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=1.5-0.5 = 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=p2-p1=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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