A block A of mass `m` is held at rest on a smooth horizontal floor. A light frictionless, small pulley is fixed at a height of `6m` from the floor. A light inextensible string of length `16m`, connected with A passes over the pulley and another identical block B is hung from the string. Initial height of B is `5m` from the floor as shown in figure

When the system is released from rest, B starts to move vertically downwards and A slides on the floor towards right.
a. If at an instant string makes an angle `theta` with horizontal, calculate relation between velocity `u` of A and `v` of B.
b. Calculate `v` when B strikes the floor. (`g=10ms^-2`).

Answer

a. `u=vsectheta`; b. `40/sqrt41ms^-1`

Explanation

Let distance of block A from foot O of pulley be x and let the depth of block B from pulley by y when string makes angle `theta` with horizontal as shown in figure.
Then length, `AP=sqrt(x^2+6^2)=sqrt(36+x^2)`
But `AP+BP=16`
or `sqrt(36+x^2)+y=16`
Differetiating above equation w.r.t. time,
`(x)/(sqrt(36+x^2))*(dx)/(dt)+(dy)/(dt)=0`
`(dy)/(dt)=v` and `(dx)/(dt)=-u-u.cos theta+v=0`
`u=vsectheta`

When B strikes the floor, `y=16m`
`AP=16-6=10m`
`sintheta=6/16impliestheta=37^@`
At that instant, kinetic energy of the two blocks=loss of potential energy to B
`1/2mu^2+1/2mv^2=mgxx5`
where `u=v sec 37^@`.
Hence, `v=(40)/(sqrt41)ms^-1`