A.1 Kinematics flashcards: IB Physics SL and HL

All 75 A.1 Kinematics cards

1. DefinitionSL & HL

   Define displacement.

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   Change in position of an object (from a fixed origin), in a stated direction. Vector. SI unit: m. Contrast: distance = total path length travelled (scalar).

2. DefinitionSL & HL

   Define velocity.

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   Rate of change of displacement: v = Δs/Δt. Vector (magnitude = speed, plus direction). SI unit: m s⁻¹.

3. DefinitionSL & HL

   Define acceleration.

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   Rate of change of velocity: a = Δv/Δt. Vector. SI unit: m s⁻². Occurs when speed changes, direction changes, or both (e.g. circular motion at constant speed is accelerated).

4. DefinitionSL & HL

   Distinguish between average velocity and instantaneous velocity.

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   Average velocity = total displacement ÷ total time. Instantaneous velocity = velocity at one instant = gradient of the tangent to the displacement–time graph at that point.

5. DefinitionSL & HL

   Distinguish between a scalar and a vector, with two examples of each.

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   Scalar: magnitude only (distance, speed, mass, energy, time). Vector: magnitude AND direction (displacement, velocity, acceleration, force, momentum). Vectors add head-to-tail; components found with cosθ (adjacent) and sinθ (opposite).

6. DefinitionSL & HL

   State the conditions under which the suvat (kinematic) equations may be used.

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   Acceleration must be constant (uniform) in magnitude and direction; motion along a straight line. For projectiles, apply separately to horizontal (a = 0) and vertical (a = −g) components.

7. DefinitionSL & HLData booklet: No – memorise

   Define relative velocity of A with respect to B.

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   v_AB = v_A − v_B (vector subtraction). E.g. two cars towards each other at 20 and 30 m s⁻¹: relative speed 50 m s⁻¹.

8. EquationSL & HLData booklet: Yes

   Write the four kinematic equations for uniformly accelerated motion.

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   s = ut + ½at²; v = u + at; v² = u² + 2as; s = (u + v)t/2. u = initial velocity, v = final velocity (m s⁻¹); a = acceleration (m s⁻²); s = displacement (m); t = time (s). Common misuse: applying when a is not constant (e.g. with air resistance).

9. EquationSL & HLData booklet: Yes

   State the value and direction of the acceleration of free fall near Earth's surface.

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   g = 9.8 m s⁻², directed vertically downward (towards Earth's centre), independent of mass when air resistance is negligible. Data booklet gives g = 9.8 m s⁻². Misuse: using +9.8 when 'up' has been chosen positive.

10. EquationSL & HLData booklet: No – memorise

    How do you find the components of a velocity u at angle θ above the horizontal?

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    u_x = u cosθ (horizontal, constant with no air resistance); u_y = u sinθ (vertical initial, changes by −g per second). Not in booklet – memorise. Misuse: swapping sin and cos when θ is measured from the vertical.

11. EquationSL & HLData booklet: No – derive

    Give an expression for the maximum height and time of flight of a projectile launched at speed u, angle θ, on level ground.

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    H = u²sin²θ/(2g); T = 2u sinθ/g; range R = u²sin2θ/g (maximum at 45°). Not in booklet – derive from suvat with vertical component u sinθ. Misuse: using total u instead of u sinθ.

12. Concept/explainSL & HL

    Explain why a ball thrown vertically upward has zero velocity but non-zero acceleration at its highest point.

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    • Velocity is momentarily zero as direction reverses
    • Gravity still acts, so resultant force ≠ 0
    • Acceleration = g downward throughout (N2)
    • Velocity changing from up to down requires acceleration.
13. Concept/explainSL & HL

    Describe the motion of a projectile launched at an angle, ignoring air resistance.

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    • Horizontal: no force, constant velocity u cosθ
    • Vertical: constant acceleration g downward
    • Components independent
    • Path is a parabola
    • Time of flight fixed by vertical motion
    • Speed minimum at top, equal to horizontal component.
14. Concept/explainSL & HL

    Explain the effect of air resistance on the path of a projectile.

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    • Drag force opposes velocity, so horizontal velocity decreases
    • Reduced range and reduced maximum height
    • Path no longer symmetric: descent steeper than ascent
    • Time of flight reduced (slightly)
    • Landing speed less than launch speed.
15. Concept/explainSL & HL

    Explain why a falling object reaches terminal speed.

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    • Initially only weight acts, a = g
    • Drag increases with speed
    • Resultant force (W − F_d) decreases, so acceleration decreases
    • When F_d = W, resultant = 0
    • Acceleration zero, constant (terminal) speed.
16. Concept/explainSL & HL

    Explain why a ball dropped and a ball projected horizontally from the same height hit the ground at the same time.

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    • Vertical and horizontal motions are independent
    • Both have same initial vertical velocity (zero)
    • Both have same vertical acceleration g
    • Same vertical displacement, so same time from s = ½gt².
17. Worked problemSL & HLData booklet: Yes

    A stone is dropped from rest from a 20 m cliff. Calculate the time to reach the ground (ignore air resistance).

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    s = ut + ½at² with u = 0: 20 = ½(9.8)t² → t² = 4.08 → t = 2.0 s (2 s.f.).

18. Worked problemSL & HLData booklet: Yes

    A ball is launched at 20 m s⁻¹ at 30° above horizontal. Calculate the maximum height.

