2021 A/L Physics Past Paper · Question 49 (English Medium)

Answer: (5) $50\ \text{N}$

Given information

What we know:

Height of the water: $H = 0.25\ \text{m}$.

Diameter of the base: $0.20\ \text{m}$, so the radius is $0.10\ \text{m}$.

Volume of the water: $V = 2.5 \times 10^{-3}\ \text{m}^3$.

Density of water: $\rho = 10^3\ \text{kg m}^{-3}$. Take $\pi = 3$.

The paper does not give $g$. We take $g = 10\ \text{m s}^{-2}$, as the marking scheme does.

What the question asks

Find the force that the water exerts on the sloping surface of the flask.

The plan has three steps.

First, find the force of the water on the base.

Then find the weight of the water.

The sloping wall must carry the difference.

Relevant theory

As pressure is the force acting normally on a unit area,

When the objects interact (collide), they exert equal and oppsite forces on one another.

Required equations

$p = H\rho g$

$F_{\text{base}} = pA$

$W = \rho V g$

$F_{\text{base}} = W + F_{\text{wall}}$

Applying it to this question

The water is at rest.

The base pushes the water up.

The weight pulls it down. The sloping wall pushes it down and inwards.

The sideways pushes of the wall cancel all round the flask. Only the vertical part is left.

$0.03$

$A = \pi r^2 = 3 \times 0.10^2$

$2500$

$p = H\rho g = 0.25 \times 1000 \times 10$

$75$

$F_{\text{base}} = pA = 2500 \times 0.03$

$25$

$W = \rho V g = 1000 \times 2.5 \times 10^{-3} \times 10$

$50$

$F_{\text{wall}} = F_{\text{base}} - W = 75 - 25$

All quantities are in SI units.

Pressure in $\text{Pa} = \text{N m}^{-2}$, area in $\text{m}^2$, so the force is in newtons.

Reasoning and calculation

The sloping wall pushes the water with $50\ \text{N}$.

The water pushes the wall with an equal and opposite force.

So the water exerts $50\ \text{N}$ on the inclined surface.

The answer is (5).

Checks

$H * rho * g * 3 * r * r - rho * V * g$

Does it make sense?

The flask narrows upwards. So it holds less water than a straight column of height $0.25\ \text{m}$ over the base would. That column would weigh $75\ \text{N}$.

The base feels $75\ \text{N}$, three times the weight of the water. The sloping wall carries the extra $50\ \text{N}$.

Conclusion

$F_{\text{wall}} = H\rho g A - \rho V g = 75 - 25 = 50\ \text{N}$

The force on the base is $pA = 2500 \times 0.03 = 75\ \text{N}$.

The water weighs $25\ \text{N}$.

So the sloping wall takes $75 - 25 = 50\ \text{N}$. The answer is (5).

In a flask that is wider at the bottom, the base force can be larger than the weight of the water.

Find the base force from the pressure. Subtract the weight. The sloping wall carries the rest.

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