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?

Respuesta :

Answer:

33.06 Kgm/s

Explanation:

Given information

Initial velocity= 20 m/s

mass, m=0.9 Kg

[tex]0.5\times 0.9\times v^{2}=0.5\times 0.7\times 0.9 Kg\times 20^{2}[/tex]

[tex]v^{2}=280[/tex]

[tex]v=\sqrt {280}=16.7332 m/s[/tex]

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

Change in momentum

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

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