Physics Help Needed

[xXHavokXx]

Its been a long while since I've done this stuff. I was tutoring a friend online and he gave me this problem and for the life of me I cant figure out how they got the answers. I get the physics I'm just not seeing the math, First I was thinking straight cross product but they didnt give
v . I set it up as a series of matrices but you cant invert B, It's probably something simple.

Anyone help?

[tropical_fishy]

did you try solving for Vx or Vy and then substituting?

[xXHavokXx]

Yes, but i have to pretend i dont know what Vx and Vy are, they are the answers. He wants to know how to do it if you dont have them

[tropical_fishy]

I meant solving for the variable Vx, and taking the equation you end up with and plugging it in for Vx everywhere... you should end up with only Vy as a variable. then solve for Vy. Once you have Vy, plug that number back in and get Vx.

[ScatterPlot]

You should be able to split that up into 2 or more equations; the sum of the forces in the I direction on the left is equal to the sum of the forces in the I direction on the right, same with J and K. I -am- kinda confused as to why there is no K on the left side- maybe Ek (sum of K) on the left is zero and understood to be such? I dunno, haven't had too much physics yet and still get kinda confused with the cross product stuff.

What do you do with the "e"? Can you distribute that or is it something special? If that wasn't there I could probly see if my way worked, but it's kinda messing me up. I take it that it's not the same natural log "e" right?

[ScatterPlot]

I meant solving for the variable Vx, and taking the equation you end up with and plugging it in for Vx everywhere... you should end up with only Vy as a variable. then solve for Vy. Once you have Vy, plug that number back in and get Vx.

I'm pretty sure that will cancel out to 0=0 or something along those lines. It's a single equation with 2 variables, and you can't find the two with normal algebra. You gotta use the funny physics kind

[bleachit]

You should be able to split that up into 2 or more equations; the sum of the forces in the I direction on the left is equal to the sum of the forces in the I direction on the right, same with J and K. I -am- kinda confused as to why there is no K on the left side- maybe Ek (sum of K) on the left is zero and understood to be such? I dunno, haven't had too much physics yet and still get kinda confused with the cross product stuff.

What do you do with the "e"? Can you distribute that or is it something special? If that wasn't there I could probly see if my way worked, but it's kinda messing me up. I take it that it's not the same natural log "e" right?

I think "e" is a constant.. so you may be able to find that in the text book.

if K is not listed, then it is understood to be zero.

[deadeye9]

"Equating corresponding components" should mean you equate
i-hat coefficients from the LHS and RHS of the equation.
Same for j-hat and k-hat.

[gimp]

the i components on the right should equal the i components on the left, same with the j components. You can ignore the k. so,

(4x10^-17) = 1.6x10^-19(.03Vy - 40)
Solve that for Vy and it comes out to 7000

You can do the same thing for the Vx component and it comes out to -3500.

*edit:

A good check is to plug the results into the K component,

If you do 1.6x10^-19(.02x-3500 + .01x7000) it comes out to be zero, which is consistent with the equation.

[xXHavokXx]

yeah I went back to sleep woke up and figured it out


also the the lack of a scientific calculator mad alot of zeroes on paper.

[Maggot6]

I am a meer grade nine, And I have always thought physics would be a great course, what kind of stuff do you learn? Just how things interact with the stuff around them(yes that was broad)....

[ascetic1]

im not in pyhsics yet, next year, but i have dealt with some basic forms of it...
basically equations relating to distance, force, propulsion, time etc...

ex. time= distance/speed...

stuff like that

[gimp]

Basic physics is kind of boring to me. It's all about throwing balls and seeing where they're gonna land. Once you get into electricity and magnetism and all that stuff it gets interesting. You know, things that you can't see with the naked eye. Physics is really just applied math.

It's definetly a good course to take.

[xmetal2001]

Yea, E&M is the good part about physics -- you just have to suffer through mechanics to get to it.

[CoolHand]

I think "e" is a constant.. so you may be able to find that in the text book.

if K is not listed, then it is understood to be zero.

You are correct. "e" is the charge on the electron which according to my handy dandy sheet of constants (still stashed from my stint with phys 24) = 1.602176462 x 10^-19 C

This kind of physics is cool, but not of much practical use (or at least not as useful as Newtonian Physics). In order to be able to solve these kinds of problems, they have to be drastically over simplified, to the point that they are not of much use in the real world anymore.

Its still neat though.