{r}_i) - U(\vec{r}_f)$.  The function
$U(\vec{r})$ is called the potential energy function, and will
depend on the force $\vec{F}$.  Every force is not necessarily
a conservative one.  For example, the contact forces of friction,
tension, and air friction are not conservative.
The work done by these forces is not equal to the difference in
a potential energy function.

Note that the potential energy function is not unique.  An arbitrary constant
can be added to $U(\vec{r})$ and the difference $U(\vec{r}_i) - U(\vec{r}_f)$
is unchanged.  That is, if $U(\vec{r})$ is the potential energy function for
some force, then $U'(\vec{r}) = U(\vec{r}) + C$ is also a valid potential 
energy function:

\begin{eqnarray*}
 \int_{\vec{r}_i}^{\vec{r}_f} \vec{F}_g \cdot d\vec{r} & = & 
                U_g(\vec{r}_i) - U_g(\vec{r}_f)\\
     & = &  (U_g(\vec{r}_i) + C) - (U_g(\vec{r}_f) + C)\\
     & = & U_g'(\vec{r}_i) - U_g'(\vec{r}_f)
\end{eqnarray*}

\noindent The arbitrary constant $C$ is chosen so that the potential energy
is zero at some reference point.  For the case of the constant gravitational
force, the reference point of zero potential energy is usually where one chooses
$y=0$, i.e. $U_g(\vec{r}) = mgy$.  To conclude this section, we mention the
potential energy functions for some of the conservative forces that you will encounter
during your first year of physics.

\bigskip

\noindent{\it Linear restoring force (ideal spring)}

\vspace{4mm}
For an ideal spring, the force that the spring exerts on an object is
proportional to the displacement from equilibrium.  If the spring
acts along the x-axis and $x=0$ is the equilibrium position, then
the force the spring exerts if the end is displaced a distance
$x$ from equilibrium is approximately

\begin{equation}
  F_x = -k x
\end{equation}

\noindent The minus sign sigifies that the force is a restoring force,
i.e. the force is in the opposite direction as the displacement.  If
$x>0$, then the force is in the negative direction towards $x=0$.  If
$x<0$, then the force is in the positive direction towards $x=0$.  The
constant $k$ is called the spring constant.  The
work done by this force from $x_i$ to $x_f$ along the x-axis is

\begin{eqnarray*}
  W_{spring} & = & \int_{x_i}^{x_f} -kx dx\\
             & = & {k \over 2} x_i^2 - {k \over 2} x_f^2
\end{eqnarray*}

\noindent Thus, the potential energy function for the ideal spring is
$U_{spring} = kx^2/2 +C$.  The constant $C$ is usually chosen to be zero.
In this case the potential energy of the spring is zero when the end is at
$x=0$, i.e. the spring is at its equilibrium position.

\begin{equation}
  U_{spring} = {k \over 2} x^2
\end{equation}

\bigskip

\noindent{\it Universal gravitational force}

\vspace{4mm}
Newton's law of universal gravitation describes the force between
two "point" objects: $\vec{F}_{12} = - (G m_1 m_2/r^2) \hat{r}_{12}$ where
$m_1$ and $m_2$ are the masses of object $1$ and $2$, $r$ is the distance
between the particles, $\vec{F}_{12}$ is the force {\it on object two} due to object
one, $\hat{r}_{12}$ is a unit vector from object one to object two.  $G$ is
a constant equal to $6.67 \times 10^{-11} NM^2/kg^2$.  The minus sign means that
the force is always attractive, since $m_1$ and $m_2$ are positive.

