Inside Collection (Book): Accessible Physics Concepts for Blind Students
Summary: This module explains momentum, impulse, and the conservation of momentum in a format that is accessible to blind students.
This module is part of a collection (see http://cnx.org/content/col11294/latest/ ) of modules designed to make physics concepts accessible to blind students. The collection is intended to supplement but not to replace the textbook in an introductory course in high school or college physics.
This module explains momentum, impulse, and the conservation of momentum in a format that is accessible to blind students.
In addition to an Internet connection and a browser, you will need the following tools (as a minimum) to work through the exercises in these modules:
The minimum prerequisites for understanding the material in these modules include:
I recommend that you open another copy of this document in a separate browser window and use the following links to easily find and view the figures and listings while you are reading about them.
I recommend that you also study the other lessons in my extensive collection of online programming tutorials. You will find a consolidated index at www.DickBaldwin.com .
This module will describe and discuss some scenarios which, never having been seen, may be hard for a blind student to imagine. Some of those scenarios can be difficult to believe even when you can see them.
The behavior of Newton's cradle is somewhat difficult to believe even when you see it in operation. Newton's cradle is a gadget that is often found in novelty shops. It typically consists of an open wood or metal frame with about five steel balls suspended by strings on parallel beams that run from one end to the other along the top.
Several steel balls in a row
Each ball is suspended by two strings so that the ball forms the lower vertex of a triangle and the two equallength strings form the sides of the triangle. Each string is attached at the upper end to a beam. The purpose of suspending each ball by two strings instead of suspending them on a single string is to cause all of the balls to swing back and forth along the same straight line.
A collision
When the system is in equilibrium, the balls are lined up in a row and each ball barely touches the one next to it. If you pull one of the balls at the end back and then release it, allowing it to swing down, it will strike its neighbor on the downswing.
Only the ball on the other end appears to move
Surprisingly, the neighboring ball doesn't appear to move when struck, nor does its neighbor, nor does that neighbor's neighbor. The only ball that appears to move is the ball at the far end of the line. That ball will be sent off in an upswing.
The process is reversed
When that ball reaches the top of its upswing, it will reverse direction, swing back down, and collide with its neighbor. This causes the ball that was used to start the process to be sent off in an upswing.
The process continues
Left alone, this process will continue until all of the energy in the system has been dissipated, which can be many minutes or even hours later.
Newton's cradle illustrates the conservation of momentum. You will find an interesting article that explains some of the technical details at http://en.wikipedia.org/wiki/Newton%27s_cradle.
You learned in an earlier module that momentum is the product of the mass of an object and the velocity of the object. Because velocity is a vector quantity, momentum is also a vector quantity. The direction of the momentum vector is the same as the direction of the underlying velocity vector.
A common symbol for momentum is p
A common symbol for momentum is p . Momentum is a derived item in the SI system of units. In the SI system, momentum is defined as kg*m/s.
In equation form , therefore:
Facts worth remembering  Momentum
The units of momentum are kg*m/s
Mass in action
Momentum can also be thought of as "mass in motion." Since all objects have mass, if an object is moving, its mass is in motion. Therefore, the object has momentum.
Proportional to both mass and velocity
From the above equation , it can be seen that an object can have a large momentum if either its mass or its velocity is large. Both variables are of equal importance in determining the momentum of an object.
A car and a tennis ball
Consider the case of a car and a tennis ball rolling down the street at the same speed. Because the car has greater mass, it has more momentum than the tennis ball. However, if the car stops and the tennis ball continues to roll, the tennis ball then has the greater momentum.
Momentum is zero at rest
The momentum of any object at rest is zero. Objects at rest do not have momentum because their mass is not in motion.
The quantity of momentum
The quantity of momentum possessed by an object depends on:
For example, a small mass moving very fast can have the same momentum as a large mass moving slowly. You sometimes hear about the major damage that a very small piece of space junk moving at a very high speed could do if it were to strike the International Space Station.
A bullet shot from a firearm has a very small mass, but it has a very high velocity. Consequently, it probably has more momentum than a baseball pitched from second base to home plate, even though the baseball has much more mass.
What happened to the dinosaurs?
Similarly, you may have heard that an asteroid with a mass that was small relative to the mass of the earth but with an extremely high velocity led to the extinction of the dinosaurs about 160 million years ago when it collided with the earth in the Gulf of Mexico.
