A pendulum consists of a bob of mass m=0.1 ~kg and a massless...

The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is,

Angular momentum about top point $L = mV \times 0 . 9$

$L = 0 . 2 \times 0 . 9$

$⇒ L = 0 . 18kg{m}^{- 2}s$

Now after the bob slides for some distance, the string becomes taut and the bob lifts off the floor so when string get taut an impulse act by string on particle.

Due to impulse, component of $V$ along string get change and becomes zero but component of $V$ perpendicular to string remain same so,

${V}_{1} = Vsin\theta = 0$

Angular momentum will be conserved about the fixed point,

${L}_{i} = {L}_{f} ⇒ 0 . 18 = 0 . 1 \times \left(Vcos\theta\right) \times 1 ⇒ Vcos\theta = 1 . 8m{s}^{- 1} = {V}_{net ( Final )}$

Now the kinetic energy just after lift up is given by,

$KE = \dfrac{1}{2}m{v}^{2} = \dfrac{1}{2} \times 0 . 1 \times 1 . {8}^{2} ⇒ KE = 0 . 162J$

Hence, $K = 0 . 162$