दो कण A तथा B स्थिर वेग क्रमश: overrightarrow | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Let\,the\,particles\,A\,and\,B\,collide\
at\,time\,t.\,For\,their\,collision,\,the\,position\
v

Vedclass · Accepted Answer

$\frac{{\overrightarrow {{r_1}} - \overrightarrow {{r_2}} }}{{\left| {\overrightarrow {{r_1}} - \overrightarrow {{r_2}} } \right|}} = \;\frac{{\overrightarrow {{v_2}} - \overrightarrow {{v_1}} }}{{\left| {\overrightarrow {{v_2}} - \overrightarrow {{v_1}} } \right|}}\;\;\;$

Vedclass · Answer

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Let\,the\,particles\,A\,and\,B\,collide\
at\,time\,t.\,For\,their\,collision,\,the\,position\
vectors\,of\,both\,particles\,sholud\,be\,\
same\,at\,time\,t,\,i.e.,\
{{\vec r}_1} + {{\vec v}_1}t = {{\vec r}_2} + {{\vec v}_2}t\,;\,{{\vec r}_1} - {{\vec r}_2} = {{\vec v}_2}t - {{\vec v}_1}t\
 = \left( {{{\vec v}_2} - {{\vec v}_1}} ight)t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i ight)
\end{array}$

$\begin{array}{l}
Also,\,\left| {\,{{\vec r}_1} - {{\vec r}_2}} ight| = \left| {{{\vec v}_2} - {{\vec v}_1}} ight|t\,\,or\,t = \frac{{\left| {\,{{\vec r}_1} - \vec r} ight|}}{{\left| {{{\vec v}_2} - {{\vec v}_1}} ight|}}\
Substituting\,this\,value\,of\,t\,in\,eqn.\,\left( i ight),\
we\,get\,\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\vec r}_1} - {{\vec r}_2} = \left( {{{\vec v}_2} - {{\vec v}_1}} ight)\frac{{\left| {\,{{\vec r}_1} - \vec r} ight|}}{{\left| {{{\vec v}_2} - {{\vec v}_1}} ight|}}\
or\,\frac{{\left| {\,{{\vec r}_1} - {{\vec r}_2}} ight|}}{{\left| {\,{{\vec r}_1} - {{\vec r}_2}} ight|}} = \frac{{\left( {{{\vec v}_2} - {{\vec v}_1}} ight)}}{{\left| {{{\vec v}_2} - {{\vec v}_1}} ight|}}
\end{array}$

Similar Questions