Figure shows a block `A` of mass `6 m` having a smooth semicircular groove of radius a placed on a smooth horizontal surface. A block `B` of mass `m` is released from a position in groove where its radius is horizontal. Find the speed of the bigger block when the smaller block reaches its bottom most position.

Explanation

Let the smaller block is moving with speed `v_(1)` relative to the bigger block when it reaches the bottommost 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_(1)^(2)`
...........ii
solving eqn i and ii we get
`2ga-36v_(2)^(2)+6v_(2)^(2)`
`v_(2)=sqrt((ga)/21)`