Answer:
Change in momentum of the stone is 3.673 kg.m/s.
Explanation:
Given:
Mass of the ball on the horizontal the surface, m = 0.10 kg
Velocity of the ball with which it hits the stone, v = 20 m/s
According to the question it rebounds with 70% of the initial kinetic energy.
We have to find the change in momentum i.e Δp
Before that:
We have to calculate the rebound velocity with which the object rebounds.
Lets say that the rebound velocity be "v1" and KE remaining after the object rebounds be "KE1".
⇒
⇒ 
⇒ 
⇒
Joules (J).
Rebound velocity "v1".
⇒ 
⇒ 
⇒ 
⇒ 
⇒
m/s ...as it rebounds.
Change in momentum Δp.
⇒ 
⇒ 
⇒ 
⇒ 
⇒
Kg.m/s
The magnitude of the change in momentum of the stone is 3.673 kg.m/s.
The overall force of gravity becomes stronger and stronger. This is because the gravity depends on the inverse of the square of the distance:

and since the distance r between the molecules becomes smaller, the gravity increases more and more.
I would rather be hit by the deflated ball because it wouldn't hurt as bad because it wouldn't have a lot of weight to hurt me in anyway