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AP Physics 2 – Practice Set

Fluids, thermodynamics, electricity, magnetism, optics, modern physics.

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12 questions  ·  41 marks  ·  answers included  ·  AP Physics 2 (algebra-based)
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This worksheet covers:

  • Free-response-style practice covering fluids
  • Thermodynamics
  • Electricity
  • Magnetism
  • Optics
  • Modern physics

A calculator is allowed; use $g=9.81\,\text{m s}^{-2}$, $k=8.99\times10^{9}\,\text{N m}^2\,\text{C}^{-2}$ and $R=8.31\,\text{J mol}^{-1}\,\text{K}^{-1}$ unless stated otherwise.

Questions

Section A – Fluids and thermodynamics

1

A hydraulic lift has a small piston of area $0.0050\,\text{m}^2$ and a large piston of area $0.045\,\text{m}^2$. A force of $250\,\text{N}$ is applied to the small piston.Calculate the force produced on the large piston.

Show answer
By Pascal’s principle, pressure is equal: $F_2=F_1\dfrac{A_2}{A_1}=250\times\dfrac{0.045}{0.0050}=2250\,\text{N}$.
[3]
2

A block of density $700\,\text{kg m}^{-3}$ and volume $0.020\,\text{m}^3$ floats in water of density $1000\,\text{kg m}^{-3}$.(a) Calculate the fraction of the block’s volume that is submerged.(b) Calculate the buoyant force acting on the block.

Show answer
(a) At equilibrium, buoyant force = weight, so $\rho_{\text{water}}V_{\text{sub}}g=\rho_{\text{block}}Vg \Rightarrow \dfrac{V_{\text{sub}}}{V}=\dfrac{700}{1000}=0.70$ (70 percent submerged).(b) $F_B=\rho_{\text{block}}Vg=700\times0.020\times9.81=137\,\text{N}$.
[4]
3

An ideal gas consists of $2.0\,\text{mol}$ confined to a volume of $0.050\,\text{m}^3$ at a temperature of $300\,\text{K}$.Calculate the pressure of the gas.

Show answer
$PV=nRT \Rightarrow P=\dfrac{nRT}{V}=\dfrac{2.0\times8.31\times300}{0.050}=9.97\times10^{4}\,\text{Pa}$.
[3]
4

A gas absorbs $500\,\text{J}$ of heat while doing $200\,\text{J}$ of work on its surroundings as it expands.(a) Calculate the change in internal energy of the gas.(b) State whether the internal energy of the gas increases or decreases, with a reason.

Show answer
(a) First law: $\Delta U=Q-W=500-200=300\,\text{J}$.(b) The internal energy increases, because the heat added exceeds the work done by the gas.
[3]

Section B – Electricity and magnetism

5

Two point charges, $q_1=+3.0\,\mu\text{C}$ and $q_2=+5.0\,\mu\text{C}$, are separated by $0.20\,\text{m}$ in air.Calculate the magnitude of the electrostatic force between them.

Show answer
$F=\dfrac{kq_1q_2}{r^2}=\dfrac{8.99\times10^{9}\times3.0\times10^{-6}\times5.0\times10^{-6}}{0.20^{2}}=3.4\,\text{N}$.
[3]
6

A $12\,\text{V}$ battery is connected to a $10\,\Omega$ resistor in series with a parallel combination of two $20\,\Omega$ resistors.(a) Calculate the total resistance of the circuit.(b) Calculate the total current supplied by the battery.(c) Calculate the current through each $20\,\Omega$ resistor.

Show answer
(a) Parallel pair: $\dfrac{1}{R_p}=\dfrac{1}{20}+\dfrac{1}{20} \Rightarrow R_p=10\,\Omega$. Total $R=10+10=20\,\Omega$.(b) $I=\dfrac{V}{R}=\dfrac{12}{20}=0.60\,\text{A}$.(c) Voltage across parallel pair $=I\times R_p=0.60\times10=6.0\,\text{V}$; current in each branch $=6.0/20=0.30\,\text{A}$.
[5]
7

A straight wire of length $0.30\,\text{m}$ carries a current of $4.0\,\text{A}$ perpendicular to a uniform magnetic field of strength $0.50\,\text{T}$.Calculate the magnitude of the magnetic force on the wire.

Show answer
$F=BIL=0.50\times4.0\times0.30=0.60\,\text{N}$.
[3]
8

A circular loop of area $0.020\,\text{m}^2$ is in a magnetic field that decreases uniformly from $0.40\,\text{T}$ to $0.10\,\text{T}$ in $0.50\,\text{s}$.Calculate the magnitude of the EMF induced in the loop.

Show answer
$\varepsilon=\left|\dfrac{\Delta\Phi}{\Delta t}\right|=\dfrac{|(0.10-0.40)\times0.020|}{0.50}=1.2\times10^{-2}\,\text{V}$.
[3]

Section C – Optics and modern physics

9

Light travelling in air ($n=1.00$) strikes a glass surface ($n=1.50$) at an angle of incidence of $30^{\circ}$.Calculate the angle of refraction in the glass.

Show answer
Snell’s law: $n_1\sin\theta_1=n_2\sin\theta_2 \Rightarrow \sin\theta_2=\dfrac{\sin30^{\circ}}{1.50}=0.333 \Rightarrow \theta_2=19.5^{\circ}$.
[3]
10

An object is placed $20\,\text{cm}$ from a converging lens of focal length $15\,\text{cm}$.(a) Calculate the image distance.(b) Calculate the magnification and describe the image (real/virtual, upright/inverted).

Show answer
(a) $\dfrac{1}{f}=\dfrac{1}{d_o}+\dfrac{1}{d_i} \Rightarrow \dfrac{1}{d_i}=\dfrac{1}{15}-\dfrac{1}{20}=\dfrac{1}{60} \Rightarrow d_i=60\,\text{cm}$.(b) $m=-\dfrac{d_i}{d_o}=-\dfrac{60}{20}=-3.0$; the image is real, inverted and magnified 3 times.
[4]
11

Light of wavelength $600\,\text{nm}$ passes through two slits separated by $0.20\,\text{mm}$ and forms an interference pattern on a screen $2.0\,\text{m}$ away.Calculate the fringe spacing on the screen.

Show answer
$\Delta y=\dfrac{\lambda L}{d}=\dfrac{600\times10^{-9}\times2.0}{0.20\times10^{-3}}=6.0\times10^{-3}\,\text{m}=6.0\,\text{mm}$.
[3]
12

The work function of a metal surface is $2.3\,\text{eV}$. Light of wavelength $400\,\text{nm}$ is incident on the surface.(a) Calculate the energy of an incident photon in $\text{eV}$ (use $hc=1240\,\text{eV nm}$).(b) Calculate the maximum kinetic energy of an emitted photoelectron.

Show answer
(a) $E=\dfrac{hc}{\lambda}=\dfrac{1240}{400}=3.1\,\text{eV}$.(b) $KE_{\max}=E-\phi=3.1-2.3=0.80\,\text{eV}$.
[4]

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