A ball of mass m 1 = 8.0 × 10−2 kg starts from rest and falls vertically downward from a height of 3.0 m. After colliding with the ground, it bounces up to a height of 2.0 m. The collision takes place over a time interval of t = 5.0 ×10−3 s. Calculate (i) the momentum of the ball immediately before and immediately after the collision, (ii) average force exerted by the ground on the ball and (iii) impulse imparted to the ball?

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To find momentums we need the speeds before and after collision. The simplest way to find these speeds is to use the energy conservation law.

Before fall, a ball had only potential energy `mgh_1.` Just before collision it had only kinetic energy m `V_1^2/2,` and they was equal. Therefore `V_1=sqrt(2gh_1),` where `h_1=3m.`

The same consideration gives that the speed after the collision is `V_2=sqrt(2gh_2),` where `h2=2m.` Now we can answer (i) and (iii).

(i) the momentum before collision is `mV_1 approx 0.61(m/s),` after is `mV_2 approx 0.50(m/s).`

(iii) note that impulse is a vector, it has the same direction as velocity. The momentum before collision is directed downwards and after -- upwards. So the difference of moments is the sum of their magnitudes, i.e. 1.11m/s.

For (ii) we have to know that a force may be expressed as the derivative of a momentum (Newton's Second law, actually). Therefore the integral of force is the difference of momentums. And the average force is this difference divided by a time. Force is upwards all the time.

(ii) average force is `1.11/(5*10^(-3)) approx 222(N).`

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