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    u_y = 20 sin30° = 10 m s⁻¹. At top v_y = 0: v² = u² + 2as → 0 = 100 − 2(9.8)H → H = 100/19.6 = 5.1 m.

19. Worked problemSL & HLData booklet: Yes

    Same ball (20 m s⁻¹ at 30°): calculate the horizontal range on level ground.

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    Time of flight: 0 = 10t − 4.9t² → t = 2.04 s. u_x = 20 cos30° = 17.3 m s⁻¹. R = 17.3 × 2.04 = 35 m (2 s.f.). Check: u²sin60°/g = 35 m.

20. Worked problemSL & HLData booklet: Yes

    A car travelling at 30 m s⁻¹ brakes uniformly to rest in 90 m. Calculate the deceleration and the stopping time.

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    v² = u² + 2as: 0 = 900 + 2a(90) → a = −5.0 m s⁻² (deceleration 5.0 m s⁻²). v = u + at: 0 = 30 − 5t → t = 6.0 s.

21. Worked problemSL & HL

    A boat heads directly across a river at 4.0 m s⁻¹ relative to the water; the river flows at 3.0 m s⁻¹. Find the boat's velocity relative to the bank.

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    Vectors perpendicular: v = √(4² + 3²) = 5.0 m s⁻¹. Direction: tanθ = 3/4 → θ = 37° downstream from the straight-across direction.

22. Worked problemSL & HLData booklet: Yes

    A ball thrown vertically upward at 15 m s⁻¹. Find the time to return to the thrower's hand.

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    Up positive: s = 0 = 15t − ½(9.8)t² → t(15 − 4.9t) = 0 → t = 3.1 s. (Or time up = 15/9.8 = 1.53 s, doubled.)

23. Graph/diagramSL & HL

    On a displacement–time graph, what do gradient, curvature and a horizontal section represent?

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    Gradient = velocity (tangent gives instantaneous velocity). Curved line = changing velocity (accelerating). Horizontal = stationary. Negative gradient = moving back toward origin. A straight sloped line = constant velocity.

24. Graph/diagramSL & HL

    On a velocity–time graph, what do gradient and area represent?

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    Gradient = acceleration (tangent for instantaneous). Area between line and time axis = displacement; area below the axis counts as negative displacement. Total distance = sum of |areas|. Curved line = changing acceleration.

25. Graph/diagramSL & HL

    On an acceleration–time graph, what does the area represent?

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    Area under a–t graph = change in velocity Δv. A horizontal line = uniform acceleration; a = 0 line = constant velocity.

26. Graph/diagramSL & HL

    Describe the velocity–time graph of a ball thrown vertically upward and caught at the same height (up positive, no air resistance).

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    Straight line, constant negative gradient −9.8 m s⁻². Starts at +u, crosses zero at maximum height (half-way in time), ends at −u. Areas above and below axis equal (total displacement zero).

27. Graph/diagramSL & HL

    Describe the velocity–time graph for a skydiver from jump to terminal speed.

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    Starts at origin with gradient g = 9.8 m s⁻². Gradient decreases continuously as drag grows. Curve levels off to a horizontal line at terminal speed (gradient zero). If parachute opens: sharp drop in velocity, then a new lower terminal speed.

28. Graph/diagramSL & HL

    Describe how to linearise s = ½gt² experimentally to find g.

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    Plot s (y-axis) against t² (x-axis). Straight line through origin. Gradient = g/2, so g = 2 × gradient. Intercept should be zero; a non-zero intercept indicates a systematic error (e.g. timing delay or offset in s).

29. Exam technique/trapSL & HL

    What sign convention should you state and how does it affect g?

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    Always state a positive direction (e.g. 'up positive'). Then a = −9.8 m s⁻² for free fall, displacements below the start are negative, and velocities downward are negative. Keep the convention consistent throughout the calculation.

30. Exam technique/trapSL & HL

    Common trap: 'The velocity of a projectile at its highest point is zero.' Correct this.

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    Only the VERTICAL component is zero. The horizontal component u cosθ is unchanged, so speed at the top = u cosθ (minimum speed on the path). Acceleration remains g downward.

31. Exam technique/trapSL & HL

    Command terms: distinguish 'Calculate', 'Determine', 'Estimate' and 'Show that'.

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    Calculate: obtain numerical answer with working. Determine: obtain the only possible answer (may need graph/data). Estimate: approximate value, order of magnitude reasoning acceptable. Show that: present full working to reach a value MORE precise than the one given (e.g. 'show that ≈ 2.0 s' → give 2.02 s).

32. Exam technique/trapSL & HL

    Experimental skill: how do you find the uncertainty in a gradient from a graph with error bars?

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    Draw best-fit line, then steepest and shallowest lines that still pass through all error bars. Δgradient = (m_max − m_min)/2. Quote gradient ± Δgradient with consistent s.f. Same method for intercept.

33. Exam technique/trapSL & HLData booklet: Yes

    Experimental skill: state the rules for combining uncertainties.

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    Adding/subtracting: add ABSOLUTE uncertainties. Multiplying/dividing: add FRACTIONAL (percentage) uncertainties. Powers: multiply fractional uncertainty by the power (Δ(x²)/x² = 2Δx/x). Constants do not change fractional uncertainty.

34. Exam technique/trapSL & HL

    Practical: outline a method to measure g by free fall and give two limitations.

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    Method: electromagnet releases steel ball, timer starts; ball breaks contact/light gate at bottom stops timer; vary height h, plot h vs t², g = 2 × gradient. Limitations: release delay (systematic, shifts intercept), air resistance (small for dense ball), measuring h to ball's centre. Prefer light gates or video over stopwatch (human reaction ≈ 0.2 s).

