Question

A 0.10 kg baseball travelling at 40 m s−1 hits straight back to the pitcher at 55 m s−1. The contact time is 0.01 seconds. What is the change in momentum of the baseball?​

Answer 1

Answer:
`-9.5 kg m/s`

Step-by-step explanation:

The change in momentum
`\Delta p` is given by:

`\Delta p=p_(2)-p_(1)` (1)

Where:

`p_(1)=m V_(1)` is the initial momentum of the baseball, being
`m=0.1 kg` and
`V_(1)=40 m/s`

`p_(2)=m V_(2)` is the final momentum of the baseball, being
`V_(2)=-55 m/s` because it is in the oppossite direction

Then (1) is rewritten as:

`\Delta p=m V_(2) - m V_(1)= m(V_(2) - V_(1))` (2)

`\Delta p=0.1 kg(-55 m/s - 40 m/s)` (3)

Finally:

`\Delta p=-9.5 kg m/s`