A system is shown in fig. Assume that the cylinder remains in contact with the two wedge. Then the velocity of cylinder is

Options

A. `sqrt(19-4sqrt(3))ms^(-1)`
B. `(sqrt(13)u)/(2)ms^(-1)`
C. `sqrt(3)ums^(-1)`
D. `sqrt(7)ms^(-1)`

Answer

D

Explanation

Method 1: Velocity components perpendicular to the contact surface remain same.
As cylinder will remain in contact with wedge A.
`V_(x)=2u`

As it also remain in contact with wedge B,
`u sin 30^(@) = V_(y) cos 30^(@)-V_(x) sin 30^(@)`
`V_(y)=V_(x)(sin 30^(@))/(cos 30^(@)) +(u sin 30^(@))/(cos 30^(@))= 3 u tan 30^(@) = sqrt(3) u`
`implies V=sqrt(V_(x)^(2)+V_(y)^(2)) = sqrt(7)u`.
Method 2: In the frame of A

`3usin 30^(@)=V_(y) cos 30^(@)`
`V_(y)= 3u tan 30^(@) = sqrt(3) u and V_(x)=2u`
`implies V=sqrt(V_(x)^(2)+V_(y)^(2))=sqrt(7)u`.