Displacement and Constant Velocity: Different Approaches

 Physics Index

Where are we going with this? The information on this page connects to standards such as:

• Investigate and evaluate the graphical and mathematical relationship (using either manual graphing or computers) of one-dimensional kinematic parameters (distance, displacement, speed, velocity, acceleration) with respect to an object's position, direction of motion, and time.

•  Algebraically solve problems involving constant velocity and constant acceleration in one- dimension.

Displacement and Constant Velocity: Different Approaches
(Hmmm… Do we need more than one approach?

So, working with motion, we end up with a couple of similar formulas.

Δd = v • Δt

This means displacement equals average velocity times time.

df = di + Δd

This means final position equals initial position plus displacement.

Combining them gives:

df = di + v • Δt

Direction matters. You choose what is positive (for example, east or right). Units must be consistent (for example, meters and seconds).

What the formulas mean (plain-English + variables)

• $d_i$: initial position (displacement) relative to some reference point (the origin you choose).

• $d_f$: final position relative to the same reference point.

• $\Delta d$: displacement during the interval (net change in position). Positive/negative shows direction.

• $\mathrm{<span\; style="font-weight:\; bold;"><i>v</i>\; : <strong\; data-end="513"\; data-start="493"\; style="display:\; inline\; !important;">average\; velocity </strong>over\; the\; interval\; (displacement\; per\; unit\; time;\; includes\; direction).<br> <br>NOTE:\; Sometimes\; average\; velocity\; is\; show\; as\; <i>v<span\; style="font-size:\; xx-small;">(ave)</span></i></span>}$

• $\Delta t$: elapsed time for the motion.

Faced with word problems, it is convenient to know which approach to take. Which formula arrives at the answer most directly! (Cause, like who want's to do extra work!)

How to know which formula to use

If you are given or want displacement without mention of any points of reference, use:

Δd = v • Δt.

If the word problem includes mention to some reference point, then you need to use this:


df = di + v∆t

Example problems and solutions

Example A – Using Δd = v • Δt

Problem: A cyclist rides west at an average velocity of –4.0 m/s for 90 s. What is the displacement?

Why this formula? We need displacement, and we are given velocity and time.

Solution: 

Δd = v • Δt
∆d= (–4.0 m/s)(90 s)
∆d= –360 m.

Answer: The displacement is –360 m (360 m west).

Example B – Using df = di + v • Δt

Problem: A boat starts 50 m west of a dock (so di = –50 m if east is positive). It cruises with an average velocity of +2.5 m/s for 3.0 minutes. Where is it at the end?

Why this formula? We need final position and we know initial position, velocity, and time.

Solution:

Convert 3.0 minutes to 180 s. 

df = di + v • Δt
df = –50 + (2.5)(180)
df = –50 + 450
df = +400 m.

Answer: 400 m east of the dock.

Angular Kinematics Overview

 Physics Index

Where are we going with this? The information on this page connects to standards such as:

• Gather evidence to defend the claim of Newton's first law of motion by explaining the effect that balanced forces have upon objects that are stationary or are moving at constant velocity.

•Recognize and communicate information about energy efficiency and/or inefficiency of machines used in everyday life.

Angular Kinematics Overview

(Wow… This sounds fancy!)

Angular kinematics is a way to describe something moving around a circle.  This is a lot like geometry!

It is pretty cool, though and NASA has this…

https://www.grc.nasa.gov/www/k-12/airplane/angdva.html

Angular Displacement

Angular displacement: The amount that the angle changes of from the initial position to the final position.

Angular displacement is sometimes represented by the letter phi φ. Angles are often represented by θ. And the ∆ is our good buddy to represent change of.

So we can say that φ is essentially ∆θ. 

We have established that ∆ anything is where did end up compared to where it started. For instance

∆T = Tf - Ti

∆d = df - di

So, 

∆θ = θf - θi

That should make sense, right? Back to the "sometimes a phi" thing:

φ = θf - θi

Keep in mind that "f" and "i" are only one way to indicate beginning and ending conditions. "0" and "1" are also seen and many sources use them for Angular displacement.

Source

So… 

φ = θf - θi

∆θ = θf - θi

is the same as

φ = θ1 - θ0

∆θ = θ1 - θ0 

Sample 1: A thing is at 30°. It moves to 50°. What is its angular displacement? (Ans. 20°)

Sample 2: A thing is at 0 rad. It moves to π rad. What is its angular displacement? (Ans  π rad)

How Far Did It Move

Geometry allows for some fun stuff when the angles are in radians.  Trust me on this!

