Calculating Force on a Bouncing Ball

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Force on a bouncing ball?

Homework Statement



this should be easy but for some reason I am making a mistake here is the problem

A ball whose mass is 0.2 kg hits the floor with a speed of 5 m/s and rebounds upward with a speed of 4 m/s. If the ball was in contact with the floor for 1.5 ms (1.5 E -3 s), what was the average magnitude of the force exerted on the ball by the floor?


The attempt at a solution

what i did was use delta(P) = Fnet * delta(t)
change in momentum equals net force times change in time

i can find change of momentum of the ball by

P3 = ((m (+-) M) / (m + M)) * P1
P1 is inital momentum of ball so just m*v .2 * 5 = 1 but would be - because movind down
m is just .2 and big M is mass of Earth
solve equation i got P3 to be 1
so i took final P minus inital P and get 2
then divided this by 1.5 E -3 to find Fnet and get some answer like 1333.333 N
that seems too big so was wondering where i messed up (or is it right)
 
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I don't understand your methodology for solving for the momentum. Both the initial and final velocities of the ball are given meaning that both momenta can be calculated directly, right?
 


BishopUser said:
I don't understand your methodology for solving for the momentum. Both the initial and final velocities of the ball are given meaning that both momenta can be calculated directly, right?

wow no for some reason i was thinking the given speeds was the acceleration. then i had to use that to figure out velocity and then momentum.

so i would use p = m*v to find final and inital
final = .2 * 4 = .8
inital = .2 * 5 = 1 (which will be -1 because it is moving towards the earth)
final - inital = 1.8 this is my change in momentum

then could i just use the change in momentum = fnet times change in time?
would my change in time just be the time it is in contact with the floor?
 


Well, the time given is the time that the ball is in contact with the floor, which is when the momentum change is taking place, so yes I would say that is the correct delta T to plug in.
 


BishopUser said:
Well, the time given is the time that the ball is in contact with the floor, which is when the momentum change is taking place, so yes I would say that is the correct delta T to plug in.

oh ok... i got it now.
im sleep deprived and my brain
is not working right now. need to
get more than 3 hrs of sleep tonight
haha
thanks for the help!