Pearson Edexcel IAL Mathematics M2.2.1 Centre of mass of discrete distributions

Practise using moments in one or two dimensions to locate the centre of mass of particles and link coordinates to a given straight line.

[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)

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M2.2.1 - Centre of mass of discrete distributions question 1

Question (a)

Question (b)