35. DefinitionSL & HL

    Define distance and speed, and state precisely how each differs from displacement and velocity.

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    Distance is the total length of path travelled; speed is distance travelled per unit time. Both are scalars (magnitude only), SI unit m and m s⁻¹. Displacement is the straight-line change in position from start to finish (vector, m) and velocity is rate of change of displacement (vector, m s⁻¹). Exam tip: marks are lost for writing "distance in a given direction" as displacement without saying it is measured from the initial position, and for calling speed "the rate of change of distance" without "travelled". Over a closed path distance > 0 but displacement = 0, so average speed > 0 while average velocity = 0.

36. DefinitionSL & HL

    Define free fall and state exactly what must be true for motion to count as free fall.

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    Free fall is motion under the influence of gravitational force only, with no other force (in particular no fluid resistance or upthrust) acting. Acceleration is then g = 9.8 m s⁻² directed vertically downward, independent of the mass, shape or initial velocity of the body. Exam tip: students lose the mark by saying "falling" or "dropped from rest" — an object thrown upward or a projectile is still in free fall while only gravity acts. Also state that free fall requires the object to be near the Earth's surface for g to be treated as constant; g is an acceleration (vector, m s⁻²), not a force.

37. DefinitionSL & HL

    Define terminal speed and state the condition that defines it.

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    Terminal speed is the constant (maximum) speed reached by a body falling through a fluid when the upward drag force plus upthrust equals the downward weight, so the resultant force and therefore the acceleration are zero. Exam tip: the mark is for "resultant force = 0 hence a = 0", not merely "drag equals weight" without the consequence; saying "the forces balance so it stops" scores zero. Terminal speed is a speed (scalar, m s⁻¹) and is approached asymptotically — it is never exactly reached in finite time, so sketch graphs must show a curve flattening, never a kink.

38. DefinitionSL & HL

    Define instantaneous acceleration as a limit, and state how it is obtained from a graph.

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    Instantaneous acceleration is the rate of change of velocity at a particular instant: a = lim(Δt→0) Δv/Δt = dv/dt = d²s/dt². It is a vector, SI unit m s⁻². Graphically it is the gradient of the tangent to the velocity–time graph at that instant. Exam tip: the common error is defining acceleration as "change in velocity" (that is Δv) or as "change in speed per unit time" — a body moving in a circle at constant speed has non-zero acceleration. State that a is in the direction of Δv, not of v: a body slowing down has a antiparallel to v.

39. DefinitionSL & HL

    Define a frame of reference and explain what makes one inertial.

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    A frame of reference is a coordinate system (origin, axes and clock) with respect to which positions, velocities and times are measured. An inertial frame is one in which Newton's first law holds: a body with no resultant force moves with constant velocity; equivalently the frame is not accelerating. Exam tip: for full marks say measurements of displacement and velocity are frame-dependent while, in Galilean relativity, time intervals, accelerations and forces are the same in all inertial frames. Students lose marks by calling the ground "the correct frame" — no inertial frame is privileged; a frame fixed to a braking car is non-inertial.

40. DefinitionSL & HL

    Define fluid resistance (drag) and list the factors that determine its magnitude.

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    Fluid resistance is the force exerted on a body by a fluid (liquid or gas) opposing its motion relative to that fluid; it acts antiparallel to the velocity of the body relative to the fluid. Its magnitude increases with speed, with cross-sectional area presented to the flow, with fluid density and with a shape-dependent drag coefficient. Exam tip: say "opposes relative motion", not "opposes motion" — a body falling in a rising airstream still experiences upward drag. Drag is a force (vector, N). Do not confuse drag with upthrust, which is a buoyancy force acting upward independently of speed.

41. DefinitionSL & HL

    Define uniform acceleration and explain the difference between deceleration and negative acceleration.

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    Uniform (constant) acceleration means the velocity changes by equal amounts in equal time intervals, so a is constant in both magnitude and direction. Deceleration means the speed is decreasing, i.e. acceleration is antiparallel to velocity. Negative acceleration means a is negative with respect to the chosen positive direction, which is not the same thing. Exam tip: a ball falling downward with up taken positive has a = −9.8 m s⁻² and is speeding up, so it has negative acceleration but is not decelerating. Always define the positive direction before assigning signs; unit m s⁻², vector.

42. DefinitionSL & HL

    Define the trajectory, range and time of flight of a projectile.

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    The trajectory is the path followed by the projectile through space (a parabola when fluid resistance is negligible). The range is the horizontal displacement from launch to the point where the projectile returns to the launch level (or reaches the specified landing level), unit m. Time of flight is the total time in the air between launch and landing, unit s. Exam tip: the frequent error is quoting "range" for launch and landing at different heights using R = u²sin2θ/g, which applies only to level ground. Range is a horizontal displacement, not the length of the curved path travelled.

43. DefinitionSL & HL

    Define average velocity and average speed and show, with a numerical example, that they are not simply related.

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    Average velocity = total displacement ÷ total time (vector, m s⁻¹); average speed = total distance travelled ÷ total time (scalar, m s⁻¹). Example: a runner completes one lap of a 400 m track in 50 s — average speed = 8.0 m s⁻¹, average velocity = 0 because the displacement is zero. Exam tip: students habitually average the initial and final speeds. (u + v)/2 gives the average velocity only for uniform acceleration; for a journey of two equal distances at different speeds the correct average speed is the harmonic mean 2v₁v₂/(v₁ + v₂), not (v₁ + v₂)/2.

