Sugar on Scales

question physics

Question

Sugar is dropped from rest onto an electric balance from a fixed height $h$, and constant rate $\dot{m}$ without bouncing. The flow of sugar is cut off at the source when the balance shows the exact weight required. What will be the final reading on the scale, once the sugar in the air has settled on the pan?


Hints

There are two effects on the reading: the weight of the sugar in the air, and the false reading due to the change in momentum of the falling sugar. Show that these two effects exactly cancel out.


Answer

False reading due to change in momentum:
$$ R = Mg + \dot{m} v $$ Using suvat (or conservation of energy) $$ gh = \frac{1}{2} v^2 $$ $$ v = \sqrt{2gh} $$ $$ R = Mg + \dot{m} \sqrt{2gh} $$ $$ \text{force due to change in momentum} = \dot{m} \sqrt{2gh} $$

Weight of sugar in the air:
Let $T$ be the time taken to fall $$ \text{mass in air} = M_{air} = \dot{m} T $$ $$ \text{final velocity} = v = \text{acceleration} * T $$ $$ \sqrt{2gh} = gT $$ $$ T = \sqrt{\frac{2h}{g}} $$ $$ M_{air} = \dot{m} \sqrt{\frac{2h}{g}} $$ $$ \text{weight in air} = M_{air} g = \dot{m} \sqrt{2gh} $$

The increase in the reading due to change in momentum is therefore equal to the decrease in the reading due to the weight of sugar still in the air.