2019 A/L Physics Past Paper · Question 08 (English Medium)

Answer: (4) Diagram (4)

What the question asks

We want: the change in velocity $\Delta\vec v$ from $A$ to $B$, its size and its direction.

Then we choose the diagram that shows it.

Relevant theory

Velocity is a vector quantity and thus has an associated direction.

However along a circular path the direction of the velocity is continuously changing.

When anyone of magnitude or direction is changing it means that the velocity is changing.

Applying it to this question

Take right as $+x$ and up as $+y$.

In the figure the object comes down on the left towards $A$.

So it goes from $A$ on to $B$.

At $A$, the bottom of the circle, the object moves to the right.

$$\vec v_A = v\,\hat{x}$$

At $B$, on the right, it moves straight up.

$$\vec v_B = v\,\hat{y}$$

Both have the same size $v$. They are at right angles.

Reasoning and calculation

A change is always final minus first.

$$\Delta\vec v = \vec v_B - \vec v_A = \vec v_B + (-\vec v_A)$$

$-\vec v_A$ points to the left.

$\Delta\vec v = (0 - v)\,\hat{x} + (v - 0)\,\hat{y}$

$|\Delta v| = \sqrt{(-v)^2 + v^2}$

$|\Delta v| = \sqrt{2}\,v$

So $\Delta\vec v$ has $v$ to the left and $v$ upwards.

The two parts are equal. So it points up and to the left, at $45°$.

Its size is $\sqrt{2}\,v$.

Reading the options

Now look at the diagrams.

(3) and (4) both point up and to the left at $45°$.

But only (4) has the size $\sqrt{2}\,v$.

So the answer is diagram (4).

Why the other options fail

This option says: $\Delta\vec v$ is $v$, straight up.

But that is only the velocity at $B$. We must also take away the velocity at $A$.

So it is wrong.

This option says: $\Delta\vec v$ is $v$, to the left.

But that is only $-\vec v_A$. The velocity at $B$ must be added to it.

This option says: $\Delta\vec v$ is $v$, up and to the left.

The direction is right. But two arrows of size $v$ at right angles give $\sqrt{2}\,v$, not $v$.

This option says: $\Delta\vec v$ is $\sqrt{2}\,v$, down and to the right.

The size is right. But the direction is turned round. That is $\vec v_A - \vec v_B$, first minus final.

Checks

$sqrt(2) * v$

Does it make sense?

In circular motion the velocity keeps turning towards the centre $O$.

So the change should point into the circle.

From halfway between $A$ and $B$, the centre is up and to the left. That matches $\Delta\vec v$.

Conclusion

$|\Delta v| = \sqrt{2}\,v$

$$\Delta\vec v = \vec v_B - \vec v_A$$

Its size is $\sqrt{2}\,v$.

It points up and to the left, at $45°$ to the horizontal.

So the answer is diagram (4).

Thinking the velocity does not change because the speed does not change.

Velocity has a direction.

When the direction changes, the velocity changes, even at the same speed.

For a quarter circle, $|\Delta v| = \sqrt{2}\,v$.