Question

A ball of 0.900 kg slides horizontally over a frictionless surface and bumps into a heavy stone at a speed of 20.0 m / s. The ball returns with 70.0% of its initial kinetic energy. How big is the change in the impulse of the stone?

Answer 1

33.06 Kgm/s

Step-by-step explanation:

Given information

Initial velocity= 20 m/s

mass, m=0.9 Kg

`0.5* 0.9* v^(2)=0.5* 0.7* 0.9 Kg* 20^(2)`

`v^(2)=280`

`v=\sqrt {280}=16.7332 m/s`

Here, v is -16.7332 m/s in opposite direction

Change in momentum

`\triangle p= m(v-u)=0.9(-16.7332-20)=-33.0599 Kgm/s\approx -33.06 Kgm/s`

Therefore, the magnitude of change in momentum is 33.06 Kgm/s