A 2.5 g ice flake is released from the edge of a hemispherical bowl whose radius r is 40.0 cm....

Question:

A 2.5 g ice flake is released from the edge of a hemispherical bowl whose radius r is 40.0 cm. The flake-bowl contact is frictionless.

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

b) If a second flake with twice the mass was substituted, what would its speed be?

Potential energy:

The term potential energy can be defined as the saved energy in the object due to the position corresponding to the other objects. The potential energy of a body in a vertical motion directly changes with height. The standard unit of the potential energy is Joules.

Given data:

• The mass of the ice flake is {eq}m = 2.5\,{\rm{g}} {/eq}
• The radius is {eq}r = 40.0\,{\rm{cm}} = 0.04\,{\rm{m}} {/eq}

(a)

The expression for the speed of the flake by using the energy conservation is

{eq}p.e = k.e {/eq}

• Here {eq}p.e = mgh {/eq} is the potential energy.
• Here {eq}k.e = \dfrac{1}{2}m{v^2} {/eq} is the kinetic energy.

Substituting the values in the above equation as,

{eq}\begin{align*} mgh &= \dfrac{1}{2}m{v^2}\\ v &= \sqrt {2gr} \\ v &= \sqrt {2\left( {9.8} \right)\left( {0.04} \right)} \\ v &= 0.88543\,{\rm{m/s}} \end{align*} {/eq}

Thus the speed of the flake is {eq}v = 0.88543\,{\rm{m/s}} {/eq}

(b)

The speed remains same because the velocity is independent of the mass.