(a) State the principle of conservation of momentum. (b) An object of mass $2m$ is travelling at a speed of $5.0\text{ m s}^{-1}$ in a straight line. It collides with an object of mass $3m$ which is initially stationary, as shown in Fig. 3.1. After the collision, the object of mass $2m$ moves with velocity $v$ at an angle of $30^\circ$ to its original direction of motion. The object of mass $3m$ moves with velocity $w$ also at an angle of $30^\circ$, as shown in Fig. 3.2. By…

(a) Total momentum is constant in an isolated system (no resultant external force). · (b) v = 2.89 m s^-1, w = 1.92 m s^-1 · (c) 1.5 × 10^3 N

9702/21 May/June 2024 — Physics Solved Past Paper

Q1MediumPhysical Quantities and UnitsDynamics

The drag force $F_D$ acting on an object falling through air is given by

F_D = \frac{1}{2} C\rho Av^2

where $A$ is the cross-sectional area of the object,
$v$ is the velocity of the object in the air,
$\rho$ is the density of the air and
$C$ is a constant called the drag coefficient.

(a)

Use SI base units to show that the drag coefficient has no units.

3M Similar

DifficultyMedium-Easy

Worked solution

Working

$[F_D] = \text{N} = \text{kg m s}^{-2}$ .

$\left[\rho\right] = \text{kg m}^{-3}$ , $[A] = \text{m}^2$ , $[v^2] = (\text{m s}^{-1})^2 = \text{m}^2\text{ s}^{-2}$ .

So

\left[\rho A v^2\right] = (\text{kg m}^{-3})(\text{m}^2)(\text{m}^2\text{ s}^{-2}) = \text{kg m s}^{-2} = \text{N}

Hence $\left[\frac{1}{2}C\rho A v^2\right] = C\,\text{N}$ must equal $\text{N}$ , so $C$ has no units.

Answer

$C$ is dimensionless (no units).

Final answer

C is dimensionless (no units).

Detailed explanation

Background Concept

Dimensional analysis (or checking homogeneity) uses the fact that any valid physical equation must have the same dimensions (units) on both sides. If an equation is

F_D = \frac{1}{2}C\rho Av^2

then the dimensions of the right-hand side must match the dimensions of force. If all other quantities on the right already have the dimensions of force, the remaining factor $C$ must be dimensionless.

Key SI base/derived units used here:

Force: $1\ \text{N} = 1\ \text{kg m s}^{-2}$
Density: $\rho$ has units $\text{kg m}^{-3}$
Area: $A$ has units $\text{m}^2$
Speed: $v$ has units $\text{m s}^{-1}$

Understanding the Question

You are told the drag force depends on $\rho$ , $A$ and $v^2$ (and constants). The question asks you to show, using SI base units, that the drag coefficient $C$ must have no units.

Approach

Write the SI units for $F_D$ , $\rho$ , $A$ , and $v$ .
Find the units of the combination $\rho A v^2$ .
Compare with the units of $F_D$ .
Conclude what units $C$ must have for the equation to be consistent.

Step-by-Step Reasoning

Force:

[F_D] = \text{N} = \text{kg m s}^{-2}

Other quantities:

[\rho] = \text{kg m}^{-3}, \quad [A] = \text{m}^2, \quad [v] = \text{m s}^{-1} \Rightarrow [v^2] = \text{m}^2\text{ s}^{-2}

Now multiply:

[\rho A v^2] = (\text{kg m}^{-3})(\text{m}^2)(\text{m}^2\text{ s}^{-2}) = \text{kg m s}^{-2} = \text{N}

So the right-hand side becomes

\left[\frac{1}{2}C\rho A v^2\right] = [C]\,\text{N}

But this must equal $\text{N}$ , therefore $[C]=1$ (no units).

Key Takeaways

A valid physics equation must be dimensionally homogeneous.
If the units of all other factors already match the left-hand side, the remaining constant is dimensionless.

Common Mistakes

Using $\text{N}$ without converting it to base units $\text{kg m s}^{-2}$ when the question explicitly asks for SI base units.
Forgetting to square the units of $v$ when handling $v^2$ .
Treating $\frac{1}{2}$ as affecting units (pure numbers do not change units).

Things to Be Careful About

Keep track of powers of metres: $\text{m}^{-3}\times\text{m}^2\times\text{m}^2 = \text{m}^{+1}$ .
State the conclusion clearly: “ $C$ has no units” or “ $C$ is dimensionless.”

Techniques used

write each quantity in SI base unitssubstitute dimensions into an equationcompare dimensions on both sides to deduce unitlessness

(b)

Fig. 1.1 shows a sphere falling at terminal velocity in air.

Fig. 1.1

Assume that the upthrust on the sphere is negligible.

On Fig. 1.1, draw and label arrows to show the directions of the two forces acting on the sphere.

2M Similar

DifficultyEasy

Worked solution

Answer

diagram-1

Weight $mg$ acts vertically downwards.

Drag force $F_D$ acts vertically upwards (opposite to the motion).

Final answer

Weight (mg) downwards; drag force (FD) upwards.

Detailed explanation

Background Concept

For an object moving through a fluid (air), two common vertical forces are:

Weight $W = mg$ acting downward due to gravity.
Drag force $F_D$ acting opposite to the direction of motion through the fluid.

At terminal velocity, the object continues to move but its acceleration is zero, so the resultant force is zero. (This is used in later parts, but the force directions are the same regardless.)

Understanding the Question

A sphere is falling downward at terminal velocity. Upthrust is stated to be negligible, so you only include two forces. You must draw and label arrows for the directions of these two forces on the given diagram.

Approach

Decide the direction of motion (downwards).
Weight always acts downwards.
Drag acts opposite to motion, so it acts upwards.
Draw two arrows on the sphere, one up and one down, and label them correctly.

Step-by-Step Reasoning

The sphere is falling downwards, so:

Draw a downward arrow from the sphere labelled $mg$ (or weight $W$ ).
Drag is resistive, so it opposes the downward motion; draw an upward arrow labelled $F_D$ (or drag).

Because upthrust is negligible, you do not draw an upthrust arrow.

diagram-1

Key Takeaways

Weight acts toward the centre of the Earth (downward).
Drag force always opposes the direction of motion relative to the fluid.
If upthrust is neglected, only these two forces act vertically.

Common Mistakes

Drawing drag in the same direction as the motion (it must oppose motion).
Including an upthrust force despite being told it is negligible.
Labelling drag as “friction” without making clear it is the air resistance $F_D$ .

Things to Be Careful About

The question asks for directions and labels: both arrows must be shown and correctly labelled.
Keep arrows clearly vertical (since only vertical forces are considered).

Techniques used

identify forces acting on a moving object in a fluiduse the direction of motion to determine the direction of dragdraw a free-body force diagram

(c)

The mass of the sphere is $49\text{ g}$ .

Calculate the drag force $F_D$ acting on the sphere.

$F_D$ = ______ $\text{N}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

At terminal velocity, resultant force $=0$ .

Upthrust negligible, so $F_D = mg$ .

F_D = 0.049 \times 9.81 = 0.481\ \text{N}

Answer

$F_D = 0.48\ \text{N}$

Final answer

0.48 N

Detailed explanation

Background Concept

Terminal velocity occurs when an object falls through a fluid and reaches a constant speed. Constant speed means acceleration $a=0$ , so by Newton’s second law the resultant force is zero:

\sum F = ma = 0

For a falling object with negligible upthrust, the forces are weight $mg$ downward and drag $F_D$ upward. If the resultant is zero, their magnitudes are equal.

Understanding the Question

You are given the mass of the sphere ( $49\ \text{g}$ ) and told it is falling at terminal velocity, with upthrust negligible. You must calculate the drag force $F_D$ .

Because it is at terminal velocity, you don’t need the speed or the drag formula yet; you just use force balance.

Approach

Convert mass to kilograms.
Use terminal velocity condition: drag balances weight (since upthrust is negligible).
Compute $F_D = mg$ using $g \approx 9.81\ \text{m s}^{-2}$ .

Step-by-Step Reasoning

Convert mass:

49\ \text{g} = 49 \times 10^{-3}\ \text{kg} = 0.049\ \text{kg}

At terminal velocity, $\sum F = 0$ . With upthrust neglected:

F_D = mg

Calculate:

F_D = 0.049 \times 9.81 = 0.481\ \text{N}

Rounded appropriately:

F_D \approx 0.48\ \text{N}

Key Takeaways

Terminal velocity implies zero acceleration and zero resultant force.
If upthrust is negligible, drag equals weight: $F_D = mg$ .
Always convert grams to kilograms before using $W = mg$ .

Common Mistakes

Using $49$ as kilograms instead of converting from grams.
Forgetting that at terminal velocity acceleration is zero, so forces balance.
Giving the answer without units.

