Two particles having position verctors `vecr_(1)=(3hati+5hatj)` metres and `vecr_(2)=(-5hati-3hatj)` metres are moving with velocities `vecv_(1)=(4hati+3hatj)m//s and vecv_(2)=(alphahati+7hatj)m//s`. If they collide after 2 seconds, the value of `alpha` is

Options

A. 2
B. 4
C. 6
D. 8

Answer

D

Explanation

It is clear from figure that the displacement vector `Delta vec(r )` between particles `p_(1) and p_(2)` is `Delta vec(r )= vec(r )_(2)- vec(r )_(1)= -8 hat(i) - 8hat(j)`

`|Delta vec(r )| = sqrt((-8)^(2) + (-8)^(2))= 8 sqrt2` ….(i)
Now, as the particles are moving in same direction (`therefore vec(v)_(1) and vec(v)_(2)` are +ve), the relative velocity is given by
`vec(r)_("rel")= vec(v)_(2) -vec(v)_(1)= (alpha -4) hat(i) + 4hat(j)`
`vec(v)_("rel") = sqrt((alpha-4)^(2) + 16)` ....(ii)
Now, we know `|vec(v)_("rel")| = (|Delta vec(r )|)/(t)`
Substituting the values of `vec(v)_("rel") and |Delta vec(r )|` from equation (i) and (ii) and t= 2s, then on solving we get `alpha= 8`