IB Physics HL Fast Paper 2 Shortcuts

Waves, Electricity, Fields, Nuclear Physics, and Thermal Physics

This premium IB Physics HL lesson brings together a set of high-value Paper 2 shortcuts that help students solve common questions more quickly and more accurately. These methods are not substitutes for understanding. Instead, they are exam-speed frameworks built on correct physical reasoning.

The lesson focuses on the kinds of questions that repeatedly appear in IB Physics:

standing waves and microwave ovens
block-on-block friction problems
electricity and circuit shortcuts
field relationships
radioactive decay and ionization current
calorimetry and phase change

⭐ Key Concepts

In standing waves, adjacent nodes or adjacent antinodes are separated by $\frac{\lambda}{2}$ .
When two blocks move together, the upper block is accelerated by static friction.
In circuit problems, choosing the correct form of the power equation saves time.
In electric and gravitational field questions, begin with the force-field relationship.
In half-life questions, use the fraction form $\left(\frac12\right)^n$ rather than repeated subtraction.
In ionization chamber questions, current is found from charge produced per second.
In calorimetry, always identify every thermal stage in order before writing the energy balance.

📘 IB Command Terms

State – give a specific answer without explanation.

Determine – obtain an answer using the relevant information.

Calculate – work out a numerical answer, showing the main steps.

Explain – give a detailed account using correct physics ideas.

Outline – give a brief summary.

Suggest – propose a reasonable answer based on scientific understanding.

1. Standing Waves Shortcut

Standing waves are one of the fastest places to gain marks if the spacing rules are remembered clearly.

Core Rule

For a standing wave:

node to node $= \frac{\lambda}{2}$

antinode to antinode $= \frac{\lambda}{2}$

node to adjacent antinode $= \frac{\lambda}{4}$

This is especially useful in microwave oven questions, where melted spots indicate antinodes.

Example 1

The distance between adjacent melted spots in a microwave oven is $6.0\,\text{cm}$ .

Determine the frequency of the microwaves.

Step 1: Melted spots are antinodes, so

$\frac{\lambda}{2}=0.060$

$\lambda=0.120\,\text{m}$

Step 2: Microwaves are electromagnetic waves, so

$c=3.0\times10^8\,\text{m s}^{-1}$

Step 3: Use

$f=\frac{c}{\lambda}$

$f=\frac{3.0\times10^8}{0.120}=2.5\times10^9\,\text{Hz}$

Answer: $2.5\times10^9\,\text{Hz}$

Examiner Tip: In microwave questions, students often treat the distance between melted spots as one full wavelength. It is usually half a wavelength.

Paper 1 Shortcut: If the separation between adjacent antinodes is given as $d$ , then immediately write $\lambda = 2d$ .

Paper 2 Strategy: State clearly that the melted regions correspond to antinodes because antinodes have maximum amplitude and therefore maximum intensity.

📊 IB Markscheme-style marks [3]

[1] identifies that adjacent melted spots are separated by $\frac{\lambda}{2}$
[1] finds $\lambda = 0.12\,\text{m}$
[1] calculates $f = 2.5\times10^9\,\text{Hz}$

⚠ Common mistakes

using $\lambda = 0.060\,\text{m}$ instead of $0.120\,\text{m}$
using the speed of sound instead of the speed of light
confusing nodes and antinodes

2. Friction Between Two Blocks Shortcut

In block-on-block questions, the key idea is that the upper block is accelerated by friction. If the blocks move together, friction is static, not kinetic.

Core Rule

At the limiting case before slipping:

$f_{\max}=\mu_s N$

Since friction provides the acceleration of the top block:

$f=ma$

So:

$ma=\mu_s mg$

$a_{\max}=\mu_s g$

Example 2

A $15\,\text{kg}$ block sits on a $45\,\text{kg}$ block on a frictionless floor. The coefficient of static friction is $0.60$ .

Determine the largest horizontal force that can be applied without slipping.