Momentum can be changed by a force
An object with momentum can be stopped if a force is applied against it for a given amount of time. For example, when a car approaches a red traffic light, the driver applies the brakes. The friction of the tires on the pavement applies a force to the car, which eventually reduces the car's velocity to zero. When the velocity goes to zero, the momentum also goes to zero.
Therefore, the momentum of an object can be changed by applying a force to the object over a given period of time.
Unbalanced forces cause acceleration
As you learned in earlier modules, an unbalanced force always accelerates an object, either causing the object to speed up or causing the object to slow down. Either way, the application of an unbalanced force to an object will change the velocity of the object. When the velocity of the object is changed, the momentum of the object is changed as well.
The impulse
Let's use what we know from Newton's second law to derive a concept known as impulse .
The product of mass and acceleration
You learned in an earlier module that force is equal to the product of mass and acceleration:
F = m * a
where
The rate of change of velocity
You also learned that acceleration is the time rate of change of velocity, or
a = (v2  v1)/t
where
Combine and rearrange equations
Therefore, by substitution we can write:
F = m * (v2  v1)/t
Multiplying both sides by t gives
F * t = m * (v2  v1), or
F*t = m*v2  m*v1
where
A change in momentum
At this point, you should recognize that the product of mass and a change in velocity is a change in momentum.
In physics, the product of force and time is given the name impulse . It follows, therefore, that
impulse = F*t = change in momentum
This equation is often referred to as the impulsemomentum change equation .
Facts worth remembering  The impulsemomentum change equation
impulse = F*t = change in momentum
F*t = m*v2  m*v1
where
One area of physics where momentum plays a large part is the physics of collisions. Momentum and possible changes in momentum are involved in the interactions among all moving objects, even when the changes in momentum are not visually obvious.
For example, the passage of a moving iron object through the magnetic field of a moving magnet will cause changes in the momentum of both the iron object and the magnet, but the result may not be obvious.
Collisions may be more obvious
The type of interaction that we call a collision may be the type of interaction that is the most familiar to us. Collisions between objects happen all the time. This module will discuss several examples that involve collisions.
Collisions in everyday life
I may be wrong, but I suspect that as a blind student, you may be more attuned to collisions in everyday life than your fellow sighted students. For example, each time the end of your cane touches an object, a collision has occurred. Regardless of how light the touch, you probably recognize that collision and take appropriate action.
The impulsemomentum change equation
A law associated with the impulsemomentum change equation may be expressed in the following way (see http://www.physicsclassroom.com/Class/momentum/U4l1b.cfm ):
In a collision, an object experiences a force for a specific amount of time that results in a change in momentum. The result of the force acting for the given amount of time is that the object's mass either speeds up or slows down (or changes direction). The impulse experienced by the object equals the change in momentum of the object.
In equation form,
F * t = m * (v2  v1)
All objects in a collision experience an impulse
When a collision occurs, each object involved in the collision experiences an impulse. The impulse is equal to the change in momentum.
A hypothetical collision with a punching bag
In gyms where students practice boxing, there is usually a large object called a punching bag. Some punching bags are large cylindrical containers made of leather or some other pliable material filled with something like sand. Typically, they hang from the ceiling or from an overhead beam.
A void below the punching bag
Often, the punching bag is not attached to the floor and the bottom of the punching bag is often several feet above the floor. This leaves a void between the bottom of the punching bag and the floor.
A walk through the gym
Assume that you, as a blind student are walking through a gym that contains a punching bag of the type described above. Because of the void, you may not detect the presence of the punching bag with your cane and you might walk directly into the punching bag.
A force over a short period of time
Unlike a solid wall, the punching bag probably wouldn't stop your forward progress instantly. Instead, there would probably be a period of time during which the bag would exert a force on you and you would exert an equal but opposite force on the bag.
Initially, both you and the heavy punching bag would probably move in the direction that you are walking. Shortly thereafter, your forward velocity would probably go to zero.
Then the punching bag would probably push you backwards giving you a negative velocity as the bag swings like a pendulum. However the collision plays itself out, you would experience an impulse that would change your momentum and the momentum of the punching bag as well.
An ideal assumption
If we assume (ideally) that the force exerted on you by the bag is constant at ten newtons for a total of five seconds, the impulse would be equal to 50*N*s.
