# A handball of mass 45 grams strikes a hard wall perpendicularly with an initial speed of 46m/s. The handball is in contact with the wall for 0.078 seconds, and then rebounds straight back with a speed of 39m/s. What average force was applied to the ball?

## Use impulse and momentum to solve.

Apr 5, 2018

I tried this:

#### Explanation:

Here I would use the Theorem of Impulse and Change in Momentum:

$I = \Delta p$

${F}_{a v} \Delta t = m {v}_{f} - m {v}_{i}$

${F}_{a v} = \frac{m {v}_{f} - m {v}_{i}}{\Delta t} = \frac{0.045 \left(- 39\right) - 0.045 \left(46\right)}{0.078} = - 49 N$

Where I considered the signs corresponding to the directions of the velocities: