Edexcel M2 2008 January — Question 2 9 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2008
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeFinding when moving in specific direction
DifficultyModerate -0.3 This is a straightforward M2 question testing standard techniques: differentiation of position vectors for velocity (routine), solving when j-component equals zero (algebraic manipulation), and applying impulse-momentum theorem (direct formula application). All parts are textbook exercises requiring no problem-solving insight, though the multi-step nature and impulse calculation elevate it slightly above pure recall.
Spec1.10a Vectors in 2D: i,j notation and column vectors3.02a Kinematics language: position, displacement, velocity, acceleration6.03f Impulse-momentum: relation

At time \(t\) seconds \((t \geq 0)\), a particle \(P\) has position vector \(\mathbf{p}\) metres, with respect to a fixed origin \(O\), where $$\mathbf{p} = (3t^2 - 6t + 4)\mathbf{i} + (3t^3 - 4t)\mathbf{j}.$$ Find
  1. the velocity of \(P\) at time \(t\) seconds, [2]
  2. the value of \(t\) when \(P\) is moving parallel to the vector \(\mathbf{i}\). [3]
When \(t = 1\), the particle \(P\) receives an impulse of \((2\mathbf{i} - 6\mathbf{j})\) N s. Given that the mass of \(P\) is 0.5 kg,
  1. find the velocity of \(P\) immediately after the impulse. [4]

Part (a)
AnswerMarks Guidance
\(\dot{\mathbf{p}} = (6t - 6)\mathbf{i} + (9t^2 - 4)\mathbf{j}\) (ms\(^{-1}\))M1 A1 (2 marks)
Part (b)
AnswerMarks Guidance
\(9t^2 - 4 = 0\), \(t = \frac{2}{3}\)M1 DM1 A1 (3 marks)
Part (c)
\(t = 1 \Rightarrow \dot{\mathbf{p}} = 5\mathbf{j}\)
(+/-) \(2\mathbf{i} - 6\mathbf{j} = 0.5(\mathbf{v} - 5\mathbf{j})\)
AnswerMarks Guidance
\(\mathbf{v} = 4\mathbf{i} - 7\mathbf{j}\) (ms\(^{-1}\))B1 ft M1 M1 A1 (4 marks) [9 marks total]
Fits their \(\dot{\mathbf{p}}\)
**Part (a)**
$\dot{\mathbf{p}} = (6t - 6)\mathbf{i} + (9t^2 - 4)\mathbf{j}$ (ms$^{-1}$) | M1 A1 | (2 marks)

**Part (b)**
$9t^2 - 4 = 0$, $t = \frac{2}{3}$ | M1 DM1 A1 | (3 marks)

**Part (c)**
$t = 1 \Rightarrow \dot{\mathbf{p}} = 5\mathbf{j}$

(+/-) $2\mathbf{i} - 6\mathbf{j} = 0.5(\mathbf{v} - 5\mathbf{j})$

$\mathbf{v} = 4\mathbf{i} - 7\mathbf{j}$ (ms$^{-1}$) | B1 ft M1 M1 A1 | (4 marks) [9 marks total]

Fits their $\dot{\mathbf{p}}$

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At time $t$ seconds $(t \geq 0)$, a particle $P$ has position vector $\mathbf{p}$ metres, with respect to a fixed origin $O$, where

$$\mathbf{p} = (3t^2 - 6t + 4)\mathbf{i} + (3t^3 - 4t)\mathbf{j}.$$

Find

\begin{enumerate}[label=(\alph*)]
\item the velocity of $P$ at time $t$ seconds, [2]

\item the value of $t$ when $P$ is moving parallel to the vector $\mathbf{i}$. [3]
\end{enumerate}

When $t = 1$, the particle $P$ receives an impulse of $(2\mathbf{i} - 6\mathbf{j})$ N s. Given that the mass of $P$ is 0.5 kg,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find the velocity of $P$ immediately after the impulse. [4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2008 Q2 [9]}}