The graviational force is a "central" force, it's direction is along the line
connecting the two objects.  This property makes the force conservative.  Work
is only done by this force when there is a change in $r$  The force does no
work if the path is circular, i.e. constant $r$.  The work done by the 
universal gravity force for paths starting at a separation distance of $r_i$ to a 
separation distance of $r_f$ is

\begin{eqnarray*}
  W_{U.G.} & = & \int_{r_i}^{r_f} - {{G m_1 m_2} \over {r^2}} dr\\
           & = & {{G m_1 m_2} \over r}|_{r_i}^{r_f}\\
           & = & {{G m_1 m_2} \over {r_f}} - {{G m_1 m_2} \over {r_i}}\\
           & = & -{{G m_1 m_2} \over {r_i}} - (- {{G m_1 m_2} \over {r_f}})
\end{eqnarray*}

\noindent Thus, the gravitational potential energy is $U_{U.G.} = - G m_1 m_2/r +C$.
The constant $C$ is usually taken to be zero, which sets the potential energy to
zero at $r = \infty$.

\begin{equation}
   U_{U.G.} = - {{G m_1 m_2} \over r}
\end{equation}

\bigskip

\noindent {\it Electrostatic interaction}

\vspace{4mm}
Coulomb's law describes the electrostatic force between two point objects that
have charge: $\vec{F}_{12} = (k q_1 q_2/r^2) \hat{r}_{12}$ where
$q_1$ and $q_2$ are the charges of object $1$ and $2$, $r$ is the distance
between the particles, $\vec{F}_{12}$ is the force {\it on object two}
due to object one, $\hat{r}_{12}$ is a unit vector from object one to object
two.  The constant $k$ equals $9 \times 10^9 NM^2/C^2$.  If the objects
have the same sign of charge, the product $q_1 q_2$ is positive and the
force is repulsive.  If the objects have opposite charge, the product
$q_1 q_2$ is negative and the force is attractive.  The force has the same
form as the universal gravitational force.  Charge is the source of the
force instead of mass.  Similar to the gravitation case, the work done 
by the electrostatic force for paths starting at a separation 
distance of $r_i$ to a separation distance of $r_f$ is

\begin{eqnarray*}
  W_{electrostatic} & = & \int_{r_i}^{r_f} {{k q_1 q_2} \over {r^2}} dr\\
           & = & -{{k q_1 q_2} \over r}|_{r_i}^{r_f}\\
           & = & -{{k q_1 q_2} \over {r_f}} - (-{{k q_1 q_2} \over {r_i}})\\
           & = & {{k q_1 q_2} \over {r_i}} - {{k q_1 q_2} \over {r_f}}
\end{eqnarray*}

\noindent Thus, the electrostatic potential energy is $U_{electrostatic} = k q_1 q_2/r +C$.
The constant $C$ is usually taken to be zero, which sets the potential energy to
zero at $r = \infty$.

\begin{equation}
 W_{electrostatic} = {{k q_1 q_2} \over r}
\end{equation}

The reason for using the word "conservative" to describe these forces is the following.
Suppose that the only forces that do work on an object are ones that are
conservative, that is, the work done by these forces is equal to the difference
in a potential energy function.  Consider the case in which the only forces that
do work on a particle are two conservative
forces.  From the work-energy theorum we have:

\begin{eqnarray*}
  \int_{\vec{r}_i}^{\vec{r}_f} \vec{F}_{net} \cdot d\vec{r} & = & {m \over 2} v_f^2 - {m \over 2} v_i^2\\ 
  \int_{\vec{r}_i}^{\vec{r}_f} \vec{F}_1 d\vec{r} + 
  \int_{\vec{r}_i}^{\vec{r}_f} \vec{F}_2 d\vec{r} +  & = & {m \over 2} v_f^2 - {m \over 2} v_i^2\\ 
  (U_1(\vec{r}_i) - U_1(\vec{r}_f)) + 
  (U_2(\vec{r}_i) - U_2(\vec{r}_f))  & = & {m \over 2} v_f^2 - {m \over 2} v_i^2 
\end{eqnarray*}

\noindent Rearrainging terms we have:

\begin{equation}
 U_1(\vec{r}_i) + U_2(\vec{r}_i) + {m \over 2} v_i^2 =
          U_1(\vec{r}_f) + U_2(\vec{r}_f) + {m \over 2} v_f^2 
\end{equation}