(In reality, the force would vary during the time interval and the computation of the impulse would be somewhat more complicated. For a timevarying force, the impulse is the area under a graph of force versus time. The average of the force over the given time interval cal also be used to compute the impulse when the force is not constant.)
The effect of the time interval
The impulse, or the change in momentum, is equal to the product of the force and the time. Therefore, a large force over a short period of time can produce the same change in momentum as is produced by a smaller force over a longer period of time.
The effect on the object experiencing the change in momentum can be quite different for the two cases. This is particularly true when the human body experiences a change in momentum. Among other things, air bags in cars are designed to lengthen the time over which the change in momentum occurs for a human body involved in a collision. In addition, the air bag can also spread the force over a larger portion of the body, which can reduce the damage to the body caused by the change in momentum.
Another representation of units
In any event, if we substitute the units for the newton in the above expression, we get
50*N*s = 50*(kg*m/s^2)*s = 50*kg*m/s
Two ways to represent the units of an impulse
Therefore, we can represent the units for an impulse as either N*s or as kg*m/s. You should recognize the latter as the product of mass and velocity, which are the same units as momentum.
In a collision, the impulse experienced by an object is always equal to the change in momentum experienced by the object.
A Super Ball
Consider the case of a Super Ball bouncing off of a solid concrete floor. (Super Ball is a brand name and registered trademark of WhamO Incorporated.)
The main characteristic of a Super Ball is its ability to bounce almost as high as the height from which it was released when dropped onto a solid surface. As a result, when the Super Ball collides with the floor, it demonstrates a strong rebound effect . The greater the rebound effect, the greater will be the acceleration, momentum change, and impulse in a collision.
A rebound collision
A rebound collision involves a change in direction in addition to a change in speed. Because the direction changes, there is a large velocity change even if the magnitude of the velocity stays the same.
Elastic collision
Collisions in which objects rebound with the same speed (and thus, the same momentum and kinetic energy) as they had prior to the collision are known as elastic collisions .
Stated differently, an elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies after the collision is equal to their total kinetic energy before the collision. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms.
Facts worth remembering  Elastic collision
An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies after the collision is equal to their total kinetic energy before the collision.
Energy conversion
A future module will explain kinetic energy and other forms of energy, such as potential energy in detail. Briefly, kinetic energy is energy possessed by a moving object simply because it is moving. For example, it hurts more to be hit by a fast moving baseball than to be hit by a slow moving baseball, simply because the collision with a fast moving baseball imparts more energy into your body. In other words, the kinetic energy possessed by the fastmoving baseball is converted into pain in your body.
Characteristics of an elastic collision
An elastic collision is typically characterized by a large velocity change, a large momentum change, a large impulse, and a large force.
While the case of a Super Ball bouncing on a solid concrete floor isn't a perfect elastic collision, it comes very close. The amount of kinetic energy that is converted into other forms of energy during each bounce is very low, and the ball will continue bouncing for a very long time with the height of each bounce being almost as high as the height of the previous bounce.
Two objects collide when they make contact while one or both are moving. As is the case with all interactions involving two or more moving objects, a collision results in a force being applied to all of the objects involved in the collision. The behavior of such collisions is governed by Newton's laws of motion.
Newton's third law
One paraphrased version of Newton's third law (see http://www.physicsclassroom.com/Class/momentum/u4l2a.cfm ) reads:
... in every interaction, there is a pair of forces acting on the two interacting objects. The size of the force on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs  equal and opposite actionreaction force pairs.
According to Newton's third law, when two objects are involved in a collision, the two objects experience forces that are equal in magnitude and opposite in direction.
In most cases, the collision will cause one object to gain momentum and the other object to lose momentum. This, in turn, will cause one object to speed up and the other object to slow down.
Railroad cars
As a child, I grew up living next to a very large railroad yard. I was accustomed to hearing the sounds of controlled collisions between railroad cars.
How railroad couples work
The devices that hold railroad cars together in a train are activated by a controlled collision. While one railroad car is either standing still, or moving at a slow speed, another railroad car purposely collides with the first car. When that happens, the two railroad cars become fastened together (coupled).
The distribution of momentum
Prior to the collision, each car possesses a given amount of momentum, which can be zero for a car at rest or nonzero for a car in motion. After the collision, the momentum of each car will have changed.
