WEBVTT
1
00:00:00.440 --> 00:00:02.940 A:middle L:90%
in this question, Asteroid A was moving to the
2
00:00:02.940 --> 00:00:05.750 A:middle L:90%
right with a velocity of 40 m per second.
3
00:00:05.759 --> 00:00:10.039 A:middle L:90%
When it collides with asteroid, be after that collision
4
00:00:10.050 --> 00:00:12.869 A:middle L:90%
. Asteroid A is moving with a new velocity V
5
00:00:12.880 --> 00:00:16.920 A:middle L:90%
A with an angle off 30 degrees with respect tweets
6
00:00:16.920 --> 00:00:21.370 A:middle L:90%
Traditional trajectory. Andi Asteroid B is now moving with
7
00:00:21.370 --> 00:00:24.670 A:middle L:90%
a velocity V. Be making an angle off 45
8
00:00:24.670 --> 00:00:28.260 A:middle L:90%
degrees with the original trajectory off the asteroid A.
9
00:00:28.739 --> 00:00:30.750 A:middle L:90%
In the first item, we have to find the
10
00:00:30.750 --> 00:00:34.299 A:middle L:90%
speed off each asteroid after the collision. That is
11
00:00:34.310 --> 00:00:37.469 A:middle L:90%
, we have to find V A M v B
12
00:00:37.840 --> 00:00:40.149 A:middle L:90%
. So for that you have to remember that the
13
00:00:40.149 --> 00:00:44.329 A:middle L:90%
net momentum is conserved both before and after the collision
14
00:00:44.679 --> 00:00:50.310 A:middle L:90%
. So the net momentum pew net before is equals
15
00:00:50.310 --> 00:00:55.770 A:middle L:90%
the net momentum after pew a net. Now we
16
00:00:55.770 --> 00:00:59.359 A:middle L:90%
have to deal with two dimensions because we have two
17
00:00:59.359 --> 00:01:02.950 A:middle L:90%
directions off movement here. The movement off Asteroid A
18
00:01:02.960 --> 00:01:06.510 A:middle L:90%
happens to the right and happens up on the movement
19
00:01:06.519 --> 00:01:08.829 A:middle L:90%
off Asteroid be happens to the right and downwards the
20
00:01:08.829 --> 00:01:12.459 A:middle L:90%
reform. We are working with two dimensions. So
21
00:01:12.540 --> 00:01:15.219 A:middle L:90%
in this question, my reference frame will be this
22
00:01:15.219 --> 00:01:19.450 A:middle L:90%
one. Everything that is pointing upwards will be positive
23
00:01:19.540 --> 00:01:21.920 A:middle L:90%
and everything that is pointing to the right will also
24
00:01:21.920 --> 00:01:25.650 A:middle L:90%
be positive. Let me call this horizontal axis the
25
00:01:25.650 --> 00:01:27.750 A:middle L:90%
X axis and the vertical one, the Y axis
26
00:01:29.439 --> 00:01:32.769 A:middle L:90%
. Okay, so what happens is that the law
27
00:01:32.780 --> 00:01:36.359 A:middle L:90%
of conservation of momentum must hold in both off these
28
00:01:36.359 --> 00:01:38.719 A:middle L:90%
axes. So these equation that I wrote right here
29
00:01:38.730 --> 00:01:42.819 A:middle L:90%
actually is equivalent to two equations, one for each
30
00:01:42.819 --> 00:01:49.459 A:middle L:90%
direction. So the net momentum in the X direction
31
00:01:49.040 --> 00:01:53.900 A:middle L:90%
before the collision must be equals to the net momentum
32
00:01:53.609 --> 00:01:57.799 A:middle L:90%
in the X direction after the collision and at the
33
00:01:57.799 --> 00:02:00.349 A:middle L:90%
same time we must have. That's the net momentum
34
00:02:01.510 --> 00:02:06.109 A:middle L:90%
in the Y direction before the collision should be equals
35
00:02:06.109 --> 00:02:08.699 A:middle L:90%
to the net momentum in the Y direction after the
36
00:02:08.699 --> 00:02:14.080 A:middle L:90%
collision. So we actually have two equations toe work
37
00:02:14.090 --> 00:02:16.750 A:middle L:90%
. And now let us use these equations. So
38
00:02:16.750 --> 00:02:19.830 A:middle L:90%
we know that to evaluate the momentum, we have
39
00:02:19.830 --> 00:02:22.900 A:middle L:90%
to multiply the mass by the velocity. Then to
40
00:02:22.900 --> 00:02:24.729 A:middle L:90%
evaluate the momentum in the X direction, we multiply
41
00:02:24.729 --> 00:02:28.370 A:middle L:90%
the mass by the velocity in the X direction.
