Question

A very hard rubber ball (m = 0.5 kg) is falling vertically at 4 m/s just before it bounces on the floor. The ball rebounds back at essentially the same speed. If the collision with the floor lasts 0.05 s, what is the average force exerted by the floor on the ball?

Answer 1

Answer:

The force exerted by the floor is 80 N.

Explanation:

Given that,

Mass of ball = 0.5 kg

Velocity= 4 m/s

Time t = 0.05 s

When the ball rebounds then the kinetic energy is

$K.E =\dfrac{1}{2}mv^2$

Where, m = mass of ball

v = velocity of ball

Put the value into the formula

$K.E=\dfrac{1}{2}\times0.5\times(4)^2$

$K.E = 4\ J$

The average force exerted by the floor on the ball = change in kinetic energy over collision time

$F = \dfrac{4}{0.05}$

$F=80\ N$

Hence, The force exerted by the floor is 80 N.