\noindent The left side of the equation contains only quantities at position
$\vec{r}_i$, and the right side only quantities at position $\vec{r}_f$.  Since
$\vec{r}_f$ is arbitrary, the quantity $U_1(\vec{r}) + U_2(\vec{r}) +
mv^2/2$ is a constant of the motion, it is a conserved quantity!  The sum of
the potential energy plus kinetic energy, which we call the {\bf total mechanical
energy}, is conserved.  This result will hold for any number of forces, as long as
the only work done on the system is by conservative forces.

If non-conservative forces act on the object, such as frictional forces, the total mechanical
energy as defined above is not conserved.  However, frictional forces are really electro-magnetic
forces at the microscopic level.  The electro-magnetic interaction is conservative if the energy of the
electromagnetic field is included.  At present, we believe there are only three fundamental
interactions in nature: gravitational, electro-magnetic-weak, and strong, and they all
are conservative.  Frictional forces are non-conservative because of the limited description 
used in analyzing the system of particles.  Since all the fundamental forces of nature (gravity, electro-
magnetic, and strong) are conservative, at the microscopic level total energy is conserved.  We
refer to the energy of the all the atoms and molecules of a material as it's {\bf internal energy}.

\bigskip

\centerline{\bf Momentum and Conservation of Momentum}

\bigskip
\noindent {\it An example to motivate the concept of momentum}

\vspace{4mm}
There is a physics professor holding on to a cart.  The physics professor,
mass $m_1$, is wearing frictionless roller skates, and the cart (mass $m_2$)
has frictionless wheels.  The professor pushes the cart with a constant force
$F$, he goes off to
the left (negative direction) and the cart goes off to the right (positive
direction).  How is the professor's final speed, $|v_1|$, related to the final
speed of the cart, $|v_2|$?  From {\bf Newton's Third Law}, the force
the cart feels due to the professor, $F_{CP}$, is equal in magnitude and opposite in
direction to the force the professor feels due to the cart, $F_{PC}$:

\begin{equation}
  -F_{PC} = F_{CP}
\end{equation}

\noindent The time that the force acts, $t$, is the same for both objects.  That
is, the time the force acts on the professor is the same as the time for force
acts on the cart:

\begin{equation}
  -F_{PC} t  = F_{CP} t
\end{equation}

\noindent Using Newton's second law, $\vec{F}_{net} = m \vec{a}$, we have

\begin{equation}
  -m_1 a_1 t = m_2 a_2 t
\end{equation}

\noindent Since the objects both stated from rest $v = a t$ for each object, gives

\begin{equation}
  - m_1 v_1 = m_2 v_2
\end{equation}

\noindent Wow, what a simple result.  The magnitude of $m v$ for the professor
to the left equals the magnitude of $m v$ for the cart to the right.  This result
seems to work for any force and any time $t$.  The quantity mass times velocity
appears to be quite special, and deserves a special name.  We call the
product of mass times velocity the momentum of the particle, $\vec{p}$.
For example, the {\bf momentum} of particle $1$ is

\begin{equation}
   \vec{p}_1 \equiv m_1 \vec{v}_1
\end{equation}

\noindent Note that momentum is a vector!

\bigskip

\noindent {\bf Some properties of the momentum of a particle}

\vspace{4mm}
Newton's law of motion for a particle can be expressed in terms of the
momentum of the particle:

\begin{eqnarray*}
  \vec{F}_{net} & = & m \vec{a}\\
                & = & m {{d \vec{v}} \over {dt}}\\
                & = & {{d (m \vec{v})} \over {dt}}\\
  \vec{F}_{net} & = & {{d \vec{p}} \over {dt}}
\end{eqnarray*}

If one multiplies both sides by $dt$ and integrate from time $t_1$ to 
time $t_2$, we have

\begin{equation}
  \int_{t_1}^{t_2} \vec{F}_{net} dt = \vec{p}_2 - \vec{p}_1
\end{equation}

\noindent This equation is called the {\bf "impulse-momentum" theorum}.
It is similar to the work-energy theorum.  Loosely speaking:
{\it force times time equals the change in momentum} of a particle,
and {\it force times distance equals the change in kinetic energy} of
a particle.

