1
00:00:03,390 --> 00:00:04,920
Let's look at some
examples of one

2
00:00:04,920 --> 00:00:09,144
dimensional elastic collisions
with no external forces

3
00:00:09,144 --> 00:00:10,060
between two particles.

4
00:00:10,060 --> 00:00:12,740
So suppose I have
particle 1 and particle 2,

5
00:00:12,740 --> 00:00:15,750
and we have them moving
on a frictionless surface.

6
00:00:15,750 --> 00:00:19,380
And let's choose
a reference frame

7
00:00:19,380 --> 00:00:21,090
in which we'll
call-- the laboratory

8
00:00:21,090 --> 00:00:26,070
frame-- in which the initial
velocity of particle 2 0.

9
00:00:26,070 --> 00:00:32,650
So our frame is called
the laboratory frame.

10
00:00:32,650 --> 00:00:38,530
And in this frame we're thinking
of particle 2 as our target.

11
00:00:38,530 --> 00:00:39,970
And the target is at rest.

12
00:00:39,970 --> 00:00:44,050
And particle 1 is moving in
with some initial velocity.

13
00:00:44,050 --> 00:00:47,920
And so here is our
initial picture.

14
00:00:47,920 --> 00:00:50,410
And then in our final
state, both the target

15
00:00:50,410 --> 00:00:52,780
and the particles can be moving.

16
00:00:52,780 --> 00:00:55,630
So we'll just indicate v1 final.

17
00:00:55,630 --> 00:01:00,640
And we'll have target was
also moving with v2 final.

18
00:01:00,640 --> 00:01:02,016
And we have i hat.

19
00:01:02,016 --> 00:01:04,150
So let's now look
at some examples

20
00:01:04,150 --> 00:01:06,850
where we're analyzing
this type of collision

21
00:01:06,850 --> 00:01:11,050
in which both energy and
momentum are constant.

22
00:01:11,050 --> 00:01:15,820
And for a particular example
that we want to look at,

23
00:01:15,820 --> 00:01:22,360
we'll have m2 is twice m1.

24
00:01:22,360 --> 00:01:24,850
Now, for our two equations that
we're going to write about,

25
00:01:24,850 --> 00:01:29,260
we'll choose components, and
we'll write our energy equation

26
00:01:29,260 --> 00:01:39,910
as 1/2 m1 v1 x initial
squared plus 1/2 m2--

27
00:01:39,910 --> 00:01:43,090
so now we're going
to write 2 m1.

28
00:01:43,090 --> 00:01:45,910
But object 2 is at
rest, so we don't have

29
00:01:45,910 --> 00:01:48,259
to worry about that one yet.

30
00:01:48,259 --> 00:01:50,740
And so, our initial
kinetic energy

31
00:01:50,740 --> 00:01:53,170
is just equal to the
final kinetic energy--

32
00:01:53,170 --> 00:01:57,920
1/2 m1 v1 x final squared.

33
00:01:57,920 --> 00:02:02,590
And now I'm going to substitute
in 2m1 for the mass of m2

34
00:02:02,590 --> 00:02:07,310
v2 x final squared.

35
00:02:07,310 --> 00:02:10,240
And that's our energy condition.

36
00:02:10,240 --> 00:02:12,340
And then our
momentum condition is

37
00:02:12,340 --> 00:02:15,490
that the incoming
momentum, 1/2 x initial

38
00:02:15,490 --> 00:02:20,800
is equal to the outgoing
momentum, m1 v1 x final.

39
00:02:20,800 --> 00:02:23,920
And again, I'm going
to substitute for m2.

40
00:02:23,920 --> 00:02:29,050
That's 2m1 v2 x final.

41
00:02:29,050 --> 00:02:34,329
And these two
equations represent

42
00:02:34,329 --> 00:02:36,950
a system of two equations
with two unknowns.

43
00:02:36,950 --> 00:02:46,030
We'll treat this as givens
along with the initial velocity

44
00:02:46,030 --> 00:02:47,230
of particle 1.

45
00:02:47,230 --> 00:02:58,050
And our target here is to
solve for the 1x final and v2 x

46
00:02:58,050 --> 00:02:58,690
final.

47
00:02:58,690 --> 00:03:01,630
So we have two equations
and two unknowns.

48
00:03:01,630 --> 00:03:03,580
Now it's a quadratic equation.

49
00:03:03,580 --> 00:03:06,070
So we have to identify-- from
a problem solving strategy,

50
00:03:06,070 --> 00:03:07,930
we have to identify
which quantity

51
00:03:07,930 --> 00:03:10,050
we're going to solve first.

52
00:03:10,050 --> 00:03:12,760
And so, let's solve
for v1 x final, which

53
00:03:12,760 --> 00:03:15,700
means we need to
eliminate v2 x final.

54
00:03:15,700 --> 00:03:17,740
And the way I'll
eliminate v2 x final

55
00:03:17,740 --> 00:03:20,300
is I'll use the
momentum equation.

56
00:03:20,300 --> 00:03:22,360
And notice that
the m's will cancel

57
00:03:22,360 --> 00:03:23,980
in the momentum equation.

