Question

A ball of mass 4.5 kg moving with speed of 2.2 m/s in the +x-direction hits a wall and bounces back with the same speed in the x direction. what is the change of the momentum of the ball?

Answer 1

Answer:

The change of the momentum of the ball is $-19.8\, \frac{mkg}{s}$

Explanation:

We should find $\varDelta\overrightarrow{p}=\overrightarrow{p_{f}}-\overrightarrow{p_{i}}$ (1)with $\overrightarrow{p_{i}}$ the initial momentum and $\overrightarrow{p_{f}}$ the final momentum. Linear momentum is defined as $\overrightarrow{p}=m\overrightarrow{v}$, using that on (1):

$\varDelta\overrightarrow{p}=m \overrightarrow{v_{f}}-m \overrightarrow{v_{i}}$ (2)

It's important to note that momentum and velocity are vectors and direction matters, so if +x direction is the direction towards the wall and the -x direction away the wall $\overrightarrow{v_{i}}=+2.2\, \frac{m}{s}$ and $\overrightarrow{v_{f}}=-2.2\, \frac{m}{s}$ so (2) becomes:

$\varDelta\overrightarrow{p}=m(-2.2- (+2.2))=-(4.5)(4.4)$

$\varDelta\overrightarrow{p}=-19.8\, \frac{mkg}{s}$