OCR MEI M1 — Question 1 6 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeResultant of three coplanar forces
DifficultyModerate -0.8 This is a straightforward M1 mechanics question requiring basic resolution of forces into components using trigonometry, then finding magnitude and direction of the resultant. It involves standard techniques (resolving at angles, Pythagoras, inverse tan) with no problem-solving insight required—purely procedural application of well-practiced methods. Easier than average A-level due to being early mechanics content with clear diagram guidance.
Spec1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication
1 Fig. 2 shows two forces acting at A . The figure also shows the perpendicular unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) which are respectively horizontal and vertically upwards. The resultant of the two forces is \(\mathbf { F } \mathbf { N }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{94f23528-931c-47b6-89aa-4b6edd25cc30-1_264_918_584_663} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure}
  1. Find \(\mathbf { F }\) in terms of \(\mathbf { i }\) and \(\mathbf { j }\), giving your answer correct to three significant figures.
  2. Calculate the magnitude of \(\mathbf { F }\) and the angle that \(\mathbf { F }\) makes with the upward vertical.
Question 1
(i)
\(F = (10 - 8\cos 50°)i + 8\sin 50°j\)
\(= 4.85769\ldots i + 6.128355\ldots j\)
\(= 4.86i + 6.13j\) (3 s.f.)
M1 – Resolution. Accept significant case. Condone resolution in only one direction.
A1 – Award for a vector with either component correct or consistent sc error is only mistake in the vector. Need not be evaluated.
A1 – cao. Must be in \(ai + bj\) or column format. Must be correct to 3 s.f.
(ii)
AnswerMarks Guidance
\(F = \sqrt{4.85769\ldots^2 + 6.12835\ldots^2} = 7.820101\ldots\)
\(= 7.82\) (3 s.f.)
B1 – FT their \(F\)
\(\text{angle} = \arctan\left(\frac{6.128\ldots}{4.857\ldots}\right)\)
\(= 38.40243\ldots\)
\(= 38.4°\) (3 s.f.)
M1 – Or equivalent. FT their \(F\). Accept \(\arctan\left(\frac{6.128\ldots}{4.857\ldots}\right)\). Accept complementary angle and \(\pm\) signs.
F1 – FT only their \(F\).
# Question 1

## (i)

$F = (10 - 8\cos 50°)i + 8\sin 50°j$

$= 4.85769\ldots i + 6.128355\ldots j$

$= 4.86i + 6.13j$ (3 s.f.)

**M1** – Resolution. Accept significant case. Condone resolution in only one direction.

**A1** – Award for a vector with either component correct or consistent sc error is only mistake in the vector. Need not be evaluated.

**A1** – cao. Must be in $ai + bj$ or column format. Must be correct to 3 s.f.

## (ii)

$|F| = \sqrt{4.85769\ldots^2 + 6.12835\ldots^2} = 7.820101\ldots$

$= 7.82$ (3 s.f.)

**B1** – FT their $F$

$\text{angle} = \arctan\left(\frac{6.128\ldots}{4.857\ldots}\right)$

$= 38.40243\ldots$

$= 38.4°$ (3 s.f.)

**M1** – Or equivalent. FT their $F$. Accept $\arctan\left(\frac{6.128\ldots}{4.857\ldots}\right)$. Accept complementary angle and $\pm$ signs.

**F1** – FT only their $F$.
1 Fig. 2 shows two forces acting at A . The figure also shows the perpendicular unit vectors $\mathbf { i }$ and $\mathbf { j }$ which are respectively horizontal and vertically upwards.

The resultant of the two forces is $\mathbf { F } \mathbf { N }$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{94f23528-931c-47b6-89aa-4b6edd25cc30-1_264_918_584_663}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}

(i) Find $\mathbf { F }$ in terms of $\mathbf { i }$ and $\mathbf { j }$, giving your answer correct to three significant figures.\\
(ii) Calculate the magnitude of $\mathbf { F }$ and the angle that $\mathbf { F }$ makes with the upward vertical.

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