Step 1: Maximum acceleration before slipping:

$a_{\max}=\mu_s g=0.60\times9.8=5.88\,\text{m s}^{-2}$

Step 2: Total mass of the system:

$m_{\text{total}}=45+15=60\,\text{kg}$

Step 3: Use $F=ma$ for the whole system:

$F_{\max}=60\times5.88=352.8\,\text{N}$

Answer: $3.5\times10^2\,\text{N}$

Examiner Tip: Friction does not always oppose motion. It opposes relative slipping. Here, it acts to the right on the upper block.

Paper 1 Shortcut: For no slipping, jump directly to $F_{\max}=(m_{\text{total}})\mu_s g$ .

Paper 2 Strategy: In explanation questions, say that the top block would otherwise lag behind, so static friction acts forwards on it.

📊 IB Markscheme-style marks [3]

[1] uses $a_{\max}=\mu_s g$ or equivalent friction argument
[1] uses total mass $60\,\text{kg}$
[1] obtains $3.5\times10^2\,\text{N}$

⚠ Common mistakes

using kinetic friction instead of static friction
using only the lower block mass in $F=ma$
stating friction acts left on the top block

3. Electricity Shortcuts

Electricity questions often become much faster when students choose the shortest valid relationship rather than the longest route.

Core Rules

For resistors in series:

$V_1:V_2=R_1:R_2$

For electrical power:

$P=VI$

$P=I^2R$

$P=\frac{V^2}{R}$

For resistors in parallel, always check that:

the equivalent resistance is less than the smallest individual resistor.

Example 3

Two resistors of $2\,\Omega$ and $6\,\Omega$ are connected in series across a supply.

Determine the ratio of the potential differences across them.

For series resistors:

$V_1:V_2=R_1:R_2$

So:

$V_1:V_2=2:6=1:3$

Answer: $1:3$

GDC Tip: If you already know $V$ and $R$ , use $P=\frac{V^2}{R}$ directly instead of first calculating current.

Paper 2 Strategy: In long circuit questions, first decide whether the current is the same everywhere or whether the potential difference is the same everywhere. That usually reveals the fastest route.

📊 IB Markscheme-style marks [2]

[1] identifies that potential difference in series is proportional to resistance
[1] obtains $1:3$

⚠ Common mistakes

calculating current unnecessarily
forgetting that current is the same in series
giving a parallel equivalent resistance larger than the smallest resistor

4. Fields Shortcuts

Field questions often become much easier once the correct force relation is written first.

Core Rules

Electric field:

$F=qE$

So:

$E=\frac{F}{q}$

Gravitational field strength:

$g=\frac{GM}{r^2}$

Near Earth:

$g\approx9.8\,\text{m s}^{-2}$

Example 4

A charge of $2.0\times10^{-6}\,\text{C}$ experiences an electric force of $8.0\times10^{-3}\,\text{N}$ .

Determine the electric field strength.

Use:

$E=\frac{F}{q}$

$E=\frac{8.0\times10^{-3}}{2.0\times10^{-6}}=4.0\times10^3\,\text{N C}^{-1}$

Answer: $4.0\times10^3\,\text{N C}^{-1}$

Examiner Tip: In field questions, students often know the definition of field but forget that the force equation usually gives the quickest markscheme path.

📊 IB Markscheme-style marks [2]

[1] uses $E=\frac{F}{q}$
[1] obtains $4.0\times10^3\,\text{N C}^{-1}$

⚠ Common mistakes

rearranging the formula incorrectly
ignoring powers of ten
mixing up field strength and potential difference

5. Nuclear Physics Shortcuts

Nuclear physics contains some of the fastest marks on the paper because the changes in alpha and beta decay follow very fixed rules.

Core Rules

Alpha decay: mass number decreases by 4, proton number decreases by 2

Beta minus decay: mass number stays the same, proton number increases by 1

Gamma emission: no change in mass number or proton number

For half-life:

$\text{remaining fraction}=\left(\frac12\right)^n$

Example 5

Write the alpha decay equation for americium-241.

