A small block of mass m is pushed on a smooth track from position A with a velocity `2sqrt5` times the minimum velocity required to reach point D. The block will leave the contact with track at the point where normal force between them becomes zero.

At what angle `theta` with horizontal does the block gets separated from the track?

Options

A. (a) `tan^-1(1/2)`
B. (b) `tan^-1(2)`
C. (c) `sin^-1(2/3)`
D. (d) The block never reaches point B.

Answer

A

Explanation

`v_1=2/sqrt5sqrt(5gR)=2sqrt(gR)`
Now, `v_2^2=v_1^2-2g(R+R sin theta)=2gR-2gR sin theta`
`N+mg sin theta=(mv_2^2)/(R)`
Put `N=0`
`implies mg sin theta=m2g(1-sin theta)`
`implies sin theta=2/3`
`implies theta=sin^-1(2/3)`

`v^2=v_1^2-2gRimpliesv^2=4gR-2gR`
`impliesv=sqrt(2gR)`
`a_c=v^2/R=2g`
`a_t=g`
`tan alpha=a_t/a_c=(g)/(2g)=1/2impliesalpha=tan^-1(1/2)`