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AP Physics 1 Study Guide

AP Physics 1 Study Guide

Comprehensive study guide for AP Physics 1 (Algebra-Based), aligned with the College Board Course and Exam Description. Covers mechanics and rotational motion with emphasis on conceptual understanding and problem-solving.

1. Kinematics

Displacement, Velocity, and Acceleration

QuantityTypeDefinitionSI Unit
DisplacementVectorChange in position, Δr\Delta \vec{r}m
DistanceScalarTotal path lengthm
VelocityVectorRate of change of displacementm/s
SpeedScalarRate of change of distancem/s
AccelerationVectorRate of change of velocitym/s2^2

One-Dimensional Kinematics

Constant acceleration equations:

v=v0+atv = v_0 + at x=x0+v0t+12at2x = x_0 + v_0 t + \frac{1}{2}at^2 v2=v02+2a(xx0)v^2 = v_0^2 + 2a(x - x_0) x=x0+12(v0+v)tx = x_0 + \frac{1}{2}(v_0 + v)t

Two-Dimensional Motion

In 2D, resolve all vectors into perpendicular components. The xx and yy components of motion are independent of each other (superposition principle).

Projectile Motion

For a projectile launched with initial speed v0v_0 at angle θ\theta above the horizontal (ignoring air resistance):

v0x=v0cosθ,v0y=v0sinθv_{0x} = v_0 \cos\theta, \quad v_{0y} = v_0 \sin\theta

Horizontal (constant velocity):

x(t)=v0xtx(t) = v_{0x}\, t

Vertical (constant acceleration ay=ga_y = -g):

y(t)=v0yt12gt2y(t) = v_{0y}\, t - \frac{1}{2}gt^2 vy(t)=v0ygtv_y(t) = v_{0y} - gt

Time of flight: t=2v0ygt = \dfrac{2v_{0y}}{g}

Maximum height: H=v0y22gH = \dfrac{v_{0y}^2}{2g}

Range: R=v02sin2θgR = \dfrac{v_0^2 \sin 2\theta}{g}

Free Fall

An object in free fall experiences only the force of gravity. All objects in free fall have the same acceleration g9.8 m/s2g \approx 9.8\ \text{m/s}^2 near Earth’s surface, regardless of mass.

Graphical Analysis

  • Position-time graph: Slope gives velocity
  • Velocity-time graph: Slope gives acceleration; area under the curve gives displacement
  • Acceleration-time graph: Area under the curve gives change in velocity

2. Dynamics

Newton’s Laws

First Law (Inertia): An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by a net external force.

Second Law: The net force on an object equals its mass times its acceleration:

Fnet=ma\vec{F}_{\text{net}} = m\vec{a}

This is a vector equation — apply separately in each direction.

Third Law: For every action force there is an equal and opposite reaction force. Forces always come in pairs acting on different objects.

Free-Body Diagrams

  1. Isolate the object of interest
  2. Draw arrows for every force acting on the object
  3. Forces act on the object; the object’s forces act on other things
  4. Common forces: weight (mgmg), normal force (NN), tension (TT), friction (ff), applied force (FF)

Friction

Static friction: fsμsNf_s \leq \mu_s N (prevents motion; maximum value is μsN\mu_s N)

Kinetic friction: fk=μkNf_k = \mu_k N (opposes motion; constant magnitude)

where μs\mu_s and μk\mu_k are coefficients of static and kinetic friction respectively, and NN is the normal force.

Inclined Planes

On an incline at angle θ\theta to the horizontal:

a=g(sinθμkcosθ)a = g(\sin\theta - \mu_k \cos\theta)

The component of weight parallel to the plane is mgsinθmg\sin\theta; the component perpendicular (which equals the normal force) is mgcosθmg\cos\theta.

Atwood Machine

Two masses m1m_1 and m2m_2 (m2>m1m_2 > m_1) connected by a string over a frictionless pulley:

a=(m2m1)gm1+m2a = \frac{(m_2 - m_1)g}{m_1 + m_2} T=2m1m2gm1+m2T = \frac{2m_1 m_2 g}{m_1 + m_2}

Uniform Circular Motion

An object moving in a circle of radius rr at constant speed vv has a centripetal acceleration directed towards the centre:

ac=v2ra_c = \frac{v^2}{r}

The net inward force providing this acceleration is:

Fc=mv2r=mω2rF_c = \frac{mv^2}{r} = m\omega^2 r

where ω\omega is the angular velocity in rad/s.

