Thank you both. I got it a little while after your post Simon, but my working was a mess and Ivy's post really clarified what it was that I had done. I latexed up the tidied version.
(ds)^2=(dx)^2+(dy)^2
(ds)^2=(dx)^2+\frac{(dx)^2(dy)^2}{(dx)^2}
(ds)^2=(dx)^2(1+\frac{(dy)^2}{(dx)^2})...
Hiya.
I got to an interesting bit in a calculus book, but as usual I'm stumped by a (probably simple) algebraic step.
The author goes from:
(ds)^2=(dx)^2+(dy)^2
to:
ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx
I understand moving the square root across, but I don't understand how the...
The following link has a pretty nice summary. It suggests the answer is YES.
http://www.mi.sanu.ac.rs/vismath/rojas/index.html
However according to this picture from wiki, neither end of an ovoid is usually a full hemisphere. I don't know if these geometric versions of an egg shape really...
Thanks alot, that's great.
\theta=43.2^\circ
v_o=2.56*10^6ms^{-1}
I see why your way would have been quicker and how I could have spotted it by looking more carefully at the unknowns in the original equations.
1=tan^2\theta+\frac{m_p}{m_ocos^2\theta}
Could multiply through by cos^2\theta to get:
cos^2\theta=cos^2\theta tan^2\theta+\frac{m_p}{m_o}
That's equivalent to:
cos^2\theta=sin^2\theta+\frac{m_p}{m_o}
And you might have guessed that I really don't know what to do with this equation. I...
Thanks that was a lot of help. I drew this diagram.
https://www.physicsforums.com/attachment.php?attachmentid=18186&d=1238173851
Momentum first:
1. vertical: 0 = m_pv_2 + m_ov_osin\theta
2. horizontal: m_pv_1 = m_ov_ocos\theta
Kinetic Energy:
3...
Homework Statement
a) State the law of conservation of linear momentum.
b) A proton of mass 1.6*10-27kg traveling with a velocity of 3*107ms-1 collides with a nucleus of an oxygen atom of mass 2.56*10-26kg (which may be assumed to be at rest initially) and rebounds in a direction at 90 degrees...