Question

please solve the question given in the attachment

Answer 1

Let velocity of third particle = V
here 3rd particle moves in x = √3y
e.g y = (1/√3)x = tan30°x
hence, line inclination with positive x - axis is 30°
so,
Velocity of 3rd particle = Vcos30° i + Vsin30° j
= √3V/2 i + V/2 j

velocity of 1st particle = V1 i
Velocity of 2nd particle = V2 j

question says that all are colinear .
so,
(V2j - V1 i) × {√3V/2i + V/2j - V1 i) = 0
(V2j - V1 i) × {( √3V/2 - V1)i + V/2j } = 0
-V2(√3V/2 - V1)k -VV1/2 K = 0

-√3V2.V/2 + V1V2 = VV1/2
V1.V2 = V.V1/2 + √3V.V2/2
V1.V2 =V ( V1+ √3V2)/2

V = 2V1,V2/(V1 + √3V2)
hence , option (B) is correct .