\bigskip

\noindent {\bf Total Momentum of a system of particles}

\vspace{4mm}
Often in physics we are interested in how particles interact with each other.
The number of interacting particles can be two or more, and we refer to the
collection of particles as a system of particles.  An important quantity
to consider for a system of particles is the {\bf total momentum of the
particles}, $\vec{P}_{tot}$.  The total momentum of the system of particles
is just the vector sum of the momenta of the individual particles:

\begin{equation}
  \vec{P}_{tot} \equiv \vec{p}_1 + \vec{p}_2 + \vec{p}_3 + ...
\end{equation}

\noindent where the subscripts $1$, $2$, etc. refer to particle number $1$, $2$, 
etc.  How does the total momentum of the system of particles change in time?  Consider
the case of two particles:

\begin{eqnarray*}
 {{d \vec{P}_{tot}} \over {dt}} & = & {{d \vec{p}_1} \over {dt}} + {{d \vec{p}_2} \over {dt}}\\
                              & = & \vec{F}_{1net} + \vec{F}_{2net}
\end{eqnarray*}

\noindent We can separate the net force on each particle into a part due to the particles within
the system and a part due to influences outside the system.  In the case of two particles, the
net force on particle one equals the force on particle one due to particle two, $\vec{F}_{12}$,
and the force due to objects outside the system, which we call external forces: $\vec{F}_{1ext}$.
The same is true for object 2:

\begin{eqnarray*}
  \vec{F}_{1net} & = & \vec{F}_{12} + \vec{F}_{1ext}\\
  \vec{F}_{2net} & = & \vec{F}_{21} + \vec{F}_{2ext}
\end{eqnarray*}

\noindent Substituting the net force equations into the change of total momentum equation
above gives:

\begin{eqnarray*}
 {{d \vec{P}_{tot}} \over {dt}} & = & \vec{F}_{12} + \vec{F}_{1ext} +  \\
                                &   & \vec{F}_{21} + \vec{F}_{2ext}
\end{eqnarray*}

\noindent Two terms on the right side of the equation 
($\vec{F}_{12}$ and $\vec{F}_{21}$) cancel each other.  From
Newton's third law, $\vec{F}_{12} = - \vec{F}_{12}$.  The force particle {\it one}
feels due to particle {\it two} is equal in magnitude and opposite in direction to the
force that particle {\it two} feels from particle {\it one}.  Although $\vec{F}_{12}$ and
$\vec{F}_{21}$ act on different particles, when all the forces of the whole
system are added up they cancel each other out.  Thus we have

\begin{equation}
  {{d \vec{P}_{tot}} \over {dt}} = \vec{F}_{1ext} + \vec{F}_{2ext}
\end{equation}

If the system contains more than two particles, the result will be similar.  In general,
we have:

\begin{equation}
  {{d \vec{P}_{tot}} \over {dt}} = \vec{F}_{1ext} + \vec{F}_{2ext} + ...  
\end{equation}

The time rate of change of the total momentum of a system of particles equals
the sum of the external forces.  {\bf If there are no external forces acting
on the system}:

\begin{equation}
    {{d \vec{P}_{tot}} \over {dt}} = 0
\end{equation}

\noindent That is, the total momentum of the system does not change in time, it is conserved.

