A horizontal....

ohk.......lemme ans...

drawing fbd diagrams......

then write all possible dynamics eq..

N₁=2mg
f₁=kN₁=2kmg
therefore
T-f₁=2mw₁
or T-2kmg=2mw₁

according to prob....w₁=w
therefore T-2kmg=2mw...............i

N₂ =mg
f₂ =kmg
T-kmg=mw₂....................ii
subtractiing eq ii from i

T-2kmg=2mw
T-kmg=mw₂

we get
-kmg=2mw-mw₂

w₂ =2w +kg

now use kinematical relations

let t be the time after which they collide
s₁=distance which bar travelled in time t=12 wt²
s₂=distance which motor body travelled in time t=12w₂t²

For collision to occur
s₁ + s₂=l

l=12 wt² +12 w₂t²
putting value of w₂ from eq iii ...in above eq we get
t==√2l/(3w+kg)