July 29, 2008

Q. 1.16 - "Problems in General Physics" by I.E. Irodov


\begin{justify}<br />\textbf{Question}: Two particles, 1 and 2, move with constant v...<br />...ce between the particles become the smallest? What is it equal to?<br />\end{justify}

\begin{justify}<br />\noindent\textbf{Answer}:<br />At any given time $t$, the $1^{st}$\...<br />...rom $O$ and the<br />$2^{nd}$ particle is at distance $l_2 - v_2 t$.<br />\end{justify}

img3.gif

\begin{justify}<br />At this time the distance between the two particles is<br />\begin{...<br />...c{(v_1 l_2 - v_2 l_1)}{\sqrt{v_1^2 + v_2^2}} \vert<br />\end{eqnarray*}\end{justify}

Posted by Apurva Sharan at 10:55 PM
Labels: ,

2 comments:

Unknown said...

these answers nicely explained please upload answers of other chapters

August 4, 2015 at 10:39 AM
Apurva Sharan said...

Hi Utkarsh,

Are you looking for answers to a specific question? Please let me know.

Regards,
Apurva

August 4, 2015 at 10:09 PM

Post a Comment

Subscribe to: Post Comments (Atom)