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?

Answer 1

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)

Answer 2

A ball is held at the top of a table. The person holding the ball drops it, and the ball is allowed to fall toward Earth. Answer the following questions about the ball.

a. When the ball is held at the top of the table (before being dropped), what type of energy does the ball have?

(1 point)
Responses

motion energy
motion energy

electrical energy
electrical energy

kinetic energy
kinetic energy

potential energy