42
00:02:28.439 --> 00:02:30.409 A:middle L:90%
So the first equality goes as follows. So in
43
00:02:30.409 --> 00:02:34.659 A:middle L:90%
the X direction before the collision we had on Lee
44
00:02:34.659 --> 00:02:37.770 A:middle L:90%
the asteroid a which has a mass m A which
45
00:02:37.770 --> 00:02:39.129 A:middle L:90%
is equal to the mass off asteroid be. But
46
00:02:39.129 --> 00:02:42.580 A:middle L:90%
we had on Lee the asteroid a moving to the
47
00:02:42.580 --> 00:02:45.960 A:middle L:90%
right with the velocity off 40 m per second.
48
00:02:46.009 --> 00:02:50.460 A:middle L:90%
Then its momentum waas the mass off asteroid a times
49
00:02:50.460 --> 00:02:53.159 A:middle L:90%
the initial velocity off 40 when it was moving to
50
00:02:53.159 --> 00:02:55.319 A:middle L:90%
the right. So we have a positive sign and
51
00:02:55.319 --> 00:02:59.669 A:middle L:90%
then after the collision, what we have is both
52
00:02:59.669 --> 00:03:04.490 A:middle L:90%
asteroids moving with some velocity in the X direction that
53
00:03:04.490 --> 00:03:07.310 A:middle L:90%
we don't know. So what happens is that after
54
00:03:07.310 --> 00:03:08.870 A:middle L:90%
the collision, the net momentum is given by the
55
00:03:08.870 --> 00:03:13.870 A:middle L:90%
mass off the asteroid A times the velocity off Asteroid
56
00:03:13.879 --> 00:03:16.840 A:middle L:90%
eight after the collision in the X direction, plus
57
00:03:16.849 --> 00:03:20.770 A:middle L:90%
the mass off Asteroid B which is equal to the
58
00:03:20.770 --> 00:03:23.189 A:middle L:90%
mass off asteroid A. So I'm writing a again
59
00:03:23.199 --> 00:03:27.990 A:middle L:90%
times the velocity off asteroid be in the X direction
60
00:03:28.000 --> 00:03:30.919 A:middle L:90%
after the collision. Then we can simplify the masses
61
00:03:30.919 --> 00:03:32.849 A:middle L:90%
. That is, we can divide by m a
62
00:03:32.860 --> 00:03:35.960 A:middle L:90%
, both the left hand side on the right hand
63
00:03:35.960 --> 00:03:38.900 A:middle L:90%
side and we end up with the following v A
64
00:03:38.909 --> 00:03:42.930 A:middle L:90%
. In the X direction plus VB in the X
65
00:03:42.930 --> 00:03:46.460 A:middle L:90%
direction is the cause to 40 now, what is
66
00:03:46.460 --> 00:03:49.319 A:middle L:90%
the relation off v A in the X direction with
67
00:03:49.319 --> 00:03:52.810 A:middle L:90%
V A and VB in the X direction with Phoebe
68
00:03:52.819 --> 00:03:54.699 A:middle L:90%
. For that, we have to decompose vectors as
69
00:03:54.699 --> 00:03:59.689 A:middle L:90%
follows. So here is our reference frame. This
70
00:03:59.689 --> 00:04:01.889 A:middle L:90%
is our X axis and this is our Y axis
71
00:04:01.900 --> 00:04:05.020 A:middle L:90%
. Now let me draw the velocity V A after
72
00:04:05.020 --> 00:04:08.759 A:middle L:90%
the collision. So it is something like that.
73
00:04:08.770 --> 00:04:12.120 A:middle L:90%
This is V A. We know that the angle
74
00:04:12.129 --> 00:04:15.860 A:middle L:90%
that v a makes with the horizontal so it makes
75
00:04:15.860 --> 00:04:18.660 A:middle L:90%
with our X axis is 30 degrees than here.
