A 10 kgball moving at 20 ms perpendicular to a wall rebounds

A. 1.0 kg-ball moving at 2.0 m/s perpendicular to a wall rebounds from the wall at 1.5 m/s. What is the change in the momentum of the ball? A 64-kg woman stands on frictionless level ice with a 0.10-kg tibe at ger feet. She kicks the stone with her foot so that she acquires a velocity of 0.017 m/s, in the forward direction. What is the velocity acquired by the stone?

Solution

1. Momentum of the ball before collision = 1.0*2.0 = 2.0 kg-m/s

                after collision                           = 1.0* -1.5 = -1.5 kg-m/s

chnage in momentum = -1.5 - 2.0 = -3.5 kg-m/s

2. let v be the velocity of the stone.

     momentum before collision =0

momentum after kick (collision) = 64*0.0017 + 0.10*v =0

velocity of the stone v = -64*0.0017/0.10 = -1.088 m/s

-ve sign indicates the velocity of the stone is opposite to that of the woman.