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Michael Fowler, U. Va. Physics.
At this point, we introduce some further conceptsthat will prove useful in describing motion. The first of these, momentum,was actually introduced by the French scientist and philosopher Descartesbefore Newton. Descartes’ idea is best understood by considering asimple example: think first about someone (weighing say 45 kg) standingmotionless on high quality (frictionless) rollerskates on a level smooth floor.A 5 kg medicine ball is thrown directly at her by someone standing in front ofher, and only a short distance away, so that we can take the ball’sflight to be close to horizontal. She catches and holds it, and becauseof its impact begins to roll backwards. Notice we’ve chosen herweight so that, conveniently, she plus the ball weigh just ten times what theball weighs by itself. What is found on doing this experiment carefullyis that after the catch, she plus the ball roll backwards at just one-tenth thespeed the ball was moving just before she caught it, so if the ball was thrownat 5 meters per second, she will roll backwards at one-half meter per secondafter the catch. It is tempting to conclude that the “total amountof motion” is the same before and after her catching the ball, since weend up with ten times the mass moving at one-tenth the speed.
Considerations and experiments like this led Descartes to invent the conceptof “momentum”, meaning “amount of motion”, and to statethat for a moving body the momentum was just the product of the mass of thebody and its speed. Momentum is traditionally labeled by the letter p,so his definition was:
momentum = p =mv
for a body having mass m and moving at speed v. It is thenobvious that in the above scenario of the woman catching the medicine ball,total “momentum” is the same before and after the catch. Initially,only the ball had momentum, an amount 5x5 = 25 in suitable units, since itsmass is 5kg and its speed is 5 meters per second. After the catch, thereis a total mass of 50kg moving at a speed of 0.5 meters per second, so thefinal momentum is 0.5x50 = 25, the total final amount is equal to the totalinitial amount. We have just invented these figures, of course, but theyreflect what is observed experimentally.
There is however a problem here—obviously one can imagine collisionsin which the “total amount of motion”, as defined above, isdefinitely not the same before and after. What about two people onrollerskates, of equal weight, coming directly towards each other at equal butopposite velocities—and when they meet they put their hands together andcome to a complete halt? Clearly in this situation there was plenty ofmotion before the collision and none afterwards, so the “total amount ofmotion” definitely doesn’t stay the same! In physics language, itis “not conserved”. Descartes was hung up on this problem along time, but was rescued by a Dutchman, Christian Huygens, who pointed outthat the problem could be solved in a consistent fashion if one did not insistthat the “quantity of motion” be positive.
In other words, if something moving to the right was taken to havepositive momentum, then one should consider something moving to the left tohave negative momentum. With this convention, two people of equalmass coming together from opposite directions at the same speed would havetotal momentum zero, so if they came to a complete halt after meeting,as described above, the total momentum before the collision would be the sameas the total after—that is, zero—and momentum would beconserved.
Of course, in the discussion above we are restricting ourselves to motionsalong a single line. It should be apparent that to get a definition ofmomentum that is conserved in collisions what Huygens really did was to tellDescartes he should replace speed by velocity in his definition ofmomentum. It is a natural extension of this notion to think of momentumas defined by
momentum = mass x velocity
in general, so, since velocity is a vector, momentum is also a vector,pointing in the same direction as the velocity, of course.
It turns out experimentally that in any collision between two objects(where no interaction with third objects, such as surfaces, interferes), thetotal momentum before the collision is the same as the total momentum after thecollision. It doesn’t matter if the two objects stick together oncolliding or bounce off, or what kind of forces they exert on each other, soconservation of momentum is a very general rule, quite independent of detailsof the collision.
Momentum Conservation and Newton’s Laws
As we have discussed above, Descartes introduced the concept of momentum,and the general principle of conservation of momentum in collisions, before Newton’s time. However, it turns out that conservation of momentum can bededuced from Newton’s laws. Newton’s laws in principle fullydescribe all collision-type phenomena, and therefore must contain momentumconservation.
To understand how this comes about, consider first Newton’s Second Lawrelating the acceleration a of a body of mass m with an externalforce F acting on it:
F = ma,or force = mass x acceleration
Recall that acceleration is rate of change of velocity, so we can rewritethe Second Law:
force = mass x rateof change of velocity.
Now, the momentum is mv, mass x velocity. This means for anobject having constant mass (which is almost always the case, of course!)
rate of change ofmomentum = mass x rate of change of velocity.
This means that Newton’s Second Law can be rewritten:
force = rate ofchange of momentum.
Now think of a collision, or any kind of interaction, between two objects Aand B, say. From Newton’s Third Law, the force Afeels from B is of equal magnitude to the force B feels from A,but in the opposite direction. Since (as we have just shown) force = rateof change of momentum, it follows that throughout the interaction process therate of change of momentum of A is exactly opposite to the rate ofchange of momentum of B. In other words, since these are vectors,they are of equal length but pointing in opposite directions. This meansthat for every bit of momentum A gains, B gains the negative ofthat. In other words, B loses momentum at exactly the rate Agains momentum so their total momentum remains the same. Butthis is true throughout the interaction process, from beginning to end. Therefore,the total momentum at the end must be what it was at the beginning.