Gravitational Force

Newton’s law of universal gravitation:

Fg=Gm1m2r2F_g = \frac{Gm_1 m_2}{r^2}

where G=6.674×1011 Nm2/kg2G = 6.674 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2.

The gravitational field strength at distance rr from mass MM:

g=GMr2g = \frac{GM}{r^2}

3. Work and Energy

Work

Work done by a constant force:

W=FdcosθW = Fd\cos\theta

where θ\theta is the angle between the force and the displacement. Work is positive when the force has a component in the direction of displacement and negative when opposing displacement.

For a variable force:

W=abF(x)dxW = \int_a^b F(x)\, dx

Work-Energy Theorem

The net work done on an object equals its change in kinetic energy:

Wnet=ΔK=KfKiW_{\text{net}} = \Delta K = K_f - K_i

Kinetic Energy

K=12mv2K = \frac{1}{2}mv^2

Potential Energy

Gravitational potential energy (near Earth’s surface):

Ug=mghU_g = mgh

Elastic (spring) potential energy:

Us=12kx2U_s = \frac{1}{2}kx^2

where kk is the spring constant and xx is the displacement from equilibrium.

Conservation of Energy

In the absence of non-conservative forces (e.g., friction):

Ki+Ui=Kf+UfK_i + U_i = K_f + U_f

When friction is present:

Ki+Ui=Kf+Uf+Elost to frictionK_i + U_i = K_f + U_f + E_{\text{lost to friction}}

where Elost to friction=fkdE_{\text{lost to friction}} = f_k d.

Power

Average power is the rate at which work is done:

P=WΔtP = \frac{W}{\Delta t}

Instantaneous power:

P=FvcosθP = Fv\cos\theta

SI unit: watt (W), where 1 W=1 J/s1\ \text{W} = 1\ \text{J/s}.

4. Momentum

Impulse and Linear Momentum

Momentum:

p=mv\vec{p} = m\vec{v}

Impulse:

J=FavgΔt=Δp=mvfmvi\vec{J} = \vec{F}_{\text{avg}}\,\Delta t = \Delta\vec{p} = m\vec{v}_f - m\vec{v}_i

The impulse-momentum theorem states that the impulse on an object equals its change in momentum. Graphically, impulse equals the area under a force-time graph.

Conservation of Momentum

When the net external force on a system is zero, the total momentum of the system is conserved:

m1v1i+m2v2i=m1v1f+m2v2fm_1\vec{v}_{1i} + m_2\vec{v}_{2i} = m_1\vec{v}_{1f} + m_2\vec{v}_{2f}

This applies independently to each component direction.

Collisions

Elastic collision: Both momentum and kinetic energy are conserved.

For a 1D elastic collision between masses m1m_1 and m2m_2:

v1f=(m1m2)v1i+2m2v2im1+m2v_{1f} = \frac{(m_1 - m_2)v_{1i} + 2m_2 v_{2i}}{m_1 + m_2} v2f=(m2m1)v2i+2m1v1im1+m2v_{2f} = \frac{(m_2 - m_1)v_{2i} + 2m_1 v_{1i}}{m_1 + m_2}

Inelastic collision: Momentum is conserved but kinetic energy is not.

Perfectly inelastic collision: The objects stick together after collision.

(m1+m2)vf=m1v1i+m2v2i(m_1 + m_2)v_f = m_1 v_{1i} + m_2 v_{2i}

Centre of Mass

The centre of mass of a system of particles:

xcm=m1x1+m2x2+m1+m2+x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2 + \cdots}{m_1 + m_2 + \cdots}

The centre of mass of a system moves as if all external forces act on a single particle of total mass located at the centre of mass.

