A 60 g ball is dropped from rest from a height of 2.4 m. It bounces off the floor and rebounds to a maximum height of 1.9 m. If the ball is in contact with the floor for 0.034 s, what is the magnitude (in N) and direction of the average force exerted on the ball by the floor during the collision

The force is 1.34 newtons and its direction is upward.

Explanation:

Choosing positive direction pointing towards the floor in this collision we're going to use the momentum-impulse theorem that states:

(1)

with the change in the momentum and J the impulse, with pi the initial momentum that is the momentum just before the collision and pf the final momentum th is the momentum just after the collision. The impulse J is also defined as:

(2)

with the average force and the time the collision lasts

We can equate expressions (2) and (1):

Using the definition of linear momentum as mass (m) time velocity (v):

We can solve for Favg:

(3)

Now we should find the velocities vf and vi, we should do this using conservation of energy:

For the velocity the ball has just before reaches the floor:

With Ui the initial potential energy (there is not initial kinetic energy) and Kf the final kinetic energy (there is not final potential energy), then:

solving for vi:

For the velocity the ball has just after bounces the floor:

There is not initial potential energy because it's a floor level at this instant, and the there is not final kinetic energy because the ball has instantly zero velocity at its maximum height (hm), then:

solvig for vf:

Using vf and vi on (3):

The negative sign indicates the direction of the force is pointing away the floor

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