2023 H2 Physics Paper 2 Q1 — A Beam, a Block and a Beaker of Water

The model, live

WeightForce from the waterStringBalanced

Part (a)

2 marks

Explain where the upthrust on the submerged block comes from.

Approach

Upthrust is not a separate kind of force. It is what is left over when you add up the pressure forces of the water on every face of the block. The two marks are for the cause (pressure increases with depth) and the consequence (the push up on the bottom face beats the push down on the top face).

The pressure in a liquid increases with depth, $p = \rho g h$ . The bottom face of the block is deeper than the top face, so the water pushes up on the bottom face with a larger force than it pushes down on the top face. The forces on the sides cancel in pairs.

Pressure in the water increases with depth, so the pressure on the bottom face of the block is greater than the pressure on the top face. The upward force on the bottom face is therefore larger than the downward force on the top face, and the resultant of these pressure forces is an upward force: the upthrust.

Note

Each pressure force grows as the block goes deeper, but they grow by the same amount, so their difference does not change. That is why the upthrust on a fully submerged block does not depend on how deep it is.

Part (b)(i)

2 marks

Show that the upthrust on the block, of volume $27.8\,\text{cm}^3$ , is about $0.27\,\text{N}$ , explaining the working.

Approach

"Explain your working" means the reason has to be written, not just the calculation. The reason is that the upthrust equals the weight of the liquid displaced, and a fully submerged block displaces its own volume of water.

The block is completely under water, so the volume of water displaced is the volume of the block, $27.8\,\text{cm}^3 = 27.8\times10^{-6}\,\text{m}^3$ . The upthrust equals the weight of that water:

U = \rho V g = (1.0\times10^{3})(27.8\times10^{-6})(9.81) = 0.273\,\text{N}

To check your answer:

Work it from the faces instead. For any block with vertical sides of area $A$ and height $\ell$ , the difference in pressure force is $\rho g (h+\ell)A - \rho g h A = \rho g \ell A = \rho g V$ . The two routes give the same expression, and the model's readout shows $0.273\,\text{N}$ at every depth.

Upthrust = weight of water displaced $= \rho V g$ $= (1.0\times10^{3})(27.8\times10^{-6})(9.81)$ $= 0.273\,\text{N}$ , about $0.27\,\text{N}$ , since the fully submerged block displaces its own volume of water.

Watch

Convert the volume before you multiply. $1\,\text{cm}^3 = 10^{-6}\,\text{m}^3$ , not $10^{-2}$ or $10^{-3}$ .

Part (b)(ii)

3 marks

Determine the weight $F$ of the block.

Approach

There are two unknowns, $W$ and $F$ , and two balanced beams, so take moments about the pivot in each. In water the string no longer carries the full weight of the block: the block is in equilibrium under $F$ down, the upthrust $U$ up and the tension up, so the tension is $F - U$ .

Fig. 1.1, moments about the pivot (anticlockwise = clockwise):

W(8.3) = F(18.0) \quad\Longrightarrow\quad W = 2.169F

Fig. 1.2, the string now pulls down on the beam with $T = F - U$ :

W(7.8) = (F - 0.273)(19.0)

Substituting for $W$ :

\begin{aligned}(2.169F)(7.8) &= 19.0F - 5.18\\ 16.92F &= 19.0F - 5.18\\ 2.08F &= 5.18\\ F &= 2.49\,\text{N}\end{aligned}

To check your answer:

With $F = 2.49\,\text{N}$ , $W = 2.169 \times 2.49 = 5.39\,\text{N}$ . In Fig. 1.2 the anticlockwise moment is $5.39 \times 7.8 = 42.1\,\text{N cm}$ and the clockwise moment is $(2.49 - 0.27)\times 19.0 = 42.1\,\text{N cm}$ . Both beams balance.

$F = 2.5\,\text{N}$

Note

The answer is sensitive to rounding. The bracket $19.0 - 16.92$ is a small difference of two large numbers, so carry at least three significant figures in $W/F$ and in $U$ until the last line.

What this question was really testing

Upthrust from pressure. The bottom face is deeper, so it is pushed harder. Name the depth dependence and the difference in force; both are needed for the marks.
Archimedes in one line. Upthrust equals the weight of fluid displaced, $\rho V g$ , and a fully submerged body displaces its own volume.
Free-body thinking before moments. In water the string carries $F - U$ , not $F$ . Getting the force on the beam right comes before writing any moment equation.
Two conditions, two unknowns. Each balanced beam gives one equation. Neither alone fixes $F$ .

A beam balanced twice, and the weight that only one value can satisfy

Part (a)

Part (b)(i)

Part (b)(ii)

What this question was really testing

This is how we mark, and how we teach.