Americium has proton number $95$ .

For alpha decay:

mass number $241-4=237$

proton number $95-2=93$

Element 93 is neptunium.

$^{241}_{95}\text{Am}\rightarrow\,^{237}_{93}\text{Np}+\,^{4}_{2}\alpha$

Example 6

A sample undergoes 4 half-lives.

Determine the fraction remaining.

$\left(\frac12\right)^4=\frac1{16}$

Answer: $\frac1{16}$

Paper 1 Shortcut: For half-life questions, calculate the number of half-lives first, then use $\left(\frac12\right)^n$ immediately.

📊 IB Markscheme-style marks

alpha decay equation [2]: one mark for correct nucleon numbers, one mark for correct proton numbers or nuclides
half-life fraction [2]: one mark for method, one mark for answer

⚠ Common mistakes

subtracting 2 from the mass number in alpha decay instead of 4
changing the mass number in beta decay
using repeated subtraction rather than the half-life fraction rule

6. Ionization Current Shortcut

This is one of the most useful one-line methods in IB nuclear physics.

Core Rule

For ionization chamber questions:

$I=Af\frac{E}{W}e$

where

$A$ = activity
$f$ = fraction entering the chamber
$E$ = particle energy
$W$ = energy needed for one ion pair
$e$ = elementary charge

Example 7

A source has activity $4.2\times10^4\,\text{Bq}$ . Only $33\%$ of particles enter the chamber. Each particle has energy $5.5\times10^6\,\text{eV}$ . The energy to form one ion pair is $15\,\text{eV}$ .

Calculate the current.

$I=(4.2\times10^4)(0.33)\left(\frac{5.5\times10^6}{15}\right)(1.6\times10^{-19})$

$I\approx8.1\times10^{-10}\,\text{A}$

Answer: $8.1\times10^{-10}\,\text{A}$

Examiner Tip: The three most common errors are forgetting to convert kBq to Bq, forgetting to convert MeV to eV, and forgetting the fraction entering the chamber.

Paper 2 Strategy: Explain briefly that current is charge per second, and that each ion pair contributes one electron to the current.

📊 IB Markscheme-style marks [3]

[1] determines number of ion pairs per second or equivalent expression
[1] converts to charge per second
[1] obtains $8.1\times10^{-10}\,\text{A}$

⚠ Common mistakes

using kBq without converting to Bq
using MeV directly instead of eV
omitting the fraction of particles entering the chamber

7. Thermal Physics and Calorimetry Shortcut

For calorimetry questions, the fastest reliable method is to list the thermal processes in order, then write one energy balance equation.

Core Rules

Temperature change:

$Q=mc\Delta T$

Phase change:

$Q=mL$

Mixing process:

$Q_{\text{lost}}=Q_{\text{gained}}$

In phase-change questions, always ask: what does the cold object do first, next, and after that?

Example 8

$0.030\,\text{kg}$ of solid paraffin at $42^\circ\text{C}$ is mixed with $0.150\,\text{kg}$ of liquid paraffin at $240^\circ\text{C}$ .

Given:

$c_{\text{solid}}=0.7\times10^3\,\text{J kg}^{-1}\text{K}^{-1}$
$c_{\text{liquid}}=2.13\times10^3\,\text{J kg}^{-1}\text{K}^{-1}$
$L_f=220\times10^3\,\text{J kg}^{-1}$
melting point $=47^\circ\text{C}$

Determine the equilibrium temperature.

Cold paraffin:

heat from $42^\circ\text{C}$ to $47^\circ\text{C}$

melt

heat from $47^\circ\text{C}$ to $T$

Hot paraffin:

cool from $240^\circ\text{C}$ to $T$

So:

$(0.030)(0.7\times10^3)(5)+(0.030)(220\times10^3)+(0.030)(2.13\times10^3)(T-47) = (0.150)(2.13\times10^3)(240-T)$

Solving gives:

$T\approx190^\circ\text{C}$

Answer: $190^\circ\text{C}$

Examiner Tip: The latent heat term is often the difference between a full-mark answer and a mid-band answer.

