Answer to Question #104271 in Mechanics | Relativity for Neha Singh

Question

A ball of mass

0.1kg,

starting from rest, falls a height of 4.0 m and then collides with

the ground. After the collision, the ball bounces up to a height of 2.0 m. The collision

with the ground takes place over a time

4.0 10 s.

3



Determine (i) the momentum of

the ball immediately before the collision and immediately after the collision and (ii) the

average force exerted by the ground on the ball. Take

10.0 ms .

Answer 1

$p_1=m\sqrt{2gh_1}$ , $p_2=m\sqrt{2gh_2}$ , $\bar{F}=\frac{p_1+p_2}{t}$ ,hence $p_1=0.1\sqrt{2\sdot 10\sdot 4}\approx0.90kg\sdot m/s$ , $p_2=0.1\sqrt{2\sdot 10\sdot 2}$ $\approx0.63kg\sdot m/s$ , $\bar{F}=\frac{0.9+0.63}{0.1}\approx15.3N$