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

Answer:

33.06 Kgm/s

Explanation:

Given information

Initial velocity= 20 m/s

mass, m=0.9 Kg

$0.5\times 0.9\times v^{2}=0.5\times 0.7\times 0.9 Kg\times 20^{2}$

$v^{2}=280$

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

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