You may be thinking at this point: so what? We already know that Newton’s laws are obeyed throughout, so why dwell on one special consequence of them?The answer is that although we know Newton’s laws are obeyed, thismay not be much use to us in an actual case of two complicated objectscolliding, because we may not be able to figure out what the forces are. Nevertheless,we do know that momentum will be conserved anyway, so if, for example,the two objects stick together, and no bits fly off, we can find their finalvelocity just from momentum conservation, without knowing any details of thecollision.
The word “work” as used in physics has a narrower meaning thanit does in everyday life. First, it only refers to physical work, ofcourse, and second, something has to be accomplished. If you lift up abox of books from the floor and put it on a shelf, you’ve done work, asdefined in physics, if the box is too heavy and you tug at it until you’reworn out but it doesn’t move, that doesn’t count as work.
Technically, work is done when a force pushes something and the object movessome distance in the direction it’s being pushed (pulled is ok, too).Consider lifting the box of books to a high shelf. If you lift the box ata steady speed, the force you are exerting is just balancing off gravity, theweight of the box, otherwise the box would be accelerating. (Of course,initially you’d have to exert a little bit more force to get it going,and then at the end a little less, as the box comes to rest at the height ofthe shelf.) It’s obvious that you will have to do twice as muchwork to raise a box of twice the weight, so the work done is proportional tothe force you exert. It’s also clear that the work done depends onhow high the shelf is. Putting these together, the definition of work is:
work = force xdistance
where only distance traveled in the direction the force is pushing counts.With this definition, carrying the box of books across the room from one shelfto another of equal height doesn’t count as work, because even thoughyour arms have to exert a force upwards to keep the box from falling to thefloor, you do not move the box in the direction of that force, that is, upwards.
To get a more quantitative idea of how much work is being done, we need to havesome units to measure work. Defining work as force x distance, as usualwe will measure distance in meters, but we haven’t so far talked aboutunits for force. The simplest way to think of a unit of force is in termsof Newton’s Second Law, force = mass x acceleration. The natural “unitforce” would be that force which, pushing a unit mass (one kilogram) withno friction of other forces present, accelerates the mass at one meter persecond per second, so after two seconds the mass is moving at two meters persecond, etc. This unit of force is called one newton (as wediscussed in an earlier lecture). Note that a one kilogram mass, whendropped, accelerates downwards at ten meters per second per second. Thismeans that its weight, its gravitational attraction towards the earth, must beequal to ten newtons. From this we can figure out that a one newton forceequals the weight of 100 grams, just less than a quarter of a pound, a stick ofbutter.
The downward acceleration of a freely falling object, ten meters per secondper second, is often written g for short. (To be precise, g= 9.8 meters per second per second, and in fact varies somewhat over the earth’ssurface, but this adds complication without illumination, so we shall alwaystake it to be 10.) If we have a mass of m kilograms, say, we know itsweight will accelerate it at g if it’s dropped, so its weight is aforce of magnitude mg, from Newton’s Second Law.
Now back to work. Since work is force x distance, the natural “unitof work” would be the work done be a force of one newton pushing adistance of one meter. In other words (approximately) lifting a stick ofbutter three feet. This unit of work is called one joule,in honor of an English brewer.
Finally, it is useful to have a unit for rate of working, also called“power”. The natural unit of “rate of working” ismanifestly one joule per second, and this is called one watt. Toget some feeling for rate of work, consider walking upstairs. A typicalstep is eight inches, or one-fifth of a meter, so you will gain altitude at,say, two-fifths of a meter per second. Your weight is, say (put in yourown weight here!) 70 kg. (for me) multiplied by 10 to get it in newtons, so it’s700 newtons. The rate of working then is 700 x 2/5, or 280 watts. Mostpeople can’t work at that rate for very long. A common English unitof power is the horsepower, which is 746 watts.
Energy is the ability to do work.
For example, it takes work to drive a nail into a piece of wood—aforce has to push the nail a certain distance, against the resistance of thewood. A moving hammer, hitting the nail, can drive it in. Astationary hammer placed on the nail does nothing. The moving hammer hasenergy—the ability to drive the nail in—because it’s moving.This hammer energy is called “kinetic energy”. Kineticis just the Greek word for motion, it’s the root word for cinema,meaning movies.
Another way to drive the nail in, if you have a good aim, might be to simplydrop the hammer onto the nail from some suitable height. By the time thehammer reaches the nail, it will have kinetic energy. It has this energy,of course, because the force of gravity (its weight) accelerated it as it camedown. But this energy didn’t come from nowhere. Work had tobe done in the first place to lift the hammer to the height from which it wasdropped onto the nail. In fact, the work done in the initial lifting,force x distance, is just the weight of the hammer multiplied by the distanceit is raised, in joules. But this is exactly the same amount of work asgravity does on the hammer in speeding it up during its fall onto the nail.Therefore, while the hammer is at the top, waiting to be dropped, it can bethought of as storing the work that was done in lifting it, which is ready tobe released at any time. This “stored work” is called potentialenergy, since it has the potential of being transformed into kineticenergy just by releasing the hammer.