The conservation of momentum  a preview
As I will explain later, (except for the conversion of some kinetic energy into other forms, such as sound energy) the two cars coupled together will possess the total amount of momentum that was possessed by the individual cars prior to the collision. This typically means that one car speeds up and the other car slows down.
A change in momentum
During the time frame surrounding the instant of the collision, each car experiences a change in momentum. With sufficiently accurate measuring devices, you could measure and record the rate of change of momentum during that short time frame.
Accelerations are not necessarily equal
Although the forces experienced by the objects are equal in magnitude, the changes in velocity (accelerations) experienced by the two objects are not necessarily equal.
The acceleration is equal to...
As we learned in an earlier module, the acceleration experienced by an object is proportional to the applied force and inversely proportional to the mass of the object. Therefore, different masses experiencing the same magnitude of force will experience different magnitudes of acceleration.
Kicking a lightweight aluminum can
Consider what happens when someone leaves an empty lightweight aluminum drink can on the floor and you accidently kick it with your bare foot while walking briskly across the room. The can exerts a force on your foot, which fortunately doesn't hurt too much because of the small mass of the can. Your foot exerts an equal and opposite force on the can that probably sends it flying across the room.
The velocity of the can changes by a large amount, going from zero to a velocity that sends it flying. Thus, the acceleration of the can is large.
The velocity of your foot, on the other hand, changes very little at the moment of impact as a result of its large mass (although your muscular reaction might be such as to slow the foot down shortly thereafter). Thus, the magnitude of the acceleration of your foot due solely to the collision is small.
Kicking a heavy can
Now consider the same scenario except that this time, the can contains your brother's collection of rocks. In this case, the acceleration of the can due to the collision with your foot would probably be quite small, and the (negative) acceleration of your foot due to the collision might be much larger than with the lightweight can.
The can of rocks probably wouldn't go flying across the room, but might simply turn over, spilling the rocks on the floor.
The negative acceleration experienced by your foot might result in some broken toes.
Once again, the forces would be equal in magnitude and opposite in direction, but the acceleration of each object would be inversely proportional to the mass of the object.
Equal and opposite forces
When you kick the can, your foot would exert a force on the can in the direction of motion of your foot. The can would exert a force on your foot that is equal in magnitude but opposite in direction.
The forward force on the can would cause it to gain velocity in the direction that your foot is moving. The backward force on your foot would cause your foot to slow down.
Equal acceleration
In the unlikely event that the mass of the can is exactly equal to the mass of your foot, the negative acceleration experienced by your foot would be equal to the positive acceleration exerted on the can. That is the one case where not only the magnitudes of the forces, but also the magnitudes of the accelerations would be equal.
For collisions between equalmass objects, each object experiences the same acceleration.
When two objects interact in an isolated system , the total momentum of the two objects before the interaction is equal to the total momentum of the two objects after the interaction. The momentum lost by one object is gained by the other object.
Facts worth remembering  Isolated system
An isolated system is a system that is free from the influence of a net external force that alters the momentum of the system.
The total momentum of a collection of objects in a system is conserved. The total amount of momentum is constant.
Many forms of interaction are possible
There are many ways that two objects can interact. For example, when a car pulls away from a stoplight, it gains momentum by exerting frictional forces on the surface of the earth. When that happens, the earth loses an equal amount of momentum. (Fortunately, this represents a very small fraction of the earth's momentum, so the loss of momentum isn't noticeable.)
When the driver applies the brakes and stops the car at the next stop light (by exerting frictional forces on the surface of the earth), the car loses all of its momentum and the earth gains an equal amount of momentum. (Once again, this represents a very small fraction of the earth's momentum, so the gain of momentum isn't noticeable.)
Railroad cars and controlled collisions
Earlier in this module, I described a process where controlled collisions are used to couple railroad cars together.
While one railroad car is either standing still, or moving at a slow speed, another railroad car purposely collides with that car. When that happens, the two railroad cars become fastened together (coupled).
Distribution of momentum
Prior to the collision, each car possesses a given amount of momentum, which can be zero for a car at rest or nonzero for a car in motion. After the collision, the momentum of each car will have changed.
Except for the conversion of some kinetic energy into other forms, such as sound energy, the two cars coupled together will possess the total amount of momentum that was possessed by the individual cars prior to the collision.
