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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 - ,

    H=1.5-0.5 = 1 m, h=0.03 m
    v1=sqrt(2gH) = sqrt(2•9.8•1)= 4.43 m/s.
    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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