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## Physics library

### Unit 6: Lesson 1

Momentum and Impulse- Apply: conservation of momentum
- Introduction to momentum
- Impulse and momentum dodgeball example
- What are momentum and impulse?
- What is conservation of momentum?
- Bouncing fruit collision example
- Momentum: Ice skater throws a ball
- 2-dimensional momentum problem
- 2-dimensional momentum problem (part 2)
- What are two dimensional collisions?
- Force vs. time graphs

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# 2-dimensional momentum problem (part 2)

We finish the 2-dimensional momentum problem. Created by Sal Khan.

## Video transcript

Welcome back. When I left off I was rushing
at the end of this problem because I tend to rush at the
end of problems when I am getting close to the YouTube
10 minute limit. But I just wanted to review the
end of it because I feel like I rushed it. And then, actually continue with
it and actually solve for the angle and then, introduce
a little bit of-- a little more trigonometry. So just to review what we did,
we said momentum is conserved and in two dimensions that means
momentum is conserved in each of the dimensions. So we figured out what the
initial momentum of the entire system was and we said, well,
in the x direction, the initial momentum-- and all the
momentum was coming from the ball A right. Because ball B wasn't moving,
so its velocity was 0. So its momentum was 0. So ball A in the x direction and
it was only moving in the x direction. So it's momentum in the x
direction was 3 meters per second times 10 kilogram
meters per second. And we got 30 kilogram
meters per second. And then there was no momentum
in the y direction. And then we knew that well after
they hit each other, ball A kind of ricochets off
at a 30 degree angle at 2 meters per second. We used that information to
figure out the x and y components of A's velocity. So A's velocity in the y
direction was 1 meter per second and A's velocity
in the x direction was square root of 3. And we used that information to
figure out A's momentum in each direction. We said well, the momentum in
the y direction must be 1 meter per second times A's mass,
which is 10 kilogram meters per second. Which I wrote-- what
I wrote here. And then we figured out A's
momentum in the B direction and we said well, that's
just going to be square root of 3 times 10. And that's 10 square
root of 3. And then we used that
information to solve for B's momentum. Because we said well, B's
momentum plus A's momentum in the x direction has
to add up to 30. This was the x direction
before. And we knew that B's momentum
plus A's momentum in the y direction had to add
up to 0, right? And so, since y's momentum going
upwards was 10 kilogram meters per second, we knew
that B's momentum going downwards would also
have to be 10 kilogram meters per second. Or you could even say
it's negative 10. And we figure that out based
on the fact that B had half the mass. That its velocity going down
was 2 meters per second. And similarly, we knew that
A's momentum in the x direction, which was 10 square
root of 3 kilogram meters per second, plus B's momentum
in the x direction is equal to 30. And then we just subtracted out
and we got B's momentum in the x direction. And then we divided by B's
mass to get its velocity. Which we got as 2.54. So that's where I left off
and we were rushing. And already, this gives you a
sense of what B is doing. Although it's broken up into
the x and y direction. Now if we wanted to simplify
this, if we wanted to kind of write B's new velocity the same
way that the problem gave us A's velocity, right? They told us A's velocity was
2 meters per second at an angle of 30 degrees. We now have to use this
information to figure out B's velocity and the angle of it. And how do we do that? Well this is just straight up
trigonometry at this point, or really just straight
up geometry. Let me clear all of this. And let's remember these two
numbers, 2.54 and minus 2. So B, we learned that in the x
direction its velocity-- this is all for B-- is equal to 2.54
meters per second and then y direction, it
was moving down. We could write this
as minus 2. But I'll just write this as 2
meters per second downwards. Right? Same thing. Minus 2 up is the same thing as
2 meters per second down. So the resulting vector's
going to look something like this. When you add two vectors you
just put them-- put the one's end at the beginning of the
other-- put them front to end, like we did here. And then you add them
together and this is the resulting vector. And I think you're used
to that at this point. And now we have to figure out
this angle and this side. Well this side is easy because
this is a right angle, so we use Pythagorean theorem. So this is going to be the
square root of 2.54 squared plus 2 squared. And what's 2.54 squared? 2.54 times-- whoops. 2.54 times 2.54 is
equal to 6.45. So that's the square root of
6.45 plus 4, which equals the square root of 10.45. And take the square
root of that. So that's 3.2, roughly. So the resulting velocity in
this direction, whatever angle this is, is 3.2 meters
per second. And I just used Pythagorean
theorem. So now all we have to do is
figure out the angle. We could use really any of the
trig ratios because we know all of the sides. So I don't know, let's
use one that you feel comfortable with. Well let's use sine. So sine of theta is
equal to what? SOH CAH TOA. Sine is opposite over
hypotenuse. So the opposite side is the y
direction, so that's 2, over the hypotenuse, 3.2. So 2 divided by 2 divided by 3.2
is equal to 0.625, which equals 0.625. So sine of theta equals 0.625. And maybe you're not familiar
with arcsine yet because I don't think I actually have
covered yet in the trig modules, although I
will eventually. So we know it's just the inverse
function of sine. So sine of theta is
equal to 0.625. Then we know that theta is equal
to the arcsine of 0.625. This is essentially saying,
when you say arcsine, this says, tell me the angle whose
sine is this number? That's what arcsine is. And we can take out Google
because it actually happens that Google has a-- let's see. Google actually-- it's an
automatic calculator. So you could type in arcsine
on Google of 0.625. Although I think the
answer they give you will be in radians. So I'll take that answer that
will be in radians and I want to convert to degrees, so I
multiply it times 180 over pi. That's just how I convert
from radians to degrees. And let's see what I get. So Google, you see, Google
says 38.68 degrees. They multiplied the whole
thing times 180 and then divided by pi, but that should
be the same thing. So roughly 38.7 degrees
is theta. Hope you understand that. You could pause it here if
you don't, but let me just write that down. So it's 38 degrees. So theta is equal
to 38.7 degrees. So then we're done. We figured out that
ball B gets hit. This is ball B and it
got hit by ball A. Ball A went off in that
direction at a 30 degree angle, at a 30 degree angle
at 2 meters per second. And now ball B goes at 38.--
or we could say roughly 39 degrees below the horizontal
at a velocity of 3.2 meters per second. And does this intuitively
make sense to you? Well if you remember the problem
from before-- and I know I erased everything. Ball A had a mass of 10
kilograms while ball B had a mass of 5 kilograms.
So it makes sense. So let's think about just
the y direction. Ball A, we figured out, the y
component of its velocity was 1 meter per second. And ball B's y component is 2
meters per second downwards. And does that makes sense? Well sure. Because their momentums
have to add up to 0. There was no y component of the
momentum before they hit each other. And in order for B to have the
same momentum going downwards in the y direction as A going
upwards, its velocity has to be essentially double, because
its mass is half. And a similar logic, although
the cosine-- it doesn't work out exactly like that. But a similar logic would mean
that its overall velocity is going to be faster than the-
than A's velocity. And so what was I
just-- oh yeah. My phone was ringing and
I got caught up. My brain starts to
malfunction. But anyway, as I was
saying, so just intuitively it makes sense. B has a smaller mass than A, so
it makes sense that-- one, B will be going faster and
that it gets deflected a little bit more as well. The reason why it seems like
it gets deflected more is because its y component
is more. But anyway, that last piece is
just to kind of hopefully give you a sense of what's happening
and I will see you in the next video.