44. EquationSL & HLData booklet: No – derive

    State the range equation for a projectile on level ground and the conditions under which it is valid.

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    R = u²sin(2θ)/g. R = horizontal range (m); u = launch speed (m s⁻¹); θ = launch angle above the horizontal (°/rad); g = 9.8 m s⁻² (magnitude of free-fall acceleration). Valid only when: fluid resistance is negligible, g is constant, and launch and landing are at the SAME height. Data booklet: not given — derive from R = u cosθ × t with t = 2u sinθ/g. Common misuse: applying it to a projectile fired from a cliff. Sanity check: u = 20 m s⁻¹, θ = 45° gives R = 400 × 1/9.8 ≈ 41 m; θ = 30° and θ = 60° give the same R ≈ 35 m since sin60° = sin120°.

45. EquationSL & HLData booklet: No – derive

    Write the equation of the trajectory y(x) of a projectile launched at speed u and angle θ from the origin.

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    y = x tanθ − gx²/(2u²cos²θ). y = vertical displacement (m); x = horizontal displacement (m); u = launch speed (m s⁻¹); θ = launch angle; g = 9.8 m s⁻². Obtained by eliminating t between x = u cosθ·t and y = u sinθ·t − ½gt². It is of the form y = ax − bx², i.e. a parabola, valid only with negligible fluid resistance and constant g. Data booklet: not given — derive. Common misuse: substituting the total speed u where u cosθ is required in the x²-term. Check: at θ = 0 it reduces to y = −gx²/(2u²), the horizontal-launch case.

46. EquationSL & HLData booklet: No – memorise

    Write the vector equation for relative velocity and explain how to apply it in two dimensions.

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    v_AB = v_A − v_B, the velocity of A relative to B, where v_A and v_B are the velocities of A and B measured in the same frame (m s⁻¹, vectors). In two dimensions resolve into perpendicular components and subtract component by component, or draw v_A and −v_B tip-to-tail and measure the resultant. Note v_BA = −v_AB, so the two have equal magnitude and opposite direction. Data booklet: not given — memorise. Common misuse: adding speeds for objects moving at an angle; only for antiparallel motion does the magnitude become v_A + v_B. Check: two cars both at 25 m s⁻¹ in the same direction give v_AB = 0.

47. EquationSL & HLData booklet: No – derive

    Write expressions for the horizontal and vertical velocity components of a projectile at time t after launch (u, θ, no fluid resistance).

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    vₓ = u cosθ (constant); v_y = u sinθ − gt (taking up as positive). vₓ, v_y in m s⁻¹; u = launch speed (m s⁻¹); θ = launch angle; g = 9.8 m s⁻²; t = time since launch (s). Speed at time t is v = √(vₓ² + v_y²) and the direction is at angle arctan(v_y/vₓ) to the horizontal. Valid only while fluid resistance is negligible. Data booklet: the underlying suvat equations are given; these component forms are derived. Common misuse: applying v = u + at to the horizontal direction — aₓ = 0, so vₓ never changes. Check: v_y = 0 at the top gives t = u sinθ/g.

48. EquationSL & HLData booklet: No – memorise

    State the drag-force model F_d = ½C_dρAv² and derive the expression for terminal speed from it.

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    F_d = ½C_dρAv²: F_d = drag force (N); C_d = drag coefficient (dimensionless, shape-dependent); ρ = fluid density (kg m⁻³); A = cross-sectional area normal to the flow (m²); v = speed relative to the fluid (m s⁻¹). At terminal speed the resultant force is zero, so mg = ½C_dρAv_t² giving v_t = √(2mg/(C_dρA)). Valid for turbulent flow at high Reynolds number; at very low speeds drag is instead ∝ v. Data booklet: not given — values are supplied in the question. Common misuse: assuming v_t ∝ m; in fact v_t ∝ √m, so doubling the mass raises terminal speed by only √2 ≈ 1.4.

49. EquationSL & HLData booklet: No – memorise

    Express instantaneous velocity and acceleration in calculus form and state how each is obtained from motion graphs.

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    v = ds/dt and a = dv/dt = d²s/dt². s = displacement (m); v = velocity (m s⁻¹); a = acceleration (m s⁻²); t = time (s). Conversely s = ∫v dt and v = ∫a dt, which is why the area under a v–t graph is displacement and the area under an a–t graph is change in velocity. Valid for any motion, uniform acceleration or not — unlike suvat. Data booklet: not given in this form; the suvat set is given. Common misuse: using suvat on non-uniform acceleration (e.g. a skydiver before terminal speed) instead of taking gradients and areas from the graph.

50. Concept/explainSL & HL

    Explain why the horizontal and vertical motions of a projectile can be treated independently.

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    • The only force acting (with negligible fluid resistance) is the weight, which is vertical
    • therefore the horizontal component of the resultant force is zero, so aₓ = 0 and the horizontal velocity is constant
    • the vertical component of the resultant force is mg, so a_y = g downward at all times
    • acceleration is a vector, and perpendicular components of a vector are independent of one another
    • the two motions share only the single variable time t, which links them
    • so suvat is applied separately in each direction, with t as the bridge. Exam tip: incomplete answers assert independence as a rule without identifying that gravity has no horizontal component; note that once drag acts the motions are coupled through the speed and are no longer independent.
51. Concept/explainSL & HL

    Explain why an object moving at constant speed may still be accelerating, using two contrasting examples.