In general, any angle θ measured in radians is defined by the formula

θ = s/r

where s is arc length and r is radius

that means that

∆θ rad = ∆s/r                    (This is probably pretty useful!)

Of course you can convert from rad to degrees. 

Multiply degrees by π/180 to get radians. 

Multiply radians by 180/π to get degrees.

But, that would be annoying! Google it, if you want.

Sample 3: A thing is at a radius of 2 from a point and is moving in a circle. If it moves such that it's arc length is 4, what is its angular displacement? (Ans 2 rad)

It follows that, in a case where something is measured from a reference point (a line or angle or something), if that something moves a final angular displacement can be found (by mathing it) as…

∆θ = θf - θi

θf  = θi  + ∆θ

This should remind you of one of the liner displacement formulas!

df = di + ∆d

Angular Velocity

Angular velocity: The amount that the angle changes in some period of time. That is to say, it is the angular displacement / elapsed time.

For instance, something with an angular velocity of 20° per second would turn 20° in 1 second, 40° in 2 seconds, 60° in 3 seconds… etc.

Other ways to describe angular velocity is to discuss revolutions per some period of time. The rate that car engines turn is reported in RPM or revolutions per minute. Revolutions per second are also normal.

Officially (SI UNITS) the units on angular velocity are radians per second  (rad/s)

It is handy to know that a circle is 360° around or 2π radians.

We need a formula!

Because of something to do with something or for some reason… Never mind.

The symbol for angular velocity is the Greek omega (not W or w!) ω

Recall that the change in time  (elapsed time) is

∆t = tf - ti

    or

∆t = t1 - t0

and angular displacement is

∆θ = θf - θi

    or

∆θ = θ1 - θ0

so we can say that 

ω = ∆θ/∆t

This can appear in other forms that mean the same thing…

ω = (θf - θi)/(tf - ti)

ω = (θ1 - θ0)/(t1 - t0)

Sample 1: A thing rotates 12° in 4 seconds. What is its angular velocity? (Ans. 3°/s)

Sample 2: A thing rotates 12 rad in 4 seconds. What is its angular velocity? (Ans. 3 rad/s)

Sample 2: A thing rotates 360 times in 1 minute (360 RPM). What is its angular velocity in degrees per second? (Ans 21,600°/s)

360 RPM is 60 rotations per second. Each rotation is 360°. Thus 

∆θ = 60 • 360°

∆t = 1

ω = ∆θ/∆t

The formula below can be rearranged to find the change in the angle:

ω = ∆θ/∆t

∆t•ω = ∆θ

Then, rearranged…

∆θ = ω∆t

This should remind you of a linear motion formula!

∆d = v∆t

Angular Displacement Part II

What if we took our angular displacement formula and did some math? That could be fun!

θf  = θi  + ∆θ

∆θ = ω∆t

Let the magic happen!

θf  = θi  + ω∆t

Wanna know a secret (that really isn't a secret)? Go to the bottom of this page!

Angular Acceleration

Angular acceleration: The amount that the rate at which angular velocity changes over some period of time. That is to say, it is the change in angular velocity / elapsed time.

Since we are using a for linear acceleration let's use a Greek a for angular acceleration—α

Where 

α is angular acceleration

∆ω is change in angular veloicty

∆t is elapsed time

α = ∆ω / ∆t

Let's do that math magic!

α = ∆ω / ∆t

∆ω = α•∆t

Pssttt… It's not REALY magic… 

If something has angular velocity to start with and it gets more, then the final angular velocity is…

ωf = ωi + ∆ω

and then…

ωf = ωi + ∆ω

∆ω = α•∆t

ωf = ωi + α•∆t

You might be thinking of the linear kinematic equation for final velocity!

Summary

Angular displacement: The amount that the angle changes of from the initial position to the final position.

Angular displacement is sometimes represented by the letter phi φ. Angles are often represented by θ.

∆θ = θf - θi

φ = θf - θi 

θf  = θi  + ∆θ 

θf  = θi  + ω∆t

In general, any angle θ measured in radians is defined by the formula

θ = s/r

where s is arc length and r is radius

that means that

∆θ rad = ∆s/r

Angular velocity: The amount that the angle changes in some period of time. That is to say, it is the angular displacement / elapsed time.

