Impulse-steel ball dropped on steel slab

[herbally]

1. Homework Statement

mass of steel ball = 40g = .04 kg
h1 = 2.00 m
h2 = 1.60 m
A steel ball is dropped from height of 2.00 m onto horizontal steel slab, rebounds to height of 1.60m. Calculate impulse of ball during impact.


2. Homework Equations

impulse J = F*dt
J = P2 - P1
K₀ + U₀ + W_other = K_f + U_f

The Attempt at a Solution

I tried to get velocities using conservation of energy in two steps; from drop to impact; from impact to rebound.

.04kg(9.8m/s²)(2m) = 1/2(.04kg)v²
v = 6.26m/s

1/2(.04kg)v² = .04kg(9.8m/s²)(1.6m)
v = 5.6m/s

then I used ΔP = .04kg(5.6m/s - 6.26m/s) = -.0264

Book shows answer to be .47N*s

Not sure where I'm botching this up. Any help is appreciated.

 

[TSny]

Hi, Herbally. This is a common problem that is easy to slip up on. Remember that momentum and velocity are vector quantities (with direction). If you take upward as the positive direction, then what is the sign of the initial velocity (momentum) just before the collision?

 

[herbally]

Naturally the initial velocity is negative. I'm failing to see the correlation. Then again, I've been doing physics for about 3 hours straight so all the numbers are beginning to run together.

Perhaps another nudge? Thanks.

 

[herbally]

WOW. NM. I'm rum-dumb at this point. I got it. Thanks a ton.

 

[TSny]

OK. So, the initial velocity is -6.26 m/s. You will have to subtract this negative number from 5.6 m/s. What is 5.6-(-6.26)? [Edit: I see you got it! Great!]