Answer
$a.\quad 0.133 \mathrm{k}\mathrm{g}\cdot \mathrm{m}/\mathrm{s}$
$ b.\quad$ greater
Work Step by Step
a.
Taking $+y$ as the upward direction, let the marble hit the floor with speed $-v_{1},$ and with initial speed of 0, rebound upward with speed $+v_{2}$.
We find $v_{1}$ and $v_{2}$ using eq.2-12:$\quad v_{1}^{2}=v_{0}^{2}-2g\Delta y$
$v_{1}=-\sqrt{v_{0}^{2}-2g\Delta y}=-\sqrt{0^{2}-2(9.81\mathrm{m}/\mathrm{s}^{2})(-1.44\mathrm{m})}=-5.32$ m/s
$v_{2}=\sqrt{v^{2}+2g\Delta y}=\sqrt{0^{2}+2(9.81\mathrm{m}/\mathrm{s}^{2})(0640\mathrm{m})}=3.54 $m/s
Now, apply $ \vec{\mathrm{I}}=\vec{\mathrm{F}}_{\mathrm{a}\mathrm{v}}\Delta t=\Delta\vec{\mathrm{p}}\qquad $9-6
$I=m\Delta v=m(v_{2}-v_{1})$
$=(0.0150\mathrm{k}\mathrm{g})[3.54\mathrm{m}/\mathrm{s}-(-5.32\mathrm{m}/\mathrm{s})]$
$=0.133 \mathrm{k}\mathrm{g}\cdot \mathrm{m}/\mathrm{s}$
$b.$
If the marble bounces to a greater height, $v_{2}$ is larger than above, and $\Delta v$ is larger, so the impulse $m\Delta v$ is greater than the impulse found above.
$a.\quad 0.133 \mathrm{k}\mathrm{g}\cdot \mathrm{m}/\mathrm{s}$
$ b.\quad$ greater
