1
00:00:03,720 --> 00:00:06,180
So we're going to return to
our one-dimensional elastic

2
00:00:06,180 --> 00:00:08,070
collision with no
external forces.

3
00:00:08,070 --> 00:00:14,520
So we have object 1 moving with
velocity V1 initial and object

4
00:00:14,520 --> 00:00:19,620
2 maybe it's moving this
way with V2 initial i hat,

5
00:00:19,620 --> 00:00:21,990
again, on a
frictionless surface.

6
00:00:21,990 --> 00:00:26,280
And we'll call that
our initial state.

7
00:00:26,280 --> 00:00:28,980
And here you can imagine we're
going to use a ground reference

8
00:00:28,980 --> 00:00:29,580
frame.

9
00:00:29,580 --> 00:00:31,350
So both objects are moving.

10
00:00:31,350 --> 00:00:36,060
And our final state
has object 1-- well,

11
00:00:36,060 --> 00:00:38,340
we don't know, again,
which way it's going.

12
00:00:38,340 --> 00:00:41,970
We can just say it bounced back.

13
00:00:41,970 --> 00:00:45,036
And object 2 also bounced back.

14
00:00:45,036 --> 00:00:46,660
But the goal of our
problem, of course,

15
00:00:46,660 --> 00:00:49,800
is to determine these vectors.

16
00:00:49,800 --> 00:00:52,500
And by knowing the vectors,
we know which way they go.

17
00:00:52,500 --> 00:00:55,290
Now because energy and
momentum are constant,

18
00:00:55,290 --> 00:00:56,885
let's write down
our two equations.

19
00:00:56,885 --> 00:00:58,260
And I'm going to
write them down,

20
00:00:58,260 --> 00:01:01,240
again, in terms of components.

21
00:01:01,240 --> 00:01:10,030
So we have 1/2 m1 V1x
initial squared plus 1/2

22
00:01:10,030 --> 00:01:20,940
m2 V2x initial squared equals
1/2 m1 V1x final squared

23
00:01:20,940 --> 00:01:27,520
plus 1/2 m2 V2x final squared.

24
00:01:27,520 --> 00:01:30,504
Now we're going to do some
algebraic manipulations here.

25
00:01:30,504 --> 00:01:31,920
So the first thing
I'm going to do

26
00:01:31,920 --> 00:01:36,280
is just eliminate these halves
because it's not necessary.

27
00:01:36,280 --> 00:01:38,880
And I don't want to
rewrite this equation.

28
00:01:38,880 --> 00:01:42,789
And this is our fact that
our energy is constant.

29
00:01:42,789 --> 00:01:46,495
And now our condition
that momentum is constant,

30
00:01:46,495 --> 00:01:48,870
we'll write this-- now, I'm
going to leave a little space

31
00:01:48,870 --> 00:01:50,300
here intentionally.

32
00:01:50,300 --> 00:01:52,110
And our condition that
momentum is constant

33
00:01:52,110 --> 00:02:03,840
is m1 Vx initial plus m2
V2x initial equals m1 V1x

34
00:02:03,840 --> 00:02:09,030
final plus m2 V2x final.

35
00:02:09,030 --> 00:02:13,050
Now this energy equation
can be factored in

36
00:02:13,050 --> 00:02:17,310
by bringing all the m1 terms
to one side and the M2 terms

37
00:02:17,310 --> 00:02:18,850
to the other side.

38
00:02:18,850 --> 00:02:21,060
So when I write that,
I'll need a little room.

39
00:02:21,060 --> 00:02:28,680
I have m1 V1x initial squared
minus V1x final squared.

40
00:02:28,680 --> 00:02:33,420
And that's equal to
m2 V2x final squared

41
00:02:33,420 --> 00:02:37,177
minus V2x initial squared.

42
00:02:37,177 --> 00:02:39,510
So I've just brought those
terms over to the other side.

43
00:02:39,510 --> 00:02:44,550
Now likewise I'll do the
same thing down here.

44
00:02:44,550 --> 00:02:53,070
I have m1 V1x initial
minus V1x final.

45
00:02:53,070 --> 00:03:00,630
And that's equal to m2 V2x
final minus V2x initial.

46
00:03:00,630 --> 00:03:03,370
Now here comes the
algebraic trick

47
00:03:03,370 --> 00:03:06,180
in which I'm going to
linearize these systems.

48
00:03:06,180 --> 00:03:08,950
This is a squared
minus b squared,

49
00:03:08,950 --> 00:03:12,270
which factors into a
plus b times a minus b.

50
00:03:12,270 --> 00:03:14,070
So let's give ourselves
a little room.

51
00:03:14,070 --> 00:03:19,220
m1 V1x initial plus-- let's
put the minus sign first.

52
00:03:22,820 --> 00:03:32,500
minus V1x final times V1x
initial plus V1x final.

