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Date	May 2019	Marks available	2	Reference code	19M.3.SL.TZ2.9
Level	Standard level	Paper	Paper 3	Time zone	2
Command term	Show that	Question number	9	Adapted from	N/A

Question

The moment of inertia of a solid sphere is $I = \frac{2}{5}m{r^2}$ where m is the mass of the sphere and r is the radius.

Show that the total kinetic energy E_k of the sphere when it rolls, without slipping, at speed v is ${E_{\text{K}}} = \frac{7}{{10}}m{v^2}$ .

[2]

A solid sphere of mass 1.5 kg is rolling, without slipping, on a horizontal surface with a speed of 0.50 m s^-1. The sphere then rolls, without slipping, down a ramp to reach a horizontal surface that is 45 cm lower.

Calculate the speed of the sphere at the bottom of the ramp.

[3]

Markscheme

E_k = E_k linear + E_k rotational

${E_{\text{k}}} = \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}$ ✔

$= \frac{1}{2}m{v^2} + \frac{1}{2} \times \frac{2}{5}m{r^2} \times {\left( {\frac{v}{r}} \right)^2}$ ✔

« $= \frac{7}{{10}}m{v^2}$ »

Answer is given in the question so check working is correct at each stage.

Initial ${E_K} = \frac{7}{{10}} \times 1.50 \times {0.5^2}$ «=0.26J» ✔

Final ${E_K} = 0.26 + 1.5 \times 9.81 \times 0.45$ «=6.88J» ✔

$v = « \sqrt {\frac{{10}}{7} \times \frac{{6.88}}{{1.5}}} =» 2.56$ «m s^–1» ✔

Examiners report

The derivation of the formula for the total kinetic energy of a rolling ball was well answered.

Although there were many correct answers, many candidates forgot to include the initial kinetic energy of the ball at the top of the ramp. The process followed to obtain the answer was too often poorly presented, candidates are encouraged to explain what is being calculated rather than just writing numbers.

Syllabus sections

Option B: Engineering physics » Option B: Engineering physics (Core topics) » B.1 – Rigid bodies and rotational dynamics

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