Answer:
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:
[tex]J=\Delta p [/tex] (1)
with [tex]\Delta p=p_f-p_i [/tex] 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:
[tex]J=F_{avg}\Delta t [/tex](2)
with [tex] F_{avg} [/tex] the average force and [tex] \Delta t[/tex] the time the collision lasts
We can equate expressions (2) and (1):
[tex]\Delta p=p_f-p_i=F_{avg}\Delta t [/tex]
Using the definition of linear momentum as mass (m) time velocity (v):
[tex]mv_f-mv_i=F_{avg}\Delta t [/tex]
We can solve for Favg:
[tex]F_{avg}=\frac{m(v_f-v_i)}{\Delta t} [/tex] (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:
[tex]U_i=K_f [/tex]
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:
[tex]mgh=\frac{mv_i^2}{2} [/tex]
solving for vi:
[tex] v_i=\sqrt{2gh}=\sqrt{2*9.81*2.4}=6.86\frac{m}{s}[/tex]
For the velocity the ball has just after bounces the floor:
[tex]K_i=U_f [/tex]
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:
[tex]\frac{mf_i^2}{2}=mgh_m [/tex]
solvig for vf:
[tex]v_f=\sqrt{2gh_m}=\sqrt{2*9.81*1.9}=6.10\frac{m}{s} [/tex]
Using vf and vi on (3):
[tex]F_{avg}=\frac{(0.06)(6.10-6.86)}{0.034}=-1.34 N [/tex]
The negative sign indicates the direction of the force is pointing away the floor