Things to Be Careful About

State the terminal-velocity condition (resultant force zero) to justify $F_D = mg$ .
Use a sensible value of $g$ consistent with Cambridge convention ( $9.8$ or $9.81\ \text{m s}^{-2}$ ).

Techniques used

use the condition for terminal velocity that resultant force is zeroequate drag force to weight when upthrust is negligiblecalculate weight using W = mg

(d)

The sphere is falling in air at a terminal velocity of $25$ in SI base units.
The density of the air is $1.2$ in SI base units.
The diameter of the sphere is $0.060$ in SI base units.

Use your answer in (c) to calculate the drag coefficient $C$ for the sphere.

$C$ = ______

3M Similar

DifficultyMedium

Worked solution

Working

Rearrange

F_D = \frac{1}{2}C\rho Av^2 \;\Rightarrow\; C = \frac{2F_D}{\rho A v^2}

A = \pi \left(\frac{0.060}{2}\right)^2 = \pi(0.030)^2 = 2.83 \times 10^{-3}\ \text{m}^2

C = \frac{2(0.481)}{(1.2)(2.83\times 10^{-3})(25^2)} = 0.454

Answer

$C = 0.45$

Final answer

0.45

Detailed explanation

Background Concept

A common model for turbulent drag on an object moving through a fluid is

F_D = \frac{1}{2} C \rho A v^2

where:

$F_D$ is drag force (opposes motion),
$C$ is the drag coefficient (dimensionless),
$\rho$ is the fluid density,
$A$ is the cross-sectional area perpendicular to the motion,
$v$ is the speed relative to the fluid.

To find $C$ , you rearrange the formula and substitute values. The cross-sectional area of a sphere (as seen from the direction of motion) is the area of a circle:

A = \pi r^2

Understanding the Question

You are given:

terminal speed $v = 25\ \text{m s}^{-1}$ ,
air density $\rho = 1.2\ \text{kg m}^{-3}$ ,
sphere diameter $d = 0.060\ \text{m}$ ,
and from part (c) the drag force $F_D$ .

You must use these to calculate the drag coefficient $C$ .

Approach

Rearrange the drag equation to make $C$ the subject.
Compute the cross-sectional area $A$ from the diameter.
Substitute $F_D$ , $\rho$ , $A$ , and $v$ into the rearranged expression.
Round to a sensible number of significant figures.

Step-by-Step Reasoning

Start with

F_D = \frac{1}{2}C\rho Av^2

Rearrange for $C$ :

C = \frac{2F_D}{\rho A v^2}

Find $A$ . The radius is half the diameter:

r = \frac{d}{2} = \frac{0.060}{2} = 0.030\ \text{m}

So

A = \pi r^2 = \pi(0.030)^2 = \pi(9.0\times 10^{-4}) = 2.83\times 10^{-3}\ \text{m}^2

Now substitute ( $F_D = 0.481\ \text{N}$ from part (c)):

C = \frac{2(0.481)}{(1.2)(2.83\times 10^{-3})(25^2)}

Calculate $25^2 = 625$ and evaluate the denominator:

(1.2)(2.83\times 10^{-3})(625) \approx 2.12

So

C \approx \frac{0.962}{2.12} = 0.454 \approx 0.45

Key Takeaways

For a sphere, use $A = \pi r^2$ where $r$ is half the diameter.
Rearranging to $C = \dfrac{2F_D}{\rho A v^2}$ is the quickest route.
Using consistent SI units throughout avoids unit errors.

Common Mistakes

Using the surface area of a sphere ( $4\pi r^2$ ) instead of the cross-sectional area ( $\pi r^2$ ).
Forgetting to square the speed ( $v^2$ ).
Using diameter as the radius in $A = \pi r^2$ .
Substituting $F_D$ in grams-force or using mass instead of force.

Things to Be Careful About

Check all quantities are already in SI base units (here they are).
Keep enough significant figures during intermediate steps (e.g. keep $F_D = 0.481\ \text{N}$ until the end).
Your final $C$ should be unitless; if you end up with units, a conversion or formula step is wrong.

Techniques used

calculate cross-sectional area from diameterrearrange an equation to make the coefficient the subjectsubstitute SI values consistently and evaluate

Q2MediumKinematics

(a)

Define velocity.

1M Similar

(b)

A student throws a ball over a vertical wall of height $h$ , as shown in Fig. 2.1.

Fig. 2.1

The ball leaves the hand of the student at a height of $1.2\text{ m}$ above the horizontal ground.
The ball has an initial velocity of $22\text{ m s}^{-1}$ at an angle of $40^\circ$ to the horizontal.
The wall is a horizontal distance of $36\text{ m}$ from where the student releases the ball.

Air resistance is negligible.

6M

(i)

Determine the time taken for the ball to reach the wall.

$\text{time taken}$ = ______ $\text{s}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Horizontal speed:

v_x = 22\cos 40^\circ = 16.9\ \text{m s}^{-1}

Time to reach wall:

t = \frac{36}{16.9} = 2.14\ \text{s}

Answer

$2.14\ \text{s}$

Final answer

2.14 s

Detailed explanation

Background Concept

In projectile motion with negligible air resistance:

horizontal acceleration is zero, so horizontal velocity is constant
vertical motion has constant acceleration $-g$

So you can treat the motion as two independent 1D motions:

Horizontal:

\text{distance} = v_x t

Vertical:

\text{displacement} = v_y t - \frac{1}{2}gt^2

Understanding the Question

The ball is thrown with speed $22\ \text{m s}^{-1}$ at $40^\circ$ above the horizontal. The wall is $36\ \text{m}$ horizontally from the release point. You need the time when the ball reaches that horizontal position.

Given:

horizontal distance to wall $x = 36\ \text{m}$
initial speed $22\ \text{m s}^{-1}$
launch angle $40^\circ$
air resistance negligible $\Rightarrow$ horizontal speed constant

Unknown: $t$ to reach the wall.

Approach

Find the horizontal component of the initial velocity: $v_x = 22\cos 40^\circ$ .
Use $x = v_x t$ to get $t = x/v_x$ .

Step-by-Step Reasoning

Resolve horizontally:

v_x = 22\cos 40^\circ

Horizontal motion has zero acceleration, so $v_x$ stays constant.
Use constant-speed relation:

36 = v_x t \quad \Rightarrow \quad t = \frac{36}{v_x}

Substitution gives $t \approx 2.14\ \text{s}$ .

Key Takeaways

For projectiles, horizontal motion is uniform (if air resistance is negligible).
Time to reach a horizontal position comes from $t = x/(u\cos\theta)$ .

Common Mistakes

Using $22\sin 40^\circ$ instead of $22\cos 40^\circ$ for horizontal speed.
Trying to use vertical equations to find the time (unnecessary here).
Forgetting that $36\ \text{m}$ is horizontal distance, not range or path length.

Things to Be Careful About

Keep units consistent (metres and seconds).
Calculator angle mode must be degrees.
Use sufficient significant figures in $v_x$ so $t$ is not rounded too early.

Techniques used

resolve the initial velocity into horizontal and vertical componentsuse constant horizontal velocity to find time from distance

(ii)

Calculate the vertical component $u$ of the initial velocity of the ball.

$u$ = ______ $\text{m s}^{-1}$

1M Similar

(iii)

The ball just goes over the wall.

Calculate the height $h$ of the wall.

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

3M Similar

DifficultyMedium

Worked solution

Working

At the wall, $t = 2.14\ \text{s}$ and $u_y = 14.1\ \text{m s}^{-1}$ .

Vertical displacement from release point:

\Delta y = u_y t - \frac{1}{2}gt^2

\Delta y = (14.1)(2.14) - \frac{1}{2}(9.81)(2.14)^2 = 7.83\ \text{m}

Height above ground at wall:

h = 1.2 + \Delta y = 1.2 + 7.83 = 9.03\ \text{m}

Answer

$9.0\ \text{m}$

Final answer

9.0 m

Detailed explanation

Background Concept

Projectile motion can be split into independent components:

Horizontal: constant velocity (no acceleration)
Vertical: constant acceleration downward of magnitude $g$

For vertical motion with upward positive:

\Delta y = u_y t + \frac{1}{2}at^2

Here $a = -g$ , so:

\Delta y = u_y t - \frac{1}{2}gt^2

The height above ground at time $t$ is then:

\text{height} = \text{initial height} + \Delta y

Understanding the Question

“The ball just goes over the wall” means that when the ball is horizontally level with the wall (at $x=36\ \text{m}$ ), its vertical position equals the top of the wall.

Given:

initial height above ground $= 1.2\ \text{m}$
launch speed $22\ \text{m s}^{-1}$ at $40^\circ$
wall is $36\ \text{m}$ away horizontally
negligible air resistance

You already found the time to reach the wall in (i), and the initial vertical component in (ii). Use these to find the vertical position at that time.