76
00:04:18.740 --> 00:04:23.389 A:middle L:90%
We haven't angle off 30 degrees. Now we want
77
00:04:23.389 --> 00:04:26.720 A:middle L:90%
to evaluate what is the X component off the velocity
78
00:04:26.720 --> 00:04:29.079 A:middle L:90%
v a. What to do? Well, it's
79
00:04:29.089 --> 00:04:30.850 A:middle L:90%
easy if we draw it. So this is the
80
00:04:30.860 --> 00:04:33.589 A:middle L:90%
X component off v A. On this is the
81
00:04:33.589 --> 00:04:38.089 A:middle L:90%
white component which will also use. So here we
82
00:04:38.089 --> 00:04:40.949 A:middle L:90%
have V A. Why component. And here we
83
00:04:40.949 --> 00:04:44.529 A:middle L:90%
have V a X component. Now note the following
84
00:04:44.670 --> 00:04:46.259 A:middle L:90%
In order to deter mined the relation between V a
85
00:04:46.269 --> 00:04:48.699 A:middle L:90%
X component and V A, we can use the
86
00:04:48.699 --> 00:04:51.730 A:middle L:90%
co sign off 30 because of the following. In
87
00:04:51.730 --> 00:04:56.949 A:middle L:90%
this context, the co sign off 30 is given
88
00:04:56.949 --> 00:05:00.230 A:middle L:90%
by the address and side off triangle, which is
89
00:05:00.230 --> 00:05:02.279 A:middle L:90%
V a X component divided by the high party news
90
00:05:02.290 --> 00:05:05.980 A:middle L:90%
which is V a. The Reform V a X
91
00:05:05.980 --> 00:05:10.480 A:middle L:90%
component is equal to the hypotenuse v a times that
92
00:05:10.480 --> 00:05:14.069 A:middle L:90%
co sign off 30 degrees. And now we can
93
00:05:14.069 --> 00:05:16.069 A:middle L:90%
use this results in this expression to get the following
94
00:05:16.740 --> 00:05:23.730 A:middle L:90%
V eight times the co sign off 30 degrees plus
95
00:05:23.949 --> 00:05:27.230 A:middle L:90%
Now VB also follows the same idea. So we
96
00:05:27.230 --> 00:05:31.009 A:middle L:90%
have vb times. The co sign off 45 degrees
97
00:05:31.129 --> 00:05:34.060 A:middle L:90%
is equals to 40 and this is how you know
98
00:05:34.069 --> 00:05:36.319 A:middle L:90%
up to this point now we go to the Y
99
00:05:36.319 --> 00:05:41.019 A:middle L:90%
direction in the Y direction we have the following Before
100
00:05:41.019 --> 00:05:43.829 A:middle L:90%
the collision, there was no movement at all in
101
00:05:43.829 --> 00:05:46.509 A:middle L:90%
the vertical axis, so the net momentum before the
102
00:05:46.509 --> 00:05:49.089 A:middle L:90%
collision was equals to zero. But after the collision
103
00:05:49.209 --> 00:05:53.569 A:middle L:90%
, both asteroids are also moving in the Y direction
104
00:05:53.839 --> 00:05:58.000 A:middle L:90%
. So we have m a times V a y
105
00:05:58.000 --> 00:06:00.399 A:middle L:90%
direction plus M b which is equals to m A
106
00:06:00.410 --> 00:06:03.529 A:middle L:90%
. So I'm writing m a times vb in the
107
00:06:03.529 --> 00:06:06.829 A:middle L:90%
Y direction again. We can divide both the left
108
00:06:06.829 --> 00:06:10.259 A:middle L:90%
hand side and the right hand side by m a
109
00:06:10.259 --> 00:06:14.000 A:middle L:90%
to get the following v a in the Y direction
110
00:06:14.189 --> 00:06:16.850 A:middle L:90%
plus VB in the Y direction is equal to zero
111
00:06:17.540 --> 00:06:20.769 A:middle L:90%
Now what is the relation between V A in the
112
00:06:20.769 --> 00:06:24.990 A:middle L:90%
Y direction and V A and VB in the right
113
00:06:24.990 --> 00:06:29.069 A:middle L:90%
direction and vb For that we can do the following
114
00:06:29.439 --> 00:06:30.879 A:middle L:90%
again. Take a look at this triangle Now we
115
00:06:30.879 --> 00:06:34.000 A:middle L:90%
want a relation between V A Y component and V
116
00:06:34.000 --> 00:06:36.639 A:middle L:90%
A. As you can see, we can use
117
00:06:36.639 --> 00:06:41.089 A:middle L:90%
the sign off 30 degrees because the sign off 30
118
00:06:41.089 --> 00:06:44.939 A:middle L:90%
degrees in this context is given by the following.
