Question

A ball of mass 0.150 kg is dropped from rest froma height of 1.25m .it rebounds from the floor to reach a height of 0.960m .what impulse was given to the ball by floor?

Answer 1

Impulse experienced on a body is defined by the change in momentum of the body.

Initial velocity is defined as

$v_1 = \sqrt{2qH_i} = \sqrt{2 \times 9.8 \times 1.25} = 4.95 \;m/s$

The final velocity of the body (note at hieghest point velocity is zero, thus from newtons equations of motion)

$v_2 = \sqrt{2qH_f} = \sqrt{2 \times 9.8 \times 0.96} = 4.34 \;m/s$

Impulse

$I = FΔt = mΔv=m(v_1-v_2) = 0.15(4.95-4.34) = 91.5 \times 10^{-3} \;kg \cdot m/s$