Pearson Edexcel IAL Mathematics M2.2 Centres of mass Question Bank

Practise finding centres of mass for particles and composite laminas, then using the centre of mass to solve hanging, pivoting and force problems.

Exam points

set up moment equations for particles or templates to find a centre of mass distance
use the vertical through the centre of mass to calculate hanging angles or forces in equilibrium

Covered syllabus points

M2.2.1 - Centre of mass of discrete distributions M2.2.2 - Centre of mass of uniform plane M2.2.3 - Simple cases of equilibrium

[Maximum number: 6]

Three particles of masses 2 m, 3 m and 4 m are placed at the points with coordinates (-2,5),(2,-3) and (3 k, k) respectively, where k is a constant.
The centre of mass of the three particles is at the point $(\bar{x}, \bar{y})$ .

(a)

Show that $\bar{x}=\frac{2+12 k}{9}$

The centre of mass of the three particles lies at a point on the straight line with equation x+2 y=3

[ 2 ]

M( x axis)

M1

Need all terms. Dimensionally consistent. Condone if m missing throughout.

Accept as part of a vector equation

$\begin{aligned} 2 m \times-+\times+\times=\times(2) 3 m 24 m 3 k \quad 9 m \quad x x=\frac{2+12 k}{9} * \end{aligned}$

A1*

Obtain given result

2

(b)

Find the value of k.

[ 4 ]

M(y axis)

M1

Need all terms. Dimensionally consistent. Might be seen as part of a vector equation in (a). It does not score any marks until referred to in part (b). Condone if m missing throughout.

$\begin{aligned} 2 m \times+\times-+\times=\times 53 m(3) 4 m k \quad 9 m \quad y \square-14+k \square \square y=\square \square \end{aligned}$

A1

Correct unsimplified equation. Allow if m missing
throughout.

Form and solve equation in k
$(2+12 k++=2 \quad 8 k \quad 27)$

DM1

Use their y ānd $x \overline{+}^{-}=2 y^{-} \quad 3$

Dependent on the two preceding M marks

$k=\frac{23}{20}(1.15)$

A1

Correct answer only

4

(6)

[Maximum number: 11]

(a)

Show that $d=\frac{4 a}{\pi}$

[ 5 ]

Masses proportional to areas: $8\pi a^2, 2\pi a^2, 2\pi a^2$

Distances from AC: $\frac{16a}{3\pi}, \frac{8a}{3\pi}, 8a-\frac{8a}{3\pi}$

8\pi a^2\left(\frac{16a}{3\pi}\right)-2\pi a^2\left(\frac{8a}{3\pi}\right)-2\pi a^2\left(8a-\frac{8a}{3\pi}\right)=8\pi a^2 d

\frac{32a^3}{\pi}=8a^2 d

A1*

d=\frac{4a}{\pi}

(b)

The template T is freely suspended from A and hangs in equilibrium with AC at an angle $\theta$ to the downward vertical.

Find the exact value of $\tan\theta$

[ 6 ]

Moments about a perpendicular axis through A give x=5a

x=5a

\tan\theta=\frac{d}{x}

\tan\theta=\frac{\frac{4a}{\pi}}{5a}

\tan\theta=\frac{4}{5\pi}

All question bank results loaded

Covered syllabus points

M2.2 - Centres of mass question 1

Question (a)

Question (b)

M2.2 - Centres of mass question 2

Question (a)

Question (b)