# Example on impulse

A rubber ball of mass $$50g$$ falls from a height of $$1m$$ and rebounds to a height of $$0.5m$$. Find the impulse and the average force between the ball and the ground if the time for which they are in contact was $$0.1s$$

The way it was solved in my book, they worked out the final velocity for each case (upward and downward motion of the ball) and finally determined the change in momentum. Once the impulse was out, the average force was resolved. My question is: As the ball was acting under the force of gravity, so it hits the ground with force $$F=mg$$; the ground should react with the same force, so can't we just multiply $$g$$ and given mass of the body to find $$F$$?

The fact that the ball bounced back by an amount which is only half of the height from which is was dropped hints at the fact that a part of the initial kinetic energy (also the corresponding part of the initial momentum) has been absorbed in the collision (by the ball or by the ground or both).

So: No. The momentum of the ball is not the same in the upward movement, because part of the energy is missing from the inelastic collision.

... it hits the ground with force $$F=mg$$ ...

This is incorrect.

To fall $$1$$ metre takes about $$0.44$$ seconds. During this time the ball is accelerating at a rate of $$g$$. During the next $$0.1$$ seconds the ball is decelerated and its velocity is partly reversed. So this requires an average deceleration greater than $$g$$ and hence an average force greater than $$mg$$.

This is a comment, I don't have reputation.

If you place a 10-lbm object on your hand against the floor, you feel the 10-lbf load (at 1 g). Whereas, if you drop the same 10-lbm object from a height of 20-feet onto your hand the sensation is completely different. It is entirely momentum ($$m \cdot v$$) driven.

Try,

$$F \cdot t = m \cdot v$$

conversely,

$$\delta F \cdot t = m \cdot \delta v$$

If I apply "the above" to the velocity "directly" before and "directly" after impact, I get (average force) $$\delta F$$ = x.78-N, where "x" (one of the digits of the answer) is for you to solve for.

The velocities achieved were calculated from: $$v^2 = v_0^2 +2 \cdot a \cdot \delta s$$

..it hits the ground with force $$F=mg$$; the ground should react with the same force.

No, the ball does not hit the ground with force $$F=mg$$. It is just the force of gravity acting on the ball. In the situation mentioned, the ball is suffering net deceleration (or acceleration in the upward direction) upon impact with the ground, that too for a very short period of time which suggests a large upward force acting on the ball. This is possible only if the normal force (acting upward) exerted by the ground is very large as compared to the force of gravity during the impact. So, the force exerted by the ground cannot be equal to $$mg$$ and hence the ball does not hit the ground with a force $$F=mg$$ as that would mean zero acceleration of the ball upon impact and the ball would continue moving downward with the same velocity.

Hope it helps!

...So, it hits the ground with force $$F = mg$$.

The fact is that we do not know the exact value of force(Remember Force cannot be seen, only its effects can be felt). When two objects are in contact, the charged constituents of the atoms of the two bodies exert Electromagnetic forces on each other. These interactions are so complicated that simply by using $$\frac{kq_1q_2}{r^2}$$(or any other formula you might possible think of) we cannot calculate the force acting on one object. Since these forces are self adjusting in nature, it is primarily by observing the effects that we determine the force. When a ball rests on a surface, the net force acting on the ball must be zero. Therefore, we make up a conclusion that the surface exerts a force on the ball equal to the force exerted by earth on the ball but opposite in direction. When the ball hits the surface with some velocity, some deformation takes place(click here to visualize). The elastic property of the ball makes it to exert a "restoring" force on the surface to regain its original shape. Equal amount of force is exerted on the ball.This force is a function of the elastic property. And hence we cannot say that it hits the ground with force $$F = mg$$.