58
00:03:23,980 --> 00:03:28,270
And so, if I divide through
by 2m in the momentum equation

59
00:03:28,270 --> 00:03:31,570
and bring this term
over to the other side,

60
00:03:31,570 --> 00:03:38,950
equation 2a becomes
v2 x final equals--

61
00:03:38,950 --> 00:03:48,790
I'm dividing through by 2--
1/2 v1 x initial minus 1/2 v1 x

62
00:03:48,790 --> 00:03:50,170
final.

63
00:03:50,170 --> 00:03:53,860
So this gives me a
target-- v2 x final--

64
00:03:53,860 --> 00:03:55,510
I can substitute into there.

65
00:03:55,510 --> 00:03:57,310
This equation, again,
can be cleaned up.

66
00:03:57,310 --> 00:03:58,531
Let's clean it up before.

67
00:03:58,531 --> 00:03:59,740
We can get rid of the halves.

68
00:03:59,740 --> 00:04:01,420
We can get rid of the m1's.

69
00:04:01,420 --> 00:04:04,360
And so the quadratic
equation that we're

70
00:04:04,360 --> 00:04:08,350
going to be working with
is v1 x initial squared

71
00:04:08,350 --> 00:04:13,160
equals v1 x final squared.

72
00:04:13,160 --> 00:04:15,220
And I'm dividing through by 2.

73
00:04:15,220 --> 00:04:22,070
So I have the factor 2
plus 2 v2 x final squared.

74
00:04:22,070 --> 00:04:25,120
Now that's our energy equation.

75
00:04:25,120 --> 00:04:27,620
And here's our
momentum equation.

76
00:04:27,620 --> 00:04:31,180
So if we substitute
this in and square it,

77
00:04:31,180 --> 00:04:36,340
we can now substitute and we
get v1 x initial squared equals

78
00:04:36,340 --> 00:04:39,200
v1 x final squared plus 2.

79
00:04:39,200 --> 00:04:41,409
Now we make our substitution.

80
00:04:41,409 --> 00:04:48,790
So that's 1/2 v1 x initial
minus 1/2 v1 x final,

81
00:04:48,790 --> 00:04:50,300
quantity squared.

82
00:04:50,300 --> 00:04:54,520
So let's now expand this.

83
00:04:54,520 --> 00:04:58,000
And we have to be careful
not to make any mistakes.

84
00:04:58,000 --> 00:05:03,460
v1 x initial squared
equals v1 x final squared.

85
00:05:03,460 --> 00:05:10,240
Now if I pull the 2 out,
I got a quarter in front,

86
00:05:10,240 --> 00:05:12,310
and divide that by 1/2.

87
00:05:12,310 --> 00:05:22,870
So I get 1/2 v1 x initial
squared plus another 1

88
00:05:22,870 --> 00:05:27,460
plus a 1/2 v1 x final squared.

89
00:05:27,460 --> 00:05:30,310
Those represent the--
squaring out those two terms.

90
00:05:30,310 --> 00:05:35,740
Now the cross-term will
have a factor of 1/2

91
00:05:35,740 --> 00:05:38,150
in the front but a factor of 2.

92
00:05:38,150 --> 00:05:45,120
So we have a simple cross-term
of v1 x initial v1 x final.

93
00:05:51,580 --> 00:05:56,170
Now, when we--
let's collect terms.

94
00:05:56,170 --> 00:05:58,450
Let's bring everything
over to this side

95
00:05:58,450 --> 00:06:07,060
so we have a 0 equals 3/2 v1
x final squared minus 1/2 v1

96
00:06:07,060 --> 00:06:15,370
x initial squared and minus
v1 x initial v1 x final.

97
00:06:15,370 --> 00:06:19,030
Now I always like to just
write this up in a simple way

98
00:06:19,030 --> 00:06:21,350
to use the quadratic formula.

99
00:06:21,350 --> 00:06:24,010
So I'm going to
divide through by 2/3.

100
00:06:24,010 --> 00:06:31,300
And I get minus 1/3 v1 x
initial squared minus 2/3 v1

101
00:06:31,300 --> 00:06:35,260
x initial v1 x final.

102
00:06:35,260 --> 00:06:37,960
And this is now a
simple application

103
00:06:37,960 --> 00:06:40,150
of the quadratic formula.

104
00:06:40,150 --> 00:06:47,060
Negative b is plus 2/3 v1
x initial plus or minus.

105
00:06:47,060 --> 00:06:49,570
And we're going to interpret
those two roots in a moment.

106
00:06:49,570 --> 00:06:55,656
We have to factor 2/3 v1 x
initial squared minus 4ac.

107
00:06:59,230 --> 00:07:02,700
So that's another
minus sign with a plus.

108
00:07:02,700 --> 00:07:08,320
So that's plus 4/3
v1 x initial squared,

109
00:07:08,320 --> 00:07:13,210
all to the square root, and
everything divided by 2.

110
00:07:13,210 --> 00:07:15,370
Now here, let's just
look at this factor.

