Questions

# Two particles having position vectors $\stackrel{\to }{{\mathrm{r}}_{1}}=\left(3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}\right)$ meters and $\stackrel{\to }{{\mathrm{r}}_{2}}=\left(-5\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}\right)$ meters are moving with velocities ${\stackrel{\to }{\mathrm{v}}}_{1}=\left(4\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}\right)\text{\hspace{0.17em}}\mathrm{m}/\mathrm{s}$ and  ${\stackrel{\to }{\mathrm{v}}}_{2}=\left(\mathrm{\alpha }\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+7\stackrel{^}{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$. If they collide after 2 seconds, the value of  $\text{'}\mathrm{\alpha }\text{'}$ is

a
2
b
4
c
6
d
8

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