Two bodies of equal mass are at rest, side by side. A constant force F is applied on the first body while at the same instant an impulsive force, producing an impulse I is applied in the same direction on the second body. Determine the time taken by them, in terms of F and I, to be side by side again.

Two bodies of equal mass are at rest, side by side. A constant force F

1.	Two bodies of equal mass are at rest, side by side. A constant force F is applied on the first body while at the same instant an impulsive force, producing an impulse I is applied in the same direction on the second body. Determine the time taken by them, in terms of F and I, to be side by side again.
Answer» Solution :Let the MASS of each body be m, and the times when they are side by 0 and t s. Acceleration of the first body due to F, a = `(F)/(m)`. `therefore` DISPLACEMENT of the first body in time t, starting from rest, `s_(1) = (1)/(2)cdot (F)/(m)cdot t^(2) = (Ft^(2))/(2m)` The change in momentum of the second body due to impulse, I = change in momentum = mv - mu = mv - m `XX` 0= mv or,`v = (I)/(m)` As no other force acts on the second body, it moves with this constant velocity (v) for t s, and covers a distance `s_(2)`. `therefore "" s_(2) = vt = (I t)/(m)` According to the problem `s_(1) = s_(2)` `therefore ""(Fr^(2))/(2m) = (I t)/(m) `[ from (1) and (2) ] `therefore "" t = (2I)/(F)` .