DP Physics (last assessment 2024)

Question 18M.3.SL.TZ2.1b

Date	May 2018	Marks available	[Maximum mark: 4]	Reference code	18M.3.SL.TZ2.1b
Level	SL	Paper	3	Time zone	TZ2
Command term	Calculate	Question number	b	Adapted from	N/A

[Maximum mark: 4]

18M.3.SL.TZ2.1b

To determine the acceleration due to gravity, a small metal sphere is dropped from rest and the time it takes to fall through a known distance and open a trapdoor is measured.

M18/4/PHYSI/SP3/ENG/TZ2/01

The following data are available.

$\begin{array}{*{20}{l}} {{\text{Diameter of metal sphere}}}&{ = 12.0 \pm 0.1{\text{ mm}}} \\ {{\text{Distance between the point of release and the trapdoor}}}&{ = 654 \pm 2{\text{ mm}}} \\ {{\text{Measured time for fall}}}&{ = 0.363 \pm 0.002{\text{ s}}} \end{array}$

(b)

Using the following equation

${\text{acceleration due to gravity}} = \frac{{2 \times {\text{distance fallen by centre of mass of sphere}}}}{{{{{\text{(measured time to fall)}}}^{\text{2}}}}}$

calculate, for these data, the acceleration due to gravity including an estimate of the absolute uncertainty in your answer.

[4]

Markscheme

«a = $\frac{{2s}}{{{t^2}}} = \frac{{2 \times 0.642}}{{{{0.363}^2}}}$ » = 9.744 «ms^–2»

fractional uncertainty in distance = $\frac{2}{{642}}$ AND fractional uncertainty in time = $\frac{{0.002}}{{0.363}}$

total fractional uncertainty = $\frac{{\Delta s}}{s} + 2\frac{{\Delta t}}{t}$ «= 0.00311 + 2 × 0.00551»

total absolute uncertainty = 0.1 or 0.14 AND same number of decimal places in value and uncertainty, ie: 9.7 ± 0.1 or 9.74 ± 0.14

Accept working in % for MP2 and MP3

Final uncertainty must be the absolute uncertainty

[4 marks]

Syllabus sections

Core

Core » Topic 1: Measurements and uncertainties » 1.2 – Uncertainties and errors

Core » Topic 1: Measurements and uncertainties

Question 18M.3.SL.TZ2.1b

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