2-dimensional momentum problem (part 2) We finish the 2-dimensional momentum problem
2-dimensional momentum problem (part 2)
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- 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.
- 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.
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