# In the figure, a 3.7 g ice flake is released from the edge of a hemispherical bowl whose radius r...

## Question:

In the figure, a {eq}3.7 g {/eq} ice flake is released from the edge of a hemispherical bowl whose radius {eq}r {/eq} is {eq}17 cm {/eq}. The flake-bowl contact is friction less.

a. What is the speed of the flake when it reaches the bottom of the bowl?

b. If we substituted a second flake with {eq}6 {/eq} times the mass, what would its speed be?

c. If, instead, we gave the flake an initial downward speed {eq}1.3 m/s {/eq} along the bowl, what would the answer be?

## Potential energy and kinetic energy

The energy stored in potential energy from gravity is

{eq}\displaystyle U = mgh {/eq}

The kinetic energy is given by

{eq}\displaystyle K = \cfrac{1}{2} mv^2 {/eq}

The given information is standard SI units is

{eq}\displaystyle r = 0.17 \,\mathrm{m} \quad m = 0.0037 \,\mathrm{kg} {/eq}

a.) Gravitational potential energy is converted into kinetic energy

{eq}\displaystyle \frac{1}{2} mv^2 = mgr \to v = \sqrt{ 2gr} = 1.82 \,\mathrm{m/2} {/eq}

b.) The size of the mass does not determine the velocity at the bottom. So the speed would be the same for a different mass.

c.) If you give it more kinetic energy initially it will be going faster at the bottom.

Let

{eq}\displaystyle v_0 = 1.3\,\mathrm{m/s} {/eq}

{eq}\displaystyle \cfrac{1}{2}mv_0^2 + mgr = \cfrac{1}{2}mv_f^2 \to v_f = \sqrt{ v_0^2 + 2gr} = 2.24 \,\mathrm{m/s} {/eq}

Not surprisingly if you give it more speed at the beginning the final speed is larger.