Question

Not sure how to begin?

Answer 1

See the diagram below,how to break the weight of the blocks into two perpendicular components,one of which `mg sin 30` is along the inclined plane,which is the component of its weight trying pull the mass downwards along the plane,and the other `mg cos 30` is responsible for the normal reaction.

So,acceleration of both the blocks are directed downwards along the plane and the value is `(mg sin 30)/m=g/2`

Suppose,both the blocks will collide after time `t`, so in that time,if the block going up moves by a distance `x`,then we can write, `x=v_o t - 1/2 (g/2) t^2` ...1

Similarly,in this time,the block coming down will move by a distance of `s-x`, so,

`s-x = 1/2 (g/2) t^2`...2

So,putting the value of `x` from 1 in 2 we get,

`s= v_o t`

So, `t= s/v_o = 4/6=0.67 s`

So,in this time distance travelled by the block going down is `1/2 (g/2) (0.67)^2=2.20 m`

So,collision will occur `4-2.20=1.8 m` above from the bottom of the plane.(distance along the plane)