This typically means that one car speeds up and the other car slows down. During the time frame surrounding the instant of the collision, each car experiences a change in momentum.
A logical proof of the conservation of momentum
We already know that when two objects interact, each object exerts a force on the other object. The two forces are equal in magnitude and opposite in direction.
We can probably agree that when two objects interact, the amount of time that objectA interacts with objectB is the same as the amount of time that objectB interacts with objectA.
Therefore, if the forces are equal and opposite, and the times are the same, we can write
Fa*t = Fb*t
where
Impulses acting on objects A and B
You should recognize these terms as the impulses acting on objects A and B. Therefore, the impulses acting on the two objects are equal and opposite.
Impulses are equal and object
You learned earlier that the impulse acting on an object is equal to the change in momentum of the object. If the impulses acting on the two objects are equal and opposite, then the change in momentum experienced by the two objects must be equal and opposite.
We can express this in equation form as
ma*(va2  va1) = mb*(vb2  vb1)
where
The law of conservation of momentum
This equation is a statement of the law of conservation of momentum. The change in momentum experienced by objectA is equal to and opposite of the change in momentum experienced by objectB.
Stated differently, the momentum lost by objectA is gained by objectB, or viceversa.
That being the case, the total momentum possessed by the system containing objectA and objectB remains unchanged by the interaction of the two objects. The total momentum of the system is conserved.
Facts worth remembering  The law of conservation of momentum
For a collision occurring between objectA and objectB in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. (By total momentum we mean the vector sum of the individual momenta of the objects.)
Bowling ball and bowling pins
Bowling is a game where the players roll a heavy ball down a long smooth wooden platform in an attempt to knock down ten heavy wooden objects (bowling pins) arranged in a triangle at the end of the platform. The bowling pins are shaped something like a flower vase that is larger at the bottom than at the top. Thus, each bowling pin has a low center of gravity.
When the ball strikes the cluster of bowling pins, there are eleven objects involved in a conservation of momentum process. The momentum that is lost by the bowling ball is distributed among the ten bowling pins, but the momentum is probably not distributed evenly among the bowling pins. There is a lot of chaos at the time of impact with the bowling ball colliding with the pins, pins colliding with other pins, etc. Some of the energy is also converted to sound.
Interactions between parts of a system transfer momentum between the parts, but do not change the total momentum of the system. We can define a point called the center of mass that serves as an average location of a system of parts.
The center of mass need not necessarily be at a location that is either in or on one of the parts. For example, the center of mass of a pair of heavy rods connected at one end so as to form a "V" shape is somewhere in space between the two rods.
Having determined the center of mass for a system, we can treat the mass of the system as if it were all concentrated at the center of mass.
For a system composed of two masses, the center of mass lies somewhere on a line between the two masses. The center of mass is a weighted average of the positions of the two masses.
Facts worth remembering  Center of mass for two objects
For a pair of masses located at two points along the xaxis, we can write
xcm = (m1*x1/M) + (m2*x2/M)
where
Multiple masses in three dimensions
When we have multiple masses in three dimensions, the definition of the center of mass is somewhat more complicated.
Facts worth remembering  Center of mass for many objects
Vector form:
rcm = sum over all i(mi*ri / M)
Component form:
xcm = sum over all i(mi*xi / M)
ycm = sum over all i(mi*yi / M)
zcm = sum over all i(mi*zi / M)
where
It can be shown that in an isolated system, the center of mass must move with constant velocity regardless of the motions of the individual particles.
It can be shown that in a nonisolated system, if a net external force acts on a system, the center of mass does not move with constant velocity. Instead, it moves as if all the mass were concentrated there into a fictitious point particle with all the external forces acting on that point.
This section contains explanations and computations involving momentum, impulse, action and reaction, and the conservation of momentum.
This section contains several examples involving momentum
Use the Google calculator to compute the momentum of a 70kg sprinter running 30 m/s at 0 degrees.
Answer: 2100 kg*m/s at 0 degrees
Use the Google calculator to compute the momentum in kg*m/s of a 2205lb truck traveling 33.6 miles per hour at 0 degrees when the changes listed below occur:
Answers:
1. Enter the following into the Google calculator and press Enter to produce the results shown.
Therefore, the initial momentum = 15000*kg*m/s at 0 degrees
2. 1000 kg*30 m/s = 30000 m kg / s at 0 degrees
3. 2000 kg*15 m/s = 30000 m kg / s at 0 degrees
4. 2000 kg*30 m/s = 60000 m kg / s at 0 degrees
A car with a weight of 10000 newtons is moving in a direction of 90 degrees at 40 m/s. After going around a curve in the road, the car is moving in a direction of 0 degrees at 20 m/s. What is the change in momentum of the car?
