I'm trying to solve the velocities v1 and v2 for these particles. It should be basic vector shit in a nuclear fission setting, but the problem is that my equations give out scalars. Like, the last line shows that v1 is the momentum for p divided by a scalar. Considering that p1 should have momentum in the y direction, unlike p, it can't be right.
I have no idea and this is the wrong board. I'm not gonna sit here and work through this aids problem, but have you tried ratios? Some shit with the ratios of mass equaling the ratio of their velocity? Saw that on my physics test in college once
I've seen that kinematics equation for years and I still don't know why anyone would need v squared for anything.
Just split into components. Cos(30)|P1|+Cos(-60)|P2|=4*10^-19 for X component, Sin(30)|P1|+Sin(-60)|P2|=0 for Y component. Two equations, two variables, solve.
One tip you should remember, for problems like these don't memorize the equations. Just apply your common sense
In this particular case make sure the x momentum is the same before and after And also make sure the y momentum is the same and after You'll get 2 equations so just solve them
I'm not saying this is the most efficient way to do the problem, but you need to stop memorizing stuff and just go with your common sense
My bad, scrap that shit I said. I didn't even read the problem. Use the conservation of momentum.
Another thing, your equation 2nd to the bottom is wrong. It doesn't make any sense to divide by a vector -- it isn't even defined. You gotta take the magnitude. So you'll have to have equations for magnitudes of the x component and y component of the momentum, but it'll have squares and sqrts which seem it may be nasty so I again refer to this
what if I told you that every component of momentum is conserved separately (even the time component of 4-momentum, that gives energy conservation law) we are living in an isotropic space-time after all so drop that stupid trigonometry shit and calculate it like an intelligent being pic unrelated
It looks like they are p1 and p2 times P"emo" x which is a scalar It's just p = (p "emo" x, 0) and you're dropping symbols First off we need to break down what information we are given. Judging from the picture, the only thing which looks apparent to me is that the triangle is isoceles, which clearly isn't enough to determine the angle
This made me laugh. Talk about overkill. Dude, you realize that even general relativity is questionable after a certain level. Like we run into paradoxes.
None of mankind's theories seem fundamentally "right". All of them have issues and we just iron them out and backwards engineer this shit to get the best answers we can
Oh, I see what information we are given. The segments BC, CD, DE, EA are all of the same length and the triangle is isosceles. Interesting problem.
I'm gonna mess around with this a moment
Can you just pay attention in class? Also ask your teacher dumbass. They're there for a reason
In , |P1| and |P2| refer to the magnitude of the vectors. Since you already know the direction(the angles in your diagram) knowing the magnitude gives you the vector. If you need to express it by components it is .
Here is my thought process: Information >isoceles >m = BC = CD = DE = EA
Q: Is angle A a function of m, or is it independent of m? A: it must be independent of m, because if we double m then the triangle which contains segments of AE, ED, DC, BC would just double in size without changing proportions so angle A would always be the same.
Q: what piece of information is equivalent to knowing angle A? A: if I knew the length of segment ED, I could determine the angle A
Q: Is AED isoceles? A: Yes, AE = ED. So angel A = angle D. Add that to the information list.
Okay, after screwing around a while and looking for easy information, it's time to actually plan our course of action and solve it.
Since AD = EB, and both have ED in common, in principle we can probably solve for ED in terms of AD. But now we gotta ask ourselves, can we put AD in terms of m? The answer is yes. Since the triangle is isoceles we can use BC and angle A to determine AD. So we'll have an equation for angle A in terms of m, which is what we wanted.
however we also said in theory angle A should be independent of length m, so in the end it should just drop out and give us angel A = yada yada
So now we've basically planned how to solve the problem without doing any calculations. I'm not gonna do it, but that's the sort of stuff you gotta learn to solve problems in general
Lmao I also have vector homwork OP What school or state are you in? Major?
So in a right triangle the cosine is the ratio of the lengths of the adjacent and hypotenuse. The length of a side is analogous to the magnitude of the vector in your diagram. So it won't work for the vector but will for the magnitude of the vector.
you are dividing by a vector It makes no sense physically or logically
You can encounter OP*s issue even without the triangle. If you add up all the x components and y components and make equations, you still have to make every vector their magnitude. If you keep them in triple form (x,y,z) you will fail.