111
00:07:15,370 --> 00:07:19,480
This is 4/9 plus 4/3.

112
00:07:19,480 --> 00:07:22,930
4/3 is 12/9, so
that's 16/9, which

113
00:07:22,930 --> 00:07:25,600
is very convenient, because when
you take the square root that's

114
00:07:25,600 --> 00:07:26,900
4/3.

115
00:07:26,900 --> 00:07:35,060
So we get equal to 2/3 v1 x
initial plus or minus 4/3 v1 x

116
00:07:35,060 --> 00:07:39,260
initial divided by 2.

117
00:07:39,260 --> 00:07:42,370
Now we see that there's
two different roots.

118
00:07:42,370 --> 00:07:49,180
So when you add them, you get
2 v1 x initial divided by 2.

119
00:07:49,180 --> 00:07:52,120
So there's one solution.

120
00:07:52,120 --> 00:07:58,030
And when you subtract them,
you're getting 2/3 minus 4/3.

121
00:07:58,030 --> 00:08:01,270
That's negative
2/3 divided by 2.

122
00:08:01,270 --> 00:08:06,550
So that is another solution--
v1 x final equals negative 1/3

123
00:08:06,550 --> 00:08:09,850
v1 x initial.

124
00:08:09,850 --> 00:08:12,610
Now let's think about
the meaning of these two

125
00:08:12,610 --> 00:08:15,370
possible solutions.

126
00:08:15,370 --> 00:08:20,500
This solution has v1 x
final equals v1 x initial.

127
00:08:20,500 --> 00:08:23,710
So that's the
initial conditions.

128
00:08:23,710 --> 00:08:24,880
Just repeat it.

129
00:08:24,880 --> 00:08:26,900
And that will
always be the case.

130
00:08:26,900 --> 00:08:29,380
One solution will describe
the initial state,

131
00:08:29,380 --> 00:08:31,810
and the other solution will
define the final state.

132
00:08:31,810 --> 00:08:34,000
You can check for
yourself that if you just

133
00:08:34,000 --> 00:08:38,380
put vx1 final into this--
equal to vx1 in the initial--

134
00:08:38,380 --> 00:08:41,650
into this momentum
equation, then v2 x final

135
00:08:41,650 --> 00:08:45,020
is 0, which just repeats
the initial conditions.

136
00:08:45,020 --> 00:08:51,970
So this solution is
the initial state.

137
00:08:51,970 --> 00:08:56,890
And this solution, here,
is the final state.

138
00:08:56,890 --> 00:08:59,020
Now just to complete
the picture,

139
00:08:59,020 --> 00:09:05,010
v2 x final-- well,
that's equal to 1/2 v1

140
00:09:05,010 --> 00:09:11,680
x initial minus 1/2
times v1 x final.

141
00:09:11,680 --> 00:09:14,890
But v1 x final is negative 1/3.

142
00:09:14,890 --> 00:09:19,720
So we have a 1/2 minus
1/2 half times minus 1/3.

143
00:09:19,720 --> 00:09:26,800
So that's 1/2 plus 1/6, which
is 4/6 or 2/3 v x initial.

144
00:09:26,800 --> 00:09:33,160
So that's 2/3 the v1 x initial.

145
00:09:33,160 --> 00:09:37,970
And that represents the solution
to this particular problem.

146
00:09:37,970 --> 00:09:41,530
Now, of course what
you want to do is,

147
00:09:41,530 --> 00:09:43,810
you want to check-- we know
the momentum condition is

148
00:09:43,810 --> 00:09:45,010
already satisfied.

149
00:09:45,010 --> 00:09:46,930
But just as a check,
you would like

150
00:09:46,930 --> 00:09:48,910
to put it into the
energy condition

151
00:09:48,910 --> 00:09:52,150
just to make sure that when
you square these things out

152
00:09:52,150 --> 00:09:53,680
you get the right terms.

153
00:09:53,680 --> 00:09:56,420
However, we're very
confident of our result,

154
00:09:56,420 --> 00:09:59,410
because we've already reproduced
the initial conditions.

155
00:09:59,410 --> 00:10:02,990
And that wouldn't happen if we
made some algebraic mistake.

156
00:10:02,990 --> 00:10:05,380
So that, by itself,
is a sufficient check

157
00:10:05,380 --> 00:10:07,210
that this is a correct solution.

158
00:10:07,210 --> 00:10:09,520
One thing we should sit
back and think about

159
00:10:09,520 --> 00:10:12,070
is that when we use
the energy and momentum

160
00:10:12,070 --> 00:10:15,430
that we're inevitably
dealing with quadratics.

161
00:10:15,430 --> 00:10:19,420
And so we expect to use-- to
solve a quadratic equation

162
00:10:19,420 --> 00:10:20,590
at one point.

163
00:10:20,590 --> 00:10:22,732
And here it was right here.

164
00:10:22,732 --> 00:10:24,190
Now we're going to
find another way

165
00:10:24,190 --> 00:10:26,770
to do this by linearizing
the system, which

166
00:10:26,770 --> 00:10:28,535
will be a lot easier.