The symbol for angular velocity is the Greek omega (not W or w!) ω

ω = ∆θ/∆t 

 

∆θ = ω∆t 

 

ωf = ωi + α•∆t

Angular acceleration: The amount that the rate at which angular velocity changes over some period of time. That is to say, it is the change in angular velocity / elapsed time.

Since we are using a for linear acceleration let's use a Greek a for angular acceleration—α

α = ∆ω / ∆t 

 

∆ω = α•∆t

Not Really A Secret!

 So abut that final displacement thing…

Where X is an amount thing (displacement, velocity, cookies, money, pizza )and ∆t is time and r is rate the thing changes…

and where the subscript f is final amount and i is initial amount…

Xf = Xi + r∆t

EXAMPLE: Bob has 4 things. Every minute he gets 2 more things.  How many things will Bob have after 5 minutes.

Where X = things…

Find Xf where…

Xi = 4 things

r = 2 things/minute

∆t = 5 minutes

Xf = Xi + r∆t

Xf = 4 + 2•5

Xf = 4 + 10

Xf = 14

    

 

 

Strength of Chemical Bonds

 General Chemistry Index

Where are we going with this? This page will give the ability to use laboratory observations and data to compare and contrast ionic, covalent, network, metallic, polar, and non-polar substances with respect to constituent particles, strength of bonds, melting and boiling points, and conductivity; provide examples of each type.

Strength of Chemical Bonds / Bond Energy
Are all chemical bonds the same… 

Compounds are formed when two or more elements combine chemically in fixed, specific ratios. Combine… What?

Elements combine to form compounds by bonding to other elements—in fixed, specific ratios. They do this in several ways, both intramolecularly (strong) and intermolecularly (weaker).  (See also.)

Intramolecular bonds occur as either ionic or covalent (and some covalent bonds are polar).

When considering the strength of chemical bonds, it is convenient to rank bond strength by type of bond. However do thing this is a drastic over-simplification.

Aside from saying that intramolecular bonds are stronger than intermolecular bonds, we cannot say always this or that (it is chemistry, after all!).

"When a bond is strong, there is a higher bond energy because it takes more energy to break a strong bond" (Source, 2024). Thus, bond strength is actually a way of comparing bond energy. And bond energy is far from simple!

The following table shows the bond energy of various bonds:

Table 1: Average Bond Energies (kJ/mol)

Single Bonds						Multiple Bonds
H—H	432	N—H	391	I—I	149	C = C	614
H—F	565	N—N	160	I—Cl	208	C ≡ C	839
H—Cl	427	N—F	272	I—Br	175	O = O	495
H—Br	363	N—Cl	200			C = O*	745
H—I	295	N—Br	243	S—H	347	C ≡ O	1072
		N—O	201	S—F	327	N = O	607
C—H	413	O—H	467	S—Cl	253	N = N	418
C—C	347	O—O	146	S—Br	218	N ≡ N	941
C—N	305	O—F	190	S—S	266	C ≡ N	891
C—O	358	O—Cl	203			C = N	615
C—F	485	O—I	234	Si—Si	340
C—Cl	339			Si—H	393
C—Br	276	F—F	154	Si—C	360
C—I	240	F—Cl	253	Si—O	452
C—S	259	F—Br	237
		Cl—Cl	239
		Cl—Br	218
		Br—Br	193
*C == O(CO2) = 799

 (Source, 2024)

A cursory look will indicate that a lot of the ionic bonds are stronger than a lot of the single covalent bonds. But looking closer will reveal that not all ionic bonds are stronger than all covalent bonds, 

For instance, compare the Si-O bond (covalent) to the H-Cl (ionic) bond.

FURTHER, most of the double and triple covalent bonds are stronger (have higher bond energy) than most of the ionic bonds (c.f. C ≡ C ).

So what?

To rank bond strength by bond type proves to be a pretty bad idea. While there are tendencies, there is no all-fitting rule. In the end, the convenience of such an attempt must be sacrificed to accuracy.  

 Physics Index

Where are we going with this? The information on this page is related to basic astronomy and astrophysics.

Astronomy: A Group-sourced Collection of

Concepts, Information, and Ideas

The following information was compiled by students (group-sourced) as an in-class project.

