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00:00:04,080 --> 00:00:07,460
So suppose we have a
point mass particle
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of mass m moving with
a velocity vector,
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v. We can introduce
a quantity we call
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00:00:15,540 --> 00:00:17,120
the momentum of that particle.
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I'll label it with the symbol p.
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And it's equal to the product
of the mass times the velocity.
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This is something you've
undoubtedly seen before.
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Now, let's think
about the dimensions
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of momentum for a moment.
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00:00:33,600 --> 00:00:37,010
So dimensionally,
momentum has units
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of mass times the velocity.
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00:00:38,980 --> 00:00:45,130
And so that's in SI units, the
units of mass are kilograms
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00:00:45,130 --> 00:00:52,490
and then the units of velocity
are meters per second.
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00:00:52,490 --> 00:00:54,800
You can also express
momentum dimensionally
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as a product of a
force times time.
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00:00:58,920 --> 00:01:02,060
And so, again in SI units,
units force are the Newton
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00:01:02,060 --> 00:01:04,670
and the units of
time is the second.
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So these are the SI units for
momentum, two different ways
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of writing the same dimensions.
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00:01:10,970 --> 00:01:20,330
Now, we've seen that
for a single particle,
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we can write Newton's
second law as the force is
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equal to the mass
times the acceleration.
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00:01:29,550 --> 00:01:31,190
Or equivalently,
we can write that
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as the mass times
the time derivative
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00:01:34,630 --> 00:01:37,970
of the acceleration, dv/dt.
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Now, if m is a constant,
then I can rewrite this as F
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00:01:51,350 --> 00:02:00,960
is equal to the time derivative
of the mass times the velocity,
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or equivalently as the time
derivative of the momentum,
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since mv is just equal to p.
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So I actually want
to stress this much--
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I'm going to put it
in a box because it's
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very important that I can
write Newton's second law,
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instead of F equals ma,
as F equals the time
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00:02:17,520 --> 00:02:20,084
derivative of the momentum.
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00:02:20,084 --> 00:02:21,500
And this is
absolutely where we'll
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00:02:21,500 --> 00:02:23,600
see the momentum
becomes very useful.
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00:02:29,400 --> 00:02:32,760
Because it turns out that this
form of Newton's second law
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is actually the most
general form of the equation
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because it's applicable not
just to a single point mass,
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but also to a more
complicated system.
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A system consisting
of many masses
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or a system where
the masses changing
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or the masses flowing,
as in a fluid.
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In all of those cases, this
form of Newton's second law
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00:02:52,079 --> 00:02:52,940
is correct.
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00:02:52,940 --> 00:02:55,710
F equals ma, which is
probably more familiar to you,
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is actually a special case
of this law for the case
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of a single point mass.
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So this is where we'll
see that momentum is quite
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00:03:02,230 --> 00:03:04,650
a useful concept, especially
as we start considering
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00:03:04,650 --> 00:03:08,840
more complicated systems, as
we'll get to a little later
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in the course.
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What I want to do now though,
is to take a closer look
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at this equation, force is
equal to the time derivative
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of the momentum.
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Whenever we have a
relation involving
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a derivative like this,
we can always also
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00:03:23,990 --> 00:03:27,156
rewrite it in an equivalent
integral form, which
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can be very useful and give
us a different way of looking
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at the same information.
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So let's take a look at that.
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So if I take this equation
and integrate both sides
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with respect to time,
then I can write
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that as the integral of
F with respect to time
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is equal to the integral
of the right-hand side,
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00:03:52,210 --> 00:03:56,542
dp/dt with respect to time.
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00:03:56,542 --> 00:03:58,250
Now, let's make this
a definite integral.
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I'll go from time t1 to
time t2 on both sides here.
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00:04:08,420 --> 00:04:12,530
Now, this right-hand
side is just--
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so the integral of dp/dt
with respect to time
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is just p at time2
minus p at time1.
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And that is just the change
in the momentum vector
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00:04:29,860 --> 00:04:31,310
going from time1 to time2.
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00:04:34,220 --> 00:04:36,830
Now, this integral on
the left-hand side,
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we give a special name.
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We call this the impulse.
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This name, impulse, calls to
mind a short, sharp, shock
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of some sort.
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But it can also
refer to a weak force
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acting over a long interval.
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00:04:52,590 --> 00:04:55,750
And notice here the
function F, the force F,
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00:04:55,750 --> 00:04:58,014
is in general a
function of time.
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00:04:58,014 --> 00:04:59,930
So this doesn't necessarily
mean a constant f.
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00:04:59,930 --> 00:05:03,160
This could mean a force
that's varying in time.
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00:05:03,160 --> 00:05:05,450
And what this
equation tells us is
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that the change in the
momentum of the system
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doesn't depend on the
detailed time dependence of F,
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00:05:11,440 --> 00:05:14,890
but rather just on
the integral of F.
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00:05:14,890 --> 00:05:23,970
And so suppose I were to graph
the force as a function of time
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00:05:23,970 --> 00:05:28,590
going from time0
to a time delta t.
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00:05:28,590 --> 00:05:33,200
And suppose I had some
complicated function that
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00:05:33,200 --> 00:05:34,020
looked like that.
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The impulse is just the
area under this curve.
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It's the integral
of this function.
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That's the impulse.
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00:05:46,590 --> 00:05:49,080
And the change in
the momentum depends
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00:05:49,080 --> 00:05:51,350
only on the area
under this curve
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00:05:51,350 --> 00:05:56,440
and not on the detailed
shape of the curve.
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00:05:56,440 --> 00:06:02,570
So what that means is that I
can define an average force
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by choosing a
constant force that
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has the same area as
this example on the left.
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So suppose I calculated that.
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And there is some
constant force here.
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I'll call this F average.
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00:06:22,660 --> 00:06:24,610
Going over the
same time interval.
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The average force is
that constant force
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00:06:31,730 --> 00:06:37,380
which has the same area as
the area under my F of t.
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00:06:37,380 --> 00:06:42,440
So in other words, F
average times delta t,
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which is the area on the
right-hand side here,
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is equal to the integral
of F of t dt integrated
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from 0 to delta t,
which is the area
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00:06:57,370 --> 00:06:58,700
under this right-hand curve.
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And so my average force is
just that integral, F of t dt,
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divided by delta t.
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00:07:10,000 --> 00:07:12,830
And this is integrated
from 0 to delta t.
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So that's my average force.