Question

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?

Answer 1

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

Step-by-step 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.