Flashcards · IB Diploma Programme Physics · Topic 1

A. Space, time and motion flashcards

47 cards on Topic 1, subtopics A.1 to A.5. Pick a subtopic or study them all.

Study mode needs JavaScript. Every card is listed below, so you can still cover the answers and test yourself.

All 47 cards

For a quick read-through or printing. Cover the right-hand side to test yourself.

A.1 Kinematics

Displacement vs distance: Displacement: the change in position, a vector; it can be zero or negative. Distance: the total path length, a scalar; never negative.
A trolley moves 60 m east, then 20 m west, in 16 s. Average speed and average velocity?: Average speed = 80 ÷ 16 = 5.0 m s⁻¹. Average velocity = 40 ÷ 16 = 2.5 m s⁻¹ east.
Average vs instantaneous velocity on a displacement–time graph: Average: the gradient of the chord joining two points. Instantaneous: the gradient of the tangent at that instant.
What do the gradient and the area of a velocity–time graph give?: Gradient: acceleration. Area: displacement (area below the time axis counts as negative).
Equations for uniformly accelerated motion: s = (u + v)t / 2 v = u + at s = ut + ½at² v² = u² + 2as Valid only while the acceleration is constant.
A car at 30 m s⁻¹ brakes at 6.0 m s⁻². Stopping time and distance?: t = 30 ÷ 6.0 = 5.0 s s = 30² ÷ (2 × 6.0) = 75 m
Projectile with no fluid resistance: horizontal and vertical motion: Horizontal: zero acceleration, constant velocity uₓ = u cos θ. Vertical: acceleration g downward, starting from u_y = u sin θ. At the top only the vertical velocity is zero.
Launch at 24 m s⁻¹, 40° above level ground. Time of flight and range?: uₓ = 18.4 m s⁻¹, u_y = 15.4 m s⁻¹ T = 2u_y ÷ g = 3.15 s Range = uₓT = 57.8 m
Qualitative effects of fluid resistance on a projectile: Smaller range and maximum height; the path is no longer a parabola and is steeper on the way down; the horizontal velocity keeps falling; the acceleration is not constant.
Explain how a falling body reaches terminal speed: As speed rises, drag rises, so the resultant force and the acceleration fall. When drag equals weight the acceleration is zero and the speed is constant.

A.2 Forces and momentum

Newton’s three laws of motion: First: a body stays at rest or at constant velocity unless a resultant force acts. Second: F = ma, or generally F = Δp/Δt. Third: the force of A on B is equal and opposite to the force of B on A.
Static and dynamic friction: Static: Ff ≤ μs FN. Dynamic (sliding): Ff = μd FN.
Elastic restoring force and buoyancy: Hooke’s law: FH = −kx. Buoyancy: Fb = ρVg (ρ the fluid’s density, V the volume displaced).
Linear momentum and impulse: p = mv. Impulse J = FΔt, equal to the change in momentum. Momentum is conserved unless a resultant external force acts.
A 0.5 kg ball at 4 m s⁻¹ is hit back at 6 m s⁻¹; contact time 0.02 s. Average force?: Δp = 0.5 × 6 − 0.5 × (−4) = 5 kg m s⁻¹ F = 5 ÷ 0.02 = 250 N
Elastic vs inelastic collisions: Momentum is conserved in both (no external force). Kinetic energy is conserved only in an elastic collision.
2 kg at 3 m s⁻¹ hits a stationary 1 kg trolley; they move off together at 2 m s⁻¹. Elastic?: Momentum: 6 = 6, conserved. KE before 9 J, after 6 J, so the collision is inelastic.
Centripetal acceleration: a = v²/r = ω²r = 4π²r/T², directed towards the centre. v = 2πr/T = ωr.
What provides the centripetal force?: Not a new force: it is whichever real forces (tension, gravity, friction, normal force) give the resultant towards the centre.

A.3 Work, energy and power

Work done by a constant force: W = Fs cos θ, θ the angle between force and displacement. Negative if θ > 90°; zero at 90°.
A 20 N force acts at 30° to a 4.0 m displacement. Work done?: W = 20 × 4.0 × cos 30° = 69.3 J
Kinetic, gravitational and elastic potential energy: Ek = ½mv² = p²/2m ΔEp = mgΔh (near the Earth’s surface) Ee = ½k(Δx)²
When is mechanical energy conserved?: When no frictional or other resistive forces act. Otherwise the change in mechanical energy equals the work done by the non-conservative forces.
Work done by the resultant force: Equals the change in kinetic energy, W = ΔEk.
Two expressions for power: P = ΔW/Δt, and P = Fv for a constant force at constant velocity.
Efficiency: η = useful energy output ÷ total energy input = useful power output ÷ total power input. No unit.
Energy density of a fuel: The energy released per unit volume of fuel (for example MJ per litre). Efficiency then says how much of it becomes useful output.
Sankey diagram: A diagram of energy flow in which arrow widths are proportional to energy; the outputs add up to the input.

A.4 Rigid body mechanics (HL only)

Torque: The turning effect of a force about an axis, τ = Fr sin θ, in N m. (HL only)
Rotational equilibrium: Zero resultant torque. The body can still rotate at constant ω; it need not be at rest. (HL only)
Moment of inertia: How the distribution of mass about an axis resists a change in rotation, in kg m². For point masses I = Σmr². (HL only)
Newton’s second law for rotation: τ = Iα, where τ is the resultant torque and α the angular acceleration. (HL only)
Angular momentum, and when it is conserved: L = Iω, in kg m² s⁻¹. Conserved if no resultant external torque acts. (HL only)
Angular impulse: τΔt, equal to the change in angular momentum. (HL only)
Rotational kinetic energy: Ek = ½Iω² = L²/2I. (HL only)
A spinning body’s I falls to a third, with no external torque. What happens to ω and Ek?: L is conserved, so ω triples, and Ek = L²/2I also triples. The extra energy comes from work done on the body. (HL only)
A hoop (I = mr²) rolls without slipping from rest down a drop of 0.80 m. Speed at the bottom?: mgh = ½mv² + ½mv² = mv², so v = √(gh) = √(9.81 × 0.80) = 2.80 m s⁻¹. (HL only)

A.5 Galilean and special relativity (HL only)

Inertial frame: A frame of reference that is not accelerating. (HL only)
The two postulates of special relativity: The laws of physics are the same in every inertial frame. The speed of light in a vacuum is the same in all inertial frames, whatever the motion of source or observer. (HL only)
Galilean transformation: x′ = x − vt, t′ = t, u′ = u − v. It works at low speeds but fails for light. (HL only)
Lorentz factor and Lorentz transformations: γ = 1/√(1 − v²/c²) x′ = γ(x − vt) t′ = γ(t − vx/c²) (HL only)
Proper time interval: The time between two events measured in the frame where both happen at the same place. It is the shortest interval; Δt = γΔt₀. (HL only)
Proper length: The length of an object measured in the frame where it is at rest. It is the longest; L = L₀/γ along the motion. (HL only)
A clock in a spacecraft ticks every 3.0 μs; the craft moves at 0.80c. Interval measured outside?: γ = 1/√(1 − 0.64) = 1.667, so Δt = 3.0 × 1.667 = 5.0 μs. (HL only)
Relativistic velocity addition: u′ = (u − v) / (1 − uv/c²). The result is always below c. (HL only)
Space-time interval: (Δs)² = (cΔt)² − (Δx)². It is invariant: the same in all inertial frames. (HL only)
Muon evidence for special relativity: In the Earth frame, the muon lifetime is dilated. In the muon frame, the distance is contracted. Both lead to the same result. (HL only)