A 0.150-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wall and rebounds with 70.0% of its initial kinetic energy. What is the magnitude of the change in momentum of the stone?

The magnitude of the change in momentum of the stone is 5.51kg*m/s.

Explanation:

the final kinetic energy = 1/2 (0.15) v^2

1/2 (0.15) v^2 = 70%*1/2 (0.15) (20) ^2

v^2 = 21/0.075

v^2 = 280

v = 16.73 m. s

if u is the initial speed and v is the final speed, then:

u = 20 m/s and v = - 16.73m/s

change in momentum = m (v-u)

= 0.15 ( - 16.73-20)

= - 5.51 kg*m. s

Therefore, The magnitude of the change in momentum of the stone is 5.51kg*m/s.