Refer to the cited sources for additional information.

Torque: Types of Levers

 Physics Index

Where are we going with this? The information on this page connects to standards such as:

• Gather evidence to defend the claim of Newton's first law of motion by explaining the effect that balanced forces have upon objects that are stationary or are moving at constant velocity.

•Recognize and communicate information about energy efficiency and/or inefficiency of machines used in everyday life.

Torque: Types of Levers

(Hmm… I thought torque was something in car engines!)

Give me a lever long enough and a fulcrum on which to place it, and I shall move the world.

~Archimedes

(As if!)

But, he's not wrong. Mathematically, that is.

Let's dig into this a little!

The concept of levers and torque are directly connected to motion that rotates around a pivot point. This motion can be considered to be clockwise around the point or counterclockwise (anticlockwise).

Every lever system has three parts: the lever, the effort and the load.

And… every lever has two parts. 

We are going to call one of those parts the lever arm. The lever arm is the beam or whatever that is used. More on this after we define other parts.

The second part of the lever is a fulcrum. The fulcrum (sometimes called the pivot) is the thing around which the lever arm rotates.

The effort is a force that is applied to the lever arm at some distance from the fulcrum.

The load is also a force that will have an effect on the lever arm. 

Generally, the effort is acting on the load. 

Circling back to lever arm…

The lever arm is the part of a lever system that sits on and rotates around the fulcrum and to which is applied the effort so that it can act on the load.

The effort is trying to rotate the lever arm around the fulcrum. The load is resisting this motion.

It is incorrect to say that if the effort is greater than the load the lever will rotate as desired. Force alone is not enough to determine what will happen.

What determines the rotation is something called torque.

Torque is the potential to rotate the lever and is the product of the applied force and the distance from the fulcrum.

Caveat: 

• it is ONLY the vector component of the force that is perpendicular to the lever arm. So, sometimes, you have to deconstruct the force to find the perpendicular component. 

Hence, where T is torque, F is force and d is distance from the fulcrum…

T = Fd

Torque adds up in a lever system. Clockwise torque and counterclockwise torque have opposite signs.

If you think of a positive torque as attempting to cause clockwise rotation around the fulcrum, then negative torque will be attempting to cause counterclockwise rotation.

Where T is torque, the net torque will be the signed sum of all the torques acting on the lever, so that…

T(net) = T1 + T2 + T3…

If you say that clockwise is positive…

…if the result is positive, then you have clockwise rotation.

…if the result is negative, then you have counterclockwise rotation.

Now, if you want to complicate the math, you can say that the fulcrum is at distance zero, then measure left as positive and right as negative, then say that forces "up" are negative and "down" are positive, then, plug all those in so that…

    T(net) = F1d1 + F2d2  + F3d3 …

It will work out…

A more "tangible" approach is to keep up logically with the torques as counterclockwise and clockwise. 

Sometimes, you are looking for a solution that results in equilibrium. That is to say that there is either no rotation or that the rotation is occuring at a constant rate.

This would mean that

T(cw) = T(ccw)

In such a case, it's easy and logical to sort the various forces to the appropriate side of the equation.

Types of Levers

Though there are three types of levers, the math on them is essentially the same:

T(net) = T1 + T2 + T3…

The types of levers are categorized according to the configuration of load, effort, and fulcrum.

Class 1 Lever

"For the Class 1 lever the pivot lies between the effort and load. A see saw in a playground is an example of a Class 1 lever where the effort balances the load."

In the above image, if the effort is counterclockwise, then the load is clockwise. So, if the product of effort and distance is greater than the product of load and distance, the lever arm will rotate counterclockwise.

Class 2 Lever

"For the Class 2 lever the load is between the pivot and the effort (like a wheelbarrow). The effort force needed is less than the load force, so there is a mechanical advantage."

In this image, load would be clockwise and effort counterclockwise.  Since torque is distance times force (effort), a small effort would move a larger load.

Class 3 Lever

"For a Class 3 lever the load is further away from the pivot than the effort. There is no mechanical advantage because the effort is greater than the load. However this disadvantage is compensated with a larger movement. This type of lever system also gives us the advantage of a much greater speed of movement."

In this image, load would be clockwise and effort counterclockwise.  Since torque is distance times force (effort), a large effort would be needed to move a smaller load.

