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IB Physics: A · Space, time and motion

Kinematics, forces, momentum, energy, rigid bodies and relativity (HL). 5 subtopics, 28 key terms, full free lesson below.

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In this topic: A.1 · A.2 · A.3 · A.4 · A.5

A.1Kinematics

Key terms

Displacement (s)
Distance moved in a stated direction from a reference point. A vector, in m.
Velocity (v)
Rate of change of displacement. A vector, in m s⁻¹.
Speed
Rate of change of distance. A scalar.
Acceleration (a)
Rate of change of velocity. A vector, in m s⁻².
Projectile
An object moving only under gravity (and possibly air resistance) after being launched.
\[v=u+at\qquad s=ut+\tfrac12at^{2}\qquad v^{2}=u^{2}+2as\qquad s=\frac{(u+v)}{2}\,t\]

Worked example

A ball is thrown horizontally at 12 m s⁻¹ from the top of a 20 m cliff. Ignoring air resistance, find the time of flight and how far from the cliff it lands.

  1. Vertical: \(u_y=0,\ s=20\text{ m},\ a=9.8\text{ m s}^{-2}\), so \(20=\tfrac12(9.8)t^{2}\).
  2. \(t=\sqrt{\dfrac{40}{9.8}}=\mathbf{2.02\text{ s}}\).
  3. Horizontal: \(s=u_xt=12\times2.02=\mathbf{24.2\text{ m}}\).

Exam tip: In projectile questions, write two columns, horizontal and vertical, and never mix values between them except time.

A.2Forces and momentum

Key terms

Newton's first law
An object stays at rest or moves with constant velocity unless a resultant force acts.
Newton's second law
Resultant force equals rate of change of momentum: F = Δp/Δt (F = ma for constant mass).
Newton's third law
If A exerts a force on B, B exerts an equal and opposite force of the same type on A.
Momentum (p)
p = mv. A vector, in kg m s⁻¹ (= N s).
Impulse (J)
Change in momentum, J = FΔt: the area under a force–time graph.
Centripetal force
The resultant force towards the centre that keeps an object in circular motion.
\[F=ma\qquad F=\frac{\Delta p}{\Delta t}\qquad p=mv\qquad J=F\Delta t=\Delta p\]
\[F_f\le\mu_sF_N\qquad F_f=\mu_dF_N\qquad F_H=-kx\qquad F_d=6\pi\eta rv\qquad F_b=\rho Vg\]
\[v=\frac{2\pi r}{T}=\omega r\qquad a=\frac{v^{2}}{r}=\omega^{2}r=\frac{4\pi^{2}r}{T^{2}}\qquad F=\frac{mv^{2}}{r}=m\omega^{2}r\]

Worked example

A 2.0 kg trolley moving at 3.0 m s⁻¹ hits a stationary 1.0 kg trolley and they stick together. Find their common velocity and the kinetic energy lost.

  1. Momentum before \(=2.0\times3.0=6.0\text{ kg m s}^{-1}\).
  2. \(6.0=3.0\,v\Rightarrow v=\mathbf{2.0\text{ m s}^{-1}}\).
  3. \(E_k\) before \(=\tfrac12(2.0)(3.0)^{2}=9.0\text{ J}\); after \(=\tfrac12(3.0)(2.0)^{2}=6.0\text{ J}\).
  4. Kinetic energy lost \(=\mathbf{3.0\text{ J}}\): the collision is inelastic.

Exam tip: Momentum is a vector: choose a positive direction first and give velocities in the opposite direction a minus sign.

A.3Work, energy and power

Key terms

Work done (W)
Energy transferred by a force: W = Fs cos θ, in J.
Kinetic energy
Energy of motion: ½mv².
Gravitational potential energy
Energy stored due to position in a gravitational field: mgΔh near Earth.
Elastic potential energy
Energy stored in a stretched or compressed object: ½kx².
Power
Rate of doing work or transferring energy, in W (J s⁻¹).
Efficiency
Useful energy (or power) out ÷ total energy (or power) in.
\[W=Fs\cos\theta\qquad E_k=\tfrac12mv^{2}=\frac{p^{2}}{2m}\qquad \Delta E_p=mg\Delta h\qquad E_H=\tfrac12k(\Delta x)^{2}\]
\[P=\frac{\Delta W}{\Delta t}=Fv\qquad \text{efficiency}=\frac{E_{\text{useful}}}{E_{\text{input}}}=\frac{P_{\text{useful}}}{P_{\text{input}}}\]

Worked example

A 60 kg student runs up stairs of vertical height 3.0 m in 4.0 s. Find the gain in potential energy and the student's useful power output.

