physics
posted by lim on .
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?

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)