Edexcel M5 2011 June — Question 1 4 marks

Exam BoardEdexcel
ModuleM5 (Mechanics 5)
Year2011
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeWork done by constant force - vector setup
DifficultyModerate -0.8 This is a straightforward application of the work done formula W = F·d, requiring only calculation of the displacement vector AB and then a dot product. It's a standard M5 question testing basic recall of the work-energy principle with no problem-solving or conceptual challenges beyond routine vector arithmetic.
Spec1.10a Vectors in 2D: i,j notation and column vectors6.02a Work done: concept and definition

A particle moves from the point \(A\) with position vector \((3\mathbf{i} - \mathbf{j} + 3\mathbf{k})\) m to the point \(B\) with position vector \((\mathbf{i} - 2\mathbf{j} - 4\mathbf{k})\) m under the action of the force \((2\mathbf{i} - 3\mathbf{j} - \mathbf{k})\) N. Find the work done by the force. [4]

AnswerMarks
\(\mathbf{A B} = (\mathbf{i} - 2\mathbf{j} - 4\mathbf{k}) - (3\mathbf{i} - \mathbf{j} + 3\mathbf{k}) = (-2\mathbf{i} - \mathbf{j} - 7\mathbf{k})\)M1 A1
\((2\mathbf{i} - 3\mathbf{j} - \mathbf{k}) \cdot (-2\mathbf{i} - \mathbf{j} - 7\mathbf{k}) = -4 + 3 + 7 = 6\text{ J}\)M1 A1

Total: 4 marks

$\mathbf{A B} = (\mathbf{i} - 2\mathbf{j} - 4\mathbf{k}) - (3\mathbf{i} - \mathbf{j} + 3\mathbf{k}) = (-2\mathbf{i} - \mathbf{j} - 7\mathbf{k})$ | M1 A1 |
$(2\mathbf{i} - 3\mathbf{j} - \mathbf{k}) \cdot (-2\mathbf{i} - \mathbf{j} - 7\mathbf{k}) = -4 + 3 + 7 = 6\text{ J}$ | M1 A1 |

Total: 4 marks

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A particle moves from the point $A$ with position vector $(3\mathbf{i} - \mathbf{j} + 3\mathbf{k})$ m to the point $B$ with position vector $(\mathbf{i} - 2\mathbf{j} - 4\mathbf{k})$ m under the action of the force $(2\mathbf{i} - 3\mathbf{j} - \mathbf{k})$ N. Find the work done by the force.
[4]

\hfill \mbox{\textit{Edexcel M5 2011 Q1 [4]}}