Approach

Use $t$ from part (i).
Use $u_y$ from part (ii).
Calculate vertical displacement from release point using

\Delta y = u_y t - \frac{1}{2}gt^2

Add the release height $1.2\ \text{m}$ to get the wall height $h$ .

Step-by-Step Reasoning

diagram-biii-1

Time to wall from horizontal motion: $t \approx 2.14\ \text{s}$ .
Vertical component of initial velocity: $u_y \approx 14.1\ \text{m s}^{-1}$ .
Take upward as positive. Acceleration is downward, so $a=-g=-9.81\ \text{m s}^{-2}$ .
Vertical displacement from the release point after time $t$ :

\Delta y = u_y t + \frac{1}{2}at^2 = u_y t - \frac{1}{2}gt^2

Substitute values:

\Delta y \approx (14.1)(2.14) - \frac{1}{2}(9.81)(2.14)^2 \approx 7.83\ \text{m}

This is the rise above the release point.

The ball started $1.2\ \text{m}$ above ground, so its height above ground at the wall is:

h = 1.2 + \Delta y \approx 1.2 + 7.83 \approx 9.03\ \text{m}

Since it “just goes over” the wall, this equals the wall height.

Key Takeaways

Use horizontal motion to find time, then substitute that time into vertical kinematics.
Always add the initial height if the projectile is launched above ground.
Sign convention: upward positive gives $a=-g$ .

Common Mistakes

Using $+\frac{1}{2}gt^2$ (wrong sign if upward is positive).
Forgetting to add the initial $1.2\ \text{m}$ height.
Using the horizontal component instead of the vertical component in the vertical equation.

Things to Be Careful About

Use the same time $t$ for both components (it’s the same motion).
Do not round $t$ and $u_y$ too aggressively before the final step.
Quote $h$ with sensible significant figures and include the unit $\text{m}$ .

Techniques used

use the time from horizontal motion in the vertical displacement equationapply constant-acceleration kinematics in the vertical directionadd the initial height to the vertical displacement to obtain wall height

Q3MediumDynamicsPhysical Quantities and UnitsKinematics

(a)

State the principle of conservation of momentum.

2M Similar

(b)

An object of mass $2m$ is travelling at a speed of $5.0\text{ m s}^{-1}$ in a straight line. It collides with an object of mass $3m$ which is initially stationary, as shown in Fig. 3.1.

Fig. 3.1

After the collision, the object of mass $2m$ moves with velocity $v$ at an angle of $30^\circ$ to its original direction of motion.

The object of mass $3m$ moves with velocity $w$ also at an angle of $30^\circ$ , as shown in Fig. 3.2.

Fig. 3.2

By considering the conservation of momentum in two dimensions, calculate the magnitudes of $v$ and $w$ .

$v$ = ______ $\text{m s}^{-1}$
$w$ = ______ $\text{m s}^{-1}$

4M Similar

DifficultyMedium

Worked solution

Working

Take $x$ along the original direction.

Initial momentum:

p_x = (2m)(5.0)=10m, \qquad p_y = 0

After collision:

p_x = (2m)v\cos 30^\circ + (3m)w\cos 30^\circ

p_y = (2m)v\sin 30^\circ - (3m)w\sin 30^\circ

Conservation of momentum in $y$ :

(2m)v\sin 30^\circ = (3m)w\sin 30^\circ \Rightarrow 2v=3w \Rightarrow v=\frac{3}{2}w

Conservation of momentum in $x$ :

(2m)v\cos 30^\circ + (3m)w\cos 30^\circ = 10m

Cancel $m$ and substitute $v=\frac{3}{2}w$ :

\cos 30^\circ \left(2\times \frac{3}{2}w + 3w\right)=10

6w\cos 30^\circ = 10 \Rightarrow w = \frac{10}{6\cos 30^\circ}=1.92\ \text{m s}^{-1}

v = \frac{3}{2}w = 2.89\ \text{m s}^{-1}

Answer

$v = 2.89\ \text{m s}^{-1}$

$w = 1.92\ \text{m s}^{-1}$

Final answer

v = 2.89 m s^-1, w = 1.92 m s^-1

Detailed explanation

Background Concept

Momentum is a vector, so in two dimensions you must conserve momentum separately in perpendicular directions (usually $x$ and $y$ ):

\sum p_x\,(\text{before}) = \sum p_x\,(\text{after}), \qquad \sum p_y\,(\text{before}) = \sum p_y\,(\text{after})

For an object of mass $m$ moving with speed $v$ at an angle $\theta$ to the $x$ -axis, its momentum components are:

p_x = mv\cos\theta, \qquad p_y = mv\sin\theta

Understanding the Question

Before collision:

mass $2m$ moves to the right at $5.0\ \text{m s}^{-1}$ ,
mass $3m$ is stationary.
So the initial momentum is entirely in the $x$ -direction.

After collision:

$2m$ moves at speed $v$ at $+30^\circ$ above the original line,
$3m$ moves at speed $w$ at $-30^\circ$ below the original line.

You must find the magnitudes $v$ and $w$ by conserving momentum in $x$ and $y$ .

Approach

Choose axes: $x$ along the original direction, $y$ perpendicular.
Write initial total momentum components.
Write final total momentum components using $\cos 30^\circ$ and $\sin 30^\circ$ .
Apply conservation in $x$ and in $y$ to get two equations.
Solve the simultaneous equations for $v$ and $w$ .

Step-by-Step Reasoning

diagram-3b-1

1. Initial momentum
Only the $2m$ object moves initially:

p_{x,\text{initial}} = (2m)(5.0)=10m

p_{y,\text{initial}} = 0

2. Final momentum components
For the $2m$ object at $+30^\circ$ :

p_{x1} = (2m)v\cos 30^\circ, \qquad p_{y1} = (2m)v\sin 30^\circ

For the $3m$ object at $-30^\circ$ (downwards $y$ component is negative):

p_{x2} = (3m)w\cos 30^\circ, \qquad p_{y2} = -(3m)w\sin 30^\circ

3. Conserve momentum in $y$
Because initial $y$ momentum is zero, the final $y$ components must cancel:

(2m)v\sin 30^\circ - (3m)w\sin 30^\circ = 0

Cancel $m\sin 30^\circ$ :

2v = 3w \Rightarrow v = \frac{3}{2}w

This is a common pattern: if the scattering is symmetric in angle, the $y$ -momentum balance directly relates the speeds.

4. Conserve momentum in $x$

(2m)v\cos 30^\circ + (3m)w\cos 30^\circ = 10m

Cancel $m$ and factor out $\cos 30^\circ$ :

\cos 30^\circ (2v + 3w)=10

Substitute $v = \frac{3}{2}w$ :

\cos 30^\circ \left(2\times \frac{3}{2}w + 3w\right) = \cos 30^\circ (6w) = 10

So

w = \frac{10}{6\cos 30^\circ}

Using $\cos 30^\circ = \frac{\sqrt{3}}{2}$ gives $w \approx 1.92\ \text{m s}^{-1}$ .
Then

v = \frac{3}{2}w \approx 2.89\ \text{m s}^{-1}

Key Takeaways

In 2D collisions, conserve momentum separately in perpendicular directions.
Assign signs carefully: upward $y$ positive, downward $y$ negative.
Symmetry in angles often makes the $y$ -equation especially simple.

Common Mistakes

Forgetting momentum is a vector and trying to conserve speeds instead of momentum components.
Using $\sin$ and $\cos$ the wrong way round for components.
Missing the negative sign for the downward $y$ -component of the $3m$ object.
Forgetting to multiply by the masses ( $2m$ and $3m$ ) when writing momentum.

Things to Be Careful About

The factor $m$ cancels, but only after you have correctly included it in every term.
Keep consistent axis choice: angles are given to the original direction, so choose $x$ along that direction.
Quote answers with units $\text{m s}^{-1}$ and sensible significant figures (matching given data, typically 2 or 3 s.f.).

Techniques used

apply conservation of momentumresolve momentum into perpendicular componentsset up simultaneous equations in two dimensionsuse trigonometric component relationships

(c)

An object of mass $4.2\text{ kg}$ is travelling in a straight line at a speed of $6.0\text{ m s}^{-1}$ . The object is brought to rest in a distance of $0.050\text{ m}$ by a constant force.

Calculate the magnitude of this force.

$\text{force}$ = ______ $\text{N}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Use

v^2 = u^2 + 2as

0 = (6.0)^2 + 2a(0.050) \Rightarrow a = -360\ \text{m s}^{-2}

Magnitude of force:

F = ma = (4.2)(360) = 1.5 \times 10^3\ \text{N}

Answer

$\text{force} = 1.5 \times 10^3\ \text{N}$

Final answer

1.5 × 10^3 N

Detailed explanation

Background Concept

For motion with constant acceleration, the SUVAT equation relating speeds, acceleration, and distance is:

v^2 = u^2 + 2as

where $u$ is initial speed, $v$ final speed, $a$ acceleration, and $s$ displacement along the line of motion.

