2019 A/L Physics Past Paper · Question 25 (English Medium)

Answer: (4) $300\ \text{km h}^{-1}$

Given information

We know:

At $80$ km from the centre the air moves at $v_1 = 150\ \text{km h}^{-1}$.

The air moves round the centre of the eye.

What the question asks

We want:

The speed $v_2$ of the same air at $r_2 = 40$ km.

Relevant theory

The angular momentum (L) of a lamina moving with an angular velocity $\omega$ about an axis passing through O is given by $L = p.r$

The total angular momentum of a system remains constant provided no external torque acts on the system.

Required equations

$L = m v r$

$m v_1 r_1 = m v_2 r_2$

Applying it to this question

We follow one air mass $m$.

Its mass does not change.

It moves round the centre of the eye.

So we use its angular momentum:

$$L = mvr$$

The push on the air points to the centre.

So it has no turning effect.

There is no outside torque.

So $L$ does not change.

Reasoning and calculation

$L_1 = m \times 150 \times 80$

$L_2 = m \times v_2 \times 40$

$m \times 150 \times 80 = m \times v_2 \times 40$

$v_2 = \frac{150 \times 80}{40}$

$300$

$v_2 = 300$

Checks

$v1 * r1 / r2$

Check the unit.

Both distances are in km.

So we need not change km to m.

The answer comes out in $\text{km h}^{-1}$.

Does it make sense?

A skater pulls in the arms and spins faster.

In the same way the air goes faster as it comes closer to the eye.

That is why the wind is strongest near the eye.

Conclusion

$v_2 = 300\ \text{km h}^{-1}$

$mvr$ does not change.

$$v_2 = \frac{150 \times 80}{40} = 300\ \text{km h}^{-1}$$

So the answer is option (4).