Answer
a. $1.3\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\mathrm{s}$
b. $-0.14\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\mathrm{s}$
c. the result of (a)
Work Step by Step
$a.$
Use equation 9-1:
$\Delta\vec{\mathrm{p}}=\vec{\mathrm{p}}_{\mathrm{f}}-\vec{\mathrm{p}}_{\mathrm{i}}=m(\vec{\mathrm{v}}_{\mathrm{f}}-\vec{\mathrm{v}}_{\mathrm{i}})$
The motion is in the $\hat{\mathrm{y}}$ direction, so expressing velocities,
$\Delta\vec{\mathrm{p}}=(0.285\mathrm{k}\mathrm{g})[(2.0\mathrm{m}/\mathrm{s})\hat{\mathrm{y}}-(-2.5\mathrm{m}/\mathrm{s})\hat{\mathrm{y}}]$
$=(1.3\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\mathrm{s})\hat{\mathrm{y}}$
$|\Delta\vec{\mathrm{p}}|=1.3\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\mathrm{s}$
$b.$
The change in the magnitude = $p_{\mathrm{f}}-p_{\mathrm{i}}=m(\mathrm{v}_{\mathrm{f}}-\mathrm{v}_{\mathrm{i}})$
$=( 0.285 \mathrm{k}\mathrm{g} ) (2.0 \mathrm{m}/\mathrm{s}-2.5\mathrm{m}/\mathrm{s})$
$=-0.14\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\mathrm{s}$
$c.$
In terms of momentum, Newton's second law is
$\displaystyle \sum\vec{\mathrm{F}}=\frac{\Delta\vec{\mathrm{p}}}{\Delta t} \qquad$9-3
That is, the net force acting on an object is equal to the rate of change of its momentum.
So, the result of (a) is more directly related to the net force acting on the ball during its collision with the floor.
