Physics 9702/42 — October/November 2023

Cambridge A-Level · A Level Structured Questions · worked solutions for every part, with the mark scheme

questions

100

marks

120

minutes

Topics Motion in a Circle · Gravitational Fields · Ideal Gases · Thermodynamics · Oscillations · Alternating Currents · +7 more

Q1Motion in a CircleFree sample

(a)

Define the radian.

DifficultyEasy

Worked solution

Answer

One radian is the angle at the centre of a circle subtended by an arc of length equal to the radius.

Final answer

Angle at centre subtended by an arc equal in length to the radius.

Detailed explanation

Background Concept

Angles can be measured in degrees or in radians. The radian is defined in a way that links angles directly to lengths on a circle.

For a circle of radius $r$ , an angle $\theta$ (in radians) subtends an arc length $s$ given by

\theta = \frac{s}{r}.

This relationship is the reason radians are so convenient in circular motion.

Understanding the Question

The question asks for the definition of the radian (not a calculation). You must state it in terms of a circle, its radius, and the arc length it subtends.

Approach

Use the defining condition $s=r$ . Substitute this into $\theta = s/r$ to identify what $1$ radian corresponds to, and state it in words.

Step-by-Step Reasoning

If $s=r$ , then

\theta = \frac{s}{r} = \frac{r}{r} = 1.

So an angle of $1\ \text{rad}$ is the angle at the centre of a circle that subtends an arc length equal to the radius.

Key Takeaways

Radian measure is defined by $\theta = s/r$ .
$1\ \text{rad}$ corresponds to the special case $s=r$ .

Common Mistakes

Defining it using degrees (e.g. saying $57.3^\circ$ ) rather than the arc/radius definition.
Missing the phrase “at the centre” or not mentioning arc length.

Things to Be Careful About

The definition must involve a circle, the radius, and an arc length equal to the radius.
Do not confuse arc length with chord length.

Techniques used

state the geometric definition of a radianrelate arc length to radius for the definition

(b)

The minute hand of a clock revolves at constant angular speed around the face of the clock, completing one revolution every hour. A small piece of modelling clay is attached to the hand with its centre of gravity at a distance $L$ from the fixed end of the hand, as shown in Fig. 1.1.

Fig. 1.1

Calculate the angular speed $\omega$ of the minute hand.

$\omega$ = ______ $\text{rad s}^{-1}$

DifficultyMedium-Easy

Worked solution

Working

Period $T = 1\ \text{h} = 3600\ \text{s}$

\omega = \frac{2\pi}{T} = \frac{2\pi}{3600} = 1.75 \times 10^{-3}\ \text{rad s}^{-1}

Answer

$1.75 \times 10^{-3}\ \text{rad s}^{-1}$

Final answer

1.75 × 10^-3 rad s^-1

Detailed explanation

Background Concept

For uniform circular motion, angular speed $\omega$ is the rate of change of angular displacement:

\omega = \frac{\Delta \theta}{\Delta t}.

One full revolution corresponds to an angular displacement of $2\pi$ radians. If the time for one revolution is the period $T$ , then

\omega = \frac{2\pi}{T}.

Understanding the Question

The minute hand completes one full revolution every hour. That means its period is $T = 1\ \text{hour}$ . The question asks for its angular speed $\omega$ in $\text{rad s}^{-1}$ , so we must express $T$ in seconds.

Approach

Convert $1\ \text{hour}$ to seconds.
Use $\omega = 2\pi/T$ .

Step-by-Step Reasoning

1 hour is

T = 60\ \text{min} \times 60\ \text{s min}^{-1} = 3600\ \text{s}.

Then

\omega = \frac{2\pi}{3600} = 1.745\times 10^{-3}\ \text{rad s}^{-1} \approx 1.75\times 10^{-3}\ \text{rad s}^{-1}.

Key Takeaways

One full revolution is always $2\pi$ radians.
For steady rotation, $\omega = 2\pi/T$ .

Common Mistakes

Leaving $T$ in hours and producing $\omega$ in $\text{rad h}^{-1}$ .
Using $\pi/T$ instead of $2\pi/T$ .

Things to Be Careful About

The required unit is $\text{rad s}^{-1}$ , so time must be in seconds.
Quote a sensible number of significant figures (typically 2–3).

Techniques used

use the period-angular speed relationconvert the period into secondssubstitute into \omega = 2\pi/T

(c)

During a time interval of $1400\ \text{s}$ , the centre of gravity of the piece of modelling clay in Fig. 1.1 moves through a total distance of $0.44\ \text{m}$ .

