question_answer

Two particles A and B, move with constant velocities \[{{v}_{1}}\] and \[{{v}_{2}}\]. At the initial moment, their position vectors are \[{{r}_{1}}\] and \[{{r}_{2}}\] respectively. The condition for particles A and B for their collision is   [NEET 2015 (Re)]

A) \[\frac{{{r}_{1}}-{{r}_{2}}}{\left| {{r}_{1}}-{{r}_{2}} \right|}=\frac{{{v}_{2}}-{{v}_{1}}}{\left| {{v}_{3}}-{{v}_{1}} \right|}\]

B) \[{{r}_{1}}\cdot {{v}_{1}}={{r}_{2}}\cdot {{v}_{2}}\]

C) \[{{r}_{1}}\times {{v}_{1}}={{r}_{2}}\times {{v}_{2}}\]

D) \[{{r}_{1}}-{{r}_{2}}={{v}_{1}}-{{v}_{2}}\]

Correct Answer: A

Solution :

For two particles A and B move with constant velocities \[{{v}_{1}}\] and \[{{v}_{2}}\]. Such that two particles to collide, the direction of the relative velocity of one with respect to other should be directed towards the relative position of the other particle.

i.e. \[\frac{{{r}_{1}}-{{r}_{2}}}{\left| {{r}_{1}}-{{r}_{2}} \right|}\xrightarrow{\,}\]direction of relative position of 1 w.r.t 2:

Similarly, \[\frac{{{v}_{1}}-{{v}_{2}}}{\left| {{v}_{1}}-{{v}_{2}} \right|}\xrightarrow{\,}\]direction of velocity of 2 w.r.t 1

So, for collision of A and B, we get

\[\frac{{{r}_{1}}-{{r}_{2}}}{\left| {{r}_{1}}-{{r}_{2}} \right|}\frac{{{v}_{2}}-{{v}_{1}}}{\left| {{v}_{2}}-{{v}_{1}} \right|}\]

Alternate Method As resultant displacement of a particle,

\[R={{r}_{1}}+{{v}_{1}}t={{r}_{2}}+{{v}_{2}}t\]

i.e.       \[{{r}_{1}}-{{r}_{2}}=({{v}_{2}}-{{v}_{1}})t\]

So,       \[\frac{{{r}_{1}}-{{r}_{2}}}{\left| {{r}_{1}}-{{r}_{2}} \right|}=\frac{({{v}_{2}}-{{v}_{1}})t}{\left| {{v}_{2}}-{{v}_{1}} \right|t}\]

\[\frac{{{r}_{1}}-{{r}_{2}}}{\left| {{r}_{1}}-{{r}_{2}} \right|}=\frac{({{v}_{2}}-{{v}_{1}})}{\left| {{v}_{2}}-{{v}_{1}} \right|}\]