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Remember, earlier in the course we measured the average speed of a bullet which we fired from a rifle.

That was because we had the ability of very fast timing.

In the old days, fast timing was not possible and people measured the speed of bullets in a very delicate way and all the tools that we have learned we can apply now to this device which we call the ballistic pendulum.

We have a pendulum with a very heavy object hanging here at the end —

I call it the block.

You see it here.

And this pendulum has length L.

Ours is about one meter.

I will give you the exact numbers later.

And we have a bullet of mass little m and the bullet comes in with velocity v.

It gets completely absorbed, sticks in there.

It's a completely inelastic collision, and the pendulum will then pick up the velocity v prime with the bullet inside.

The bullet is somewhere here.

Momentum is conserved, so we clearly have that, m v equals (m plus M) times v prime.

So if you could measure v prime, then you could measure the speed of the bullet, which is v.

How do we measure v prime? Well, we wait for the pendulum to come to a halt, let's say here, when the speed is zero.

When it was here, it had a speed v prime.

And we know that there was kinetic energy here —

no gravitational potential energy.

I can call this level U = 0, but right here, if this difference in height is h, then all the kinetic energy has been converted to gravitational potential energy.

So we apply the theorem, the work-energy theorem, or you could say... and it's equally valid, you could say we applied the conservation of mechanical energy.

And so this kinetic energy, which is one-half (m plus M) times v prime squared is now converted exclusively to gravitational potential energy, which equals (m plus M) times g times that h.

And we lose our (m plus M), so v prime would be the square root of 2gh.

And so all you would have to measure h, and then you know v prime, and if you know v prime, you know the speed of the bullet.

But life is not that simple.

It's very difficult to measure h, and I can make you see that.

Suppose this angle —

angle theta —

when it comes to a halt is only two degrees.

Then h, which is L times (1 minus cosine theta), is only 0.6 millimeters for the dimensions that I have chosen here for a length of one meter.

And you can't even see it —

it's invisible —

let alone that you can measure it to any degree of accuracy.

So what are we going to do now? Well, we are going to not measure h, but we are going to measure x.

I call this x = 0.

And here, when the pendulum comes to a halt, I call that x.

For a two-degree angle, x is approximately 3? centimeters.

It can easily be checked by you, of course.

So you get a huge displacement in this direction compared to h.

If you use small angle approximation —

and you better believe that two degrees is very small —

then you can prove, which is purely geometrical mathematics —

and I leave you with that proof —

that this is approximately x squared divided by 2L.

I want you to prove that.

You take the expansion of the cosine, the theories of the Taylor series, and you cut it off somewhere, and this is not so difficult to prove.

In other words, v prime squared, which is 2gh, can now be replaced by approximately 2g times x squared divided by 2L which is g times x squared divided by L.

That was because we had the ability of very fast timing.

In the old days, fast timing was not possible and people measured the speed of bullets in a very delicate way and all the tools that we have learned we can apply now to this device which we call the ballistic pendulum.

We have a pendulum with a very heavy object hanging here at the end —

I call it the block.

You see it here.

And this pendulum has length L.

Ours is about one meter.

I will give you the exact numbers later.

And we have a bullet of mass little m and the bullet comes in with velocity v.

It gets completely absorbed, sticks in there.

It's a completely inelastic collision, and the pendulum will then pick up the velocity v prime with the bullet inside.

The bullet is somewhere here.

Momentum is conserved, so we clearly have that, m v equals (m plus M) times v prime.

So if you could measure v prime, then you could measure the speed of the bullet, which is v.

How do we measure v prime? Well, we wait for the pendulum to come to a halt, let's say here, when the speed is zero.

When it was here, it had a speed v prime.

And we know that there was kinetic energy here —

no gravitational potential energy.

I can call this level U = 0, but right here, if this difference in height is h, then all the kinetic energy has been converted to gravitational potential energy.

So we apply the theorem, the work-energy theorem, or you could say... and it's equally valid, you could say we applied the conservation of mechanical energy.

And so this kinetic energy, which is one-half (m plus M) times v prime squared is now converted exclusively to gravitational potential energy, which equals (m plus M) times g times that h.

And we lose our (m plus M), so v prime would be the square root of 2gh.

And so all you would have to measure h, and then you know v prime, and if you know v prime, you know the speed of the bullet.

But life is not that simple.

It's very difficult to measure h, and I can make you see that.

Suppose this angle —

angle theta —

when it comes to a halt is only two degrees.

Then h, which is L times (1 minus cosine theta), is only 0.6 millimeters for the dimensions that I have chosen here for a length of one meter.

And you can't even see it —

it's invisible —

let alone that you can measure it to any degree of accuracy.

So what are we going to do now? Well, we are going to not measure h, but we are going to measure x.

I call this x = 0.

And here, when the pendulum comes to a halt, I call that x.

For a two-degree angle, x is approximately 3? centimeters.

It can easily be checked by you, of course.

So you get a huge displacement in this direction compared to h.

If you use small angle approximation —

and you better believe that two degrees is very small —

then you can prove, which is purely geometrical mathematics —

and I leave you with that proof —

that this is approximately x squared divided by 2L.

I want you to prove that.

You take the expansion of the cosine, the theories of the Taylor series, and you cut it off somewhere, and this is not so difficult to prove.

In other words, v prime squared, which is 2gh, can now be replaced by approximately 2g times x squared divided by 2L which is g times x squared divided by L.

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