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: $-9.5 kg m/s$

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:

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