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

The change of the momentum of the ball is
`-19.8\, (mkg)/(s)`

Step-by-step 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\, (m)/(s)` and
`\overrightarrow{v_(f)}=-2.2\, (m)/(s)` so (2) becomes:

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

`\varDelta\overrightarrow{p}=-19.8\, (mkg)/(s)`