A particle of mass $1$ kg is threaded on a smooth circular wire of radius $0.5$ m fixed in a vertical plane with centre O. It is slightly disturbed from rest at the highest point A. Taking $g = 9.81\,\mathrm{ms^{-2}}$ , find its speed in $\mathrm{ms^{-1}}$ when angle AOP is $90^\circ$ , to 2 decimal places.

Model answer

3.13

Also accepted: 3.13 m/s, 3.1, 3.132

Explanation

When angle AOP is

90^\circ

the particle is level with the centre O, so it has fallen a height equal to the radius,

0.5

m. The wire is smooth, so energy is conserved:

\dfrac12 mv^2 = mg(0.5)

, giving

v^2 = g

and

v = \sqrt{g}

. With

g = 9.81

this is

3.13\,\mathrm{ms^{-1}}

(with

g = 9.8

it is also

3.13

Derived from ZIMSEC Additional Mathematics Paper 2, Specimen Paper, Q3

Get the full paper, not just one question

Danho has every sitting for this paper, with your progress tracked question by question, offline.