In the arrangement shown in fig. a wedge of mass `m_(3) =3.45 kg` is placed on a smooth horizontal surface. Small and light pulley is connected on its top edge, as shown. A light, flexible thread passes over the pulley. Two block having mass `m_(1)=1.3kg and m_(2) =1.5 kg` are connected at the ends of the thread.

`m_(1)` is on smooth horizontal surface and `m_(2)` rests on inclined smooth surface of the wedge. the base length of wedge is 2m and inclination is `37^(@)`. `m_(2)` is initially near the top edge of the wedge.
If the whole system is released from rest, calculate:
(a) velocity of wedge when `m_(2)` reaches its bottom.
(b) velocity of `m_(2)` at that instant.
`(g=10ms^(-2))`.

Answer

`2ms^(-1); sqrt(13)ms^(-1)`

Explanation

Let acceleration of `m_(1)` be a (rightwards) and that of wedge be b(leftwards). Acceleration of `m_(2)` (relative to wedge) becomes `(a+b)`, down the plane. Therefore, resulant acceleration of `m_(2)` is vector sum of the two acceleration.
(a) `(a+b)` down the plane and
(b) b leftwards.
Hence, components of this resultant acceleration are
(a) `[(a+b) cos 37^(@)-b]=((4a-b))/(5)` horizontal rightward, and
(b) `(a+b) sin 37^(@)=3/5 (a+b)` vertically downward.
Considering free body diagrams

For horizontal force on `m_(1), T=m_(1)a` ...(i)
For horizontal force on wedge,
`T-T cos 37^(@)+N_(2) sin 37^(@)=m_(3)b`...(ii)
For horizontal force on `m_(2)`
`N_(2). sin 37^(@)-T cos 37^(@)=m_(2)(0.8a-0.2)` ...(iii)
For vertical force on `m_(2)`
`m_(2)g-N_(2) cos 37^(@)-T sin 37^(@)=m_(2)(0.6 a + 0.6b)`...(iv)
From above equations , `a=3ms^(-2), b=2ms^(-2)`
Since base angle and base length of wedge are `37^(@)` and 2m respectively, therefore, height of its vertical faces is `2tan 37^(@)=1.5m`.
Now considering vertical motion of `m_(2)` from top of bottom of the wedge, `u=0`, acceleration =`(0.6a=0.6b)=3ms^(-2)` and displacement = `1.50m`.
`t=1s`
At this instant, horizontal component of velocity of `m_(2)` is
`v_(2x)=(0.8a-0.2b)t=2ms^(-1)`
and vertical component,
`v_(2y)=(0.6a+0.6b)t=3ms^(1)`
Vertical of `m_(2)` is
`v_(2)=sqrt(v_(2x)^(2)+v_(2y)^(2))=sqrt(13)ms^(-1)`
Velocity of wedge at this instant = `bt=2ms^(-1)`.