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}

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\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}

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2 comments:

  1. UnknownAugust 4, 2015 at 10:39 AM

    these answers nicely explained please upload answers of other chapters

    ReplyDelete
  2. Apurva SharanAugust 4, 2015 at 10:09 PM

    Hi Utkarsh,

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

    Regards,
    Apurva

    ReplyDelete