Posted by lim on Saturday, July 21, 2012 at 1:08am.
One man hold a 3kg iron ball fall from 1.5m in the air. The ball then fall on to a table .The table is 0.5m tall. The ball bound up 3cm up and fall again. What is the impact(in Newton) the ball hit the table on the first time?

physics  Elena, Saturday, July 21, 2012 at 2:04am
H=1.50.5 = 1 m, h=0.03 m
m•g•H=m•v1²/2
v1=sqrt(2gH) = sqrt(2•9.8•1)= 4.43 m/s.
m•v2²/2=mgh.
v2= sqrt(2gh) = sqrt(2•9.8•0.03)= 0.77 m/s.
Δp=p2p1=mv2 –(mv1) = m(v2+v1) =3•(4.43+0.77) =15.6 kg•m/s.
If you knew time of impact “Δt”,
F•Δt= Δp,
F= Δp/ Δt (in Newtons)
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