A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor?

Question

anonymous · Answer

mass (m) = 0.120kg
initial velocity (Vo) = 0                                                      came from rest
distance traveled (s) = 1.25 m 
acceleration (g) = 9.8 m/s^2                                            due to gravity
final velocity (Vf) = ?

$$Vf ^{2} = 2gs + Vo ^{2}$$
$$Vf ^{2} = 2(9.8 m/s ^{2})(1.25 m) + 0$$
$$Vf = 7\sqrt{2}\div2 m/s$$                                                 
 $$Vf \approx 4.950 m/s$$

$$change \in momentum (\Delta P) = Impulse (J)$$
$$\Delta P = m( Vf - Vo )$$
$$\Delta P = 0.120kg(7\sqrt{2}\div2 m/s)$$
$$\Delta P \approx 0.594 kg m/s$$
kg m/s = Ns ( Newton*seconds )                             Ns is the common unit of Impulse 

Impulse is approximately 0.594 Ns, directed upward (opposing the direction of velocity)