Solution:
While this problem could be solved using the Google calculator, because of the number of steps involved, JavaScript is probably a better approach.
The solution script for this problem is shown in Listing 1 .
<! File JavaScript01.html >
<html><body>
<script language="JavaScript1.3">
//The purpose of this function is to receive the adjacent
// and opposite side values for a right triangle and to
// return the angle in degrees in the correct quadrant.
function getAngle(x,y){
if((x == 0) && (y == 0)){
//Angle is indeterminate. Just return zero.
return 0;
}else if((x == 0) && (y > 0)){
//Avoid divide by zero denominator.
return 90;
}else if((x == 0) && (y < 0)){
//Avoid divide by zero denominator.
return 90;
}else if((x < 0) && (y >= 0)){
//Correct to second quadrant
return Math.atan(y/x)*180/Math.PI + 180;
}else if((x < 0) && (y <= 0)){
//Correct to third quadrant
return Math.atan(y/x)*180/Math.PI + 180;
}else{
//First and fourth quadrants. No correction required.
return Math.atan(y/x)*180/Math.PI;
}//end else
}//end function getAngle
document.write("Start Script </br>");
var weight = 10000//N
var g = 9.8// m/s^2
//Find the mass of the car
var mass = weight/g;// kg
var ang1 = 90;//initial angle in degrees
var ang2 = 0; //final angle in degrees
var speed1 = 40;//initial speed in m/s
var speed2 = 20;//final speed in m/s
var ang1r = ang1*Math.PI/180;//initial angle in radians
var ang2r = ang2*Math.PI/180;//final angle in radians
//Remember, momentum is a vector quantity and momenta must
// be added and subtracted using vector arithmetic.
//Compute the components of the change in momentum.
var P1x = mass * speed1 * Math.cos(ang1r);
var P1y = mass * speed1 * Math.sin(ang1r);
var P2x = mass * speed2 * Math.cos(ang2r);
var P2y = mass * speed2 * Math.sin(ang2r);
var deltaPx = P2xP1x;//change in horizontal component
var deltaPy = P2yP1y;//change in vertical component
//Compute the magnitude of the change in momentum using
// the Pythagorean theorem.
var deltaPm = Math.sqrt(deltaPx*deltaPx + deltaPy*deltaPy);
//Compute the angle of the change in momentum usiing
// trigonometry.
var deltaPa = getAngle(deltaPx,deltaPy);
document.write("The givens." + "</br>");
document.write("weight = " + weight.toFixed(0)
+ " kg</br>");
document.write("speed1 = " + speed1.toFixed(0)
+ " m/s</br>");
document.write("angle 1 = " + ang1.toFixed(0)
+ " degrees</br>");
document.write("speed2 = " + speed2.toFixed(0)
+ " m/s</br>");
document.write("angle 2 = " + ang2.toFixed(0)
+ " degrees</br>");
document.write("Computed mass." + "</br>");
document.write("mass = " + mass.toFixed(0) + " kg</br>");
document.write("Components of momentum vectors." + "</br>");
document.write("P1x = " + P1x.toFixed(0) + "</br>");
document.write("P1y = " + P1y.toFixed(0) + "</br>");
document.write("P2x = " + P2x.toFixed(0) + "</br>");
document.write("P2y = " + P2y.toFixed(0) + "</br>");
document.write("Components of momentum change vectors."
+ "</br>");
document.write("deltaPx = " + deltaPx.toFixed(0) + "</br>");
document.write("deltaPy = " + deltaPy.toFixed(0) + "</br>");
document.write("Magnitude and angle of change vector."
+ "</br>");
document.write("deltaPm = " + deltaPm.toFixed(0)
+ " m kg/s</br>");
document.write("deltaPa = " + deltaPa.toFixed(0)
+ " degrees</br>");
document.write("End Script");
</script>
</body></html>
The comments in Listing 1 explain the steps involved in finding the solution.
The output produced by Listing 1 is shown in Figure 1 with the magnitude and angle of the vector that describes the change of momentum at the end.