Once $a$ is known, Newton's second law links resultant force and acceleration:

F = ma

If the question asks for the magnitude of the force, you give a positive value even if the acceleration is negative (deceleration).

Understanding the Question

Given:

mass $m = 4.2\ \text{kg}$
initial speed $u = 6.0\ \text{m s}^{-1}$
final speed $v = 0$ (comes to rest)
stopping distance $s = 0.050\ \text{m}$
constant force (so constant acceleration)

Find: magnitude of the force producing the stopping.

Approach

Use $v^2 = u^2 + 2as$ to find the acceleration $a$ .
Use $F = ma$ to find the force.
Take magnitude (ignore the negative sign of deceleration).

Step-by-Step Reasoning

1. Find the acceleration
Substitute into

v^2 = u^2 + 2as

with $v=0$ , $u=6.0$ , $s=0.050$ :

0 = 36 + 2a(0.050)

0 = 36 + 0.10a \Rightarrow a = -360\ \text{m s}^{-2}

The negative sign shows the acceleration is opposite to the motion (a deceleration).

2. Use Newton's second law

F = ma = (4.2)(-360) = -1512\ \text{N}

Magnitude:

|F| = 1512\ \text{N} \approx 1.5 \times 10^3\ \text{N}

Key Takeaways

Constant force implies constant acceleration, so SUVAT applies.
Deceleration gives a negative $a$ , but the force magnitude is positive.
Always keep units consistent (metres, seconds, kilograms).

Common Mistakes

Using $v$ instead of $v^2$ in the SUVAT equation.
Forgetting that $s = 0.050\ \text{m}$ is already in metres and accidentally converting incorrectly.
Giving a negative force when the question asks for magnitude.

Things to Be Careful About

The stopping distance is small, giving a large deceleration; this is physically plausible.
Significant figures: inputs are mostly 2 s.f., so $1.5 \times 10^3\ \text{N}$ is appropriate.

Techniques used

apply an equation of uniformly accelerated motionsolve for acceleration from kinematicsapply Newton's second law to find force magnitude

Q4Medium-EasyDeformation of Solids

(a)

Define strain.

1M Similar

(b)

A copper wire of length $4.0\text{ m}$ has a uniform cross-sectional area of $4.5 \times 10^{-7}\text{ m}^2$ .

A tensile force of $18\text{ N}$ is applied to the wire. This causes the wire to extend by $1.4\text{ mm}$ up to its limit of proportionality.

5M

(i)

Calculate the Young modulus of the wire.

$\text{Young modulus}$ = ______ $\text{Pa}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

\text{stress} = \frac{F}{A} = \frac{18}{4.5 \times 10^{-7}} = 4.0 \times 10^{7}\ \text{Pa}

\text{strain} = \frac{\Delta L}{L} = \frac{1.4 \times 10^{-3}}{4.0} = 3.5 \times 10^{-4}

E = \frac{\text{stress}}{\text{strain}} = \frac{4.0 \times 10^{7}}{3.5 \times 10^{-4}} = 1.14 \times 10^{11}\ \text{Pa}

Answer

E \approx 1.1 \times 10^{11}\ \text{Pa}

Final answer

1.1 × 10^11 Pa

Detailed explanation

Background Concept

For a wire under tension:

Stress is the force per unit cross-sectional area:

\sigma = \frac{F}{A}

Strain is the fractional extension:

\varepsilon = \frac{\Delta L}{L}

Young modulus $E$ measures stiffness and is defined (in the linear, proportional region) by:

E = \frac{\text{stress}}{\text{strain}} = \frac{\sigma}{\varepsilon}

This only applies up to the limit of proportionality (where stress is directly proportional to strain).

Understanding the Question

You are given:

$L = 4.0\ \text{m}$
$A = 4.5 \times 10^{-7}\ \text{m}^2$
$F = 18\ \text{N}$
extension $\Delta L = 1.4\ \text{mm} = 1.4 \times 10^{-3}\ \text{m}$

You must calculate the Young modulus of copper using these values.

Approach

Find the stress using $\sigma = F/A$ .
Find the strain using $\varepsilon = \Delta L/L$ .
Use $E = \sigma / \varepsilon$ .

(Equivalently, you can combine into one expression $E = FL/(A\Delta L)$ .)

Step-by-Step Reasoning

1) Stress

Convert nothing here except ensure SI units are used (they are).

\sigma = \frac{F}{A} = \frac{18}{4.5 \times 10^{-7}} = 4.0 \times 10^{7}\ \text{Pa}

2) Strain

Convert extension from mm to m:

\Delta L = 1.4\ \text{mm} = 1.4 \times 10^{-3}\ \text{m}

Then:

\varepsilon = \frac{\Delta L}{L} = \frac{1.4 \times 10^{-3}}{4.0} = 3.5 \times 10^{-4}

3) Young modulus

E = \frac{\sigma}{\varepsilon} = \frac{4.0 \times 10^{7}}{3.5 \times 10^{-4}} = 1.14 \times 10^{11}\ \text{Pa}

To suitable significant figures:

E \approx 1.1 \times 10^{11}\ \text{Pa}

Key Takeaways

Always convert to SI units before calculating $E$ .
$E$ comes from the linear region: $E = \sigma/\varepsilon$ .
For a given material, $E$ is (approximately) constant and does not depend on wire dimensions.

Common Mistakes

Forgetting to convert $1.4\ \text{mm}$ to $1.4 \times 10^{-3}\ \text{m}$ .
Using area in $\text{mm}^2$ or mixing units.
Calculating $E = \varepsilon/\sigma$ (inverted).

Things to Be Careful About

Stress has unit $\text{Pa}$ ; strain has no unit; therefore Young modulus has unit $\text{Pa}$ .
Quote the final answer to 2–3 significant figures (matching the given data).

Techniques used

calculate stress using force divided by cross-sectional areacalculate strain using extension divided by original lengthapply Young modulus as stress divided by strain

(ii)

On Fig. 4.1, draw a line to show how the stress varies with the strain for the wire up to its limit of proportionality.

Fig. 4.1

2M Similar

DifficultyMedium-Easy

Worked solution

Working

At limit of proportionality:

\sigma = \frac{F}{A} = \frac{18}{4.5 \times 10^{-7}} = 4.0 \times 10^{7}\ \text{Pa}

\varepsilon = \frac{\Delta L}{L} = \frac{1.4 \times 10^{-3}}{4.0} = 3.5 \times 10^{-4}

Answer

Draw a straight line through the origin to the point $\left(3.5 \times 10^{-4},\ 4.0 \times 10^{7}\ \text{Pa}\right)$ (i.e. $x=3.5$ on the $\text{strain}/10^{-4}$ axis and $y=4.0$ on the $\text{stress}/10^{7}\ \text{Pa}$ axis).

diagram-4bii

Final answer

Straight line through origin to (strain 3.5×10^-4, stress 4.0×10^7 Pa).

Detailed explanation

Background Concept

In the proportional (Hooke’s law) region for a metal wire:

\sigma \propto \varepsilon

So a graph of stress $\sigma$ against strain $\varepsilon$ is a straight line through the origin. The gradient of the stress–strain graph is the Young modulus:

E = \frac{\sigma}{\varepsilon}

The graph is only straight up to the limit of proportionality.

Understanding the Question

You are given the force, dimensions, and extension at the limit of proportionality. Figure 4.1 has scaled axes:

Vertical axis is stress in units of $10^{7}\ \text{Pa}$ .
Horizontal axis is strain in units of $10^{-4}$ .

You must draw the correct stress–strain line up to that limit.

Approach

Calculate the stress at the limit: $\sigma = F/A$ .
Calculate the strain at the limit: $\varepsilon = \Delta L/L$ .
Plot the point using the scaled axes.
Draw a straight line from the origin to this point.

Step-by-Step Reasoning

Compute the endpoint values:

\sigma = \frac{18}{4.5 \times 10^{-7}} = 4.0 \times 10^{7}\ \text{Pa}

\varepsilon = \frac{1.4 \times 10^{-3}}{4.0} = 3.5 \times 10^{-4}

Now convert these to the axis numbers:

On the $y$ -axis, stress is labelled as $\text{stress} / 10^{7}\ \text{Pa}$ , so $4.0 \times 10^{7}\ \text{Pa}$ corresponds to $y = 4.0$ .
On the $x$ -axis, strain is labelled as $\text{strain} / 10^{-4}$ , so $3.5 \times 10^{-4}$ corresponds to $x = 3.5$ .

So the required graph is a straight line from $(0,0)$ to $(3.5,4.0)$ .

diagram-4bii

Key Takeaways

In the proportional region, stress–strain is a straight line through the origin.
You must use the axis scaling correctly to place the endpoint.

