question_answer

Two particles having position vectors \[\overrightarrow{{{r}_{1}}}=(3\hat{i}+5\hat{j})\] metres and \[\overrightarrow{{{r}_{2}}}=(-5\hat{i}-3\hat{j})\] metres are moving with velocities \[{{\overrightarrow{v}}_{1}}=(4\hat{i}+3\hat{j})\,m/s\] and \[{{\overrightarrow{v}}_{2}}=(\alpha \,\hat{i}+7\hat{j})\] \[m/s.\] If they collide after 2 seconds, the value of \['\alpha '\] is                        [EAMCET 2003]

A)          2   

B)          4

C)          6   

D)          8

Correct Answer: D

Solution :

         It is clear from figure that the displacement vector \[\Delta \overrightarrow{r}\] between particles \[{{p}_{1}}\] and \[{{p}_{2}}\] is \[\Delta \overrightarrow{r}=\overrightarrow{{{r}_{2}}}-\overrightarrow{{{r}_{1}}}=-8\hat{i}-8\hat{j}\] \[|\Delta \overrightarrow{r}|\,=\sqrt{{{(-8)}^{2}}+{{(-8)}^{2}}}=8\sqrt{2}\]                        ?..(i) Now, as the particles are moving in same direction \[(\because \ \overrightarrow{{{v}_{1}}}\text{ and }\overrightarrow{{{v}_{2}}}\text{ are }+ve)\], the relative velocity is given by \[{{\overrightarrow{v}}_{rel}}=\overrightarrow{{{v}_{2}}}-\overrightarrow{{{v}_{1}}}=(\alpha -4)\hat{i}+4\hat{j}\] \[{{\overrightarrow{v}}_{rel}}=\sqrt{{{(\alpha -4)}^{2}}+16}\]                         ?..(ii) Now, we know \[|{{\overrightarrow{v}}_{rel}}|\,=\frac{|\Delta \overrightarrow{r}|}{t}\] Substituting the values of \[{{\overrightarrow{v}}_{rel}}\] and \[|\Delta \overrightarrow{r}|\] from equation (i) and (ii) and \[t=2s\], then on solving we get \[\alpha =8\]