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Current time:0:00Total duration:9:32

2-dimensional momentum problem (part 2)

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 fail 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 zero so its momentum was zero so ball a in the X direction and it was only moving in the X direction so its momentum in the X Direction was three meters per second times 10 kilogram meters per second we got 30 kilogram meters per second and then there was no momentum in the 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 of two meters per second we use 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 direction phase velocity in the X direction was square root of 3 and we use that information to figure out as well mentum in each direction we said well the momentum in the Y direction must be 1 meter per second times a smash which is 10 kilogram meters per second which I wrote weight right when I wrote here and then we figured out a 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 roots of 3 and then we use that information to solve for B is momentum because we said well B is momentum plus s momentum in the X Direction has to add up to 30 right this was the X direction before and we knew that B's momentum plus a zwo mentum in the Y Direction had to add up to zero right and so since Y is momentum going upwards was 10 kilogram meters per second we knew that B's momentum going downwards would also have to be 10 kilogram meter per second or you could even say it's negative 10 and we figured that out based on the fact that we had half the mass that it's velocity going down was 2 meters per second and similarly we knew that a A's A's momentum in the X direction Plus which was 10 square roots of 3 kilogram meters per second plus B's momentum in the X Direction is equal to 30 and then we just subtract it 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 is two point five four so that's where I left off and we were rushing and I mean 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 is velocity right they told us a is velocity plus two meters per second at an angle of 30 degrees we now have to use this information to figure out B's velocity and 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 so if we let me clear all of this so just remember these two numbers 2.5 4 and -2 so B we we learned that in the X direction it's velocity this is all for B is equal to 2.5 4 meters per second and then the y direction it was moving down we could write this as minus 2 we could write this as minus 2 but I'll just write this as 2 meters per second downwards down right same thing minus 2 up is the same thing as 2 meters per second down so the resulting vector is going to look something like this when you add two vectors you just put them put the one end at the beginning of the other would put 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 two squared and what's 2.54 squared 2.5 4 times ty whoops 2.5 4 times two point five four is equal to six point four five so that's the square root of six point four five plus four which equals the square root of ten point four five plus four equals ten point four five and take the square root of that so that's three point two roughly three point two so it's the resulting velocity in this direction whatever angle this is is 3.2 meters per second and I just use Pythagorean theorem so now all we have to do is figure out the angle and we could use a bunch of we could use really any of the trig ratios because we know all of the sides so I don't know let's let's use one that you you feel comfortable with well unless you sign so sine of theta sine of theta is equal to what so cut Toa sine is opposite over hypotenuse so the opposite side is the Y direction so that's 2 over the hypotenuse 3.2 so two divided by two divided by three point two is equal to 0.625 which equals 0.625 so sine of theta is equal to 0.625 and maybe you're not familiar with arc sine yet because I don't think I actually have covered it yet in the trig modules but 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 arc sine of 0.625 this is essentially saying when you say arc sine this is telling me the angle whose sine is this number that's what arc sine is and we can take out Google because it actually 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 0.625 although i think there's the answer they give you will be in radians so i'll take that answer that'll be in radians and i want to convert to degrees so i multiply 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 thirty eight point six eight degrees they multiplied the whole thing times Wendy and then divide by five but that should be the same thing so thirty eight so roughly thirty eight point seven degrees this 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 thirty eight point seven so theta is equal to thirty eight point seven 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 two meters per second and now ball B goes at thirty eight point or we could say roughly 39 degrees below the horizontal at a velocity of 3.2 meters per second and and does this does this intuitively make sense to you well if you remember the problem from before and I know I erased everything ball a was had a mass of 10 kilograms while ball B had a mass of five kilograms so it makes sense so let's think about just the Y Direction ball a we figured out its what it the y component of its velocity was one meters per second one meter per second and ball B's y component is two meters per second downwards and does that make sense well sure because they're Momentum's have to add up to 0 there was no y component of the momentum before we before they hit each other 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 it's mass is 1/2 and a similar logic although the the cosine doesn't 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 going in then a z' velocity and so most I just say oh yeah okay my phone is ringing and I got I got caught up my brain starts to starts to malfunction but anyway I was saying so just intuitively it makes sense B has a smaller mass then a so it makes sense that one B will be going faster and that it gets deflected a little bit more as well and I mean the reason why it seems like it gets deflected more because it's Y components 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