Common Mistakes

Drawing a curve (that would represent beyond the proportional limit).
Not passing through the origin.
Plotting $(3.5,4.0)$ as if the axes were unscaled (forgetting the $10^{-4}$ and $10^{7}$ factors).

Things to Be Careful About

The extension is given at the limit of proportionality, so your line should stop at that point.
Use a ruler to draw a single straight line with a clear endpoint.
Ensure the point lies at $x = 3.5$ and $y = 4.0$ on the printed axes (not at $x=0.35$ or $y=0.40$ ).

Techniques used

calculate the endpoint values of stress and strain from given dataplot a straight-line relationship through the originread and use axis scaling factors to place points correctly

(c)

A second copper wire has the same length as the wire in (b) but a larger diameter. Both wires are subjected to a tensile force of $18\text{ N}$ .

By placing a tick (✓) in each row, complete Table 4.1 to compare the stress and strain of the two wires.

Table 4.1

	greater in second wire	less in second wire	the same in both wires
stress
strain

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

For the second wire, $A$ is larger.

\sigma = \frac{F}{A} \Rightarrow \text{stress is less}

Same material so same $E$ and

\varepsilon = \frac{\sigma}{E} \Rightarrow \text{strain is less}

So:

stress: less in second wire
strain: less in second wire

Final answer

Stress: less in second wire. Strain: less in second wire.

Detailed explanation

Background Concept

Two key definitions link force, geometry, and stretching:

\sigma = \frac{F}{A}

\varepsilon = \frac{\Delta L}{L}

For a given material in the proportional region, Hooke’s law in stress–strain form is:

\sigma = E\varepsilon

So if the material is the same, $E$ is the same, and strain is directly proportional to stress:

\varepsilon = \frac{\sigma}{E}

Understanding the Question

There are two copper wires of the same length. The second wire has a larger diameter, so it has a larger cross-sectional area $A$ . Both wires experience the same tensile force $F = 18\ \text{N}$ .

You must decide whether stress and strain are greater, less, or the same in the second wire compared to the first.

Approach

Use $\sigma = F/A$ to compare stresses when $A$ changes but $F$ is the same.
Then use $\varepsilon = \sigma/E$ (same material means same $E$ ) to compare strains.

Step-by-Step Reasoning

Stress comparison

Same force $F$ , but second wire has larger area $A$ .

\sigma = \frac{F}{A}

If $A$ increases, $\sigma$ decreases. Therefore stress is less in the second wire.

Strain comparison

Both are copper, so Young modulus $E$ is the same. In the proportional region:

\varepsilon = \frac{\sigma}{E}

Since $\sigma$ is less for the second wire, $\varepsilon$ is also less for the second wire.

Key Takeaways

Increasing diameter (area) reduces stress for the same applied force.
For the same material, lower stress implies lower strain (in the linear region).

Common Mistakes

Saying stress is the same because the force is the same (stress depends on area too).
Saying strain is the same because the length is the same (strain depends on extension, which changes with stress).
Confusing stress with strain.

Things to Be Careful About

The statement “same material” is crucial: it means Young modulus $E$ is the same, linking stress and strain directly.
The question does not ask about extension directly, but note that smaller strain would also mean smaller extension because $\Delta L = \varepsilon L$ and $L$ is the same for both wires.

Techniques used

relate stress to force and cross-sectional areause Hooke's law relation between stress and strain for the same materialcompare quantities qualitatively when one geometric parameter changes

Q5MediumSuperpositionWaves

A stretched string PQ has length $1.2\text{ m}$ . One end of the string is attached to a vibration generator and the other end is attached to a wall, as shown in Fig. 5.1.

Fig. 5.1

The vibration generator is switched on and a stationary wave is formed on the string. The string is shown at one instant of time in Fig. 5.2.

Fig. 5.2

(a)

Explain how a stationary wave is formed between the vibration generator and the wall.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Waves from the generator travel along the string to the wall and are reflected back.
The incident and reflected waves (same frequency) travel in opposite directions and superpose, producing fixed nodes (zero displacement) and antinodes (maximum displacement), i.e. a stationary wave.

Final answer

Incident wave reflects at the wall; incident and reflected waves superpose to form nodes and antinodes (stationary wave).

Detailed explanation

Background Concept

A stationary wave on a string forms when two progressive waves of the same frequency (and usually similar amplitude) travel in opposite directions along the same line.

By the principle of superposition, the resultant displacement at any point is the sum of the displacements due to each wave.

At some points the two waves always cancel: nodes (zero displacement at all times).
At other points they always add or subtract to the largest possible magnitude: antinodes (maximum amplitude).

A reflection at a fixed end (the wall) produces a reflected wave that is inverted in displacement compared with the incident wave, but it has the same frequency.

Understanding the Question

The generator at one end produces a progressive wave on the stretched string. The other end is attached to a wall (a fixed boundary), so the wave reflects. The question asks for the physical mechanism (reflection + superposition) that leads to a stationary-wave pattern between the generator and wall.

Approach

State that the generator sends a progressive wave along the string.
State that the wave reflects at the wall.
Explain that the incident and reflected waves travel in opposite directions and superpose.
Conclude that repeated constructive/destructive interference makes fixed nodes and antinodes (stationary wave).

Step-by-Step Reasoning

The vibration generator produces a sinusoidal disturbance, so a progressive transverse wave travels from P to Q.
When this wave reaches the wall at Q, it cannot move the wall: Q is effectively a fixed end, so the wave is reflected back toward P.
Now there are two waves on the same string segment: one travelling right and one travelling left.
At each position, the instantaneous displacement is the algebraic sum of the two displacements.
- Where the two displacements are always equal and opposite, the resultant displacement is always zero: a node.
- Where they are always in the same direction, the resultant has maximum magnitude: an antinode.
Because this pattern stays at fixed positions, it is a stationary wave.

Key Takeaways

Stationary waves come from superposition of two waves of the same frequency travelling in opposite directions.
Reflection at a fixed end provides one of the counter-propagating waves.
Nodes and antinodes are the hallmark of stationary waves.

Common Mistakes

Saying “the wave stops” at the wall: it reflects; energy is not simply lost (though some may be absorbed).
Describing only reflection but not superposition/interference.
Confusing a stationary wave with a travelling wave (stationary pattern vs moving crests).

Things to Be Careful About

Use the correct terms incident, reflected, superposition, nodes, antinodes.
Make clear the waves travel in opposite directions and have the same frequency.

Techniques used

describe reflection of a progressive wave at a fixed endapply the principle of superposition to two oppositely travelling waveslink interference to the formation of nodes and antinodes

(b)

Calculate the wavelength of the stationary wave shown in Fig. 5.2.

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

1M Similar

DifficultyEasy

Worked solution

Working

Three loops are shown, so

L = 3\left(\frac{\lambda}{2}\right)

1.2 = \frac{3\lambda}{2} \;\Rightarrow\; \lambda = \frac{2 \times 1.2}{3} = 0.80

Answer

$\lambda = 0.80\ \text{m}$

Final answer

0.80 m

Detailed explanation

Background Concept

For a string with nodes at both ends (fixed ends), a stationary wave consists of a number of equal “loops”. Each loop is the distance between two adjacent nodes, and that distance is half a wavelength:

\text{one loop length} = \frac{\lambda}{2}

If there are $n$ loops along a length $L$ , then

L = n\left(\frac{\lambda}{2}\right)

Understanding the Question

The string length is $L = 1.2\ \text{m}$ . The diagram shows a stationary wave pattern with three distinct loops (three bulges) between P and Q. The task is to find the wavelength $\lambda$ of the underlying progressive waves.

Approach

Count the number of loops between the fixed ends.
Use $L = n(\lambda/2)$ .
Rearrange for $\lambda$ .

Step-by-Step Reasoning

From the diagram, there are $n=3$ loops between P and Q.
Each loop corresponds to a half-wavelength, so total length is

L = 3\left(\frac{\lambda}{2}\right)

Substitute $L = 1.2\ \text{m}$ :

1.2 = \frac{3\lambda}{2}

Solve for $\lambda$ :

\lambda = \frac{2 \times 1.2}{3} = 0.80\ \text{m}

Key Takeaways

In a stationary wave on a string, node-to-node distance is $\lambda/2$ .
Counting loops is often the fastest way to relate $L$ to $\lambda$ .

Common Mistakes

Treating one loop as one full wavelength (it is $\lambda/2$ ).
Miscounting loops or forgetting that both ends are nodes for a fixed end.

Things to Be Careful About

Ensure you are counting complete loops between adjacent nodes.
Quote the wavelength with unit (m).

