## [SOLVED] collision and impulse

1. The problem statement, all variables and given/known data
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

2. Relevant equations
p=m*v
deltap=J
F=m(deltav)/deltat

3. 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!
 PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks
 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.