Question

A 0.500-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 mangnitude of the change in momentum of the stone?

Answer 1

Answer:

change in momentum of the stone is 1.635 kg.m/s

Explanation:

let m = 0.500kg ball and M  be the mass of the stone, v be the velocity of the ball and V be the velocity of the stone

the  initial kinetic energy of the ball is 1/2(0.500)(20^2) = 100 J

the kinetic energy of the ball after rebounding is 70/100(100) = 70 J

Kb = 1/2mv^2

 v =  \sqrt{2k/m} = \sqrt{2(70)/0.500} = 16.73 m/s  

from the conservation of linear momentum, we know that:

mvi + MVi = mvf + MVf

MVf - MVi = mvi - mvf

MVf - MVi = (0.500)(20) - (0.500)(16.73)

                 = 1.635 kg.m/s  

therefore, the change is momentum of the stone is 1.635 kg.m/s