Answer
$a.\quad 0.69 \mathrm{m}/\mathrm{s}$
$ b.\quad$no
$c.\quad 0.40\mathrm{J}$
Work Step by Step
In a system of several objects, the total linear momentum is the vector sum of the individual momenta:
$\displaystyle \sum\vec{\mathrm{p}}=m_{1}\vec{\mathrm{v}}_{1}+m_{2}\vec{\mathrm{v}}_{2}$
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$a.$
$\displaystyle \sum\vec{\mathrm{p}}=0$
$m_{1}\vec{\mathrm{v}}_{1}+m_{2}\vec{\mathrm{v}}_{2}=0$
$m_{2}\vec{\mathrm{v}}_{2}=-m_{1}\vec{\mathrm{v}}_{1}$
$\displaystyle \mathrm{v}_{2}=-\frac{m_{1}\mathrm{v}_{1}}{m_{2}}$
$=\displaystyle \frac{(0.35\mathrm{k}\mathrm{g})(\mathrm{l}.2\mathrm{m}/\mathrm{s})}{0.61\mathrm{k}\mathrm{g}}=0.69 \mathrm{m}/\mathrm{s}$
$b.$
Both carts are in motion and have mass. Both have nonzero kinetic energies.
The kinetic energy of the system is not zero
$c.$
$\displaystyle \sum K=\frac{1}{2}m_{1}\mathrm{v}_{1}^{2}+\frac{1}{2}m_{2}\mathrm{v}_{2}^{2}$
$=\displaystyle \frac{1}{2}(0.35 \mathrm{k}\mathrm{g} )(1.2 \displaystyle \mathrm{m}/\mathrm{s})^{2}+\frac{1}{2}(0.61 \mathrm{k}\mathrm{g} )(0.69 \mathrm{m}/\mathrm{s})^{2}$
$=0.40\mathrm{J}$