GDC Tip: Enter the entire equation carefully and solve for $T$ in one step. This reduces arithmetic mistakes in long calorimetry questions.

Paper 2 Strategy: Before substituting numbers, write the stages in words first. This makes it much harder to forget a phase change or use the wrong specific heat capacity.

📊 IB Markscheme-style marks [4]

[1] correct heat balance idea
[1] includes latent heat term
[1] substitutes relevant values correctly
[1] obtains $190^\circ\text{C}$

⚠ Common mistakes

forgetting the latent heat term
using $220\,\text{kJ kg}^{-1}$ as $2.2\times10^6\,\text{J kg}^{-1}$ instead of $220\times10^3\,\text{J kg}^{-1}$
using $(T-42)$ instead of $(T-47)$ after melting
using the solid specific heat capacity after the substance has melted

📝 Paper 1 Quick-Fire Summary

Standing waves: adjacent antinodes $= \frac{\lambda}{2}$
Block friction: $a_{\max}=\mu_s g$
Series resistors: voltage ratio = resistance ratio
Power: choose from $P=VI$ , $P=I^2R$ , or $P=\frac{V^2}{R}$
Electric field: $E=\frac{F}{q}$
Half-life: remaining fraction $=\left(\frac12\right)^n$
Ionization current: $I=Af\frac{E}{W}e$
Calorimetry: $Q_{\text{lost}}=Q_{\text{gained}}$

🧾 Paper 2 Extended Response Advice

Start with the physics model, not the numbers.
State what each region or force represents before calculating.
Use short linking statements such as “the top block is accelerated by friction” or “adjacent antinodes are separated by half a wavelength”.
Keep working clear and linear so the examiner can award method marks.
Always check whether your final answer is physically reasonable.

🧪 Challenge Problems

Challenge 1: Standing Waves

The distance between adjacent antinodes in a standing wave is $8.0\,\text{cm}$ . The wave speed is $320\,\text{m s}^{-1}$ . Determine the frequency.

Show answer

$\frac{\lambda}{2}=0.080$

$\lambda=0.160\,\text{m}$

$f=\frac{320}{0.160}=2.0\times10^3\,\text{Hz}$

Challenge 2: Friction

A $20\,\text{kg}$ block sits on a $30\,\text{kg}$ block on a frictionless floor. The coefficient of static friction between them is $0.40$ . Find the maximum horizontal force so they move together.

Show answer

$a_{\max}=\mu_s g=0.40\times9.8=3.92\,\text{m s}^{-2}$

$m_{\text{total}}=50\,\text{kg}$

$F_{\max}=50\times3.92=196\,\text{N}$

Challenge 3: Ionization Current

A source has activity $1.5\times10^5\,\text{Bq}$ . Only $20\%$ of the particles enter a chamber. Each particle has energy $4.0\times10^6\,\text{eV}$ and the ionization energy is $20\,\text{eV}$ . Calculate the current.

Show answer

$I=Af\frac{E}{W}e$

$I=(1.5\times10^5)(0.20)\left(\frac{4.0\times10^6}{20}\right)(1.6\times10^{-19})$

$I=9.6\times10^{-10}\,\text{A}$

Challenge 4: Calorimetry

$0.10\,\text{kg}$ of water at $20^\circ\text{C}$ is mixed with $0.20\,\text{kg}$ of water at $80^\circ\text{C}$ . Determine the final temperature.

Show answer

$0.10(T-20)=0.20(80-T)$

$0.10T-2=16-0.20T$

$0.30T=18$

$T=60^\circ\text{C}$

✅ Self-Practice Questions

1. The distance between adjacent melted spots in a microwave is $5.8\,\text{cm}$ . Calculate the microwave frequency.

Show worked solution

$\frac{\lambda}{2}=0.058$

$\lambda=0.116\,\text{m}$

$f=\frac{3.0\times10^8}{0.116}\approx2.6\times10^9\,\text{Hz}$

2. A $10\,\text{kg}$ object rests on a $25\,\text{kg}$ block. The coefficient of static friction is $0.50$ . Find the maximum force so they move together.

