Introduction to Physics (PHSX 1010)
Adam Johnston

Course notes:
MOMENTUM

Imagine what you would do if, on your way to physics class, someone started shooting koosh balls (squishy and soft) and hard balls at you.  If all the balls are fired at the same velocity, and all have the same mass, which balls do you most want to dodge?  This probably isn't a very hard question to answer.  Obviously, you don't want to get hit by the hard balls.

But, why?  Each ball has the same mass and the same motion.  Each ball will hit you and come to rest.  So why should the hard ball hurt more?

One way to think of this is to consider the physics of the situation.  First, when you think of "pain," physics thinks of "force."  This is the cause of your pain.  In other words, you exert more force on the hard ball than on the koosh ball to get it to come to a stop.  But, you must have had to give up something else, right?  If each ball had the same motion, then there must be more to the story besides the force part.  This is where impulse comes into play.

Before we can fully understand impulse, you need to also understand momentum.  These two concepts play as a pair.  Momentum is the product of the mass and velocity (with direction -- very important!) of an object.  So, our koosh balls and hard balls all had the same momentums.  Momentum is kind of like "oomph."  You expect a cement truck to have more oomph/momentum than a tennis ball, even if they are moving at the same pace.  This is because it has more mass.  Likewise, you expect a major league pitcher to place more oomph/momentum on the same baseball that I throw across the room.  This is because his/her throw has more velocity.

Okay, so each ball (koosh and hard) has the same momentum as the other.  Getting them each to stop requires the same change in momentum.  This is where impulse comes in.  Impulse is the change in momentum, and it is also equal to the average force exerted during a collision/interaction times the time this takes.  If each impulse is the same, but the contact times are different, then the forces must also be different.  That is, if Impulse = F*t, then for two equal impulses (as in our example), the object with the smaller t (pool ball) must have the larger F (more pain), and the object with the bigger t (squish ball) must have the smaller F (less pain).  Greater impulses mean greater changes in momentum, but they can do this in different ways.  A greater impulse could be the result of a greater force; or it could be the result of a greater contact time. (You might imagine looking at replays of these collisions in slow motion.)  That's the koosh ball, and that's why it is less painful (forceful).  It takes its time.

On a test/homework question, you should analyze these kinds of problems by going back to the definition of impulse, rather than just saying something like "one is softer than the other."  This doesn't tell us much when you think about it.  Why is the "softer" part important?

There are lots of examples of how this works.  We might imagine two gymnasts jumping up and down, one on a mat and another on a concrete floor.  We also imagine two balls, one which bounces and another which does not.  Which experiences more force?  In this case, the contact times are about equal, but the changes in momentum are different.  The bouncy ball's momentum changes by twice as much (going down, then going back up), so the force exerted on it is about twice as much.  This might seem a little backwards, but, believe me, you would much rather squish than bounce, unless you can find a way to increase that contact time (say with a trampoline).

You could use the ideas of impulse and momentum to analyze why a gymnast decides to land on a spongy mat rather than on a concrete floor.  Yes, it's true to say that one is "hard" and one is "soft," but in physics we want to know what is going on in terms of the forces exerted in each case.  For which is the force greater?  Why?  What exactly is different?

In a similar way, we considered the happy and unhappy balls.  Which would you rather be?  The happy ball bounces when dropped, but the unhappy ball just sticks to the table.  Assuming that both have the same amount of contact time, which of these experiences the most force (and therefore the most pain)?

Great, but what about momentum conservation?  An impulse is fine, but what conclusions can we draw regarding a system that has no external impulses (although it might have all kinds of internal impulses)?  Such a system should conserve its momentum.  Momentum is the mass multiplied by the velocity of that mass.  A system's total momentum is the sum of all objects' momentums added together.  When you consider the fact that velocity has a direction, then momentum has a direction; and then different objects in a system can have opposite momentum, adding up to a total momentum of zero.  Two carts at rest have this momentum, and they still have zero momentum after an explosion has occurred in between them, causing them to move in opposite directions.

If two objects have exactly equal mass, using conservation of momentum to describe the system is relatively easy.  Let's say one of the masses is sitting still, and the other is moving.  Then, all the momentum of the system is in the moving mass.  If it collides in a perfectly bouncy (more on this later) collision, the first mass will come to a stop, and the second will start moving at exactly the same velocity that the first mass had originally.  Since the m*v is exactly the same as before for the system, momentum is conserved.

If the same collision happens, but the two masses stick to one another, what results?  Sticking together, the combination of objects make one object that is twice as massive as the object that was moving before the collision.  Using conservation of momentum, we can see that the resulting velocity should be half that of the original moving object.

Something that was less clearly explained in class (but a little better in the book) was why exactly the motion of objects after taking part in a collision looks so much different when it is an elastic collision (in which kinetic energy is conserved, and objects bounce off one another) versus when it is an inelastic collision (in which kinetic energy is not conserved, and objects stick to one another).  The short (but not very informative) answer is: The motion is different because energy goes to different places in the two different collisions.  We'll try to show this a little better next time.  The good news is that it will help us to understand the rack of swinging ball bearings known as Newton's Cradle: Why do two balls colliding with the rack always knock two balls off the other side?  (Conservation of momentum allows for the possibility that one ball will fly off with twice the speed, or four balls will fly off with twice the speed of the original balls.  But this never happens . . . )