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00:00:03,580 --> 00:00:06,070
So we've now drawn
pictures of the interaction

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00:00:06,070 --> 00:00:08,590
of a person jumping off
a cart in both the ground

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00:00:08,590 --> 00:00:12,940
frame and a reference frame
moving with velocity, vc, which

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00:00:12,940 --> 00:00:14,650
is the final speed of the cart.

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00:00:14,650 --> 00:00:17,500
These were pictures in the
reference frame moving with vc,

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00:00:17,500 --> 00:00:19,360
and these are the
pictures-- momentum

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00:00:19,360 --> 00:00:23,830
diagrams-- of the person and
the cart in the ground frame.

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00:00:23,830 --> 00:00:26,140
Now, we would like to apply
the momentum principle

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00:00:26,140 --> 00:00:29,200
to solve for vp and vc.

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00:00:29,200 --> 00:00:33,280
We are given that the velocity
of the person in the moving

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00:00:33,280 --> 00:00:35,470
frame, u.

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00:00:35,470 --> 00:00:39,160
We'll treat this as
a given quantity.

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00:00:39,160 --> 00:00:42,130
And we'll express
the velocity, vc,

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00:00:42,130 --> 00:00:56,530
and so we want to find vc and
vp in terms of u and p and mc.

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00:00:56,530 --> 00:01:00,670
Now, we'll use the
ground frame first,

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00:01:00,670 --> 00:01:04,949
and our assumption here is that
the person jumped horizontally

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00:01:04,949 --> 00:01:06,490
and that there are
no external forces

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00:01:06,490 --> 00:01:07,656
in the horizontal direction.

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00:01:07,656 --> 00:01:10,600
So all of the vectors
we've drawn are horizontal,

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00:01:10,600 --> 00:01:13,550
and there's no external forces
in the horizontal direction.

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00:01:13,550 --> 00:01:19,090
Then we know that the initial
momentum of the system

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00:01:19,090 --> 00:01:22,450
is equal to the final
momentum of the system.

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00:01:22,450 --> 00:01:25,420
In our initial picture,
nothing is moving.

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00:01:25,420 --> 00:01:27,340
And in our final
picture, we have

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00:01:27,340 --> 00:01:33,740
m cart v cart plus m
person v person is 0.

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00:01:33,740 --> 00:01:38,140
Now, recall that we also showed
that the velocity of the person

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00:01:38,140 --> 00:01:43,150
in the ground frame is related
to the velocity of the person

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00:01:43,150 --> 00:01:47,229
in the moving frame by
adding the relative velocity

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00:01:47,229 --> 00:01:49,450
of the two frames.

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00:01:49,450 --> 00:01:52,000
The relative velocity
of the two frames

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00:01:52,000 --> 00:01:59,350
is the velocity of the cart, vc,
so what we have is vc plus u.

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00:01:59,350 --> 00:02:03,130
So this is how vp
is related to u.

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00:02:03,130 --> 00:02:05,350
And now, I can use
this start equation

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00:02:05,350 --> 00:02:13,630
to write it as 0 equals mc
vc plus mp times vc plus u.

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00:02:13,630 --> 00:02:25,270
And by adding terms, I get that
mc plus mp vc equals minus mpu.

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00:02:25,270 --> 00:02:31,660
Or the velocity of the cart
is equal to minus mpu divided

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00:02:31,660 --> 00:02:35,620
by mc plus mp.

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00:02:35,620 --> 00:02:38,150
Now that I have the
velocity of the cart,

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00:02:38,150 --> 00:02:40,230
I can just substitute
that in here

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00:02:40,230 --> 00:02:43,720
to find the velocity of the
person, which, remember,

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00:02:43,720 --> 00:02:46,930
was the velocity
of the cart plus u.

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00:02:46,930 --> 00:02:50,710
So now, I've solved this
problem in the ground frame.

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00:02:50,710 --> 00:02:55,031
Next, I'll do the same
analysis in the moving frame.