From graph in time from `t=0` to `t=3` sec.
acceleration of object of mass `m_(1)=10kg` is
`a=(15-0)/(3)=5m//s^(2)`
`therefore` Force on object of mass `m_(1)` from `t=0` to `t=3` sec. (i)
`=10xx5=50N`
Before and after collision at `t=4` sec. the velocites of blocks are as shown.
after collision
`therefore` initial momentum of system
`=m_(1)u_(1)+m_(2)=150+25u_(2)`
final momentum of system
`=(m_(1)+m_(2))v=35xx5=175`
From conservationof momentum
`therefore 150+25u_(2)=175`
or `u_(2)=+1 m//s`
`therefore` speed of second particle just before colision is `1m//s` and before collision both blocks move in same direction.