Kinetic Energy of Pendulum Bob Just After Liftoff

To find the kinetic energy $K$ of the pendulum bob immediately following liftoff, we analyze the motion in two main stages: the horizontal motion before the string becomes taut, and the impulse delivered by the string when it becomes taut.

1. Motion Before the String Becomes Taut

Initially, the bob of mass $m = 0.1 \text{ kg}$ is at rest at the position $(0, 0)$ directly below the fixed pivot located at $(0, H)$, where $H = 0.9 \text{ m}$.

A horizontal impulse $P = 0.2 \text{ kg}\cdot\text{m/s}$ is applied to the bob. The initial velocity $v_0$ of the bob along the frictionless floor (in the $+x$-direction) is: $v_0 = \frac{P}{m} = \frac{0.2}{0.1} = 2 \text{ m/s}$

The bob slides horizontally along the line $y = 0$. The distance from the pivot $(0, H)$ to the bob at horizontal position $x$ is: $d = \sqrt{x^2 + H^2}$

The string becomes taut when $d = L = 1.0 \text{ m}$. Thus: $x^2 + H^2 = L^2 \implies x = \sqrt{L^2 - H^2} = \sqrt{1.0^2 - 0.9^2} = \sqrt{0.19} \text{ m}$

2. Velocity Decomposition Just Before Liftoff

Just before the string becomes taut, the bob is at position $\vec{r} = x \hat{i} - H \hat{j}$ relative to the pivot, moving with velocity $\vec{v}_0 = v_0 \hat{i} = 2 \hat{i} \text{ m/s}$.

The unit vector pointing along the string away from the pivot is: $\hat{r} = \frac{x}{L} \hat{i} - \frac{H}{L} \hat{j}$

We can resolve the initial velocity $\vec{v}_0$ into components along the string (radial) and perpendicular to the string (tangential):

1. Radial component (along the string): $v_r = \vec{v}_0 \cdot \hat{r} = v_0 \frac{x}{L}$

2. Tangential component (perpendicular to the string): $v_{\perp} = \sqrt{v_0^2 - v_r^2} = v_0 \sqrt{1 - \frac{x^2}{L^2}} = v_0 \frac{H}{L}$

3. Motion Immediately After Liftoff

When the string becomes taut, an impulsive tension acts along the string towards the pivot. Since the string is inextensible, it completely destroys the radial component of velocity ($v_r \to 0$), while leaving the tangential component ($v_{\perp}$) unaffected.

Additionally, because the tension impulse has an upward vertical component, the floor exerts no reaction impulse, allowing the bob to lift off the floor immediately.

The velocity of the bob just after liftoff is therefore purely tangential: $v' = v_{\perp} = v_0 \frac{H}{L}$

Substituting the given values: $v' = 2 \times \frac{0.9}{1.0} = 1.8 \text{ m/s}$

4. Calculation of Kinetic Energy $K$

The kinetic energy $K$ of the bob immediately following liftoff is: $K = \frac{1}{2} m (v')^2 = \frac{1}{2} \times 0.1 \times (1.8)^2 = 0.05 \times 3.24 = 0.162 \text{ Joules}$

Thus, the value of $K$ is $0.162$.