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    • Acceleration is the rate of change of velocity, and velocity is a vector with magnitude and direction
    • a change in direction alone is a change in velocity, so acceleration is non-zero
    • example 1: a car rounding a bend at a steady 15 m s⁻¹ accelerates toward the centre of the bend
    • example 2: a projectile at its highest point has zero vertical velocity but a = 9.8 m s⁻² downward
    • the direction of a is the direction of Δv, not of v
    • only motion in a straight line at unchanging speed has a = 0. Exam tip: the incomplete answer equates acceleration with "getting faster"; the mark scheme awards the mark specifically for identifying velocity as a vector and a change in direction as a change in velocity.
52. Concept/explainSL & HL

    Explain, in terms of forces, how the trajectory of a projectile launched at an angle changes when fluid resistance is significant.

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    • Drag acts opposite to the instantaneous velocity, so it has both horizontal and vertical components
    • the horizontal component decelerates the projectile, so vₓ is no longer constant and decreases throughout
    • the maximum height is reduced because drag adds to the retarding force on the way up
    • on the way down drag opposes gravity, so the downward acceleration is less than g and falls toward zero as terminal speed is approached
    • the range is reduced
    • the path becomes asymmetric: the descending branch is steeper and shorter horizontally than the ascending branch
    • the angle of impact is steeper than the launch angle and the impact speed is less than the launch speed. Exam tip: many students say only "lower and shorter"; the asymmetry and the loss of horizontal velocity are the discriminating marks.
53. Concept/explainSL & HL

    Explain why 45° gives maximum range on level ground with no fluid resistance, and why the optimum angle is smaller when drag acts.

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    • Range R = u²sin2θ/g, which is maximum when sin2θ = 1, i.e. 2θ = 90° so θ = 45°
    • physically, increasing θ increases the time of flight but decreases the horizontal velocity component, and 45° is the best compromise
    • angles equidistant from 45° (e.g. 30° and 60°) give equal ranges
    • with drag, the longer a projectile is in flight the more horizontal velocity it loses
    • a high-angle launch therefore wastes speed on a long flight against drag
    • the optimum angle falls (typically to ≈ 30–40° for a ball, lower still for a fast projectile). Exam tip: students quote 45° without conditions; state "level ground and negligible fluid resistance" or the mark is withheld.
54. Concept/explainSL & HL

    Explain why rain falling vertically appears to a moving cyclist to come at an angle from in front.

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    • The rain has velocity v_R vertically downward in the ground frame
    • the cyclist has velocity v_C horizontally in the ground frame
    • what the cyclist observes is the velocity of the rain relative to the cyclist, v_RC = v_R − v_C
    • subtracting v_C is equivalent to adding a horizontal component of magnitude v_C directed backward relative to the cyclist's motion
    • the resultant therefore points downward and toward the cyclist, at angle arctan(v_C/v_R) from the vertical
    • the faster the cyclist rides, the closer the apparent direction is to the horizontal
    • the cyclist tilts an umbrella forward by this angle. Exam tip: this illustrates that velocity is frame-dependent while the rain's actual motion is unchanged; incomplete answers claim the rain "really" changes direction.
55. Concept/explainSL & HL

    Explain why two skydivers of different mass but identical shape and area reach different terminal speeds.

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    • At terminal speed drag equals weight: ½C_dρAv_t² = mg
    • so v_t = √(2mg/(C_dρA)) and, with C_d, ρ and A identical, v_t ∝ √m
    • the heavier skydiver has a greater weight, so a greater drag force and hence a greater speed is needed to balance it
    • the heavier one therefore reaches a higher terminal speed and takes longer to reach it
    • doubling the mass increases terminal speed by only a factor of √2 ≈ 1.4, not 2
    • adopting a spread-eagle posture increases A and lowers v_t; diving head-first lowers A and raises v_t. Exam tip: students often say heavier objects "fall faster" generally — in free fall (no fluid) both accelerate at g regardless of mass; mass matters only because drag depends on speed and area, not mass.
56. Concept/explainSL & HL

    Explain how the signs of velocity and acceleration together determine whether a body speeds up or slows down.

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    • First define a positive direction; v and a are then signed quantities along that axis
    • if v and a have the same sign the body speeds up
    • if v and a have opposite signs the body slows down
    • if a = 0 the speed is constant
    • a negative acceleration therefore does not always mean slowing down: a ball falling with up positive has v < 0 and a < 0 and gains speed
    • a body moving in the negative direction with positive acceleration is slowing, may stop, and then reverses
    • formally, the speed increases when v·a > 0. Exam tip: the standard error is to equate "negative acceleration" with "deceleration"; state the sign convention explicitly in the answer, as the mark scheme awards a mark for it in multi-stage motion questions.
57. Concept/explainSL & HL

    Explain why suvat equations cannot be applied to a skydiver before terminal speed, and state what may be used instead.

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    • The suvat equations are derived assuming acceleration is constant in magnitude and direction
    • for a falling body in air the drag force grows as speed grows, so the resultant force and hence the acceleration continuously decrease
    • acceleration falls from g at the instant of release toward zero at terminal speed
    • using v = u + at with a = 9.8 m s⁻² therefore overestimates the speed and the distance fallen
    • instead use a velocity–time graph: gradient gives instantaneous acceleration and area under the curve gives displacement
    • or apply Newton's second law instant by instant: ma = mg − ½C_dρAv²
    • suvat becomes valid again once terminal speed is reached, with a = 0. Exam tip: students blindly quote suvat; "only for uniform acceleration" is an explicit syllabus condition.
58. Concept/explainSL & HL

    Explain the Nature of Science significance of Galileo's inclined-plane experiments for the study of free fall.

