Answer
a. same amount of sting
b. Best: II
Work Step by Step
The linear momentum of an object of mass $m$ moving with velocity $\vec{\mathrm{v}}$ is
$\vec{\mathrm{p}}=m\vec{\mathrm{v}} \qquad $9-1
By Newton's second law, the impulse delivered to an object is equal to the change in its momenta:
$\vec{\mathrm{I}}=\vec{\mathrm{F}}_{\mathrm{a}\mathrm{v}}\Delta t=\Delta\vec{\mathrm{p}}\qquad $9-6
----------
$a.$
$mv=(2m)(\displaystyle \frac{v}{2})$, so both balls have the same momentum, so, over the same amount of time, the same net force acts on your hand, causing the same amount of sting.
$b.$
In terms of kinetic energy,
the first ball has $\displaystyle \Delta K_{1}=\frac{mv^{2}}{2}$
the second: $\displaystyle \Delta K_{2}=\frac{(2m)(\frac{v^{2}}{4})}{2}=\frac{mv^{2}}{4}=\frac{1}{2}\Delta K_{1}.$
The second ball has half as much kinetic energy as the first.
The work done by the hand in stopping the ball over a distance $d$ for the first ball $F_{1}=\displaystyle \frac{\Delta K_{1}}{d},$
and for the second, $F_{2}=\displaystyle \frac{1}{2}F_{1},$ but only if the distance of stopping, d, is equal
So,
I is true only if the stopping distances are equal
II: is true (see part a.)
III: is wrong, as mass isn't the only thing to be considered (such as speed)
Best: II