To give an example, suppose we have a hammer of mass 2 kg, and we lift it upthrough 5 meters. The hammer’s weight, the force of gravity, is 20newtons (recall it would accelerate at 10 meters per second per second undergravity, like anything else) so the work done in lifting it is force x distance= 20 x 5 = 100 joules, since lifting it at a steady speed requires a liftingforce that just balances the weight. This 100 joules is now stored readyfor use, that is, it is potential energy. Upon releasing the hammer, thepotential energy becomes kinetic energy—the force of gravity pulls thehammer downwards through the same distance the hammer was originally raisedupwards, so since it’s a force of the same size as the original liftingforce, the work done on the hammer by gravity in giving it motion is the sameas the work done previously in lifting it, so as it hits the nail it has akinetic energy of 100 joules. We say that the potential energy istransformed into kinetic energy, which is then spent driving in the nail.
We should emphasize that both energy and work are measured in the sameunits, joules. In the example above, doing work by lifting just addsenergy to a body, so-called potential energy, equal to the amount of work done.
From the above discussion, a mass of m kilograms has a weight of mgnewtons. It follows that the work needed to raise it through a height hmeters is force x distance, that is, weight x height, or mgh joules.This is the potential energy.
Historically, this was the way energy was stored to drive clocks. Largeweights were raised once a week and as they gradually fell, the released energyturned the wheels and, by a sequence of ingenious devices, kept the pendulumswinging. The problem was that this necessitated rather large clocks toget a sufficient vertical drop to store enough energy, so spring-driven clocksbecame more popular when they were developed. A compressed spring is justanother way of storing energy. It takes work to compress a spring, but(apart from small frictional effects) all that work is released as the springuncoils or springs back. The stored energy in the compressed spring isoften called elastic potential energy, as opposed to the gravitationalpotential energy of the raised weight.
We’ve given above an explicit way to find the potential energyincrease of a mass m when it’s lifted through a height h,it’s just the work done by the force that raised it, force x distance =weight x height = mgh.
Kinetic energy is created when a force does work accelerating a mass andincreases its speed. Just as for potential energy, we can find thekinetic energy created by figuring out how much work the force does in speedingup the body.
Remember that a force only does work if the body the force is acting onmoves in the direction of the force. For example, for a satellite goingin a circular orbit around the earth, the force of gravity is constantlyaccelerating the body downwards, but it never gets any closer to sea level, itjust swings around. Thus the body does not actually move any distance inthe direction gravity’s pulling it, and in this case gravity does no workon the body.
Consider, in contrast, the work the force of gravity does on a stone that issimply dropped from a cliff. Let’s be specific and suppose it’sa one kilogram stone, so the force of gravity is ten newtons downwards. Inone second, the stone will be moving at ten meters per second, and will havedropped five meters. The work done at this point by gravity is force xdistance = 10 newtons x 5 meters = 50 joules, so this is the kinetic energy ofa one kilogram mass going at 10 meters per second. How does the kineticenergy increase with speed? Think about the situation after 2 seconds. Themass has now increased in speed to twenty meters per second. It hasfallen a total distance of twenty meters (average speed 10 meters per second xtime elapsed of 2 seconds). So the work done by the force of gravity inaccelerating the mass over the first two seconds is force x distance = 10newtons x 20 meters = 200 joules.
So we find that the kinetic energy of a one kilogram mass moving at 10meters per second is 50 joules, moving at 20 meters per second it’s 200joules. It’s not difficult to check that after three seconds, whenthe mass is moving at 30 meters per second, the kinetic energy is 450 joules.The essential point is that the speed increases linearly with time, but thework done by the constant gravitational force depends on how far the stone hasdropped, and that goes as the square of the time. Therefore, the kineticenergy of the falling stone depends on the square of the time, and that’sthe same as depending on the square of the velocity. For stones ofdifferent masses, the kinetic energy at the same speed will be proportional tothe mass (since weight is proportional to mass, and the work done by gravity isproportional to the weight), so using the figures we worked out above for a onekilogram mass, we can conclude that for a mass of m kilograms moving ata speed v the kinetic energy must be:
kinetic energy = ½mv²
Exercises for the reader: bothmomentum and kinetic energy are in some sense measures of the amount of motionof a body. How do they differ?
Can a body change in momentum without changing in kinetic energy?
Can a body change in kinetic energy without changing in momentum?
Suppose two lumps of clay of equal mass traveling in opposite directions atthe same speed collide head-on and stick to each other. Is momentumconserved? Is kinetic energy conserved?
As a stone drops off a cliff, both its potential energy and its kineticenergy continuously change. How are these changes related to each other?
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