IB Maths IA criteria A to E: checklist, self-grade sheet and examples

Aim: to find a model for the height of a tennis ball after n bounces, and to use it to predict when the bounces become too small to see. Section 2 describes how I measured the bounces; Section 3 fits and tests two models; Section 4 uses the better model to answer the aim; Section 5 evaluates it. Full measurements are in Appendix A.

• The aim is precise and stated first.
• The reader knows the route through the exploration.
• Raw data is in an appendix, so the body stays concise.

Let n be the bounce number (n = 0 is the drop) and hₙ the maximum height, in metres, after the nth bounce. The fitted model is hₙ ≈ 1.24 × 0.720ⁿ (parameters to 3 s.f., consistent with heights read to the nearest centimetre). Figure 2 shows the measured heights with the model; for ln hₙ against n the correlation coefficient is r ≈ −0.999.

• Variables are defined with units before use.
• ≈ and 3 s.f. are used and justified by the measurement accuracy.
• The graph is numbered and referred to in the text.

Before fitting anything I predicted that each bounce would keep the same fraction of the previous height, because the ball and the floor do not change between bounces. If that is true, the ratio hₙ₊₁/hₙ should be constant, so I calculated it for each bounce (Table 2). It was not quite constant: it fell from 0.74 to 0.69. This made me wonder whether a new ball behaves differently, so I repeated the experiment with a ball from a new tin and compared the two sets of ratios.

• A prediction is made and then tested with the student's own data.
• A surprise leads to a new question the student follows up themselves.
• The engagement drives the mathematics forward, not just the introduction.

The residuals of the exponential model are negative at the drop, positive for bounces 2 to 4 and negative again from bounce 6, so the model overestimates the later bounces. That matches the falling ratios in Table 2: the ball seems to lose a slightly larger fraction of its energy on low bounces, perhaps because the felt absorbs relatively more of a small impact. For my aim this matters: the model predicts a visible bounce (above 1 cm) up to n = 14, but if the ratio keeps falling the real answer is nearer n = 12. A model in which the ratio depends on the height would be the natural next step.

• Reflection interprets a specific feature of the results (the residual pattern).
• It links back to the aim and says how the limitation changes the answer.
• The extension follows from what was found, not from a generic list.

If each bounce keeps a fraction k of the previous height, then hₙ = h₀kⁿ, so ln hₙ = ln h₀ + n ln k: a graph of ln hₙ against n should be linear. The least-squares line through bounces 0 to 6 has gradient −0.328 and intercept 0.214, so k = e^−0.328 ≈ 0.720 and h₀ ≈ 1.24 m, close to the 1.20 m drop (the gap is a clue I return to in Section 5). Testing on bounce 7, which I did not use for the fit, the model gives 0.124 m against a measured 0.115 m.

• The model is derived from an assumption, not just chosen.
• Linearising with logarithms shows understanding of why the fit works.
• The model is tested on data not used to fit it.

Measuring from the 1.20 m drop, the total distance travelled is D = h₀ + 2h₀(k + k² + …) = h₀ + 2h₀k/(1 − k) for 0 < k < 1, so D ≈ 7.37 m. A fall from height h takes √(2h/g), so the times between bounces also form a geometric series, with ratio √k; the total time T = √(2h₀/g)(1 + 2√k/(1 − √k)) ≈ 6.0 s is finite even though the number of bounces is not. My video shows the ball at rest after about 4.9 s, which suggests the constant-k model fails for very small bounces, consistent with the residuals in Section 4.

• Sophistication: links sequences, series and kinematics.
• Rigour: the convergence condition is stated and the result justified.
• Precise: the prediction is checked against evidence and the discrepancy explained.