Edexcel M1 2002 January — Question 6 11 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2002
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNewton's laws and connected particles
TypeSingle particle, Newton's second law – vector (2D forces)
DifficultyModerate -0.8 This is a straightforward M1 mechanics question testing basic vector addition, Newton's second law, and constant acceleration kinematics. All parts follow standard procedures: (a) add force vectors, (b) use arctangent for angle, (c) apply F=ma, (d) use v=u+at then find magnitude. No problem-solving insight required, just routine application of formulas with 11 marks total spread across simple steps.
Spec1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.03a Force: vector nature and diagrams3.03d Newton's second law: 2D vectors

A particle \(P\), of mass 3 kg, moves under the action of two constant forces (6\(\mathbf{i}\) + 2\(\mathbf{j}\)) N and (3\(\mathbf{i}\) - 5\(\mathbf{j}\)) N.
  1. Find, in the form (\(a\mathbf{i}\) + \(b\mathbf{j}\)) N, the resultant force \(\mathbf{F}\) acting on \(P\). [1]
  2. Find, in degrees to one decimal place, the angle between \(\mathbf{F}\) and \(\mathbf{j}\). [3]
  3. Find the acceleration of \(P\), giving your answer as a vector. [2]
The initial velocity of \(P\) is (-2\(\mathbf{i}\) + \(\mathbf{j}\)) m s\(^{-1}\).
  1. Find, to 3 significant figures, the speed of \(P\) after 2 s. [5]

Part (a)
AnswerMarks
\(F = (6i + 2j) + (3i - 5j) = (9i - 3j)\text{ N}\)B1 (1)
Part (b)
\(\tan\theta = \frac{3}{9} \Rightarrow \theta \simeq 71.6°\)
AnswerMarks Guidance
\(\phi = 108.4°\)M1 A1 (E) A1 (3)
Part (c)
AnswerMarks
\(F = ma'' \Rightarrow a = (3i - j)\text{ m s}^{-2}\)M1 A1 (E) (2)
Part (d)
\(v = (-2\xi + j) + 2(3\xi - j) = 4\xi - j\)
AnswerMarks Guidance
Speed \(= \sqrt{(4^2 + 1^2)} \simeq 4.12\text{ m s}^{-1}\)M1, M1, A1 (a) M1 A1 (5) (11)
## Part (a)
$F = (6i + 2j) + (3i - 5j) = (9i - 3j)\text{ N}$ | B1 (1)

## Part (b)
$\tan\theta = \frac{3}{9} \Rightarrow \theta \simeq 71.6°$

$\phi = 108.4°$ | M1 A1 (E) | A1 (3)

## Part (c)
$F = ma'' \Rightarrow a = (3i - j)\text{ m s}^{-2}$ | M1 A1 (E) (2)

## Part (d)
$v = (-2\xi + j) + 2(3\xi - j) = 4\xi - j$

Speed $= \sqrt{(4^2 + 1^2)} \simeq 4.12\text{ m s}^{-1}$ | M1, M1, A1 (a) | M1 A1 (5) (11)

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A particle $P$, of mass 3 kg, moves under the action of two constant forces (6$\mathbf{i}$ + 2$\mathbf{j}$) N and (3$\mathbf{i}$ - 5$\mathbf{j}$) N.

\begin{enumerate}[label=(\alph*)]
\item Find, in the form ($a\mathbf{i}$ + $b\mathbf{j}$) N, the resultant force $\mathbf{F}$ acting on $P$. [1]
\item Find, in degrees to one decimal place, the angle between $\mathbf{F}$ and $\mathbf{j}$. [3]
\item Find the acceleration of $P$, giving your answer as a vector. [2]
\end{enumerate}

The initial velocity of $P$ is (-2$\mathbf{i}$ + $\mathbf{j}$) m s$^{-1}$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find, to 3 significant figures, the speed of $P$ after 2 s. [5]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2002 Q6 [11]}}