Question

A 1.2 kg ball drops vertically onto a floor from a height of 32 m, and rebounds with an initial speed of 10 m/s.

Answer 1

Explanation:

Given that,

Mass of the ball, m = 1.2 kg

Initial speed of the ball, u = 10 m/s

Height of the floor from ground, h = 32 m

(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 32}$

v = -25.04 m/s (negative as it rebounds)

The impulse acting on the ball is equal to the change in momentum. It can be calculated as :

$J=m(v-u)$

$J=1.2\times (-25.04-10)$

J = -42.048 kg-m/s

(b) Time of contact, t = 0.02 s

Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :

$J=\dfrac{F}{t}$

$F=J\times t$

$F=42.048\times 0.02$

F = 0.8409 N

Hence, this is the required solution.