\begin{equation}
     \vec{P}_{tot} = constant
\end{equation}

\noindent This is a very nice result, and it is straight forward to see
why it is true in the two particle case.  The key physics is Newton's
third law.  If there are no external forces, the net force on object
1 is caused only by object 2 (and visa-versa).  Whatever force object
1 feels, object 2 will feel the opposite force.  Since the time of
interaction, $\Delta t$ is the same for both particles, 
$\vec{F}_{12} \Delta t = - \vec{F}_{21} \Delta t$.  Force times time
is the change in momentum, thus the vector change in the momentum of
particle one will be opposite to the vector change in momentum of
particle 2: $\Delta \vec{p}_1 = - \Delta \vec{p}_2$. Object 1's
gain (loss) of momentum equals object 2's loss (gain).  The net
change in the total momenta of the two objects is zero.  This
same argument is true if there are more then two objects.

In lecture we will discuss several cases where there are no external
forces: collisions, explosions, etc.  When no external forces act
in a collision, the total momentum ({\bf vector sum} of the momenta of
the particles) of the system is conserved.  If kinetic energy is also
conserved, we call the collision elastic.  It is remarkable that one
doesn't need to know the details of the collision.  As long as there
are no external forces, it doesn't matter
what kind of forces are involved: the total momentum remains constant
before, during and after the collision.

In the derivation above, {\bf the conservation of total momentum} comes 
from Newton's third law, which {\bf is a result of a symmetry in nature}.  
The interaction between object 1 and object 2 is symmetric, they each
must experience the same force.  Is this a general result, that
symmetries in nature lead to conserved quantities?  We find that in
many cases this is true.  Rotational symmetry leads to conservation
of angular momentum, and momentum and energy conservation are a result of
space and time symmetry.  This is a grand idea, and helps in describing 
the physics of subatomic particles.  We do experiments to identify
conserved quantities, then develop mathematical descriptions that
have the corresponding symmetry properties.  The connection between
symmetries in nature and conserved quantities is one of the more
"beautiful" principles of physics.

\bigskip

\noindent {\bf Center of Mass of a system of particles}

\vspace{4mm}
From the total momentum of a system of particles, we can 
define another special quantity for a system of particles: 
the "center-of-mass" velocity. The center-of-mass velocity,
$\vec{V}_{cm}$, is defined as the total momentum divided by the total mass
of the system:

\begin{equation}
  \vec{V}_{cm} \equiv {{\vec{P}_{tot}} \over {M_{tot}}}
\end{equation}

\noindent where the total mass of the system, $M_{tot}$
is defined as

\begin{equation}
  M_{tot} \equiv m_1 + m_2 + ...
\end{equation}

The center-of-mass velocity is easily expressed in terms of the
individual masses and velocities of the particles that make up 
the system:

\begin{equation}
 \vec{V}_{cm} = {{m_1 \vec{v}_1 + m_2 \vec{v}_2 + ...} \over
                 {m_1 + m_2 + ...}}
\end{equation}

\noindent Note that the center-of-mass velocity is a vector.  From
the center-of-mass velocity, it is straight forward to define
a center-of-mass position and a center-of-mass acceleration:

\begin{equation}
 \vec{a}_{cm} = {{m_1 \vec{a}_1 + m_2 \vec{a}_2 + ...} \over
                 {m_1 + m_2 + ...}} 
\end{equation}

\noindent for the center-of-mass acceleration, and

\begin{equation}
 \vec{R}_{cm} = {{m_1 \vec{r}_1 + m_2 \vec{r}_2 + ...} \over
                 {m_1 + m_2 + ...}}
\end{equation}

\noindent for the center-of-mass position.  $\vec{R}_{cm}$ is
usually called center-of-mass.  The connection between these
three "kinematical" quantities is the same as with
the position, velocity, and acceleration of one particle:

\begin{eqnarray*}
   \vec{V}_{cm} & = & {{d \vec{R}_{cm}} \over {dt}}\\
   \vec{a}_{cm} & = & {{d \vec{V}_{cm}} \over {dt}}
\end{eqnarray*}

