Two particles having position vectors $\overrightarrow{{\mathrm{r}}_1}=(3\widehat{\mathrm{i}}+5\widehat{\mathrm{j}})$ meters and $\overrightarrow{{\mathrm{r}}_2}=(-5\widehat{\mathrm{i}}-3\widehat{\mathrm{j}})$ meters are moving with velocities ${\overrightarrow{\mathrm{v}}}_1=(4\widehat{\mathrm{i}}+3\widehat{\mathrm{j}})\mathrm{m}/\mathrm{s}$ and  ${\overrightarrow{\mathrm{v}}}_2=(\mathrm{\alpha }\widehat{\mathrm{i}}+7\widehat{\mathrm{j}})\mathrm{m}/\mathrm{s}$. If they collide after 2 seconds, the value of  $\text{'}\mathrm{\alpha }\text{'}$ is

detailed solution

Correct option is D

It is clear from figure that the displacement vector Δr→ between particles p1 and p2 is Δr→=r2→−r1→=−8i^−8j^ |Δr→| =(−8)2+(−8)2=82          …..(i)Now, as the particles are moving in same direction (∵ v1→ and v2→ are +ve), the relative velocity is given byv→rel=v2→−v1→=(α−4)i^+4j^v→rel=(α−4)2+16                           …..(ii)Now, we know  |v→rel| =|Δr→|tSubstituting the values of  v→rel and |Δr→| from equation (i) and (ii) and t=2s, then on solving we get  α=8