53
00:03:32,500 --> 00:03:34,040
Factored that term.

54
00:03:34,040 --> 00:03:37,500
We have the same factoring
on the other side.

55
00:03:37,500 --> 00:03:40,400
So it's just
identical, V2x final

56
00:03:40,400 --> 00:03:49,680
minus V2x initial times
V2x final plus V2x initial.

57
00:03:49,680 --> 00:03:57,870
Now let's call this equation
1a and our momentum factored

58
00:03:57,870 --> 00:03:59,250
as 2a.

59
00:03:59,250 --> 00:04:02,190
Now if you notice,
the momentum piece

60
00:04:02,190 --> 00:04:06,850
is appearing exactly
there and exactly here.

61
00:04:06,850 --> 00:04:10,500
So when I divide 1a
by 2a-- and I'll just

62
00:04:10,500 --> 00:04:15,960
symbolically represent that--
then these two pieces cancel.

63
00:04:15,960 --> 00:04:21,420
And that leads to just this
term equal to that term.

64
00:04:21,420 --> 00:04:24,370
And the significance, as
you'll see when I write it out,

65
00:04:24,370 --> 00:04:33,360
1 of x 1x final equals V2x
final plus V2x initial.

66
00:04:33,360 --> 00:04:38,190
I've solved the quad-- I've
eliminated the squared terms,

67
00:04:38,190 --> 00:04:40,240
linearized the system.

68
00:04:40,240 --> 00:04:43,440
Now I still want to write
this equation in another way.

69
00:04:43,440 --> 00:04:45,420
Another important
point to notice

70
00:04:45,420 --> 00:04:49,290
is that this equation
is independent of mass.

71
00:04:49,290 --> 00:04:51,150
Now what I want to
do is write this

72
00:04:51,150 --> 00:04:54,400
in terms of those concepts
of relative velocity we had.

73
00:04:54,400 --> 00:04:59,220
Remember just to motivate
this, V relative by definition

74
00:04:59,220 --> 00:05:01,980
was V1 minus V2.

75
00:05:01,980 --> 00:05:09,030
So let's write this in terms
of the initial and the final.

76
00:05:09,030 --> 00:05:12,120
So in order to do that, we
have to bring this initial term

77
00:05:12,120 --> 00:05:15,210
over to here and this
final term over to there.

78
00:05:15,210 --> 00:05:20,520
And so this equation, which
will give it a number 3,

79
00:05:20,520 --> 00:05:23,760
and now we'll modify
that by calling it 3a.

80
00:05:23,760 --> 00:05:30,730
We have V1x initial
minus V2x initial.

81
00:05:30,730 --> 00:05:32,270
Now notice the sign.

82
00:05:32,270 --> 00:05:34,770
I'm going to want to keep
the order of 1 and 2.

83
00:05:34,770 --> 00:05:42,690
So I have to put a minus sign,
V1x final minus V2x final.

84
00:05:45,909 --> 00:05:51,280
And when written this way,
this is the initial component

85
00:05:51,280 --> 00:05:53,290
of the relative velocity.

86
00:05:53,290 --> 00:05:56,620
And in there is the final
component of relative velocity.

87
00:05:56,620 --> 00:05:58,420
So by combining
these two equations,

88
00:05:58,420 --> 00:06:02,830
I have this remarkable result
that V relative initial

89
00:06:02,830 --> 00:06:06,025
is minus V relative final.

90
00:06:09,130 --> 00:06:13,390
And this condition is
a very powerful tool

91
00:06:13,390 --> 00:06:17,740
for simply analyzing
one-dimensional elastic and

92
00:06:17,740 --> 00:06:18,934
inelastic collisions.

93
00:06:18,934 --> 00:06:20,350
I'd like to even
give this a name.

94
00:06:20,350 --> 00:06:30,890
I'd like to call it the
energy momentum equation.

95
00:06:30,890 --> 00:06:33,140
Now there's a lot of
significant things about this.

96
00:06:33,140 --> 00:06:35,390
So let's just think
about it for a moment.

97
00:06:35,390 --> 00:06:40,100
We have that V relative
initial in magnitude

98
00:06:40,100 --> 00:06:43,060
is equal to V relative final.

99
00:06:43,060 --> 00:06:45,800
And so right away, this
gives us some insight

100
00:06:45,800 --> 00:06:46,850
into any collision.

101
00:06:46,850 --> 00:06:49,850
We can see whether
a collision, if we

102
00:06:49,850 --> 00:06:51,800
know what the relative
initial velocity is,

103
00:06:51,800 --> 00:06:56,240
we know that the final relative
velocity has the same magnitude

104
00:06:56,240 --> 00:06:58,940
but simply switches direction.

105
00:06:58,940 --> 00:07:02,030
And that's a powerful tool in
which to analyze collisions

106
00:07:02,030 --> 00:07:04,870
without doing a lot of algebra.