A small ball of mass `100 g` is attached to a light and inextensible string of length `50 cm`. The string is tied to a support `O` and the mass `m` released from point `A` which is `a'` a horizontal distance of `30cm` from the support. Calculate the speed of the ball is its lowest point of the trajectory.

Explanation

`mgxx0.4=(1)/(2)mv^(2)`
`v=2sqrt(2)m//s`
(from conservation of mechanical energy just before the string becomes taut during the application of impulse, velocity along perpendicular to the string remain same. So after impulse velocity of the ball equal to `(6sqrt(2))/(5)m//s`
again from conservation of mechanical energy
`(1)/(2)m((6sqrt(2))/(5))^(2)+mg(0.1)=(1)/(2)mv_(f)^(2)`
Now velocity at the bottommost point `v_("final")=2.2m//s`