119
00:06:44.949 --> 00:06:46.509 A:middle L:90%
It is the opposite side off the triangle, which
120
00:06:46.509 --> 00:06:49.990 A:middle L:90%
is V a Y component divided by the hypotenuse which
121
00:06:49.990 --> 00:06:54.550 A:middle L:90%
is v A to reform its true that V A
122
00:06:54.560 --> 00:06:57.970 A:middle L:90%
y component is equals to v a times this sign
123
00:06:57.980 --> 00:07:01.449 A:middle L:90%
off 30 degrees. So here we have V a
124
00:07:02.040 --> 00:07:09.649 A:middle L:90%
times the sign off 30 degrees minus vb times this
125
00:07:09.649 --> 00:07:13.490 A:middle L:90%
sign off 45 degrees and this is equals to zero
126
00:07:13.939 --> 00:07:15.290 A:middle L:90%
. Now you might want to know why there is
127
00:07:15.290 --> 00:07:18.139 A:middle L:90%
a negative sign here. And the negative sign here
128
00:07:18.149 --> 00:07:19.870 A:middle L:90%
is because, as you can see, the Y
129
00:07:19.870 --> 00:07:25.300 A:middle L:90%
component off VB will point downwards in the negative direction
130
00:07:25.310 --> 00:07:28.769 A:middle L:90%
off our vertical axis. So this is why I
131
00:07:28.769 --> 00:07:31.569 A:middle L:90%
have this minus sign here. Then using this equation
132
00:07:31.639 --> 00:07:35.540 A:middle L:90%
, you can bet remind the following V a times
133
00:07:35.540 --> 00:07:41.629 A:middle L:90%
The sign off 30 degrees is equals to vb times
134
00:07:41.629 --> 00:07:46.029 A:middle L:90%
. They sign off 45 degrees. The Reform V
135
00:07:46.029 --> 00:07:50.660 A:middle L:90%
A is equals to vb times the sign off 45
136
00:07:50.660 --> 00:07:57.490 A:middle L:90%
degrees divided by the sine off 30 degrees. And
137
00:07:57.490 --> 00:08:01.000 A:middle L:90%
now we can use this result these equations in order
138
00:08:01.000 --> 00:08:03.610 A:middle L:90%
to finally determine what is VB. So by using
139
00:08:03.610 --> 00:08:07.360 A:middle L:90%
these results, we get the following. Now we
140
00:08:07.360 --> 00:08:15.259 A:middle L:90%
have vb times the sign off 45 degrees, divided
141
00:08:15.259 --> 00:08:20.579 A:middle L:90%
by the sine off 30 degrees times the co sign
142
00:08:20.040 --> 00:08:26.300 A:middle L:90%
off 30 degrees plus vb times. The co sign
143
00:08:26.569 --> 00:08:31.110 A:middle L:90%
off 45 degrees is equals to 40. Then we
144
00:08:31.110 --> 00:08:37.080 A:middle L:90%
can factor V B to get following vb times this
145
00:08:37.080 --> 00:08:43.460 A:middle L:90%
sign off 45 degrees times the co sign off 30
146
00:08:43.460 --> 00:08:50.509 A:middle L:90%
degrees divided by the sine off 30 degrees, plus
147
00:08:50.509 --> 00:08:54.169 A:middle L:90%
the co sign off 45 degrees is a close to
148
00:08:54.169 --> 00:09:01.669 A:middle L:90%
40 and finally, VB is equals to the sign
149
00:09:01.240 --> 00:09:07.470 A:middle L:90%
off 45 degrees times the co sign off 30 degrees
150
00:09:07.129 --> 00:09:13.679 A:middle L:90%
, divided by the sine off 30 degrees, plus
151
00:09:13.929 --> 00:09:18.820 A:middle L:90%
the co sign off 45 degrees. Now those are
152
00:09:18.820 --> 00:09:22.799 A:middle L:90%
dividing 40. This results in a speed VP off
153
00:09:22.799 --> 00:09:28.269 A:middle L:90%
approximately 20.7 m per second. Now that we know
154
00:09:28.269 --> 00:09:31.110 A:middle L:90%
VB, we can use this expression to Dr Mind
155
00:09:31.110 --> 00:09:33.039 A:middle L:90%
V A. By doing that, to get the
156
00:09:33.039 --> 00:09:39.399 A:middle L:90%
following V A is equals to 20 0.7 times the
157
00:09:39.399 --> 00:09:45.740 A:middle L:90%
sign off 45 degrees divided by the sine off 30
158
00:09:45.740 --> 00:09:48.399 A:middle L:90%
degrees. By doing that, we get va being
159
00:09:48.399 --> 00:09:54.919 A:middle L:90%
approximately 29.3 m per second. And this is the
160
00:09:54.919 --> 00:09:58.309 A:middle L:90%
answer to the first item off this question. Now
161
00:09:58.309 --> 00:10:01.710 A:middle L:90%
, for the second item, we will need the
162