Techniques used

count the number of loops (segments) between fixed endsuse the relation between loop length and half-wavelengthrearrange to determine wavelength from string length

(c)

Fig. 5.3 shows the stationary wave at time $t = 0$ when all points on the wave are at their maximum displacements.

Fig. 5.3

The period of the wave is $0.16\text{ s}$ .

On Fig. 5.3, sketch the shape of the stationary wave at time $t = 0.24\text{ s}$ .

2M Similar

DifficultyMedium

Worked solution

Working

T = 0.16\ \text{s},\quad t = 0.24\ \text{s} = 1.5T

After $1.5T$ the wave is the same as after $0.5T$ , so all displacements are reversed.

diagram-5c-1

Answer

Sketch the same three-loop shape as at $t=0$ but inverted about the equilibrium line (crests become troughs, troughs become crests; nodes unchanged).

Final answer

Inverted stationary-wave shape (same as t=0 but reflected in the equilibrium line).

Detailed explanation

Background Concept

In a stationary wave, each point on the string oscillates in simple harmonic motion (SHM) with the same frequency as the driving generator, but the amplitude depends on position:

Nodes: amplitude $0$ (never move).
Antinodes: maximum amplitude.

The time dependence at any fixed position is SHM. After half a period ( $T/2$ ), the displacement at every oscillating point is reversed (because SHM is sinusoidal):

x(t + T/2) = -x(t)

So a snapshot of the string at $t = T/2$ looks like the original snapshot flipped about the equilibrium line.

Understanding the Question

At $t=0$ , the diagram shows the string when all points are at maximum displacement (so each loop is at its extreme). The period is $T = 0.16\ \text{s}$ . We must sketch the shape at $t = 0.24\ \text{s}$ .

Approach

Express $0.24\ \text{s}$ as a multiple of $T$ .
Use periodicity: adding a whole number of periods returns the same shape.
Recognise that an extra half-period inverts the displacement.
Draw the inverted pattern; keep nodes in the same positions.

Step-by-Step Reasoning

Calculate how many periods $t$ corresponds to:

\frac{t}{T} = \frac{0.24}{0.16} = 1.5

So $t = 1.5T = T + 0.5T$ .

After one full period $T$ , the wave shape repeats exactly.
The remaining extra time is $0.5T$ $0.5 T$ (half a period). In SHM, after half a period, every displacement changes sign:
- a crest becomes a trough of the same magnitude,
- a trough becomes a crest,
- nodes remain at zero.
Therefore, the required sketch is the same three-loop pattern but flipped vertically about the equilibrium line.

diagram-5c-1

Key Takeaways

Convert the given time into a fraction/multiple of the period.
A stationary-wave snapshot after $T/2$ is the inverted snapshot (nodes unchanged).

Common Mistakes

Treating $0.24\ \text{s}$ as $0.24/0.16 = 0.24T$ instead of $1.5T$ .
Redrawing a travelling wave (shifted sideways). A stationary wave does not move sideways; its nodes stay fixed.
Changing node positions: nodes must remain at the same locations.

Things to Be Careful About

Use the identity: shape repeats every $T$ , but is inverted after $T/2$ .
Ensure the number of loops is unchanged and the endpoints/nodes stay fixed.

Techniques used

convert time into a number of periodsuse the phase of SHM at a given fraction of a periodapply stationary-wave behaviour at all points simultaneously (sign reversal at half-period)

(d)

Points R and T on the string are a horizontal distance of $0.30\text{ m}$ apart and in the positions shown in Fig. 5.4.

Fig. 5.4

State the phase difference between the oscillations of points R and T.

$\text{phase difference}$ = ______ $^\circ$

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

R and T are in adjacent loops (separated by a node), so they oscillate in antiphase.

$\text{phase difference} = 180^\circ$

Final answer

180°

Detailed explanation

Background Concept

In a stationary wave on a string, all points between a pair of adjacent nodes oscillate in phase (they reach maxima and cross equilibrium together). However, neighbouring segments (on opposite sides of a node) oscillate in antiphase:

When one segment is above the equilibrium line, the neighbouring segment is below it.
This corresponds to a phase difference of $180^\circ$ .

Understanding the Question

The diagram marks two points:

R is at a crest in one loop.
T is at a trough in the next loop.
They are separated by a node between those loops. The question asks for the phase difference between their oscillations.

Approach

Decide whether R and T are in the same loop or in different loops.
Use the stationary-wave phase rule: same loop (\to 0^\circ), adjacent loops (\to 180^\circ).
State the phase difference in degrees.

Step-by-Step Reasoning

In the given stationary-wave pattern, nodes separate loops.
R lies in one loop and T lies in the neighbouring loop, with a node in between.
Adjacent loops oscillate in antiphase, so when R has maximum positive displacement, T has maximum negative displacement.
Therefore the phase difference is

180^\circ

Key Takeaways

Same segment between nodes: phase difference $0^\circ$ .
Adjacent segments across a node: phase difference $180^\circ$ .

Common Mistakes

Using progressive-wave phase rules based on separation distance (e.g. (\Delta x/\lambda)) without noticing this is a stationary wave.
Saying $360^\circ$ (which is actually in phase, not antiphase).

Things to Be Careful About

The key clue is whether there is a node between the points.
Phase difference is about timing of oscillations, not about the instantaneous shape alone (though the shape helps you see antiphase).

Techniques used

identify whether points lie in the same or adjacent stationary-wave loopsuse the phase relationship between oscillations in adjacent segmentsstate phase difference in degrees

(e)

Calculate the speed of the progressive waves on the stretched string.

$\text{speed}$ = ______ $\text{m s}^{-1}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

f = \frac{1}{T} = \frac{1}{0.16} = 6.25\ \text{Hz}

With $\lambda = 0.80\ \text{m}$ ,

v = f\lambda = 6.25 \times 0.80 = 5.0\ \text{m s}^{-1}

Answer

$5.0\ \text{m s}^{-1}$

Final answer

5.0 m s⁻¹

Detailed explanation

Background Concept

The speed $v$ of a progressive wave is related to its frequency $f$ and wavelength $\lambda$ by the wave equation:

v = f\lambda

Also, frequency and period are reciprocals:

f = \frac{1}{T}

On a string, the stationary wave pattern is produced by these progressive waves, so the wavelength found from the stationary wave is the wavelength of the progressive waves.

Understanding the Question

You are given the period $T = 0.16\ \text{s}$ of the wave and you can obtain the wavelength from the stationary-wave pattern (from part (b), $\lambda = 0.80\ \text{m}$ ). The question asks for the wave speed on the string.

Approach

Convert period to frequency using $f = 1/T$ .
Use $v = f\lambda$ (or directly $v = \lambda/T$ ).
Substitute values with correct units.

Step-by-Step Reasoning

Find the frequency:

f = \frac{1}{T} = \frac{1}{0.16\ \text{s}} = 6.25\ \text{Hz}

Use the wave equation:

v = f\lambda

Substitute $\lambda = 0.80\ \text{m}$ :

v = 6.25 \times 0.80 = 5.0\ \text{m s}^{-1}

(Equivalently, $v = \lambda/T = 0.80/0.16 = 5.0\ \text{m s}^{-1}$ .)

Key Takeaways

$v = f\lambda$ links wave speed to time information (period/frequency) and space information (wavelength).
The wavelength measured from a stationary wave belongs to the underlying progressive waves.

Common Mistakes

Using $v = \lambda f$ but substituting $T$ instead of $f$ without taking the reciprocal.
Using the wrong wavelength (e.g. confusing loop length with full wavelength).
Missing units or giving $\text{m s}^{-2}$ instead of $\text{m s}^{-1}$ .

Things to Be Careful About

Keep everything in SI units (seconds and metres are already SI here).
Quote the final answer to a sensible number of significant figures (here $5.0\ \text{m s}^{-1}$ ).

Techniques used

use frequency-period relation to find frequencyapply the wave equation v = f lambdasubstitute wavelength and period consistently in SI units

Q6MediumD.C. Circuits

(a)

State Kirchhoff’s first law.

1M Similar

(b)

A cell with internal resistance $r$ is connected to two resistors of resistances $R_1$ and $R_2$ as shown in Fig. 6.1.

Fig. 6.1

The potential differences (p.d.s) across $R_1$ and $R_2$ are $V_1$ and $V_2$ respectively.
The terminal p.d. across the cell is $V$ .
The current in the circuit is $I$ .

Use Kirchhoff’s laws to show that the total resistance $R_T$ of the external circuit is given by

R_T = R_1 + R_2

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using Kirchhoff’s second law for the external circuit,

V = V_1 + V_2

For series resistors, the current is the same $= I$ , so

V_1 = IR_1, \quad V_2 = IR_2

Hence

V = I(R_1 + R_2)

But $R_T = V/I$ , so

R_T = R_1 + R_2

Answer

$R_T = R_1 + R_2$

Final answer

RT = R1 + R2

Detailed explanation

Background Concept

Kirchhoff’s second law (KVL) states that the algebraic sum of potential differences around any closed loop is zero. In a series circuit, the same current $I$ flows through each component. Ohm’s law for a resistor is $V = IR$ .

