Question

A 0.150-kg lump of clay is dropped from a height of 1.15 m onto the floor. it sticks to the floor and does not bounce.

Answer 1

Missing question: "What is the magnitude of the impulse I imparted to the clay by the floor during the impact?"

Solution:
The impulse is equal to the variation of momentum of the object: 
$I=\Delta p=m \Delta v$
where m is the mass of the object and $\Delta v$ is the variation of velocity of the object during the impact.

By using energy conservation, we can calculate the velocity of the object before the impact. In fact, the initial potential energy of the object is all converted into kinetic energy before the impact:
$mgh= \frac{1}{2}mv^2$
from which we find v:
$v= \sqrt{2gh}= \sqrt{2 \cdot 9.81 m/s^2 \cdot 1.15 m}=4.75 m/s$

This is the velocity of the object just before the impact. After the impact, the object comes to rest: this means that the variation of velocity of the object is equal to 4.75 m/s. Therefore, the impulse is

$I=m \Delta v=(0.150 kg)(4.75 m/s)=0.71 Kg m s^{-1}$