Solution results.  


I find it interesting that the magnitude of the change in momentum is greater than the magnitude of either the initial or final momentum.
What happens if the car in the previous example changes speed but doesn't change direction.
Solution:
Change the given conditions in the script in Listing 1 to those shown at the beginning of Figure 2 .
Change in speed only.  


This change causes the car to slow down, but to continue in the same direction. As a result, the angle of the change in momentum is an angle that is opposite to the direction that the car is moving. The magnitude of the change in momentum depends entirely on the initial and final speeds.
What happens if the car in the previous example changes direction but doesn't change speed?
Solution:
Change the given conditions in the script in Listing 1 to those shown at the beginning of Figure 3 . This scenario simulates the car making a 10degree turn to the right without changing speed.
Change in direction only.  


This section contains several examples involving the impulse.
1. What is the impulse experienced by pushing a 10kg wagon that was initially at rest, with a constant force of 2 newtons for a period of 3 seconds?
Answer:
The impulse is given by the product of force and time. The mass of the wagon is superfluous for this question.
impulse = 2 N * 3 s = 6*N*s
2. What is the acceleration of the wagon in question 1 above?
Answer:
Now we do need to know the mass.
The most straightforward solution comes from the fact that we know the mass and that the force is uniform. Therefore,
F = m*a, or
a = F/m = 2N/10kg = 0.2 m/s^2
A more interesting solution comes from the fact that since
impulse = F*t, and
F = m*a, then
impulse = m*a*t, or
a = impulse/(m*t) = 6*N*s/(10*kg*3*s) = 0.2 m/s^2
3. What is the velocity at the end of the 3second interval in question 1 above.
Answer:
The impulse is equal to the change in momentum, and the initial velocity is 0.
impulse = m*(v2  v1) = m*v2, or
v2 = impulse/m = 6*N*s/(10*kg) = 0.6 m/s
We can check that answer by knowing that the acceleration is uniform at 0.3 m/s^2 for 3 s = 0.6 m/s.
A dip in the pool
You have a body mass of 70 kg. You are on your knees on an inflatable raft in a swimming pool. The raft has a mass of 1 kg. Your outstretched hands are about two meters from a safety rope that is strung across the pool.
You decide to launch yourself from the raft to catch the rope, exerting a force with a horizontal component of 5 newtons. (You assume that the vertical component of your launching force will take care of the downward pull of gravity, allowing you to fly in a parabolic arc to the rope.) Assuming uniform acceleration (which is unrealistic but we will assume that anyway), how long will it take you to fly through the air to reach the rope?
Answer:
The force that you exert on the raft will be equal and opposite to the force that the raft exerts on you. Therefore,
F = my*ay = mr*ar
where
Therefore
ay = F/my = 5N/70kg = 0.07143 m/s^2 toward the rope
ar = F/mr = 5N/1kg = 5.00000 m/s^2 away from the rope
We learned in an earlier module that given a constant acceleration, the distance traveled versus time is:
d = v0t + 0.5*a*t^2
In this case, v0 is zero, so
d = 0.5*ay*t^2, or
t = sqrt(d/(0.5*ay)), or
t = sqrt(2m/(0.5*0.07143m/s^2)) = 7.48324 seconds
I doubt that you will stay in the air long enough to reach the rope.
During that time period, the raft will travel the following distance in the opposite direction (assuming no resistance from the water).
d = 0.5*ar*t^2, or
d = 0.5*(5m/s^2)*(7.48324s)^2 = 140 meters
Railroad cars
Getting back to my example of coupling railroad cars, when the collision has been completed, the two masses have effectively been joined into a single mass and they are moving at the same velocity.
In that case, we can write the above equation as
ma*(v2  va1) = mb*(v2  vb1)
ma*v2 +mb*v2 = ma*va1 +mb*vb1
v2 = (ma*va1 +mb*vb1)/(ma + mb)
where
Then for any set of assumed mass values for the railroad cars and assumed values for the initial velocities, we can calculate the final velocity of the coupled pair of railroad cars.
Scenario #1: Assume that the two railroad cars are just alike and empty giving them the same mass. Also assume that the initial velocity for CarA is 10 m/s and the initial velocity for Carb is 0.
Question: What would be the final velocity of the coupled railroad cars?
