A string AB of length 21 is fixed at A to a point on a smooth horizontal table. A particle - of mass m attached to B is initially at a distance I from A as shown in figure. The particle is projected horizontally with speed u at right angles to AB. Find the impulsive tension in the string when it becomes taut and the velocity of the particle immediately afterwards.

Explanation

When the string b eomes thut `AB. = 2l and angle B. AB = 60^(@)`

Just before the string jerks, the particles is velocity components parallel and perpendicular to AB. are u `sin 60^(@)` and u `cos 60^(@)` respectively. When the string becomes thut the length of AB is fixed and particle cannot travel in the direction AB.. After the jerk, the velocity of the particle is, therefore, perpendicular to AB.. Using impulse = change in momentum
(a) Along `B.A , J =0- (- m u sin 60^(@)) (or) J = (sqrt3)/(2) m u`
(b) Perpendicular to B.A (no impulsive component) `0= mv - m u cos 60^(@) (or) v = u cos 60^(@) = (u)/(2)`
Therefore, the velocity of the particle just after the string becomes thut is `(u)/(2) ` perpendicular to the string.