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    • Galileo could not time free fall accurately with the water clocks available, so he "diluted" gravity by rolling balls down a gentle incline
    • this slowed the motion so that distances and times could be measured reliably
    • he found s ∝ t² for a body released from rest, consistent with uniform acceleration
    • extrapolating the incline to 90° gave the free-fall case, an early use of extrapolation and idealisation in modelling
    • friction and rolling were treated as negligible, showing how simplifying assumptions make a model tractable
    • the result contradicted Aristotle's claim that heavier bodies fall proportionately faster
    • later evidence (the vacuum tube, the Apollo 15 hammer-and-feather demonstration) confirmed it. Exam tip: credit is for the method — controlling variables and idealising — not for retelling the story.
59. Concept/explainSL & HL

    Explain why a projectile passes a given height with the same speed going up as coming down, but with a different velocity.

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    • With negligible fluid resistance the only acceleration is g downward, constant throughout
    • the horizontal component vₓ is unchanged at all times
    • applying v_y² = u_y² + 2a(Δy) with the same Δy gives the same magnitude of v_y on the ascent and descent
    • hence the speed √(vₓ² + v_y²) is identical at equal heights
    • but v_y has opposite sign, so the velocity vectors differ in direction: the path makes equal angles above and below the horizontal
    • the motion is symmetric in time about the highest point, so the time up equals the time down
    • this symmetry is destroyed once drag acts, and the descending speed is then smaller. Exam tip: a question asking for velocity requires magnitude AND direction; giving only the speed loses a mark.
60. Concept/explainSL & HL

    Explain why the time of flight of a projectile launched horizontally from a cliff does not depend on its launch speed.

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    • The vertical and horizontal motions are independent because weight has no horizontal component
    • vertically the projectile starts with u_y = 0 and accelerates at g regardless of how fast it is launched horizontally
    • the time to fall a height h is therefore found from h = ½gt², giving t = √(2h/g), which contains no term in u
    • so a stone launched at 5 m s⁻¹ and one at 50 m s⁻¹ from the same cliff land at the same instant
    • only the horizontal range x = u·t differs, being proportional to u
    • this holds only while fluid resistance is negligible, since drag couples the two motions. Exam tip: the answer must identify u_y = 0 and quote t = √(2h/g); simply asserting "the motions are independent" scores at most one mark.
61. Worked problemSL & HL

    A stone is projected horizontally at 15 m s⁻¹ from the top of a 45 m cliff. Calculate the horizontal distance travelled and the speed and direction of impact with the level ground below.

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    Take down as positive; the motions are independent. Vertically u_y = 0, a = 9.8 m s⁻², s = 45 m: 45 = ½ × 9.8 × t² → t² = 9.18 s² → t = 3.03 s. Horizontally aₓ = 0: x = vₓt = 15 × 3.03 = 45.5 ≈ 45 m. Impact vertical velocity: v_y = u_y + at = 9.8 × 3.03 = 29.7 m s⁻¹ down. Impact speed v = √(15² + 29.7²) = √(225 + 882) = √1107 = 33.3 ≈ 33 m s⁻¹. Direction: arctan(29.7/15) = 63° below the horizontal. Check/Trap: the horizontal velocity is unchanged, so never use v = u + at horizontally; the impact speed is not 15 + 29.7 but the vector sum.

62. Worked problemSL & HL

    A ball is kicked at 25 m s⁻¹ at 40° above the horizontal. Calculate its velocity 1.5 s after the kick (ignore fluid resistance).

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    Resolve the launch velocity: uₓ = 25 cos40° = 19.2 m s⁻¹, u_y = 25 sin40° = 16.1 m s⁻¹ (up positive). Horizontally aₓ = 0 so vₓ = 19.2 m s⁻¹ throughout. Vertically v_y = u_y − gt = 16.1 − 9.8 × 1.5 = 16.1 − 14.7 = 1.4 m s⁻¹ (still upward). Magnitude: v = √(19.2² + 1.4²) = √(369 + 2.0) = √371 = 19.3 ≈ 19 m s⁻¹. Direction: arctan(1.4/19.2) = 4.1° above the horizontal. Check/Trap: v_y is still positive, so the ball has not yet reached its highest point — that occurs at t = 16.1/9.8 = 1.64 s. A velocity answer must include the direction, not just 19 m s⁻¹.

63. Worked problemSL & HL

    An aircraft flies at 60 m s⁻¹ relative to the air. A wind of 15 m s⁻¹ blows from west to east. Determine the heading the pilot must take to travel due north, and the resulting ground speed.

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    The ground velocity is the vector sum: v_ground = v_air-relative + v_wind. For the resultant to point due north, the westward component of the aircraft's air velocity must cancel the 15 m s⁻¹ eastward wind. So 60 sinθ = 15, where θ is the heading west of north: sinθ = 0.25, θ = 14.5° ≈ 14° west of north. Ground speed is the northward component: v = 60 cos14.5° = 58.1 ≈ 58 m s⁻¹, or by Pythagoras √(60² − 15²) = √3375 = 58.1 m s⁻¹. Check/Trap: the ground speed is less than the airspeed even though the wind is perpendicular to the intended track — crosswinds always cost ground speed. Do not add 60 and 15 or subtract them; they are perpendicular only after the correction.

