Question

A daydreaming soccer player takes a 0.42 kg ball to the face experiencing an impact force of 1215.4 N. If the ball hit the players face with a speed of 13.8 m/s and bounces off in the opposite direction with the same speed, calculate the time of impact. Note: The time of impact will be a fraction of a second so answer with at least 5 decimal places.

Answer 1

Given data

*The given mass of the ball is m = 0.42 kg

*The given impact force is F = 1215.4 N

*The given initial speed is u = 13.8 m/s

*The ball bounces off in the same direction is v = -13.8 m/s

The formula for the time of impact is calculated by the Impulse-momentum theorem as

$\begin{gathered} F\Delta t=m\Delta v \\ \Delta t=\frac{m\Delta v}{F} \end{gathered}$

Substitute the known values in the above expression as

$\begin{gathered} \Delta t=\frac{0.42\times(13.8-(-13.8))}{(1215.4)} \\ =0.00953\text{ s} \end{gathered}$

Hence, the time of impact is 0.00953 s