See also: https://www.sciencelearn.org.nz/resources/1924-what-levers-does-your-body-use

___________________________

There can be far more complex arrangements involving multiple efforts and loads. However, the principle remains the same:

Where T is torque, the net torque will be the signed sum of all the torques acting on the lever, so that…

T(net) = T1 + T2 + T3…

Where the roadway is considered the lever arm, each support footing is a fulcrum, each cable is an upward force (effort) and the roadway and every car is a load. It is notable that the upward forces only exist in opposition to the downward forces. 

Momentum: Notes and Such

Physics Index

Where are we going with this? The information on this page connects to standards such as:

• Using experimental evidence and investigations, determine that Newton’s second law of motion defines force as a change in momentum, F = Δp/Δt.

• Develop and apply the impulse-momentum theorem along with scientific and engineering ideas to design, evaluate, and refine a device that minimizes the force on an object during a collision (e.g., helmet, seatbelt, parachute).

•  Assess the validity of the law of conservation of linear momentum (p=mv) by planning and constructing a controlled scientific investigation involving two objects moving in one-dimension.

Momentum: Notes and Such

(Wow! Now, that's a fancy title!)

What is momentum? Well… It's not a force… And it's not energy… Basically, it's a thing that helps us understand motion. And collisions.

(You gotta do better than that!)

A more formal definition would say that momentum is the quantity of motion, the product of an object's mass and its velocity, and it can be quantized as

p = mv

  

where is p is momentum, m is mass, and v is velocity.

This means that the units for momentum are

kg•m/s

Momentum is… boring… unless something changes. Just watching a football player run at a constant velocity isn't very interesting.

Where momentum becomes interesting is when something happens. Momentum concepts can then help us describe it.

The first application is when momentum changes… So, you got a thing with some momentum. Then something happens and there is less mass. Or more. Or the velocity changes.

This is sort of an abstract concept that helps us illustrate that (this is important!)…

within any closed system, momentum is conserved. Unless something different is introduced to the system, momentum is a constant.

Hence, within a closed system, any changes are limited to within that system, so the momentum can be thought of as having a first and second state (and 3rd, 4th, etc… if you so desired). So, we can formulize that to say that

p1 = p2

And since p = mv, then it follows that

m1v1 = m2v2

This will lead to questions where WITHIN THE CLOSED SYSTEM the mass or velocity will change (by means that somehow do not violate the "closed system" concept).

A far more realistic application is one in which something affects the moving object and thus causes a change to the momentum. This gives rise questions that examine the m1v1 = m2v2 in momentum from a first condition to a second (and third, fourth, fifth… if so desired).

This can be represented as…

∆p = p2 - p1

∆p =  m2v2  -  m1v1       <--- plug in                (eq. 1

Frequently, this will be seen when the velocity of an object changes. Since it is the same object

m1 = m2

So the formula can be represented as:

∆p = m(v2 - v1)                         (eq. 2

The astute observer will recognize that 

v2 - v1 = ∆v

Leading to the fancified version of equation 2 shown below

∆p = m∆v                         (eq. 3

A less frequent possibility is where mass changes (or when both mass and velocity change). Equation 1 can be a starting point for those situations.

IMPORTANT NOTE

Momentum is the product of velocity and mass, and velocity has a direction, so therefore, momentum has a direction (it is a vector). You will have to attend to the positive and negative values as directions.

Further, (LIKE ALL VECTORS), momentum need not occur on a single axis. Vectors can be deconstructed into orthogonal components, then all of the components can be recombined into resultant vectors.

Momentum In Action—A First Look: Inelastic Collisions

Key to understanding what's going when two (or more) objects collide is to know that, within a closed system momentum is conserved. 

That means that, at any point within the closed system the sum of all the momentums of all the objects in the system stays the same. You can't end up with more momentum than you start with!

So if there are 2 objects, and one has a momentum of 5 kg•m/s and the other has a momentum of 3 kg•m/s the total momentum will always be 8 kg•m/s. If the two objects interact with each other (collide) the distribution of momentum can change, but the total will always be 8.

Collisions

There are two types of collisions. 

In an elastic collision, two things collide and bounce off each other, going off separately after the collision.

In an inelastic collision, two things collide and stick to each other, going off as a single thing after the collision.

Collisions (Both types)

In a system involving a collision, we look at the two (or more) objects as they interact. We look at them in isolation, generally. That is to say, we only consider how they are interacting with each other.