Show worked solution

$a_{\max}=\mu_s g=0.50\times9.8=4.9\,\text{m s}^{-2}$

$m_{\text{total}}=35\,\text{kg}$

$F_{\max}=35\times4.9=171.5\,\text{N}$

$F_{\max}\approx1.7\times10^2\,\text{N}$

3. A source has activity $3.0\times10^4\,\text{Bq}$ . $25\%$ of particles enter a chamber. Each particle has energy $6.0\times10^6\,\text{eV}$ and the energy required to create one ion pair is $30\,\text{eV}$ . Determine the current.

Show worked solution

$I=Af\frac{E}{W}e$

$I=(3.0\times10^4)(0.25)\left(\frac{6.0\times10^6}{30}\right)(1.6\times10^{-19})$

$I=2.4\times10^{-10}\,\text{A}$

4. After 5 half-lives, what fraction of a radioactive sample remains?

Show worked solution

$\left(\frac12\right)^5=\frac1{32}$

5. $0.020\,\text{kg}$ of ice at $0^\circ\text{C}$ is added to $0.10\,\text{kg}$ of water at $30^\circ\text{C}$ . Given $L_f=3.34\times10^5\,\text{J kg}^{-1}$ and $c=4200\,\text{J kg}^{-1}\text{K}^{-1}$ , determine the final temperature.

Show worked solution

Heat needed to melt the ice:

$Q=0.020\times3.34\times10^5=6680\,\text{J}$

Heat available if warm water cools to $0^\circ\text{C}$ :

$Q=0.10\times4200\times30=12600\,\text{J}$

So all the ice melts.

Remaining heat:

$12600-6680=5920\,\text{J}$

This heats $0.12\,\text{kg}$ of water:

$5920=0.12\times4200\times T$

$T=11.7^\circ\text{C}$

Final answer: approximately $12^\circ\text{C}$

Final Summary

Fast Paper 2 formulas to remember:

Standing waves: $\text{adjacent antinodes}=\frac{\lambda}{2}$

Block friction: $a_{\max}=\mu_s g$

Electric field: $E=\frac{F}{q}$

Half-life: $\left(\frac12\right)^n$

Ionization current: $I=Af\frac{E}{W}e$

Calorimetry: $Q_{\text{lost}}=Q_{\text{gained}}$

These shortcuts are powerful because they are rooted in correct physics. When students combine them with careful reading, unit conversion, and clear written working, they save time and pick up method marks more reliably across both Paper 1 and Paper 2.

Waves, Electricity, Fields, Nuclear Physics, and Thermal Physics

⭐ Key Concepts

📘 IB Command Terms

1. Standing Waves Shortcut

Core Rule

Example 1

2. Friction Between Two Blocks Shortcut

Core Rule

Example 2

3. Electricity Shortcuts

Core Rules

Example 3

4. Fields Shortcuts

Core Rules

Example 4

5. Nuclear Physics Shortcuts

Core Rules

Example 5

Example 6

6. Ionization Current Shortcut

Core Rule

Example 7

7. Thermal Physics and Calorimetry Shortcut

Core Rules

Example 8

📝 Paper 1 Quick-Fire Summary

🧾 Paper 2 Extended Response Advice

🧪 Challenge Problems

Challenge 1: Standing Waves

Challenge 2: Friction

Challenge 3: Ionization Current

Challenge 4: Calorimetry

✅ Self-Practice Questions

Final Summary

Share this:

Like this:

Related

You may also like

Leave a ReplyCancel reply

Company

Links

Support

Recommend

alphy school publishers

Discover more from Alphy School

Customise Consent Preferences

Login with your site account