Question

A rubber ball is dropped and bounces vertically from a horizontal concrete floor. If the ball has a speed of 4.1 m/s just before striking the floor, and a speed of 3.9 m/s just after bouncing, find the average force of the floor on the ball. Assume that the ball is in contact with the floor for 0.26 s, and that the mass of the ball is 0.44 kg.

Answer 1

To solve this problem it is necessary to apply Newton's second law with the peculiarity that the expression given for velocity must be expressed in terms of the rate of change of velocity and time.

Mathematically this is,

F= ma

$F = m\frac{\Delta V}{t}$

Where,

m = mass

$\Delta V$= Change in velocity

t = Time

Since the time interval occurs when the speed hits - in the downward direction - until it bounces - in the upward direction - we will have that the change in speed is given by

$\Delta V = (3.9)-(-4.1)$

The rest of our values are given as:

m = 0.26s

t = 0.44kg

Finally replacing we will have to

$F = 0.44 \frac{(3.9)-(-4.1) }{0.26}$

$F = 13.53N$

Therefore the final average force of the floor on the ball is 13.53N