Edexcel M2 — Question 4 10 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeWork done by constant force - vector setup
DifficultyStandard +0.3 This is a straightforward M2 work-energy question requiring standard vector displacement calculation (PQ = Q - P), work done formula (F·d), and work-energy principle (work = ΔKE). All steps are routine applications of well-practiced formulas with no conceptual challenges or novel problem-solving required, making it slightly easier than average.
Spec1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02e Calculate KE and PE: using formulae

\(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane. A body of mass 1 kg moves under the action of a constant force \((4\mathbf{i} + 5\mathbf{j})\) N. The body moves from the point \(P\) with position vector \((-3\mathbf{i} - 15\mathbf{j})\) m to the point \(Q\) with position vector \(9\mathbf{i}\) m.
  1. Find the work done by the force in moving the body from \(P\) to \(Q\). [5 marks]
  2. Given that the body started from rest at \(P\), find its speed when it is at \(Q\). [5 marks]

AnswerMarks Guidance
(a) Displacement \(= 12\mathbf{i} + 15\mathbf{j}\)M1 A1 A1
Distance \(= 3\sqrt{41}\)M1
\(\mathbf{F} = \sqrt{41}\), so work done \(= 3\sqrt{41} \times \sqrt{41} = 123\) J
(b) \(a = 4\mathbf{i} + 5\mathbf{j}\)B1 M1 A1
\(12\mathbf{i} + 15\mathbf{j} = \frac{1}{2}(4\mathbf{i} + 5\mathbf{j})t^2\)M1
\(t = \sqrt{6}\)A1
\(\mathbf{v} = \sqrt{6}(4\mathbf{i} + 5\mathbf{j})\)M1
\(\mathbf{v} = \sqrt{246} = 15 \cdot 7\) ms\(^{-1}\)
(a) Displacement $= 12\mathbf{i} + 15\mathbf{j}$ | M1 A1 A1 |

Distance $= 3\sqrt{41}$ | M1 |

$|\mathbf{F}| = \sqrt{41}$, so work done $= 3\sqrt{41} \times \sqrt{41} = 123$ J | A1 |

(b) $a = 4\mathbf{i} + 5\mathbf{j}$ | B1 M1 A1 |

$12\mathbf{i} + 15\mathbf{j} = \frac{1}{2}(4\mathbf{i} + 5\mathbf{j})t^2$ | M1 |

$t = \sqrt{6}$ | A1 |

$\mathbf{v} = \sqrt{6}(4\mathbf{i} + 5\mathbf{j})$ | M1 |

$|\mathbf{v}| = \sqrt{246} = 15 \cdot 7$ ms$^{-1}$ | A1 | (10 marks)
$\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors in a horizontal plane. A body of mass 1 kg moves under the action of a constant force $(4\mathbf{i} + 5\mathbf{j})$ N. The body moves from the point $P$ with position vector $(-3\mathbf{i} - 15\mathbf{j})$ m to the point $Q$ with position vector $9\mathbf{i}$ m.

\begin{enumerate}[label=(\alph*)]
\item Find the work done by the force in moving the body from $P$ to $Q$. [5 marks]
\item Given that the body started from rest at $P$, find its speed when it is at $Q$. [5 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2  Q4 [10]}}