In this case, we have a closed system in which momentum is conserved. That means that, where case 1 is before the collision and case 2 is after, 

pinitial = pfinal            (Eq. 1

Since the total momentum of a system is the momentum of all the objects in it, for any case we can say that

ptot = p1 + p2 + …

Inelastic Collisions

In an inelastic collision, two things collide and stick to each other, going off as a single thing after the collision.

Recalling that momentum is conserved, it follows that the momentum of the two objects before the collision must be equal to the momentum of the two objects after the collision. This…

pinitial = pfinal 

So, if there are 2 objects, object 1 and object 2, then the initial momentum is

pinitial = p1 + p2 

And since momentum is found as

p = mv

we can substitute in for p1 and p2

pinitial = p1 + p2 

pinitial = m1v1 + m2v2  

Now, for inelastic collisions, m1 and m2 become stuck to each other. They go off as one thing, with only one mass and only one velocity.

Okay… so, we have 2 objects and a before (initial) and after (final) situation. We could pile up subscrips such as…

m1v1i + m2v2i …

OR… we can change the symbol for initial velocity and keep things less messy… Let's do that…

 So, what we get is…

let u be initial velocity and v be final velocity

 m1u1 + m2u2 = (m1+ m2)v            (Eq. 3

I like to think of it as…

m3 = m1 + m2 

This makes the the equation

 m1u1 + m2u2 = m3v            (Eq. 3 alt

___________________________

We will look at inelastic collisions after we examine the relationship between force, time, mass, and momentum.
___________________________

Force and Momentum

Another important consideration of momentum is the forces involved in creating momentum change.

This concept assumes that, for some period of time, a force of some magnitude and direction act on an object causing a change in velocity which causes a change in momentum. Let's have a look!

            p = mv            (Fundamental momentum equation)

∆p = m∆v         (Eq. 1

Now, what causes velocity to change? Acceleration! 

∆v = a∆t        (Eq. 2

 

Now, how is force related to acceleration?

F = ma

a = F/m         (Eq. 3

Plug a from Eq. 3 into Eq. 2 we get

∆v = (F/m)∆t        (Eq. 4

Now, plug Eq. 4 into Eq. 1 to get…

∆p = m • F/m • ∆t        

∆p = F∆t      <--- Cancel the m in above               (Eq. 5

One more version…  Substitute Eq. 1 into Eq. 5:

m∆v = F∆t            (Eq. 6

Different situations will call for the use of different equations, so look at what is given and pick accordingly.

Up to this point, we have examined momentum only from the perspective of a single object. However, the interaction of more than one object leads many interesting outcomes. 

This is usually considered collisions

Elastic Collisions

Since momentum for any object is 

p = mv

then

ptot = m1v1 + m2v2 + … (Eq. 2

If we limit our discussion to two objects, we can roll Equation 2 back into Equation 1 and get

pinitial = pfinal 

m1v1i + m2v2i = m1v1f + m2v2f

Woooo… That's a lot of subscripts!

We can make the formula easier on the eyes if we do this…

let u be initial velocity and v be final velocity

Since mass before and after does not change, then:

pinitial = pfinal 

m1u1 + m2u2 = m1v1 + m2v2            (Eq. 3

 

Equation 3 is easy to solve if you have seven of the eight values given. 

Often, though, you need to find one of the final velocities, but know the value of neither. Yikes!

Through a fairly tedious algebra process, you can arrive at the following formulas:

${\displaystyle }$v1=(m1-m2 ) / ( m1+m2) • u1    +    2m2 / (m1+m2) • u2              (Eq. 4

${\displaystyle }$v2=(m2-m1) / (m1+m2) • u2  +    (2m1 / (m1+m2) • u1                    (Eq 5

You will see this in different forms, if you do research. You might see the second arranged like this:

${\displaystyle }$v2= (2m1 / (m1+m2) • u1   +  (m2-m1) / (m1+m2) • u2                    (Eq 5 alt

It's math! You can rearrange things following the… you know… math rules!

Doing a bit more algebra we can get to simpler forms:

v1 =  (m1-m2 ) • u1  +  2m2 • u2

                       (m1+m2)                            (Eq. 6

 

v2 =  (m2-m1 ) • u2  +  2m1 • u1

                       (m1+m2)                            (Eq. 7