64. Worked problemSL & HL

    A ball is thrown vertically downward at 5.0 m s⁻¹ from a window 30 m above the ground. Calculate the time to reach the ground and the impact speed.

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    Take down as positive: u = 5.0 m s⁻¹, a = 9.8 m s⁻², s = 30 m. Use s = ut + ½at²: 30 = 5.0t + 4.9t², so 4.9t² + 5.0t − 30 = 0. t = [−5.0 + √(25 + 4 × 4.9 × 30)]/(2 × 4.9) = (−5.0 + √613)/9.8 = (−5.0 + 24.8)/9.8 = 2.02 ≈ 2.0 s (reject the negative root). Impact speed from v² = u² + 2as = 25 + 2 × 9.8 × 30 = 613, v = 24.8 ≈ 25 m s⁻¹ downward. Check/Trap: dropped from rest it would take 2.47 s, so 2.0 s is sensible. Always discard the negative time root and state why; keep the sign convention consistent throughout.

65. Worked problemSL & HL

    A projectile is launched at 20 m s⁻¹ at 30° above the horizontal from the edge of a cliff 25 m above the sea. Calculate the total time of flight and the horizontal distance from the cliff base to the landing point.

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    Components: uₓ = 20 cos30° = 17.3 m s⁻¹, u_y = 20 sin30° = 10.0 m s⁻¹. Take up as positive; the landing displacement is s_y = −25 m. Using s = ut + ½at²: −25 = 10.0t − 4.9t², so 4.9t² − 10.0t − 25 = 0. t = [10.0 + √(100 + 490)]/9.8 = (10.0 + 24.3)/9.8 = 3.50 s. Horizontal distance: x = uₓt = 17.3 × 3.50 = 60.6 ≈ 61 m. Check/Trap: the range formula R = u²sin2θ/g gives 35 m and is wrong here because launch and landing heights differ. The vertical displacement is −25 m, not +25 m or 25 m plus the maximum height; sign errors here are the single most common cause of lost marks.

66. Worked problemSL & HL

    A skydiver of total mass 80 kg falls with drag F = ½C_dρAv², where C_d = 1.0, ρ = 1.2 kg m⁻³ and A = 0.70 m². Calculate the terminal speed and the initial acceleration at the instant the speed is half of it.

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    At terminal speed the resultant force is zero: mg = ½C_dρAv_t². So v_t = √(2mg/(C_dρA)) = √(2 × 80 × 9.8/(1.0 × 1.2 × 0.70)) = √(1568/0.84) = √1867 = 43.2 ≈ 43 m s⁻¹. At v = v_t/2 the drag is ½C_dρA(v_t/2)² = mg/4 = 196 N. Newton's second law: ma = mg − F_d = 784 − 196 = 588 N, so a = 588/80 = 7.35 ≈ 7.4 m s⁻² downward. Check/Trap: drag ∝ v², so halving the speed quarters the drag — a is three-quarters of g, not half. The acceleration is not constant, so suvat cannot be used to find how long reaching v_t takes.

67. Worked problemSL & HL

    A car starts from rest and accelerates uniformly at 2.5 m s⁻² for 8.0 s, then travels at constant velocity for a further 12 s. Calculate the total distance travelled and the average velocity for the whole 20 s.

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    Stage 1: v = u + at = 0 + 2.5 × 8.0 = 20 m s⁻¹. Distance s₁ = ut + ½at² = 0 + ½ × 2.5 × 8.0² = 80 m (or use area of the triangle on the v–t graph). Stage 2: constant velocity, s₂ = vt = 20 × 12 = 240 m. Total distance = 80 + 240 = 320 m. Average velocity = total displacement/total time = 320/20 = 16 m s⁻¹ in the direction of travel. Check/Trap: do NOT apply suvat across the whole 20 s — the acceleration is not uniform over that interval. Also 16 m s⁻¹ is not the mean of 0 and 20 m s⁻¹ by coincidence only; splitting the v–t graph into a triangle plus a rectangle is the safest method.

68. Graph/diagramSL & HL

    Describe the displacement–time, velocity–time and acceleration–time graphs for a ball dropped onto a hard floor and bouncing twice (up positive, drag negligible).

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    Displacement–time: starts at the drop height, curves downward with increasing steepness to zero at the first impact, then a series of parabolic arcs of decreasing height and decreasing width. Velocity–time: straight lines of constant gradient −9.8 m s⁻² between bounces; at each impact an almost vertical jump from a large negative velocity to a smaller positive velocity, the shortening of each sawtooth showing energy loss. Acceleration–time: constant at −9.8 m s⁻² during flight, with a brief large positive spike at each impact. Extract quantities: the gradient of the v–t line gives g; the area of each triangle gives the rebound height; the ratio of successive rebound speeds gives the coefficient of restitution. If the floor were softer the impact spikes would be lower and wider.

69. Graph/diagramSL & HL

    Explain how to determine an instantaneous velocity from a curved displacement–time graph, and how uncertainty arises.

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    Axes: displacement s/m on the y-axis against time t/s on the x-axis. A curve of increasing gradient shows increasing velocity. Instantaneous velocity is the gradient of the tangent drawn to the curve at that instant: draw the tangent so it touches at one point only, extend it across as much of the grid as possible and use a large triangle, v = Δs/Δt. The larger the triangle, the smaller the fractional uncertainty in reading the coordinates. Uncertainty is estimated by drawing the steepest and shallowest reasonable tangents and halving the difference in gradient. A chord between two nearby points instead gives the AVERAGE velocity over that interval, which systematically differs from the tangent value on a curve; a horizontal tangent means instantaneously at rest.

