Edexcel M2 2014 June — Question 1 6 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2014
SessionJune
Marks6
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Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeParticles at coordinate positions
DifficultyModerate -0.8 This is a straightforward application of the centre of mass formula for particles in 2D. Students need to set up two equations (one for x-coordinates, one for y-coordinates) using the standard formula and solve simultaneously for k and a. It requires only direct substitution and basic algebra, making it easier than average but not trivial since it involves two unknowns and coordinate geometry.
Spec6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass
  1. Three particles of mass \(3 m , 2 m\) and \(k m\) are placed at the points whose coordinates are \(( 1,5 ) , ( 6,4 )\) and \(( a , 1 )\) respectively. The centre of mass of the three particles is at the point with coordinates \(( 3,3 )\).
Find
  1. the value of \(k\),
  2. the value of \(a\).
Question 1:
Part (a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Moments about \(x\) axis: \(3m \times 5 + 2m \times 4 + km \times 1 = (5+k)m \times 3\)M1 Use moments to form equation in \(k\). All terms required. Condone sign errors on LHS. Condone \(6+k\). \(m\) not required. Could be in fraction form. Correct unsimplified equation. Allow with common factor of \(g\)
\(15 + 8 + k = 3k + 15\)A1 Correct unsimplified equation
\(k = 4\)A1 cso
Part (b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Moments about \(y\) axis: \(3m \times 1 + 2m \times 6 + km \times a = (5+k)m \times 3\)M1 Use moments to form equation in \(a\) and \(k\) (or their \(k\)) only. All terms required. \(m\) not required. Could be in fraction form. Condone \(6+k\).
\(3 + 12 + 4a = 27\)A1 Correct unsimplified equation (follow their \(k\) if \(k\) substituted). Accept with common factor of \(g\).
\(a = 3\)A1 Allow after use of incorrect value for \(k\) 
# Question 1:

## Part (a)

| Answer/Working | Marks | Guidance |
|---|---|---|
| Moments about $x$ axis: $3m \times 5 + 2m \times 4 + km \times 1 = (5+k)m \times 3$ | M1 | Use moments to form equation in $k$. All terms required. Condone sign errors on LHS. Condone $6+k$. $m$ not required. Could be in fraction form. Correct unsimplified equation. Allow with common factor of $g$ |
| $15 + 8 + k = 3k + 15$ | A1 | Correct unsimplified equation |
| $k = 4$ | A1 | cso |

## Part (b)

| Answer/Working | Marks | Guidance |
|---|---|---|
| Moments about $y$ axis: $3m \times 1 + 2m \times 6 + km \times a = (5+k)m \times 3$ | M1 | Use moments to form equation in $a$ and $k$ (or their $k$) only. All terms required. $m$ not required. Could be in fraction form. Condone $6+k$. |
| $3 + 12 + 4a = 27$ | A1 | Correct unsimplified equation (follow their $k$ if $k$ substituted). Accept with common factor of $g$. |
| $a = 3$ | A1 | Allow after use of incorrect value for $k$ |

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\begin{enumerate}
  \item Three particles of mass $3 m , 2 m$ and $k m$ are placed at the points whose coordinates are $( 1,5 ) , ( 6,4 )$ and $( a , 1 )$ respectively. The centre of mass of the three particles is at the point with coordinates $( 3,3 )$.
\end{enumerate}

Find\\
(a) the value of $k$,\\
(b) the value of $a$.\\

\hfill \mbox{\textit{Edexcel M2 2014 Q1 [6]}}
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Q1 6 Q2 9 Q3 9 Q4 9 Q5 9 Q6 12 Q7 12 Q8 9