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