5. Rotational Motion

Torque

Torque is the rotational analogue of force:

τ=rFsinθ=rF\tau = rF\sin\theta = r_\perp F

where r=rsinθr_\perp = r\sin\theta is the perpendicular distance from the axis of rotation to the line of action of the force (the moment arm). Positive torque causes counter-clockwise rotation; negative causes clockwise.

Net torque and angular acceleration:

τnet=Iα\tau_{\text{net}} = I\alpha

where II is the rotational inertia and α\alpha is the angular acceleration.

Rotational Inertia (Moment of Inertia)

I=miri2I = \sum m_i r_i^2

For continuous objects: I=r2dmI = \int r^2\, dm

Common moments of inertia:

ObjectRotational Inertia
Solid disc/cylinder (axis through centre)12MR2\dfrac{1}{2}MR^2
Hoop/ring (axis through centre)MR2MR^2
Solid sphere25MR2\dfrac{2}{5}MR^2
Thin rod (axis through centre)112ML2\dfrac{1}{12}ML^2
Thin rod (axis through end)13ML2\dfrac{1}{3}ML^2
Point mass at distance rrmr2mr^2

Angular Momentum

L=IωL = I\omega

For a point mass: L=mvrL = mvr_\perp

Conservation of angular momentum: When no net external torque acts on a system, IωI\omega is constant:

Iiωi=IfωfI_i\omega_i = I_f\omega_f

Rotational Kinetic Energy

Kr=12Iω2K_r = \frac{1}{2}I\omega^2

Total kinetic energy for a rolling object:

Ktotal=12mv2+12Iω2K_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

Rolling Motion

For an object rolling without slipping:

vcm=Rωv_{\text{cm}} = R\omega acm=Rαa_{\text{cm}} = R\alpha

Acceleration of a solid sphere rolling down an incline of angle θ\theta:

a=57gsinθa = \frac{5}{7}g\sin\theta

6. Oscillations

Simple Harmonic Motion (SHM)

An object undergoes SHM when a restoring force is proportional to the displacement from equilibrium:

F=kxF = -kx

This yields sinusoidal motion:

x(t)=Acos(ωt+ϕ)x(t) = A\cos(\omega t + \phi)

where AA is the amplitude, ω\omega is the angular frequency, and ϕ\phi is the phase constant.

Mass-Spring System

ω=km,T=2πmk,f=12πkm\omega = \sqrt{\frac{k}{m}}, \quad T = 2\pi\sqrt{\frac{m}{k}}, \quad f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

where TT is the period (time for one complete oscillation) and ff is the frequency.

Simple Pendulum

For small angles (θ<15°\theta < 15°), a simple pendulum approximates SHM:

ω=gL,T=2πLg,f=12πgL\omega = \sqrt{\frac{g}{L}}, \quad T = 2\pi\sqrt{\frac{L}{g}}, \quad f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}

Note: the period of a pendulum is independent of mass.

Energy in SHM

Total mechanical energy is conserved and constant:

Etotal=12kA2=Kmax=UmaxE_{\text{total}} = \frac{1}{2}kA^2 = K_{\text{max}} = U_{\text{max}}

At any position xx:

K=12k(A2x2),U=12kx2K = \frac{1}{2}k(A^2 - x^2), \quad U = \frac{1}{2}kx^2

Period and Frequency

T=1f,f=1T,ω=2πf=2πTT = \frac{1}{f}, \quad f = \frac{1}{T}, \quad \omega = 2\pi f = \frac{2\pi}{T}

7. Key Equations

Kinematics

v=v0+atx=x0+v0t+12at2v2=v02+2aΔxv = v_0 + at \quad x = x_0 + v_0 t + \tfrac{1}{2}at^2 \quad v^2 = v_0^2 + 2a\Delta x xmax=v02sin2θ2gR=v02sin2θgx_{\text{max}} = \frac{v_0^2\sin^2\theta}{2g} \quad R = \frac{v_0^2 \sin 2\theta}{g}

Dynamics

Fnet=mafk=μkNFc=mv2rF_{\text{net}} = ma \quad f_k = \mu_k N \quad F_c = \frac{mv^2}{r}

