Figure shows a wedge `A` of mass `6 m` smooth semicircular groove of radius `a= 8.4 m` placed on a smooth horizontal surface. A small block `B` of mass `m` is released from a position in groove where its radius is horizontal. Find the speed (in `ms^(-1)`) of bigger block when smaller block reaches its bottommost position.

Answer

2

Explanation

Let smaller block is moving with speed `v_(1)` relative to bigger block when it reaches the bottom most position and at this instant bigger block is moving at `v_(2)` (say) then using conservation of momentum in horizontal direction we have

`6mv_(2)=m(v_(1)-v_(2))`..........i
Now using energy conservation
`mga=1/2m(v_(1)-v_(2))^(2)+1/2(6m)v_(2)^(2)`......ii
Solving i and ii we get
`2ag=36v_(2)^(2)+v_(2)^(2)`
`v_(2)=sqrt((ga)/21)=sqrt((10xx8.4)/21)=2ms^(-1)`