Answer
2.08 m/s
Work Step by Step
Since the contact is frictionless, the mechanical energy $E_{mec}$ is conserved and so $\Delta E_{mec} = \Delta K + \Delta U = \frac{1}{2}m(v^{2} – v_{0}^{2}) + mg\Delta h = 0$.
Since we’re looking for the final speed $v$, we isolate it:
$ \frac{1}{2}m(v^{2} – v_{0}^{2}) + mg\Delta h = 0$.
$\frac{1}{2}m(v^{2} – v_{0}^{2}) = – mg\Delta h$.
$v^{2} – v_{0}^{2} = – 2g\Delta h$.
$v^{2} = v_{0}^{2} – 2g\Delta h$.
$v = \sqrt {v_{0}^{2} – 2g\Delta h}$.
The problem gives us the mass $m$ = 2.00 g = 0.00200 kg and the radius $r$ = 22.0 cm = 0.220 m of the bowl. At the bottom of the bowl, the flake would have fallen a vertical distance equal to r, and so the change in height is $\Delta h$ = – 0.220 m. We also know that at the edge of the bowl, the flake was at rest and thus $v_{0}$ = 0. Now we just plug in:
$v = \sqrt {0^{2} – 2(9.8 m/s^{2})(–0.220 m)}$ = 2.08 m/s
