Q4.12 Kinetic and potential energy (S)
A very small ice cube is released from the inside edge of a hemispherical bowl, radius 20 cm. How fast is the cube moving at the bottom of the bowl?Solution
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![\[ K_f = \frac{1}{2}m v^2;~~~~~~U_f = 0 \]](mastermathpng-0.png)

Solution
If we assume that:
- Air resistance can be neglected
- Friction between ice and bowl can be neglected
then the total amount of energy is conserved. Energy is then exchanged between potential and kinetic forms.
We solve the problem by equating energies of the initial and final states (Key Point 3.16)
Ki+Ui=Kf+Uf
The initial kinetic energy and potential energies are:
Ki=0;~~~~~~~Ui=mgh
The final kinetic and potential energies are
![\[ K_f = \frac{1}{2}m v^2;~~~~~~U_f = 0 \]](mastermathpng-0.png)
equating the two gives
![\[ \frac{1}{2}m v^2 = m gh \]](mastermathpng-1.png)
so that