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You see here the topics that will be on your plate on Monday.

It's clearly not possible for me in one exam to cover all these topics, so I will have to make a choice Monday.

Today I also have to make a choice.

I cannot do justice to all these topics in any depth, so I will butterfly over them and some of them I won't even touch at all.

However, what is not covered today may well be on the exam.

You will get loads of equations.

Almost every equation that I could think of will be on your exam.

There's also special tutoring this weekend you can check that on the Web —

Saturday and Sunday.

Let's start with a completely inelastic collision.

Completely inelastic collision.

We have mass m1, we have mass m2.

It's a one-dimensional problem.

This one has velocity v1, and this has velocity v2.

They collide, and after the collision they are together —

because that's what it means when the collision is completely inelastic —

and they have a velocity, v prime.

If there is no net external force on the system, momentum must be conserved.

So I now have that m1 v1 plus m2 v2 must be m1 plus m2 times v prime.

One equation with one unknown —

v prime follows immediately.

You may say, "Gee, you should really have put arrows over here." Well, in the case that it is a one-dimensional collision, you can leave the arrows off because the signs automatically take care of that.

Kinetic energy is not conserved.

Before the collision, your kinetic energy equals one-half m1 v1 squared plus one-half m2 v2 squared.

After the collision, kinetic energy equals one-half m1 plus m2 times v prime squared.

And you can easily prove that this is always less than that in case of a completely inelastic collision.

There's always kinetic energy destroyed, which then comes out in the form of heat.

Let's do now an elastic collision.

And I add the word "completely" elastic, but "elastic" itself is enough because that means that kinetic energy is conserved.

I start with the same initial condition: m1 v1, m2 v2, but now, after the collision, m1 could either go this way or this way, I don't know.

So this could be v1 prime, this could be v1 prime.

m2, however, will always go into this direction.

That's clear, because if you get hit from behind by object one, after the collision, you obviously go in this direction.

Again, we don't have to put the arrows over it, because the signs take care of it in a one-dimensional case.

This will be plus, then.

That could be... you could adopt that as your convention, and if it goes in this direction, then it will be negative.

So if you find, for v1 prime, minus something, it means it's bounced back.

So now we can apply the conservation of momentum if there is no external force on the system.

Internal force is fine.

All right, so now we have m1 v1 plus m2 v2 equals m1 v1 prime plus m2 v2 prime —

conservation of momentum.

Now we get the conservation of kinetic energy, because we know it's an elastic collision.

One-half m1 v1 squared, one-half m2 v2 squared —

that's before the collision.

After the collision, one-half m1 v1 prime squared plus one-half m2 v2 prime squared.

Two equations with two unknowns and in principle, you can solve for v1 prime and for v2 prime, except that it could be time-consuming.

And so on exams, what is normally done when you get a problem like that...

It's clearly not possible for me in one exam to cover all these topics, so I will have to make a choice Monday.

Today I also have to make a choice.

I cannot do justice to all these topics in any depth, so I will butterfly over them and some of them I won't even touch at all.

However, what is not covered today may well be on the exam.

You will get loads of equations.

Almost every equation that I could think of will be on your exam.

There's also special tutoring this weekend you can check that on the Web —

Saturday and Sunday.

Let's start with a completely inelastic collision.

Completely inelastic collision.

We have mass m1, we have mass m2.

It's a one-dimensional problem.

This one has velocity v1, and this has velocity v2.

They collide, and after the collision they are together —

because that's what it means when the collision is completely inelastic —

and they have a velocity, v prime.

If there is no net external force on the system, momentum must be conserved.

So I now have that m1 v1 plus m2 v2 must be m1 plus m2 times v prime.

One equation with one unknown —

v prime follows immediately.

You may say, "Gee, you should really have put arrows over here." Well, in the case that it is a one-dimensional collision, you can leave the arrows off because the signs automatically take care of that.

Kinetic energy is not conserved.

Before the collision, your kinetic energy equals one-half m1 v1 squared plus one-half m2 v2 squared.

After the collision, kinetic energy equals one-half m1 plus m2 times v prime squared.

And you can easily prove that this is always less than that in case of a completely inelastic collision.

There's always kinetic energy destroyed, which then comes out in the form of heat.

Let's do now an elastic collision.

And I add the word "completely" elastic, but "elastic" itself is enough because that means that kinetic energy is conserved.

I start with the same initial condition: m1 v1, m2 v2, but now, after the collision, m1 could either go this way or this way, I don't know.

So this could be v1 prime, this could be v1 prime.

m2, however, will always go into this direction.

That's clear, because if you get hit from behind by object one, after the collision, you obviously go in this direction.

Again, we don't have to put the arrows over it, because the signs take care of it in a one-dimensional case.

This will be plus, then.

That could be... you could adopt that as your convention, and if it goes in this direction, then it will be negative.

So if you find, for v1 prime, minus something, it means it's bounced back.

So now we can apply the conservation of momentum if there is no external force on the system.

Internal force is fine.

All right, so now we have m1 v1 plus m2 v2 equals m1 v1 prime plus m2 v2 prime —

conservation of momentum.

Now we get the conservation of kinetic energy, because we know it's an elastic collision.

One-half m1 v1 squared, one-half m2 v2 squared —

that's before the collision.

After the collision, one-half m1 v1 prime squared plus one-half m2 v2 prime squared.

Two equations with two unknowns and in principle, you can solve for v1 prime and for v2 prime, except that it could be time-consuming.

And so on exams, what is normally done when you get a problem like that...

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