Answer to Question #102688 in Mechanics | Relativity for sakshi

Question

A ball of mass kg,.10 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 -³s 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 g = 10.0m/s².

Answer 1

Upwards is the positive direction.

i)

$$p_i=mv_i=-m\sqrt{2gh_i}$$

$$p_i=-10\sqrt{2(10)(4)}=-89.4\frac{kg\cdot m}{s}$$

$$p_f=mv_f=m\sqrt{2gh_f}$$

$$p_f=10\sqrt{2(10)(2)}=63.2\frac{kg\cdot m}{s}$$

ii)

$$F=\frac{p_f-p_i}{t}$$

$$F=\frac{63.2-(-89.4)}{0.004}=38200\ N=38.2\ kN$$