- #1

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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

K

_{0}+ U

_{0}+ 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/s

^{2})(2m) = 1/2(.04kg)v

^{2}

v = 6.26m/s

1/2(.04kg)v

^{2}= .04kg(9.8m/s

^{2})(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.