Calculating Impulse and Average Force for a Ball Dropped on a Floor

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    Collision Impulse
AI Thread Summary
The discussion focuses on calculating the impulse and average force for a ball dropped onto a floor. The ball, weighing 1.0 kg, hits the floor at 20 m/s and rebounds at 5 m/s. The correct impulse is determined by calculating the change in momentum as P_after - P_before, which corrects the initial misunderstanding. Additionally, when calculating average force, it's important to include the force of gravity alongside the impulse. The final calculations clarify the necessary adjustments to arrive at the correct answers for both impulse and average force.
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[SOLVED] collision and impulse

Homework Statement


A 1.0 kg ball drops vertically onto a floor, hitting with a speed of 20 m/s. It rebounds with an initial speed of 5 m/s.
(a) What impulse acts on the ball during the contact?
kg·m/s

(b) If the ball is in contact with the floor for 0.020 s, what is the average force exerted on the floor?
N



Homework Equations


p=m*v
deltap=J
F=m(deltav)/deltat


The Attempt at a Solution



For (a) i used p=m*v and deltap=J

so I got (1*-20)-(1*5)=-25

and for (b) i used that F=(m*deltav)/deltat

which came out to be... 1(-25)/.020=0.5 N

both answers were wrong and I'm not sure what is wrong with the formulas I used...

Any help is appreciated! Thanks!
 
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25/0.020=1250
 
For a) the impulse is P_after - P_before and not P_before - P_after.

For b) see The reply of Mikelepore. I think you need to add the force of gravity as well.
 
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