Understanding the Question

In Fig. 6.1, the two external resistors $R_1$ and $R_2$ are connected in series. Their p.d.s are $V_1$ and $V_2$ , and the terminal p.d. across the cell (i.e. across the external circuit) is $V$ . You must use Kirchhoff’s laws to show the external total resistance $R_T$ equals $R_1 + R_2$ .

Approach

Use KVL around the loop containing the external resistors to relate $V$ to $V_1$ and $V_2$ .
Use the fact that series components carry the same current, then apply Ohm’s law to write $V_1$ and $V_2$ in terms of $I$ .
Rearrange into the form $V = IR_T$ and identify $R_T$ .

Step-by-Step Reasoning

Around the external part of the circuit, the total drop across both resistors equals the terminal p.d.:

V = V_1 + V_2

Since $R_1$ and $R_2$ are in series, the same current $I$ flows through each.
Apply Ohm’s law separately:

V_1 = IR_1, \qquad V_2 = IR_2

Substitute into the KVL equation:

V = IR_1 + IR_2 = I(R_1 + R_2)

By definition, the equivalent (total) external resistance is $R_T = V/I$ , so comparing with $V = IR_T$ gives:

R_T = R_1 + R_2

Key Takeaways

In series: same current; p.d.s add.
KVL gives the sum of voltage drops around a loop.
Equivalent resistance in series is the sum of resistances.

Common Mistakes

Using Kirchhoff’s first law at a junction (there is no junction in a simple series circuit).
Writing $V_1 = V_2 = V$ for series resistors (incorrect: voltages divide).
Forgetting that $R_T$ is for the external circuit, not including internal resistance $r$ .

Things to Be Careful About

Make clear that $V$ is the terminal p.d. across the external network.
Use consistent symbols: $V_1$ across $R_1$ , $V_2$ across $R_2$ , current $I$ through both.

Techniques used

apply Kirchhoff's second law around a loopuse the same current through series componentssubstitute Ohm's law into a potential difference equationidentify total resistance as terminal p.d. divided by current

(c)

The electromotive force (e.m.f.) of the cell in Fig. 6.1 is $1.50\text{ V}$ .

The values of $R_1$ and $R_2$ are $10\ \Omega$ and $15\ \Omega$ respectively. The terminal p.d. of the cell is $1.35\text{ V}$ .

Calculate the internal resistance $r$ of the cell.

$r$ = ______ $\Omega$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

External resistance:

R_T = R_1 + R_2 = 10 + 15 = 25\ \Omega

Current:

I = \frac{V}{R_T} = \frac{1.35}{25} = 5.40 \times 10^{-2}\ \text{A}

Using $\mathcal{E} = V + Ir$ :

r = \frac{\mathcal{E} - V}{I} = \frac{1.50 - 1.35}{5.40 \times 10^{-2}} = 2.78\ \Omega

Answer

$r = 2.8\ \Omega$

Final answer

2.8 Ω

Detailed explanation

Background Concept

A real cell can be modelled as an ideal source of e.m.f. $\mathcal{E}$ in series with an internal resistance $r$ . When a current $I$ is delivered, some energy is dissipated inside the cell, giving an internal voltage drop $Ir$ .

The terminal p.d. $V$ (the p.d. across the external circuit) is then

V = \mathcal{E} - Ir

or equivalently $\mathcal{E} = V + Ir$ .

Understanding the Question

You are given:

$\mathcal{E} = 1.50\ \text{V}$
$R_1 = 10\ \Omega$ , $R_2 = 15\ \Omega$ in series, so they form the external load
terminal p.d. $V = 1.35\ \text{V}$
You must find the internal resistance $r$ .

Approach

Combine the external resistors to get the total external resistance $R_T$ .
Use the terminal p.d. across the external circuit: $I = V/R_T$ .
Use $\mathcal{E} - V = Ir$ to solve for $r$ .

Step-by-Step Reasoning

Series combination:

R_T = R_1 + R_2 = 10 + 15 = 25\ \Omega

The terminal p.d. $V$ is across the external circuit, so by Ohm’s law:

I = \frac{V}{R_T} = \frac{1.35}{25} = 0.0540\ \text{A}

The “lost volts” across the internal resistance are:

\mathcal{E} - V = 1.50 - 1.35 = 0.15\ \text{V}

Since this equals $Ir$ :

r = \frac{\mathcal{E} - V}{I} = \frac{0.15}{0.0540} = 2.78\ \Omega \approx 2.8\ \Omega

Key Takeaways

Terminal p.d. is across the external load only.
Lost volts $= \mathcal{E} - V = Ir$ .
Find $I$ from the external circuit, then deduce $r$ .

Common Mistakes

Using $I = \mathcal{E}/R_T$ instead of $I = V/R_T$ (only true if $r=0$ ).
Forgetting to add $R_1$ and $R_2$ for series.
Writing $V = \mathcal{E} + Ir$ (sign error).

Things to Be Careful About

Quote $r$ in $\Omega$ .
Use a sensible rounding: inputs are mostly 2–3 s.f., so $r \approx 2.8\ \Omega$ is appropriate.

Techniques used

calculate external total resistance for series resistorsuse terminal p.d. to find circuit currentuse the internal resistance relation E = V + Irrearrange to solve for internal resistance

(d)

A resistor of resistance $R_3$ is added to the circuit in Fig. 6.1, so that the circuit is as shown in Fig. 6.2.

Fig. 6.2

State and explain the effect, if any, of this change on:

4M

(i)

the current in the cell

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The current in the cell increases.

Adding $R_3$ in parallel reduces the total external resistance, so the total circuit resistance $(r + R_{\text{ext}})$ decreases and hence

I = \frac{\mathcal{E}}{r + R_{\text{ext}}}

increases.

Final answer

Increases

Detailed explanation

Background Concept

For a cell of e.m.f. $\mathcal{E}$ and internal resistance $r$ , supplying an external equivalent resistance $R_{\text{ext}}$ , the circuit current is

I = \frac{\mathcal{E}}{r + R_{\text{ext}}}

Adding resistors in parallel decreases the equivalent resistance because there are more paths for charge to flow.

Understanding the Question

Originally, $R_1$ and $R_2$ are in series. Then a resistor $R_3$ is added in parallel with the series combination of $R_1$ and $R_2$ (Fig. 6.2). You must state and explain what happens to the current drawn from the cell.

Approach

Compare the equivalent external resistance before and after adding $R_3$ .
Use the relationship between total resistance and current for a cell with internal resistance.

Step-by-Step Reasoning

diagram-d-i-1

Before adding $R_3$ , the external resistance is

R_{\text{ext,1}} = R_1 + R_2

After adding $R_3$ in parallel with $(R_1+R_2)$ , the new external equivalent resistance is

R_{\text{ext,2}} = \left(\frac{1}{R_1 + R_2} + \frac{1}{R_3}\right)^{-1}

A parallel combination always gives a smaller resistance than the smallest branch resistance, so

R_{\text{ext,2}} < R_{\text{ext,1}}

The total resistance seen by the e.m.f. becomes $r + R_{\text{ext}}$ . Since $R_{\text{ext}}$ decreases while $r$ is unchanged, the denominator in

I = \frac{\mathcal{E}}{r + R_{\text{ext}}}

decreases, so the current $I$ increases.

Key Takeaways

Adding a parallel path reduces equivalent resistance.
With fixed $\mathcal{E}$ and fixed $r$ , smaller total resistance means larger current.

Common Mistakes

Saying current decreases because “more components” are added (not true for parallel additions).
Treating $R_3$ as if it were in series with $R_1$ and $R_2$ .

Things to Be Careful About

The question asks for current in the cell (total current supplied), not the current in a particular branch.
Internal resistance $r$ does not change when you add external resistors.

Techniques used

determine how a parallel resistor changes equivalent resistancerelate total circuit resistance to current using I = E / (Rext + r)apply qualitative proportional reasoning between resistance and current

(ii)

the terminal p.d. of the cell.

2M Similar

DifficultyMedium

Worked solution

Answer

The terminal p.d. decreases.

With internal resistance $r$ ,

V = \mathcal{E} - Ir

Adding $R_3$ decreases external resistance so $I$ increases (part (i)), hence the drop $Ir$ increases and $V$ becomes smaller.

Final answer

Decreases

Detailed explanation

Background Concept

For a cell with internal resistance $r$ , the terminal p.d. across the external circuit is

V = \mathcal{E} - Ir

The term $Ir$ is the voltage lost inside the cell due to internal resistance (energy dissipated internally).

