A pendulum consists of a bob of mass m =0.1 kg and a massless inextensible string of length L =1.0 m . It is suspended from a fixed point at height H =0.9 m above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse P =0.2 kg - m / s is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is J kg - m 2 / s. The kinetic energy of the pendulum just after the lift-off is K Joules. The value of K is .

Question

A pendulum consists of a bob of mass m =0.1 kg and a massless inextensible string of length L =1.0 m . It is suspended from a fixed point at height H =0.9 m above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse P =0.2 kg - m / s is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is J kg - m 2 / s. The kinetic energy of the pendulum just after the lift-off is K Joules. The value of K is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

As bob lifts off due to impulse of string angular momentum will be conserved about suspension point

\Rightarrow J_{f} = J_{i}

\Rightarrow M v_{f} ℓ sin 9 0^{\circ} = 0.18

\Rightarrow v_{f} = 1.8 m / s

Kinetic energy after lifting off

\Rightarrow K_{f} = \frac{1}{2} m v_{f}^{2}

= \frac{1}{2} \times 0.1 \times (1.8)^{2} = 0.162 J

As bob lifts off due to impulse of string angular momentum will be conserved about suspension point ⇒Jf​=Ji​ ⇒Mvf​ℓsin90∘=0.18 ⇒vf​=1.8m/s Kinetic energy after lifting off ⇒Kf​=21​mvf2​ =21​×0.1×(1.8)2=0.162J