  1. \(\Delta E_p=mg\Delta h=60\times9.8\times3.0=\mathbf{1764\text{ J}}\approx1.8\text{ kJ}\).
  2. \(P=\dfrac{1764}{4.0}=\mathbf{441\text{ W}}\approx440\text{ W}\).

Exam tip: When an object falls from rest without air resistance, equate mgh to ½mv²: the mass cancels, so v = √(2gh).

A.4Rigid body mechanics HL

Key terms

Torque (τ)
Turning effect of a force: τ = Fr sin θ, in N m.
Moment of inertia (I)
Resistance of a body to angular acceleration: I = Σmr², in kg m².
Angular velocity (ω)
Rate of change of angular displacement, in rad s⁻¹.
Angular momentum (L)
L = Iω, in kg m² s⁻¹. Conserved when no external torque acts.
Rotational equilibrium
Resultant torque is zero.
\[\tau=Fr\sin\theta\qquad \tau=I\alpha\qquad I=\sum mr^{2}\qquad L=I\omega\qquad E_k=\tfrac12I\omega^{2}=\frac{L^{2}}{2I}\]
\[\omega=\omega_0+\alpha t\qquad \theta=\omega_0t+\tfrac12\alpha t^{2}\qquad \omega^{2}=\omega_0^{2}+2\alpha\theta\qquad \Delta L=\tau\Delta t\]

Worked example

A flywheel with I = 0.50 kg m² starts from rest and has a constant torque of 2.0 N m applied for 5.0 s. Find its final angular velocity and rotational kinetic energy.

  1. \(\alpha=\dfrac{\tau}{I}=\dfrac{2.0}{0.50}=4.0\text{ rad s}^{-2}\).
  2. \(\omega=\alpha t=4.0\times5.0=\mathbf{20\text{ rad s}^{-1}}\).
  3. \(E_k=\tfrac12I\omega^{2}=\tfrac12(0.50)(20)^{2}=\mathbf{100\text{ J}}\).

Exam tip: Always check that angles are in radians when you use rotational equations.

A.5Galilean and special relativity HL

Key terms

Reference frame
A coordinate system with a clock, from which an observer measures positions and times.
Inertial frame
A frame that is not accelerating; Newton's first law holds in it.
Proper time (Δt₀)
Time between two events measured in the frame where they happen at the same place.
Proper length (L₀)
Length of an object measured in its own rest frame.
Lorentz factor (γ)
γ = 1/√(1 − v²/c²); always ≥ 1.
Spacetime interval
(ct)² − x², the same for all inertial observers.
\[x'=x-vt\qquad u'=u-v\qquad\text{(Galilean)}\]
\[\gamma=\frac{1}{\sqrt{1-\dfrac{v^{2}}{c^{2}}}}\qquad x'=\gamma(x-vt)\qquad t'=\gamma\left(t-\frac{vx}{c^{2}}\right)\qquad u'=\frac{u-v}{1-\dfrac{uv}{c^{2}}}\]
\[\Delta t=\gamma\Delta t_0\qquad L=\frac{L_0}{\gamma}\qquad (ct')^{2}-(x')^{2}=(ct)^{2}-x^{2}\]

Worked example

A muon with proper lifetime 2.2 μs moves at 0.98c. Find its lifetime measured in the Earth frame and the distance it travels in that time.

  1. \(\gamma=\dfrac{1}{\sqrt{1-0.98^{2}}}=\dfrac{1}{\sqrt{0.0396}}=5.03\).
  2. \(\Delta t=\gamma\Delta t_0=5.03\times2.2=\mathbf{11\ \mu s}\).
  3. Distance \(=0.98\times3.00\times10^{8}\times1.1\times10^{-5}\approx\mathbf{3.3\text{ km}}\), compared with about 650 m without time dilation.

Exam tip: Identify which observer measures the proper time or proper length first: it is always the shorter time or the longer length.

Practise A · Space, time and motion with exam-style questions →

B · The particulate nature of matter →

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