(i)

Calculate the angle through which the minute hand moves in this time interval.

angle = ______ $\text{rad}$

DifficultyMedium-Easy

Worked solution

Working

\theta = \omega t = \left(1.75\times 10^{-3}\right)(1400) = 2.44\ \text{rad}

Answer

$2.44\ \text{rad}$

Final answer

2.44 rad

Detailed explanation

Background Concept

For constant angular speed $\omega$ , angular displacement increases linearly with time:

\theta = \omega t.

This is the rotational analogue of $s = vt$ for constant linear speed.

Understanding the Question

In $1400\ \text{s}$ the minute hand (and the clay fixed to it) rotates through some angle $\theta$ . You are given the time interval and (from part (b)) the constant angular speed of the minute hand.

Approach

Use the uniform rotation relation $\theta = \omega t$ with $t = 1400\ \text{s}$ .

Step-by-Step Reasoning

With $\omega = 1.75\times 10^{-3}\ \text{rad s}^{-1}$ and $t=1400\ \text{s}$ :

\theta = \omega t = (1.75\times 10^{-3})(1400) = 2.44\ \text{rad}.

The seconds cancel, leaving radians.

Key Takeaways

Uniform rotation: $\theta$ is found from $\omega t$ .

Common Mistakes

Using $\theta = 2\pi t/T$ but mixing units for $t$ and $T$ .
Forgetting that radians are dimensionless and still writing an incorrect unit.

Things to Be Careful About

Ensure $\omega$ is in $\text{rad s}^{-1}$ and $t$ in $\text{s}$ so the product is consistent.

Techniques used

use \theta = \omega t for constant angular speedsubstitute time and angular speed

(ii)

Determine distance $L$ .

$L$ = ______ $\text{m}$

DifficultyMedium-Easy

Worked solution

Working

s = L\theta \Rightarrow L = \frac{s}{\theta} = \frac{0.44}{2.44} = 0.180\ \text{m}

Answer

$0.18\ \text{m}$

Final answer

0.18 m

Detailed explanation

Background Concept

Arc length $s$ on a circle is related to radius $r$ and angle $\theta$ (in radians) by

s = r\theta.

This formula is only valid when $\theta$ is in radians.

Understanding the Question

The clay’s centre of gravity moves along a circular arc of radius $L$ (distance from the centre of the clock to the clay). In $1400\ \text{s}$ it travels a distance $s = 0.44\ \text{m}$ along that arc. From part (c)(i) you have the angle moved through, $\theta$ .

You are asked to determine $L$ .

Approach

Use $s = L\theta$ and rearrange to $L = s/\theta$ .

Step-by-Step Reasoning

Given $s=0.44\ \text{m}$ and $\theta = 2.44\ \text{rad}$ :

L = \frac{s}{\theta} = \frac{0.44}{2.44} = 0.180\ \text{m}.

So $L \approx 0.18\ \text{m}$ .

Key Takeaways

Use $s=r\theta$ for circular motion distances.
Always check $\theta$ is in radians.

Common Mistakes

Using degrees in $s=r\theta$ (would give a wrong value for $L$ ).
Confusing $L$ with the full length of the minute hand to the edge of the clock.

Things to Be Careful About

Use consistent significant figures; $0.44\ \text{m}$ suggests $L$ should be quoted to about 2 s.f.
$L$ is the radius of the circular path of the clay’s centre of gravity, not necessarily the hand’s full length.

Techniques used

use arc length relation s = r\thetarearrange to isolate the radiussubstitute numerical values

(iii)

Calculate the magnitude of the centripetal acceleration of the piece of modelling clay.

centripetal acceleration = ______ $\text{m s}^{-2}$

DifficultyMedium

Worked solution

Working

a = L\omega^2 = (0.180)\left(1.75\times 10^{-3}\right)^2 = 5.5\times 10^{-7}\ \text{m s}^{-2}

Answer

$5.5\times 10^{-7}\ \text{m s}^{-2}$

Final answer

5.5 × 10^-7 m s^-2

Detailed explanation

Background Concept

In uniform circular motion, the speed may be constant but the velocity changes direction, so there is an acceleration towards the centre called centripetal acceleration.

Its magnitude can be written as

a = r\omega^2

or equivalently

a = \frac{v^2}{r},

where:

$r$ is the radius of the circular path,
$\omega$ is angular speed,
$v$ is linear (tangential) speed.

Understanding the Question

The clay’s centre of gravity moves in a circle of radius $L$ about the clock’s centre. You have already found $L$ and you know the minute hand rotates at constant $\omega$ . The question asks for the magnitude of the centripetal acceleration of the clay.

Approach

Use $a = r\omega^2$ because $r=L$ is known and $\omega$ is known. Substitute values and keep units consistent.

Step-by-Step Reasoning

From earlier parts:

$L = 0.180\ \text{m}$
$\omega = 1.75\times 10^{-3}\ \text{rad s}^{-1}$

Compute $\omega^2$ :

\omega^2 = (1.75\times 10^{-3})^2 = 3.06\times 10^{-6}\ \text{s}^{-2}.

