IB Physics HL · guided topic map
Mechanics for IB Physics HL
Mechanics for IB Physics HL, organized into 1 syllabus topic and 3 mapped concept guides.
- Syllabus topics
- 1
- Mapped concept guides
- 3
- Educational level
- IB Diploma Physics Higher Level
Syllabus to lesson
Choose the exact concept
Work in order or jump to the concept named in your specification, course outline, or assignment.
A.1Kinematics
Space, time and motion
3 guides+
Kinematics
Space, time and motion
Diagrams
Mechanics as IB Physics HL draws it
The figures from the IB Physics HL practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.
01Figure 2KinematicsIB
Figure comment
Figure 2Side view of the apparatus. A vertical rod in a heavy base carries a horizontal clamp arm, and a string hangs from the clamp jaws with a small spherical bob tied to its lower end. A dimension line beside the string marks the length L, running from the point of suspension down to the centre of the bob. A dashed line shows the string displaced to one side and a dashed arc through the bob shows the path it follows as it swings.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Guided questions 5 parts
Reading cue. L runs from the jaws of the clamp to the centre of the bob, not to its top: the radius of the bob is part of the length, and the dashed arc shows the path, not an angle.
aState A student can only reach the string with a metre rule. State the two lengths that must be added together to give the L marked on the drawing.
Check answer 2 marks
- the length of string from the jaws of the clamp down to the top of the bob
- plus the radius of the bob
bOutline Outline why the timing of each swing should be started and stopped as the bob passes the lowest point of the dashed arc rather than at either end of it.
Check answer 3 marks
- the bob moves fastest at the lowest point, so the uncertainty in judging the instant of passing is smallest
- at the ends of the arc the bob is momentarily at rest, so the eye cannot fix the instant it turns
- a fiducial mark placed at the lowest point makes the judgement repeatable from swing to swing
cDetermine A student grips the string in the jaws, measures 0.788 m from the jaws to the top of a bob of radius 12 mm, and uses that as L with a measured period of 1.794 s. Determine the percentage error this introduces into g.
Check answer 3 marks
- true length L = 0.788 + 0.012 = 0.800 m
- g = 4π²L / T², so with T fixed the fractional error in g equals the fractional error in L
- error = 0.012 / 0.800 = 1.5%, and g comes out as 9.67 m s⁻² instead of 9.81 m s⁻² — too low
dDiscuss The clamp arm is carried on a vertical rod in a heavy base. Discuss the effect on the measured period if the base were light enough to let the rod rock slightly in time with the swing.
Check answer 4 marks
- the point of suspension would no longer be fixed, so the length governing the period is not the L that was measured
- rod and pendulum exchange energy, so the amplitude decays faster than air resistance alone would cause
- the period is shifted in the same direction on every trial, so the error is systematic and repeating the timing does not reduce it
- concludes that the heavy base, or clamping the base to the bench, is what makes L a valid measure of the pendulum length
Transfer challenge
A mass on a vertical spring oscillates with T = 2π√(m/k). A student times 20 complete oscillations as 15.6 s, with an uncertainty of ±0.20 s in the total time. Determine the percentage uncertainty in T, and compare it with timing a single oscillation.
Check answer 4 marks
- T = 15.6 / 20 = 0.780 s
- dividing by an exact count of 20 leaves the fractional uncertainty unchanged: 0.20 / 15.6 = 1.3%
- timing one oscillation would give 0.20 / 0.78 = 26%
- the reaction-time uncertainty is fixed per timing run, so spreading it over many oscillations is what reduces it — twenty oscillations cut it by a factor of twenty
02Figure 3Kinematics · Forces and momentumIB
Figure comment
Figure 3Side view. A building stands on level hatched ground, its height marked by a dimension line labelled 25 m running from roof level down to the ground. A small ball, labelled as having mass 0.150 kg, sits at the right-hand edge of the roof, and a horizontal arrow from the ball labelled 12 m s⁻¹ points away from the building. A faint dashed curve traces the path the ball follows from the roof edge down to the ground.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Guided questions 5 parts
Reading cue. The 12 m s⁻¹ arrow is horizontal and is the whole initial velocity, so the ball leaves the roof with zero vertical speed and 25 m is a free-fall drop, not a length of the dashed curve.
aState State the horizontal component of the ball's velocity at the instant it reaches the ground, and give a reason.
Check answer 2 marks
- 12 m s⁻¹
- no horizontal force acts while air resistance is negligible, so the horizontal component is unchanged throughout
bDetermine Determine the horizontal distance from the foot of the building to the point where the dashed path meets the ground.
Check answer 3 marks
- vertical: 25 = ½ × 9.81 × t², so t = 2.26 s
- horizontal motion is at constant velocity: x = 12 × 2.26
- x = 27 m
cDetermine Determine the angle below the horizontal at which the dashed path meets the ground.
Check answer 3 marks
- vertical component on landing: v_y = 9.81 × 2.26 = 22.1 m s⁻¹
- horizontal component is still 12 m s⁻¹, so tan θ = 22.1 / 12 = 1.85
- θ = 62° below the horizontal
dDiscuss A second ball is thrown horizontally from the same point at 24 m s⁻¹. Discuss how its time of flight, its landing distance and its landing angle each compare with those of the first ball.
Check answer 4 marks
- the vertical motion is unaffected by the horizontal velocity, so the time of flight is the same, 2.26 s
- the landing distance doubles to 54 m, since x = ut with the same t
- the vertical component on landing is still 22.1 m s⁻¹, so tan θ = 22.1 / 24 and θ = 43°
- concludes that throwing harder flattens the path and moves the landing point out, but does not keep the ball in the air any longer
Transfer challenge
A stone is released from a hot-air balloon that is rising steadily at 4.0 m s⁻¹ when it is 25 m above the ground. Determine the time the stone takes to reach the ground.
Check answer 4 marks
- the stone shares the balloon's velocity, so it leaves with u = 4.0 m s⁻¹ upwards, not from rest
- taking up as positive: −25 = 4.0t − ½ × 9.81 × t²
- 4.905t² − 4.0t − 25 = 0, so t = 2.7 s
- notes this exceeds the 2.26 s of a ball with no vertical velocity, because the stone first rises before falling