Answer: For this scenario, we have
v2 = (ma*va1 +mb*vb1)/(ma + mb), or
v2 = (ma*10)/2*ma = 5 m/s
The final velocity of the pair of coupled cars is half the initial velocity of the car that was moving.
Scenario #2: Now assume that due to loading, CarA has twice the mass of CarB, the initial velocity for CarA is 10m/s and the initial velocity for CarB is 0.
Question: What would be the final velocity of the coupled railroad cars?
Answer: For this scenario, we have
v2 = (ma*va1 +mb*vb1)/(ma + mb), or
v2 = (ma*va1)/(ma + 0.5ma), or
v2 = ma*10/1.5*ma = 6.67 m/s
When the car at rest is less massive than the car in motion, the final velocity is a little higher than when the two cars have the same mass.
Scenario #3: Finally, assume that due to loading, CarA has twice the mass of CarB, the initial velocity for CarA is 10m/s and the initial velocity for CarB is 5m/s.
Question: What would be the final velocity of each railroad car?
Answer: For this scenario, we have
v2 = (ma*va1 +mb*vb1)/(ma + mb), or
v2 = (ma*10 +0.5ma*5)/(1.5*ma), or
v2 = (10/1.5) + (0.5*5/1.5) = 8.33 m/s
When both cars are already moving in the same direction, the final velocity in that direction is greater than when one of the cars is stationary.
This section contains solutions to problems involving the center of mass.
Two objects are located on a flat lawn with the following mass values and locations:
What are the coordinates of the center of mass?
Solution:
A JavaScript script that will solve this problem is shown in Listing 2 .
<! File JavaScript02.html >
<html><body>
<script language="JavaScript1.3">
document.write("Start Script </br>");
//Create arrays for mass,xCoor, and yCoor values;
var mass = new Array(15,3);
var xCoor = new Array(1,4);
var yCoor = new Array(5,2);
//Declare a counter variable.
var cnt = 0;
//Use a loop to compute total mass by summing the individual
// mass values.
var massTotal = 0;
for(cnt = 0;cnt < mass.length;cnt++){
massTotal += mass[cnt];
}//end for loop
//Use a loop to compute xcoordinate of the center of mass
// by summing the normalized sum of products.
var cmX = 0;
for(cnt = 0;cnt < mass.length;cnt++){
cmX += mass[cnt]*xCoor[cnt]/massTotal
}//end for loop
//Use a loop to compute ycoordinate of the center of mass
// by summing the normalized sum of products.
var cmY = 0;
for(cnt = 0;cnt < mass.length;cnt++){
cmY += mass[cnt]*yCoor[cnt]/massTotal
}//end for loop
//Display the results
document.write("massTotal = " + massTotal + " kg</br>");
document.write("cmX = " + cmX + " meters</br>");
document.write("cmY = " + cmY + " meters</br>");
document.write("End Script");
</script>
</body></html>
The comments describe how the problem is solved.
The output is shown in Figure 4 .
Center of mass for two objects.  


The xcoordinate for the center of mass is 1.5 meters, and the ycoordinate for the center of mass is 4.5 meters.
Three objects are located on a flat lawn with the following mass values and locations:
What are the coordinates of the center of mass?
Use the code from Listing 2 but make the change shown in Listing 3 in order to populate the world with three objects having different mass values and different coordinates.
//Create arrays for mass,xCoor, and yCoor values;
var mass = new Array(5,5,10);
var xCoor = new Array(1,3,2);
var yCoor = new Array(1,1,3);
The output is shown in Figure 5 .
Center of mass for three objects.  


I encourage you to repeat the calculations that I have presented in this lesson to confirm that you get the same results. Experiment with the scenarios, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.
I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modules in this collection are published.
This section contains a variety of miscellaneous information.
Financial : Although the Connexions site makes it possible for you to download a PDF file for this module at no charge, and also makes it possible for you to purchase a preprinted version of the PDF file, you should be aware that some of the HTML elements in this module may not translate well into PDF.
I also want you to know that I receive no financial compensation from the Connexions website even if you purchase the PDF version of the module.
Affiliation : I am a professor of Computer Information Technology at Austin Community College in Austin, TX.
end
"Blind students should not be excluded from physics courses because of inaccessible textbooks. The modules in this collection present physics concepts in a format that blind students can read […]"