Understand the setup

• Mass of bob: $m=0.1\,\text{kg}$
• String length: $L=1.0\,\text{m}$
• Height of suspension point above floor: $H=0.9\,\text{m}$
• Horizontal impulse given: $P=0.2\,\text{kg m/s}$

Since the bob is initially on the floor vertically below the support, the vertical distance from support to floor is $H=0.9\,\text{m}$, while string length is $L=1.0\,\text{m}$.

Therefore initially the string is slack by $L-H=1.0-0.9=0.1\,\text{m}.$

Motion before the string becomes taut

The bob lies on the frictionless floor, so it can move only horizontally while on the floor.

The impulse $P$ gives the bob horizontal momentum $p=P=0.2\,\text{kg m/s}.$

Hence its speed just after impulse is $v=\frac{P}{m}=\frac{0.2}{0.1}=2\,\text{m/s}.$

Since the floor is frictionless and the string is slack, the bob continues with constant horizontal speed $2\,\text{m/s}$ until the string becomes taut.

When does the string become taut?

Let the support be at height $H=0.9$ m above the floor. While the bob is on the floor, its vertical distance below the support remains $0.9$ m.

If the bob has moved horizontally by distance $x$, then the distance from support to bob is $\sqrt{x^2+H^2}.$

The string becomes taut when this distance equals $L=1.0$ m: $\sqrt{x^2+0.9^2}=1.0.$ Squaring, $x^2+0.81=1,$ $x^2=0.19,$ $x=\sqrt{0.19}.$

Angular momentum about the point of suspension just before lift-off

Just before lift-off, the bob is still moving horizontally with momentum magnitude $p=mv=0.2\,\text{kg m/s}.$

Angular momentum about the support is $J = p \times (\text{perpendicular distance from support to line of motion}).$

The line of motion is horizontal along the floor, whose perpendicular distance from the support is simply $H=0.9$ m.

Therefore, $J = pH = 0.2 \times 0.9 = 0.18\,\text{kg m}^2/\text{s}.$

Speed just after lift-off

At the instant the string becomes taut, the impulsive tension can only remove the radial component of velocity; the tangential component remains.

Let $\theta$ be the angle the string makes with the vertical at lift-off. Then $\cos\theta = \frac{H}{L} = \frac{0.9}{1} = 0.9,$ $\sin\theta = \frac{x}{L} = \sqrt{0.19}.$

The bob's velocity just before tautening is horizontal: $v=2$ m/s.

The tangential component relative to circular motion is equal to $v_t = \frac{J}{mL}.$

Using $J=0.18$, $v_t = \frac{0.18}{0.1\times 1} = 1.8\,\text{m/s}.$

This is the speed just after lift-off.

Kinetic energy just after lift-off

Therefore, $K = \frac12 m v_t^2 = \frac12 (0.1)(1.8)^2.$ $K = 0.05 \times 3.24 = 0.162\,\text{J}.$

Hence, $K \approx 0.16\,\text{J}.$

Final answer

$\boxed{0.16}$