Homework Help: Need help

1. Apr 16, 2007

rootX

need urgent help

1. The problem statement, all variables and given/known data
Still working on the question I posted few days ago, here's the link for that question:

I kinda got confident over my answer, but I discovered another formula, and it screwed up everything.
It is:
$$V_{1,f} = \frac{m_{1}-m_{2}}{m_{1}+m_{2}} v_{1,i} + \frac{2 * m_{1}}{m_{1}+m_{2}} v_{2,i}$$
$$V_{2,f} = \frac{m_{1}-m_{2}}{m_{1}+m_{2}} v_{2,i} + \frac{2 * m_{1}}{m_{1}+m_{2}} v_{1,i}$$

Hope, I put the tex code right. Anyhow, it gives me
$$x_{1,f}=-4.37$$
$$x_{2,f}=10.03$$
$$y_{1,f}=7.01$$
$$y_{2,f}=4.48$$

But using my that previous way(that relativity method) I get
$$x_{1,f}=-1.20$$
$$x_{2,f}=9.40$$
$$y_{1,f}=13.20$$
$$y_{2,f}=5.85$$

Now, I am confused over mine(I tried that applet, but it is pretty cumbersome
to set values.

Anyhow, if anyone can tell me which answer is right?
I would be really thankful. I have this thing due tomorrow

2. Apr 16, 2007

hage567

I didn't really look that closely at your relativity method, but did you convert back to the "non-relativity" frame after you found your values? Just an idea.

3. Apr 16, 2007

rootX

Yes, I did.
And if I use my answers, then they give total final kinetic energy same as before the collisions.
However, the answer that that formula gives provide a different total kinetic energy(But I saw it in like two Physics books since then)

4. Apr 16, 2007

rootX

just one last more question

A metal tube is 2.40 m long and has a mass of 800g. It contains 200g sliding mass at its front end. There is no friction between any surface. When the sliding mass reached the end of the tube, it sticks to a magnet attached there. How long would it take the tube to travel a distance of 6.00 m, if the metal tube has an initial velocity of 6.00 m/s?

I calculated time it took before the sliding mass reached the end, and then found the distance left to cover when this happens, and find time for that distance. And, I got 1.16 s.

5. Apr 16, 2007

rootX

oops, that was supposed to go in a new thread

6. Apr 16, 2007

hage567

I think those equations you posted are intended for one-dimensional collisions.

7. Apr 16, 2007

rootX

yes, the book mentions that. However, I did break the velocities into components. So, shouldn't that make the collision 1-dimensional for each axis?