A mass m starting from A reaches B of a frictionless track. On reaching B, it pushes the track with a force equal to x times its weight, then the applicable relation is

Options

A. (a) `h=((x+5))/(2)r`
B. (b) `h=x/2r`
C. (c) `h=r`
D. (d) `h=((x+1)/(2))r`

Answer

A

Explanation

KE of blocks at `B=PE` at `A-PE` at B
`1/2mv^2=mgh-mg2r=mg(h-2r)`
`v^2=2g(h-2r)` (i)
Also, `(mv^2)/(r)=xmg+mg`
or `v^2=(x+1)rg` (ii)
Equating Eqs (i) and (ii), we get `2g(h-2r)=(x+1)gr`
or `2gh=(x+1)gr+4gr=(x+5)gr`
`h=((x+5)/(2))r`