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 at a horizontal distance of 30 cm from the support. Calculate the speed of the ball at its lowest point of the trajectory. (mark the answer in ~m / s )

Step-by-step solution

aligned & mg 0.4= 1 2 mv ^2 & v =2 2 ~m / s aligned (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 6 2 5 ~m / s again from conservation of mechanical energy 1 2 m ( 6 2 5 )^2+m g(0.1)= 1 2 m v_f^2 Now velocity at the bottommost point v _ final =2.2 ~m / s