Solution to Question 18
Step 1: Cloud Expansion
The balls bounce elastically. The maximum height $H$ attained is inversely proportional to the effective acceleration due to gravity ($g_{eff}$).
$$ H = \frac{v_0^2}{2g_{eff}} $$Initially $H_1$ corresponds to 750 ml with $g_{eff} = g$.
When the beaker accelerates down with $0.5g$, the effective gravity is $g’ = g – 0.5g = 0.5g$.
$$ H_2 = \frac{v_0^2}{2(0.5g)} = 2 \left( \frac{v_0^2}{2g} \right) = 2 H_1 $$So the new cloud of balls fills the beaker up to $2 \times 750 = 1500$ ml.
Step 2: Density Distribution
The balls are in a steady dynamic state. The flux of particles crossing any horizontal plane is constant. Since flux $\Phi = n(y) v(y)$, the number density $n(y) \propto 1/v(y)$.
Velocity at height $y$: $v(y) = \sqrt{2g_{eff}(H_{max} – y)}$.
Density: $n(y) \propto \frac{1}{\sqrt{H_{max} – y}}$.
Total number of balls $N_{total} = \int_0^{H_{max}} n(y) dy = k \int_0^{H_{max}} (H_{max}-y)^{-1/2} dy$.
$$ N_{total} = k [-2\sqrt{H_{max}-y}]_0^{H_{max}} = 2k\sqrt{H_{max}} $$Step 3: Calculating Fraction
We need the fraction of balls below the 1125 ml mark ($h = 1125$). The total height is $H_{max} = 1500$.
Number below $h$: $N(h) = \int_0^h n(y) dy = 2k (\sqrt{H_{max}} – \sqrt{H_{max}-h})$.
Fraction $f = \frac{N(h)}{N_{total}} = 1 – \sqrt{1 – \frac{h}{H_{max}}}$.
$$ \frac{h}{H_{max}} = \frac{1125}{1500} = \frac{3}{4} = 0.75 $$ $$ f = 1 – \sqrt{1 – 0.75} = 1 – \sqrt{0.25} = 1 – 0.5 = 0.5 $$Answer: 0.5