WEBVTT

00:00:03.570 --> 00:00:06.350
So we've been trying to find
the velocity of the cart

00:00:06.350 --> 00:00:08.300
and the velocity of
the person in terms

00:00:08.300 --> 00:00:11.310
of the relative
velocity of the person

00:00:11.310 --> 00:00:14.530
jumping in the reference
frame of the moving cart,

00:00:14.530 --> 00:00:16.760
the mass of the person,
and the mass of the cart.

00:00:16.760 --> 00:00:19.160
We've already solved
this in the ground frame,

00:00:19.160 --> 00:00:25.010
now I would like to solve this
in which my reference frame is

00:00:25.010 --> 00:00:25.775
the moving frame.

00:00:29.190 --> 00:00:31.440
So our pictures in
the moving frame

00:00:31.440 --> 00:00:33.910
were actually much easier.

00:00:33.910 --> 00:00:36.030
Because the moving
frame is traveling

00:00:36.030 --> 00:00:38.400
with the final
speed of the cart,

00:00:38.400 --> 00:00:45.070
at the final interaction--
after the interaction is

00:00:45.070 --> 00:00:48.370
done-- in this moving
frame, the cart is at rest.

00:00:48.370 --> 00:00:49.580
They're moving together.

00:00:49.580 --> 00:00:51.190
You're sitting in the cart.

00:00:51.190 --> 00:00:52.700
You're moving at the same speed.

00:00:52.700 --> 00:00:56.390
The person jumped off with
speed u relative to the cart.

00:00:56.390 --> 00:01:00.290
The tricky part was to
realize that, in the ground

00:01:00.290 --> 00:01:04.530
frame-- when the cart is at
rest here in the ground frame--

00:01:04.530 --> 00:01:07.410
an observer moving
with speed Vc would

00:01:07.410 --> 00:01:10.580
see the person in the cart
moving backwards in the moving

00:01:10.580 --> 00:01:14.640
train with a speed minus Vc.

00:01:14.640 --> 00:01:18.789
So now we can apply the momentum
principle to these two states.

00:01:18.789 --> 00:01:23.020
In the moving frame,
we have that momentum

00:01:23.020 --> 00:01:28.900
of the system initially equals
the momentum of the system

00:01:28.900 --> 00:01:29.770
finally.

00:01:29.770 --> 00:01:36.900
Now, the initial momentum is
just minus Mp plus M cart V

00:01:36.900 --> 00:01:41.090
cart-- notice the minus V cart.

00:01:41.090 --> 00:01:48.580
And the final momentum is
equal to just M person u.

00:01:48.580 --> 00:01:52.259
And so we see that V cart
equals minus M person

00:01:52.259 --> 00:01:57.120
u over M person plus
Mc, which is exactly

00:01:57.120 --> 00:02:00.130
the same result that we
got in the ground frame

00:02:00.130 --> 00:02:05.040
but it was actually much easier
to solve in the moving frame.