2019 A/L Physics Past Paper · Question 38 (English Medium)

Flipping a floating wooden block with a metal piece

A piece of metal is attached to the top of a rectangular wooden block that floats in water. As shown in the figure, $50\%$ of the volume of the wooden block is submerged in water. The metal piece and the wooden block have the same mass. If the wooden block with the metal piece is flipped up side down, what could be the percentage of the volume of the wooden block submerged in water?

(1) Slightly smaller than 50%
(2) Much smaller than 50%
(3) 50%
(4) Slightly larger than 50%
(5) Much larger than 50%

Answer and explanation

Answer: (1) Slightly smaller than 50%

What the question asks

We change one thing.

The metal piece goes from the top (out of the water) to the bottom (under the water).

We want to know how much of the wood is under water after that.

Relevant theory

When a body is completely or partially immersed in a fluid at rest, it experiences an upthrust equal to the weight of the fluid displaced by the body.

If a body is to float in a liquid, the forces on it should be in equilibrium.

Applying it to this question

The block floats.

So the push up equals the weight.

The weight is $2mg$ , the wood and the metal together.

The push up is the weight of the water pushed aside.

So the volume of water pushed aside is:

$V_{sub} = \dfrac{2m}{\rho_w}$

Look at what does not change.

The total mass is still $2m$ .

So the water pushed aside is still the same volume, $\dfrac{V}{2}$ .

Before the flip, the wood alone pushes aside all of it.

Reasoning and calculation

After the flip the metal is under water.

It pushes aside its own volume $V_m$ .

The total must still be $\dfrac{V}{2}$ .

So the wood needs to push aside only $\dfrac{V}{2} - V_m$ .

So less of the wood is under water.

$\rho_w\frac{V}{2} = 2m \Rightarrow \frac{m}{V} = \frac{\rho_w}{4}$

$V_m = \frac{m}{8\rho_w} = \frac{V}{32}$

$46.9$

$\frac{V}{2} - \frac{V}{32} = 0.469V$

$V_m$ is how big?

The metal is many times denser than water. The masses are equal.

So its volume is only a few percent of the wood's volume.

If the metal is 8 times denser than water, it is about 47 % instead of 50 %.

So it is a little less than 50 %.

Reading the options

Less than 50 %, but only a little.

So it is option (1).

Why the other options fail

This option says: much less than 50 %.

That needs the metal to push aside a lot of water.

But the metal is much denser than the wood, and the masses are equal. So its volume is small.

So the change is only a few percent.

So it is wrong.

This option says: 50 %.

It is true that the total weight does not change.

But after the flip the metal is under water too. It also gives some push up.

So the wood needs to push aside less water.

This option says: a little more than 50 %.

That would mean the wood sinks deeper.

But the metal under water adds to the push up. So the wood sinks less, not more.

This option says: much more than 50 %.

The metal does go down. But the total weight stays the same. The water pushed aside must stay the same too.

The metal now supplies a part of it. So the wood sinks less.

Checks

Does it make sense?

Try a metal as light as water. Its volume would be $\dfrac{V}{4}$ .

Then only 25 % of the wood would be under water. That is much less.

Real metals are many times denser. So the change is only 2 to 4 %.

Conclusion

The metal goes under water and pushes a little water aside.

So the wood sinks a little less.

The change is small.

So the answer is option (1).

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