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Momentum of a Dropped Ball

A golf ball weighing .45 N is dropped from a height 1 meter and has a perfect elastic collision with the floor.

What is time required to hit floor?

What is the instantaneous momentum of the ball before it strikes the floor?

What will be the change in momentum from the instant before it hits the ground to right after it rebounds?

If it was in contact with the floor for 4 X 10^-4 seconds then what was the average force on the ball while on the ground?

Solution Preview


t= sqrt(2*H/g)=sqrt(2*1/9.8)=0.45(s)

p = m*v = m*g*t = 0.45*0.45= (m/s) ...

Solution Summary

H = g*t*(t/2)
p = m*v = m*g*t