70. Graph/diagramSL & HL

    Sketch and describe the graphs of the horizontal and vertical velocity components of a projectile against time (no fluid resistance), and state how each changes when drag acts.

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    Axes: velocity/m s⁻¹ on the y-axis, time/s on the x-axis, up and forward taken positive. Horizontal component: a horizontal straight line at vₓ = u cosθ, gradient zero (aₓ = 0); the area under it is the horizontal range. Vertical component: a straight line of gradient −9.8 m s⁻², starting at u sinθ, crossing zero at the highest point and ending at −u sinθ on level ground; the area above the axis equals the maximum height and the total area is zero. With significant fluid resistance the horizontal line becomes a curve decaying toward zero, and the vertical line becomes a curve whose gradient magnitude decreases, tending to a constant terminal value; symmetry about the peak is lost.

71. Graph/diagramSL & HL

    Describe the acceleration–time graph for a falling body from release to terminal speed and explain how to use it.

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    Axes: acceleration/m s⁻² (down positive) on the y-axis against time/s on the x-axis. The graph starts at 9.8 m s⁻² at the instant of release, when the speed and therefore the drag are zero, and decreases as a smooth curve that flattens asymptotically toward zero as terminal speed is approached; it never becomes negative and never crosses the axis. The area under the curve between two times equals the change in velocity over that interval, so the total area from release equals the terminal speed. A more massive body of the same shape has a curve that decays more slowly to zero, giving a larger area and a higher terminal speed; a larger area A presented to the flow makes the curve decay faster to a lower terminal speed.

72. Exam technique/trapSL & HL

    Practical: design an investigation to test how the range of a projectile depends on launch angle. State apparatus, method, controls, limitations and improvements.

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    Apparatus: spring launcher clamped to a bench with an adjustable protractor scale, steel ball, metre rules, sand tray or carbon paper on the floor, video camera. Method: fix the launcher at the floor's level, set θ, fire, mark the landing point, repeat 5 times and take the mean range; vary θ in 10° steps from 10° to 80°; plot R against θ and against sin2θ. Independent variable θ, dependent variable R; control launch speed (same spring compression every time), same ball, same launch height. Expect a straight line through the origin for R against sin2θ, gradient u²/g, giving u. Limitations: launch speed drifts as the spring fatigues; air resistance lowers ranges at large θ; launch height may not equal landing height. Improvements: use video at 240 fps to measure u directly; use a light-gate pair at the muzzle.

73. Exam technique/trapSL & HL

    Exam trap: students apply R = u²sin2θ/g or "time up = time down" to a projectile launched from a cliff or from a moving platform. Explain the error and the correct method.

    Show answer

    The trap: both results are derived assuming launch and landing at the SAME height and no fluid resistance, but students memorise them as universal projectile facts. They fall for it because the symmetric level-ground case dominates textbook examples. Correct approach: always set up the vertical equation with the actual displacement, e.g. s_y = −h for a cliff, and solve the quadratic 0 = u sinθ·t − ½gt² + h for t, taking the positive root; then x = u cosθ·t. For a launch from a moving platform, add the platform velocity vectorially to the launch velocity first. Command-term guidance: "Determine" allows any valid route but demands working, "Show that" requires you to reach the printed value with one more significant figure than quoted, and "Estimate" accepts one significant figure with a stated assumption.

74. Exam technique/trapSL & HL

    Exam trap: propagating uncertainties when g is found from s = ½gt². Show the correct treatment and the usual mistake.

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    Rearranging, g = 2s/t². The fractional uncertainty rule for products and powers gives Δg/g = Δs/s + 2(Δt/t) — the power 2 multiplies the fractional uncertainty in t, which students routinely omit, so they under-report the uncertainty by a factor approaching two. Worked example: s = 1.500 ± 0.005 m, t = 0.553 ± 0.010 s gives g = 9.81 m s⁻², Δg/g = 0.0033 + 2(0.0181) = 0.039, so Δg = 0.39 and the result is g = 9.8 ± 0.4 m s⁻². Quote the absolute uncertainty to one significant figure and round the value to the same decimal place. Note reaction time is a systematic (not random) error if the operator consistently stops the clock late; repeating readings reduces random scatter but never removes systematic bias.

75. Exam technique/trapSL & HL

    Exam trap: choosing the wrong suvat equation and misusing s = (u + v)t/2. Give a reliable selection strategy.

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    The trap: s = (u + v)t/2 is valid ONLY for uniform acceleration; students apply it to multi-stage journeys or to falling bodies with drag, where the average velocity is not the mean of the end values. Strategy: for each stage, list the five quantities s, u, v, a, t, mark the three you know and the one you want, then pick the equation that omits the fifth — v = u + at omits s, s = ut + ½at² omits v, v² = u² + 2as omits t, and s = (u + v)t/2 omits a (use it when a is unknown or not required). Always define the positive direction first and keep signs consistent. Command-term guidance: "Calculate" needs a numerical answer with unit and correct significant figures (match the least precise datum, usually 2 or 3); "Sketch" needs correct shape with labelled axes but not plotted points.

Practise this topic with exam-style questions: A.1 Kinematics questions (SL) · A.1 Kinematics questions (HL) · all flashcards

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