Energy

K=12mv2Ug=mghUs=12kx2K = \tfrac{1}{2}mv^2 \quad U_g = mgh \quad U_s = \tfrac{1}{2}kx^2 W=FdcosθP=WΔt=FvcosθW = Fd\cos\theta \quad P = \frac{W}{\Delta t} = Fv\cos\theta

Momentum

p=mvJ=FavgΔt=Δpp = mv \quad J = F_{\text{avg}}\,\Delta t = \Delta p

Rotation

τ=IαL=IωKr=12Iω2\tau = I\alpha \quad L = I\omega \quad K_r = \tfrac{1}{2}I\omega^2

Oscillations

Tspring=2πmkTpendulum=2πLgT_{\text{spring}} = 2\pi\sqrt{\frac{m}{k}} \quad T_{\text{pendulum}} = 2\pi\sqrt{\frac{L}{g}}

8. Exam Tips

  1. Draw a free-body diagram for every problem. It earns points, organises your thinking, and helps identify all forces. Never skip this step on the free-response section.
  2. Check dimensions and units at every step. If your answer has units of kg·m/s when you expect joules, there is an error. Dimensional analysis catches many mistakes quickly.
  3. Define your coordinate system. State which direction is positive and where the origin is. This prevents sign errors and makes your reasoning clear to the grader.
  4. Distinguish between vector and scalar quantities. Displacement is not distance; velocity is not speed. The AP exam frequently tests this distinction.
  5. Apply conservation laws first. Before writing out Newton’s second law for every object, check whether conservation of energy or momentum solves the problem more efficiently.
  6. Explain your reasoning in words. The AP Physics 1 exam emphasises conceptual understanding. After writing equations, explain what each term represents and why the equation applies.
  7. Use graphs to support your answers. The exam often asks you to sketch or interpret graphs. Practise translating between equations, graphs, and physical descriptions.

9. Common Mistakes

  1. Confusing mass and weight. Mass is an intrinsic property measured in kg; weight is the gravitational force mgmg measured in newtons. An object has the same mass on the Moon but different weight.
  2. Including internal forces in Fnet\vec{F}_{\text{net}}. Only external forces contribute to the net force on a system. The tension in a rope between two parts of the same system is an internal force and cancels out.
  3. Forgetting the normal force is not always mgmg. On an incline, N=mgcosθN = mg\cos\theta. In an accelerating lift, N=m(g±a)N = m(g \pm a). Always derive NN from Newton’s second law.
  4. Mixing up centripetal and centrifugal forces. “Centrifugal force” is a fictitious force that appears in a rotating reference frame. In an inertial frame, only the centripetal force (directed towards the centre) acts on the object.
  5. Using the wrong collision formula. Only use elastic collision equations when kinetic energy is conserved. For perfectly inelastic collisions, use conservation of momentum alone and set final velocities equal.
  6. Incorrect rotational inertia. Do not assume I=MR2I = MR^2 for all objects. A solid disc has I=12MR2I = \tfrac{1}{2}MR^2; a solid sphere has I=25MR2I = \tfrac{2}{5}MR^2. Know the common values.
  7. Applying the small-angle approximation outside its range. The pendulum formula T=2πL/gT = 2\pi\sqrt{L/g} only holds for angles below roughly 15°. For larger amplitudes, the period is longer.

10. Summary

TopicKey Ideas
KinematicsDisplacement, velocity, acceleration, projectile motion, free fall, motion graphs
DynamicsNewton’s three laws, free-body diagrams, friction, inclined planes, circular motion
Work and EnergyWork, work-energy theorem, kinetic/potential energy, conservation of energy, power
MomentumImpulse-momentum theorem, conservation of momentum, elastic and inelastic collisions
Rotational MotionTorque, rotational inertia, angular momentum, conservation of angular momentum, rolling
OscillationsSHM, mass-spring systems, pendulums, energy in oscillations

AP Physics 1 rewards deep conceptual understanding over memorisation. Focus on building physical intuition — ask yourself why an object behaves as it does before reaching for an equation. Practise explaining your reasoning evidently in writing, as the free-response section heavily weights qualitative explanations alongside quantitative solutions.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

  • Confusing terminology or concepts that appear similar but have distinct meanings.
  • Overlooking key assumptions or boundary conditions that limit applicability.