Edexcel M1 2016 June — Question 5 10 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2016
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeForces in vector form: resultant and acceleration
DifficultyModerate -0.5 This is a straightforward M1 vector mechanics question requiring standard techniques: adding vectors, using the parallel condition (ratio of components), solving a linear equation, then applying F=ma. Part (a) is routine algebraic manipulation, and part (b) is direct application of Newton's second law with given values. Slightly easier than average due to its predictable structure and standard methods.
Spec3.03a Force: vector nature and diagrams3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors

Two forces, \(\mathbf{F}_1\) and \(\mathbf{F}_2\), act on a particle \(A\). \(\mathbf{F}_1 = (2i - 3j)\) N and \(\mathbf{F}_2 = (pi + qj)\) N, where \(p\) and \(q\) are constants. Given that the resultant of \(\mathbf{F}_1\) and \(\mathbf{F}_2\) is parallel to \((\mathbf{i} + 2\mathbf{j})\),
  1. show that \(2p - q + 7 = 0\) [5] Given that \(q = 11\) and that the mass of \(A\) is 2 kg, and that \(\mathbf{F}_1\) and \(\mathbf{F}_2\) are the only forces acting on \(A\),
  2. find the magnitude of the acceleration of \(A\). [5]

Two forces, $\mathbf{F}_1$ and $\mathbf{F}_2$, act on a particle $A$.

$\mathbf{F}_1 = (2i - 3j)$ N and $\mathbf{F}_2 = (pi + qj)$ N, where $p$ and $q$ are constants.

Given that the resultant of $\mathbf{F}_1$ and $\mathbf{F}_2$ is parallel to $(\mathbf{i} + 2\mathbf{j})$,

\begin{enumerate}[label=(\alph*)]
\item show that $2p - q + 7 = 0$
[5]

Given that $q = 11$ and that the mass of $A$ is 2 kg, and that $\mathbf{F}_1$ and $\mathbf{F}_2$ are the only forces acting on $A$,

\item find the magnitude of the acceleration of $A$.
[5]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2016 Q5 [10]}}