These center-of-mass quantities have some nice properties.  Since
$\vec{P}_{tot} = M_{tot} \vec{V}_{cm}$, we have

\begin{eqnarray*}
  M_{tot} {{d \vec{V}_{cm}} \over {dt}} & = & {{d \vec{P}_{tot}} \over {dt}}\\
                         & = & \vec{F}_{ext-net}
\end{eqnarray*}

\noindent where $\vec{F}_{ext-net}$ is the sum of the external forces.
This equation can be rewritten as

\begin{equation}
  {{d \vec{V}_{cm}} \over {dt}}  = {{\vec{F}_{ext-net}} \over {M_{tot}}}    
\end{equation}

\noindent Thus, the acceleration of the center-of-mass equals the net force
divided by the total mass.  {\bf If there are no external forces}:

\begin{eqnarray*}
  {{d \vec{V}_{cm}} \over {dt}} & = & 0\\
            \vec{V}_{cm} & = & constant
\end{eqnarray*}

\noindent Wow, another nice result.  If there are no external forces,
the center-of-mass of the system moves at a constant velocity.  If it
is initially at rest (in an inertial frame) it remains at rest.  This
result also defines a special reference frame for a system of particles:
the reference frame for which the center-of-mass does not move.

If there are no external forces acting on a system of particles,
There exists an inertial frame of reference in which the center-of-mass
$\vec{R}_{cm}$ remains at rest.  We refer to this reference frame
as {\bf the center-of-mass reference frame}, or simply center-of-mass
frame.  Although the individual particles in a system might move in a
complicated manner, there is one special position, the center-of-mass,
which moves in a simple way.  It is often easier to analyze the motion 
of a system of particles from the center-of-mass frame. 

\bigskip

\noindent {\bf Final comments on Energy and Momentum}

\vspace{4mm}
The first 4 weeks of the quarter we discussed the physics of the interaction
of particles using the "force-motion" approach: find the net force on
each object, then $\vec{a} = \vec{F}_{net}/m$.  The symmetry of interaction
was expressed via Newton's third law.  During the last 3 weeks, we expressed these
laws using the energy and momentum of the particles.  The two approaches are equivalent.  Since
we believe the fundamental forces are conservative, from the potential energy
functions the forces can be determined and visa-versa: 

\begin{eqnarray*}
  F_x = - {{\partial U} \over {\partial x}}\\
  F_y = - {{\partial U} \over {\partial y}}\\
  F_z = - {{\partial U} \over {\partial z}}
\end{eqnarray*} 

\noindent What are the advantages of using the "energy-momentum" approach verses
the "force-motion" approach?  In terms of solving problems, if one is
interested in how an objects speed changes from one point to another then
the energy approach is much simplier than solving for the acceleration.
However, {\it problem solving is not the main motivation for studying energy
and momentum}.  

The laws of physics for "small" systems (atoms, nuclei, subatomic
particles) and for relatively fast objects (special relativity) are best described
in terms of momentum and energy.  Energy and momentum transform from one
inertial reference frame to another the same way as displacements in time and
space.  The interference properties of a particle is related to the particle's
momentum ($\hbar / p$).  In the Schr\"odinger equation, which is used to describe
quantum systems, the interaction is expressed in terms of the potential energy 
of the system.  Advances in modern physics were guided by the principles
of energy and momentum conservation.

Energy plays a key role in life on earth.  If an animal needs to use
more energy obtaining food than the food supplies, then the animal starves
and species can go extinct.  Energy is an important commodity in our
daily lives: for our bodies, for our homes, and for our personal transportation.
Wars have been fought over energy, and our lifestyle will 
be determined by how successful we are in using the sun's energy.  
Energy is perhaps more important than momentum because it is a scalar quantity allowing
it to be stored and sold.  The importance of energy cannot be understated.
Next quarter, a large part of your physics course (Phy132) will be 
devoted to an understanding of internal energy and energy transfer processes.  

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