1
00:00:03,340 --> 00:00:05,020
Let's calculate
the kinetic energy

2
00:00:05,020 --> 00:00:08,800
in the center of mass frame of
two objects that are colliding.

3
00:00:08,800 --> 00:00:12,020
I'm going to use prime for the
center of mass frame velocity.

4
00:00:12,020 --> 00:00:14,180
So we have V1 prime.

5
00:00:14,180 --> 00:00:17,140
And we have V2 prime.

6
00:00:17,140 --> 00:00:21,400
And recall two things that
when we calculated V1 prime

7
00:00:21,400 --> 00:00:23,560
in the center of
mass frame, we found

8
00:00:23,560 --> 00:00:26,560
that this was equal
to the reduced mass

9
00:00:26,560 --> 00:00:31,030
divided by m1 times
the relative velocities

10
00:00:31,030 --> 00:00:36,070
of the two objects, V1
prime minus V2 prime.

11
00:00:36,070 --> 00:00:40,960
Now in the center of
mass frame, this quantity

12
00:00:40,960 --> 00:00:43,130
is a reference
frame independent.

13
00:00:43,130 --> 00:00:47,440
And so we can just
write that as V1, 2.

14
00:00:47,440 --> 00:00:55,477
It's the relative velocity is
reference frame independent.

15
00:01:02,020 --> 00:01:06,160
So we also knew
that V2 prime was

16
00:01:06,160 --> 00:01:11,980
equal to minus mu over m2 V1, 2.

17
00:01:11,980 --> 00:01:14,530
So now let's calculate
the kinetic energy

18
00:01:14,530 --> 00:01:16,300
in the center of mass frame.

19
00:01:16,300 --> 00:01:23,170
So K cm is 1/2 m1
V1 prime squared

20
00:01:23,170 --> 00:01:27,289
plus 1/2 m2 V2 prime squared.

21
00:01:27,289 --> 00:01:30,620
Now let's use our results in
terms of relative velocity.

22
00:01:30,620 --> 00:01:39,729
So we have 1/2 m1 mu
over and m1 times V1,

23
00:01:39,729 --> 00:01:43,500
2 squared-- because
we're squaring

24
00:01:43,500 --> 00:01:48,430
that-- plus 1/2 m2 mu-- the
minus sign doesn't matter

25
00:01:48,430 --> 00:01:50,800
anymore because
we're squaring it--

26
00:01:50,800 --> 00:01:54,970
V2 squared-- I could put that
in there, it wouldn't matter.

27
00:01:54,970 --> 00:01:59,350
And now what we
have is we have 1/2.

28
00:01:59,350 --> 00:02:06,160
One of the m1s cancels so we
have a mu squared over m1 V1, 2

29
00:02:06,160 --> 00:02:12,070
squared plus 1/2 another
mu square over m2 times

30
00:02:12,070 --> 00:02:14,590
V1, 2 squared.

31
00:02:14,590 --> 00:02:20,410
So I pull out the common terms,
mu square, V1, 2 squared.

32
00:02:20,410 --> 00:02:24,700
I have 1 over m1 plus 1 over m2.

33
00:02:24,700 --> 00:02:30,460
But recall that the reduced
mass was precisely 1 over m1

34
00:02:30,460 --> 00:02:32,380
plus 1 over m2.

35
00:02:32,380 --> 00:02:38,079
And so I get 1/2 mu
times V1, 2 squared

36
00:02:38,079 --> 00:02:41,650
is the kinetic energy in
the center of mass frame.

37
00:02:41,650 --> 00:02:46,090
So what that means
is a way of thinking

38
00:02:46,090 --> 00:02:48,030
if this were just a
simple reduced mass

39
00:02:48,030 --> 00:02:51,579
problem with a relative
speed of V1, 2 squared,

40
00:02:51,579 --> 00:02:53,710
I can write down
the kinetic energy.

41
00:02:53,710 --> 00:02:58,780
But that's the kinetic energy
in the center of mass frame.

42
00:02:58,780 --> 00:03:03,850
And therefore, the change in
kinetic energy in a collision

43
00:03:03,850 --> 00:03:11,140
is just 1/2 V1, 2 V
final squared minus--

44
00:03:11,140 --> 00:03:13,750
and it's the same
mu, so we'll just

45
00:03:13,750 --> 00:03:23,079
put a parentheses-- minus
V1, 2 initial squared.

46
00:03:23,079 --> 00:03:26,530
And if the collision
is elastic-- remember,

47
00:03:26,530 --> 00:03:34,000
our elastic collision
show that V1, 2 initial

48
00:03:34,000 --> 00:03:38,320
was minus V1, 2 final.

49
00:03:38,320 --> 00:03:42,079
So an elastic collision
satisfies that condition.

50
00:03:42,079 --> 00:03:47,360
And you can see directly that
delta Kcm is 0 in that case.

51
00:03:47,360 --> 00:03:49,960
And a completely
inelastic collision

52
00:03:49,960 --> 00:03:52,840
has the two objects moving
together at the end.

53
00:03:52,840 --> 00:03:55,240
So V1, 2 final is 0.

54
00:03:55,240 --> 00:04:06,820
So completely inelastic is
the condition that V1, 2 final

55
00:04:06,820 --> 00:04:10,820
is 0 because the
objects stick together.

56
00:04:10,820 --> 00:04:12,670
And then have a
very simple result

57
00:04:12,670 --> 00:04:15,310
for the change in kinetic
energy for completely

58
00:04:15,310 --> 00:04:16,600
inelastic collision.

59
00:04:16,600 --> 00:04:22,230
It's only negative 1/2 mu V1,
2 initial quantity squared.