00:10:01.710 --> 00:10:03.179 A:middle L:90%
values off the speeds, so I will keep them
163
00:10:03.179 --> 00:10:07.190 A:middle L:90%
here in the second item, we want to know
164
00:10:07.200 --> 00:10:11.899 A:middle L:90%
what fraction off the original kinetic energy off Asteroid A
165
00:10:11.909 --> 00:10:15.190 A:middle L:90%
is dissipated in the collision. So for that,
166
00:10:15.200 --> 00:10:18.620 A:middle L:90%
we can compare the kinetic energy off the system before
167
00:10:18.620 --> 00:10:20.769 A:middle L:90%
the collision which was composed off the kinetic energy off
168
00:10:20.779 --> 00:10:24.179 A:middle L:90%
asteroid A only with the kinetic energy off the system
169
00:10:24.190 --> 00:10:28.039 A:middle L:90%
after the collision. So before the collision, let
170
00:10:28.039 --> 00:10:31.190 A:middle L:90%
me call this E B. So the kinetic energy
171
00:10:31.190 --> 00:10:35.090 A:middle L:90%
before the collision is given by the kinetic energy off
172
00:10:35.090 --> 00:10:39.559 A:middle L:90%
Asteroid A only so one half off the massive asteroid
173
00:10:39.570 --> 00:10:43.750 A:middle L:90%
a times the square off its speed. So 40
174
00:10:43.759 --> 00:10:48.629 A:middle L:90%
squared after the collision. The kinetic energy is given
175
00:10:48.629 --> 00:10:52.549 A:middle L:90%
by the some off the kinetic energy off Asteroid A
176
00:10:52.559 --> 00:10:54.330 A:middle L:90%
on asteroid. Be so after the collision, we
177
00:10:54.330 --> 00:10:58.940 A:middle L:90%
have one half off a times the speed off asteroid
178
00:10:58.950 --> 00:11:05.509 A:middle L:90%
a 29.3 squared after the collision, plus one half
179
00:11:05.519 --> 00:11:07.059 A:middle L:90%
off the mass off Asteroid B, which is equal
180
00:11:07.059 --> 00:11:11.190 A:middle L:90%
to the mass off asteroid eight times the speed off
181
00:11:11.200 --> 00:11:16.049 A:middle L:90%
Asteroid B, which is 20.7 squared after the collision
182
00:11:16.639 --> 00:11:18.179 A:middle L:90%
. Okay, now we can calculate what is the
183
00:11:18.179 --> 00:11:22.090 A:middle L:90%
variation in the kinetic energy. By doing that,
184
00:11:22.100 --> 00:11:24.590 A:middle L:90%
we got the following. The variation in the kinetic
185
00:11:24.600 --> 00:11:28.320 A:middle L:90%
energy is given by the kinetic energy after the collision
186
00:11:28.330 --> 00:11:35.460 A:middle L:90%
. So one half m a factoring this times 29.3
187
00:11:35.460 --> 00:11:43.009 A:middle L:90%
squared plus 20.7 squared minus one half m a times
188
00:11:43.009 --> 00:11:46.259 A:middle L:90%
40 squared. This is a variation in the kinetic
189
00:11:46.269 --> 00:11:50.490 A:middle L:90%
energy. Then we can divide the variation in the
190
00:11:50.490 --> 00:11:56.519 A:middle L:90%
kinetic energy by the initial kinetic energy E K.
191
00:11:56.990 --> 00:11:58.460 A:middle L:90%
And this is a fraction that we want to evaluate
192
00:11:58.840 --> 00:12:03.980 A:middle L:90%
. This fraction is given by one half off M
193
00:12:03.990 --> 00:12:11.490 A:middle L:90%
A times 29.3 squared plus 20.7 squared minus 40 squared
194
00:12:11.860 --> 00:12:18.049 A:middle L:90%
, divided by one half off M a times 40
195
00:12:18.059 --> 00:12:22.210 A:middle L:90%
squared. We can simplify the factors off one half
196
00:12:22.220 --> 00:12:24.710 A:middle L:90%
and m a. And then by performing the calculation
197
00:12:24.720 --> 00:12:28.710 A:middle L:90%
, we get the fraction, which is approximately minus
198
00:12:28.730 --> 00:12:37.820 A:middle L:90%
0.196 which is equivalent to minus 19.6%. So the
199
00:12:37.820 --> 00:12:43.799 A:middle L:90%
conclusion is that 19.6% off the initial kinetic energy was
200
00:12:43.799 --> 00:12:46.049 A:middle L:90%
dissipated during the collision. We know that it was
201
00:12:46.049 --> 00:12:50.460 A:middle L:90%
dissipated because we have this minus sign here on this
202
00:12:50.460 --> 00:12:50.960 A:middle L:90%
is the answer to this question.