A bob of heavy mass m is suspended by a light string of length l . The bob is given a horizontal velocity v ₀ as shown in figure. If the string gets slack at some point P making an angle from the horizontal, the ratio of the speed v of the bob at point P to its initial speed v₀ is :

Step-by-step solution

C.O.M.E. 1 2 mv ₀^2= mg (1+ )+ 1 2 mv _ p ^2....(1) At pt P Tp + mg = mv _ p ^2 ( as T _ p =0 ) mg = mv _ p ^2 mv _ p ^2= mg .( ii ) From (i) & (ii) aligned & 1 2 m v₀^2=m g (1+ )+ 1 2 m g & v₀^2=2 g (1+ )+g & v₀= 2 g +3 g ....(iii) aligned aligned & v _ p = g & v ₀= 2 ~g +3 ~g & v _ p v ₀ = 2+3 aligned