OCR MEI M1 — Question 1 6 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeForces in vector form: equilibrium (find unknowns)
DifficultyModerate -0.8 This is a straightforward equilibrium problem requiring only direct application of the principle that forces sum to zero in equilibrium, followed by basic vector magnitude and angle calculations. It involves minimal problem-solving—just routine manipulation of vectors in component form with no conceptual challenges beyond recall of equilibrium conditions.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.03b Newton's first law: equilibrium3.03m Equilibrium: sum of resolved forces = 0

1 A particle rests on a smooth, horizontal plane. Horizontal unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) lie in this plane. The particle is in equilibrium under the action of the three forces \(( - 3 \mathbf { i } + 4 \mathbf { j } ) \mathrm { N }\) and \(( 21 \mathbf { i } - 7 \mathbf { j } ) \mathrm { N }\) and \(\mathbf { R N }\).
  1. Write down an expression for \(\mathbf { R }\) in terms of \(\mathbf { i }\) and \(\mathbf { j }\).
  2. Find the magnitude of \(\mathbf { R }\) and the angle between \(\mathbf { R }\) and the \(\mathbf { i }\) direction.

Question 1:

\(\mathbf{R} + \begin{pmatrix} -3 \\ 4 \end{pmatrix} + \begin{pmatrix} 21 \\ -7 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)
\(\mathbf{R} = \begin{pmatrix} -18 \\ 3 \end{pmatrix}\)
M1
A1 — Sum to zero. Award if seen here or in (ii) or used in (ii).
SC1 for \(\begin{pmatrix} 18 \\ -3 \end{pmatrix}\)
2 marks

\(R = \sqrt{18^2 + 3^2} = \sqrt{324 + 9} = 18.248\ldots\) so \(18.2\) N (3 s.f.)
M1 — Use of Pythagoras
A1 — Any reasonable accuracy. FT \(R\) (with 2 non-zero components)
angle is \(180° - \arctan\left(\frac{3}{18}\right) = 170.53\ldots°\) so \(171°\) (3 s.f.)
M1
A1 — Allow \(\arctan\left(\frac{\pm 3}{\pm 18}\right)\) or \(\arctan\left(\frac{\pm 18}{\pm 3}\right)\). Any reasonable accuracy. FT \(R\) provided their angle is obtuse but not \(180°\)
4 marks

Total: 6 marks

Question 1:

--- 1(i) ---

$\mathbf{R} + \begin{pmatrix} -3 \\ 4 \end{pmatrix} + \begin{pmatrix} 21 \\ -7 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

$\mathbf{R} = \begin{pmatrix} -18 \\ 3 \end{pmatrix}$

M1

A1 — Sum to zero. Award if seen here or in (ii) or used in (ii).

SC1 for $\begin{pmatrix} 18 \\ -3 \end{pmatrix}$

2 marks

--- 1(ii) ---

$R = \sqrt{18^2 + 3^2} = \sqrt{324 + 9} = 18.248\ldots$ so $18.2$ N (3 s.f.)

M1 — Use of Pythagoras

A1 — Any reasonable accuracy. FT $R$ (with 2 non-zero components)

angle is $180° - \arctan\left(\frac{3}{18}\right) = 170.53\ldots°$ so $171°$ (3 s.f.)

M1

A1 — Allow $\arctan\left(\frac{\pm 3}{\pm 18}\right)$ or $\arctan\left(\frac{\pm 18}{\pm 3}\right)$. Any reasonable accuracy. FT $R$ provided their angle is obtuse but not $180°$

4 marks

Total: 6 marks
1 A particle rests on a smooth, horizontal plane. Horizontal unit vectors $\mathbf { i }$ and $\mathbf { j }$ lie in this plane. The particle is in equilibrium under the action of the three forces $( - 3 \mathbf { i } + 4 \mathbf { j } ) \mathrm { N }$ and $( 21 \mathbf { i } - 7 \mathbf { j } ) \mathrm { N }$ and $\mathbf { R N }$.\\
(i) Write down an expression for $\mathbf { R }$ in terms of $\mathbf { i }$ and $\mathbf { j }$.\\
(ii) Find the magnitude of $\mathbf { R }$ and the angle between $\mathbf { R }$ and the $\mathbf { i }$ direction.

\hfill \mbox{\textit{OCR MEI M1  Q1 [6]}}