A0.150-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.

b. lifting dumbbell weight at 20-degree angle from the body after a 45-degree angle