Then

a = L\omega^2 = (0.180)(3.06\times 10^{-6}) = 5.5\times 10^{-7}\ \text{m s}^{-2}.

This is extremely small because the rotation is very slow (one revolution per hour).

(Alternative check: $v = s/t = 0.44/1400 = 3.14\times 10^{-4}\ \text{m s}^{-1}$ , then $a=v^2/L$ gives the same result.)

Key Takeaways

Centripetal acceleration depends on how fast the object goes around the circle: $a \propto \omega^2$ .
Slow rotation gives very small centripetal acceleration.

Common Mistakes

Using $a = r\omega$ (missing the square).
Using the distance travelled $0.44\ \text{m}$ as the radius.
Mixing up $L$ with the full minute-hand length.

Things to Be Careful About

Radians are dimensionless, so $\omega$ has unit $\text{s}^{-1}$ and $a$ ends up in $\text{m s}^{-2}$ .
Because the answer is very small, standard form is usually clearest.

Techniques used

apply centripetal acceleration formula a = r\omega^2use a previously determined radiussquare the angular speed and substitute values

(d)

Use your answer in (c)(iii) to explain why the variation with time of the magnitude of the force exerted by the minute hand on the piece of modelling clay is negligible as the minute hand undergoes one full revolution.

DifficultyMedium

Worked solution

Answer

From (c)(iii), $a$ is extremely small ( $\approx 5\times 10^{-7}\ \text{m s}^{-2}$ ), so the centripetal force needed is

F_c = ma

and hence $F_c \ll mg$ .
Therefore the force exerted by the hand on the clay is dominated by the (approximately constant) support of the clay’s weight, so any change in the magnitude during a revolution is negligible.

Final answer

Because a is extremely small, Fc = ma is negligible compared with mg, so the hand’s force magnitude is essentially constant through a revolution.

Detailed explanation

Background Concept

If an object moves in a circle at constant speed, it needs a centripetal acceleration towards the centre.

The net inward (centripetal) force required is

F_c = ma = mr\omega^2.

In a real situation, the contact force provided by a support (here, the clock hand) may also have to balance other forces such as the weight $mg$ .

The important idea for this question is comparing sizes of forces/accelerations:

if the centripetal acceleration is tiny, then the additional force needed to provide it is also tiny.

Understanding the Question

You found in (c)(iii) that the centripetal acceleration of the clay is extremely small. The question asks you to use that fact to explain why, over a full revolution, the magnitude of the force that the minute hand exerts on the clay does not change appreciably with time.

In other words: does the required centripetal effect noticeably change the contact force the hand must provide as it rotates?

Approach

Use $F_c = ma$ to link your small value of $a$ to a small required centripetal force.
Compare this with the weight $mg$ (which is of order $10\ \text{m s}^{-2}$ in acceleration terms).
Conclude that the centripetal contribution is negligible, so the magnitude of the hand’s force is approximately constant.

Step-by-Step Reasoning

From (c)(iii),

a \approx 5.5\times 10^{-7}\ \text{m s}^{-2}.

The centripetal force needed is

F_c = ma.

Even without knowing $m$ , we can compare accelerations:

centripetal acceleration is $\sim 10^{-7}\ \text{m s}^{-2}$ ,
gravitational acceleration is $g \approx 9.8\ \text{m s}^{-2}$ .

\frac{a}{g} \approx \frac{5.5\times 10^{-7}}{9.8} \approx 6\times 10^{-8}.

That means the centripetal force is about $10^{-7}$ times the weight:

\frac{F_c}{mg} = \frac{ma}{mg} = \frac{a}{g} \sim 10^{-8}.

So the force the hand must provide to cause the circular motion is utterly negligible compared with the force associated with the clay’s weight (which is essentially constant). Therefore, as the minute hand goes around, the magnitude of the force exerted by the hand changes by an amount that is far too small to matter.

diagram-d1

Key Takeaways

Use $F=ma$ to convert “small acceleration” into “small force”.
Comparing $a$ with $g$ is a powerful way to judge whether centripetal effects are significant.

Common Mistakes

Saying “the force is constant because the acceleration is constant” without using the smallness of $a$ (the question specifically asks you to use the value from (c)(iii)).
Forgetting that the direction of centripetal force changes, but the question is about the magnitude.
Not making a comparison (e.g. with $mg$ ) to justify “negligible”.

Things to Be Careful About

“Negligible variation” needs a reason: here it is because $a$ (and thus $F_c$ ) is extremely small.
If you choose to mention weight, be clear you are comparing magnitudes (using $a/g$ or $F_c/mg$ ).

Techniques used

relate force to centripetal acceleration using F = macompare magnitudes of forces to judge significancereason about whether a small additional force changes the resultant appreciably

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