Understanding the Question

After adding $R_3$ in parallel with the existing external resistors, you must describe what happens to the terminal p.d. of the cell and explain why.

Approach

Use the conclusion that the current drawn from the cell increases when external resistance decreases.
Substitute this into the terminal p.d. relation $V = \mathcal{E} - Ir$ to predict the effect on $V$ .

Step-by-Step Reasoning

diagram-d-ii-1

Adding $R_3$ in parallel reduces the equivalent external resistance, so the current in the cell increases.
The internal voltage drop is $Ir$ . If $I$ increases while $r$ is constant, then $Ir$ increases.
Since

V = \mathcal{E} - Ir,

a larger $Ir$ means a smaller terminal p.d. $V$ .

So the terminal p.d. decreases.

Key Takeaways

Terminal p.d. falls as current increases (for fixed internal resistance).
Any change that reduces external resistance tends to increase current and therefore increase “lost volts”.

Common Mistakes

Saying terminal p.d. stays equal to e.m.f. (only true when $I=0$ ).
Claiming $V$ increases because “more current flows” without considering the internal voltage drop.

Things to Be Careful About

The e.m.f. $\mathcal{E}$ is a property of the cell and is unchanged by changing the external circuit (assuming no significant depletion).
Be explicit that $r$ is internal and unchanged; it is the increase in $I$ that changes $Ir$ and hence $V$ .

Techniques used

use V = E - Ir for terminal potential differencelink increased current to increased internal voltage dropdeduce qualitative change in terminal p.d.

Q7MediumParticle Physics

Nuclei of an isotope of copper (Cu) each have 29 protons and 37 neutrons. This isotope is a $\beta^-$ emitter.

(a)

State the nuclide notation in the form ${}^A_Z X$ for this nucleus of copper.

1M Similar

(b)

The energy spectrum of the $\beta^-$ radiation emitted by a sample of this isotope is shown in Fig. 7.1.

Fig. 7.1

7M

(i)

Use Fig. 7.1 to explain why other particles apart from the $\beta^-$ particles must be emitted during this decay.

3M Similar

DifficultyMedium

Worked solution

Answer

diagram-b-i-spectrum

The $\beta^-$ particles have a range of kinetic energies from $0$ up to a maximum value (continuous spectrum).

If only the daughter nucleus and the $\beta^-$ particle were produced, conservation of energy and momentum would give the $\beta^-$ a single fixed kinetic energy.

Since the energy varies, some energy (and momentum) must be carried away by another emitted particle.

Final answer

Continuous range of beta energies implies energy/momentum shared with another emitted particle.

Detailed explanation

Background Concept

In radioactive decay, quantities that must be conserved include:

total energy (including kinetic energy of products and any rest-mass energy changes),
momentum.

For a simple two-body decay (parent at rest decays into exactly two products), conservation of momentum fixes the magnitudes of the momenta of the two products, and therefore their kinetic energies are fixed (a single value, not a spread).

In $\beta^-$ decay specifically, a neutron in the nucleus changes to a proton and emits an electron ( $\beta^-$ ). In reality, a (anti)neutrino is also emitted, allowing energy and momentum to be shared in different proportions.

Understanding the Question

Fig. 7.1 shows the number of emitted $\beta^-$ particles against their kinetic energy. The curve is not a single line at one energy; it is spread out from near $0$ up to a maximum (endpoint) energy.

The question asks you to use that fact to argue that more than just $\beta^-$ particles must be emitted.

Approach

Identify the key observation from the graph: the electron kinetic energy is continuous from $0$ to a maximum.
Use the principle: two-body decay gives a fixed kinetic energy for one product.
Conclude: at least one additional particle must be taking a variable share of the available energy and momentum.

Step-by-Step Reasoning

diagram-b-i-spectrum

From the spectrum, electrons are detected with many different kinetic energies (not one value).
Suppose only two products were emitted: the daughter nucleus and the electron.
- If the parent nucleus is initially at rest, momentum conservation requires the electron and the daughter nucleus to have equal and opposite momenta.
- With only two bodies, that momentum magnitude is fixed, so their kinetic energies would also be fixed.
- Therefore the electron would have a single discrete kinetic energy.
But the graph shows a continuous range, so the electron does not always take the same share of the available decay energy.
Hence there must be at least one other emitted particle that can carry away the “missing” energy and momentum in varying amounts.

Key Takeaways

A continuous $\beta$ spectrum is evidence that decay energy is shared between more than two products.
Conservation of momentum is the key reason a two-body decay would give a single electron energy.

Common Mistakes

Saying “energy is not conserved” (energy is conserved; it is shared among products).
Referring only to conservation of charge/nucleon number (true but doesn’t explain the spectrum shape).
Claiming the spread is due to measurement error; the spread is a real physical feature of $\beta$ decay.

Things to Be Careful About

The maximum (endpoint) energy corresponds to the case where the other particle takes almost zero kinetic energy (and the recoil is minimal), but most decays share energy differently.
The daughter nucleus recoil kinetic energy is usually small but not the main explanation for the wide continuous spectrum.

Techniques used

interpret a continuous beta energy spectrumapply conservation of energy to nuclear decayapply conservation of momentum to infer missing particles

(ii)

State the name of the other particle emitted during the decay of this isotope.

1M Similar

(iii)

The copper isotope decays to an isotope of zinc (Zn).

Give the radioactive decay equation for this decay. Include the nucleon and proton numbers of all the particles involved.

3M Similar

DifficultyMedium-Easy

Worked solution

Working

For $\beta^-$ decay, $A$ stays the same and $Z$ increases by $1$ .

Initial nucleus: $A = 66$ , $Z = 29$ .
So daughter zinc nucleus has $A = 66$ , $Z = 30$ .

Answer

{}^{66}_{29}\text{Cu} \rightarrow {}^{66}_{30}\text{Zn} + {}^{0}_{-1}e + {}^{0}_{0}\bar{\nu}_e

Final answer

${}^{66}_{29}\text{Cu} \rightarrow {}^{66}_{30}\text{Zn} + {}^{0}_{-1}e + {}^{0}_{0}\bar{\nu}_e$

Detailed explanation

Background Concept

A nuclear decay equation must conserve:

nucleon number $A$ (total number of protons + neutrons),
proton number $Z$ (charge number).

In $\beta^-$ decay, a neutron in the nucleus converts to a proton and emits an electron and an electron antineutrino:

n \rightarrow p + e^- + \bar{\nu}_e

This means for the nucleus:

$A$ stays the same (one nucleon is still one nucleon),
$Z$ increases by $1$ (a neutron becomes a proton).

Understanding the Question

The copper isotope ( ${Z=29}$ ) decays by $\beta^-$ emission into an isotope of zinc. You must write the full decay equation including nucleon and proton numbers for every particle.

Approach

Find the original $A$ and $Z$ for copper.
Use the $\beta^-$ rules: $A$ unchanged, $Z \to Z+1$ .
Write the balanced equation including $e^-$ and $\bar{\nu}_e$ with their correct ${}^A_Z$ values.

Step-by-Step Reasoning

Original copper nucleus has:

A = 29 + 37 = 66, \quad Z = 29

so it is ${}^{66}_{29}\text{Cu}$ .

After $\beta^-$ decay:
- Nucleon number unchanged: $A = 66$ .
- Proton number increases by 1: $Z = 30$ .
- Element with $Z=30$ is zinc, so the daughter is ${}^{66}_{30}\text{Zn}$ .
The emitted electron has:

{}^{0}_{-1}e

(nucleon number 0, charge number $-1$ ).

The emitted antineutrino has:

{}^{0}_{0}\bar{\nu}_e

(0 nucleons, 0 charge).

Putting this together:

{}^{66}_{29}\text{Cu} \rightarrow {}^{66}_{30}\text{Zn} + {}^{0}_{-1}e + {}^{0}_{0}\bar{\nu}_e

Check conservation:

$A$ : $66 = 66 + 0 + 0$
$Z$ : $29 = 30 + (-1) + 0$

Key Takeaways

In $\beta^-$ decay: $A$ unchanged, $Z$ increases by 1.
Always include the electron and the (anti)neutrino with correct ${}^A_Z$ labels.

Common Mistakes

Decreasing $Z$ (that would be $\beta^+$ decay).
Changing $A$ (not correct for $\beta$ decay).
Writing ${}^{0}_{+1}e$ (positron) instead of ${}^{0}_{-1}e$ .
Omitting the antineutrino when asked for “all particles involved”.

Things to Be Careful About

Use the correct nuclide symbols: Cu and Zn.
Ensure both $A$ and $Z$ are shown for every particle, including $e^-$ and $\bar{\nu}_e$ .

Techniques used

apply nucleon number conservation in nuclear equationsapply charge (proton number) conservation in nuclear equationswrite a balanced $\beta^-$ decay equation including the (anti)neutrino

Physics 9702/21 — May/June 2024

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