OCR MEI M1 2014 June — Question 2 8 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Year2014
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeForces in equilibrium (find unknowns)
DifficultyModerate -0.8 This is a straightforward M1 equilibrium question requiring basic vector addition and setting components equal to zero. Part (i) is simple addition and scalar multiplication check, part (ii) requires showing the j-component is zero, and part (iii) involves solving two simultaneous equations from equilibrium conditions. All steps are routine applications of standard techniques with no problem-solving insight required.
Spec1.10d Vector operations: addition and scalar multiplication3.03a Force: vector nature and diagrams3.03b Newton's first law: equilibrium3.03m Equilibrium: sum of resolved forces = 0
2 The unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) shown in Fig. 2 are in the horizontal and vertically upwards directions. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{63a2dc41-5e8b-4275-8653-ece5067c4306-2_132_145_1726_968} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure} Forces \(\mathbf { p }\) and \(\mathbf { q }\) are given, in newtons, by \(\mathbf { p } = 12 \mathbf { i } - 5 \mathbf { j }\) and \(\mathbf { q } = 16 \mathbf { i } + 1.5 \mathbf { j }\).
  1. Write down the force \(\mathbf { p } + \mathbf { q }\) and show that it is parallel to \(8 \mathbf { i } - \mathbf { j }\).
  2. Show that the force \(3 \mathbf { p } + 10 \mathbf { q }\) acts in the horizontal direction.
  3. A particle is in equilibrium under forces \(k \mathbf { p } , 3 \mathbf { q }\) and its weight \(\mathbf { w }\). Show that the value of \(k\) must be - 4 and find the mass of the particle.
Question 2:
Part (i)
AnswerMarks Guidance
AnswerMarks Guidance
\(\mathbf{p}+\mathbf{q}=28\mathbf{i}-3.5\mathbf{j}\)B1
\(28\mathbf{i}-3.5\mathbf{j}=k(8\mathbf{i}-\mathbf{j})\)M1 Or equivalent. \(k\) may be implied by going straight to \(3.5\)
\(k=3.5\), (So they are parallel)A1
Alternative: \(\mathbf{p}+\mathbf{q}\): \(\tan\theta=\frac{-3.5}{28}\Rightarrow\theta=-7.13°\)B1
\(8\mathbf{i}-\mathbf{j}\): \(\tan\theta=\frac{-1}{8}\Rightarrow\theta=-7.13°\), So they are parallelM1, A1 Comparing ratio of components in each vector is sufficient. Both ratios the same and correct
[3]
Part (ii)
AnswerMarks Guidance
AnswerMarks Guidance
\(3\mathbf{p}+10\mathbf{q}=(36+160)\mathbf{i}+(-15+15)\mathbf{j}=196\mathbf{i}\)B1
Zero \(\mathbf{j}\) component so horizontalB1 Or equivalent explanation. May be shown on a diagram
[2]
Part (iii)
AnswerMarks Guidance
AnswerMarks Guidance
The horizontal component must be zero, so \(12k+3\times16=0\Rightarrow k=-4\)B1 Substituting \(k=-4\) and showing \(\mathbf{i}\) component is zero is acceptable
\(\mathbf{w}=-24.5\mathbf{j}\)B1 Award for \(24.5\) seen
\(mg=24.5\Rightarrow m=2.5\), The mass is \(2.5\) kgB1 Award for \(2.5\) seen. FT from their weight
[3] 
# Question 2:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{p}+\mathbf{q}=28\mathbf{i}-3.5\mathbf{j}$ | B1 | |
| $28\mathbf{i}-3.5\mathbf{j}=k(8\mathbf{i}-\mathbf{j})$ | M1 | Or equivalent. $k$ may be implied by going straight to $3.5$ |
| $k=3.5$, (So they are parallel) | A1 | |
| **Alternative:** $\mathbf{p}+\mathbf{q}$: $\tan\theta=\frac{-3.5}{28}\Rightarrow\theta=-7.13°$ | B1 | |
| $8\mathbf{i}-\mathbf{j}$: $\tan\theta=\frac{-1}{8}\Rightarrow\theta=-7.13°$, So they are parallel | M1, A1 | Comparing ratio of components in each vector is sufficient. Both ratios the same and correct |
| **[3]** | | |

## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3\mathbf{p}+10\mathbf{q}=(36+160)\mathbf{i}+(-15+15)\mathbf{j}=196\mathbf{i}$ | B1 | |
| Zero $\mathbf{j}$ component so horizontal | B1 | Or equivalent explanation. May be shown on a diagram |
| **[2]** | | |

## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| The horizontal component must be zero, so $12k+3\times16=0\Rightarrow k=-4$ | B1 | Substituting $k=-4$ and showing $\mathbf{i}$ component is zero is acceptable |
| $\mathbf{w}=-24.5\mathbf{j}$ | B1 | Award for $24.5$ seen |
| $mg=24.5\Rightarrow m=2.5$, The mass is $2.5$ kg | B1 | Award for $2.5$ seen. FT from their weight |
| **[3]** | | |

---
2 The unit vectors $\mathbf { i }$ and $\mathbf { j }$ shown in Fig. 2 are in the horizontal and vertically upwards directions.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{63a2dc41-5e8b-4275-8653-ece5067c4306-2_132_145_1726_968}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}

Forces $\mathbf { p }$ and $\mathbf { q }$ are given, in newtons, by $\mathbf { p } = 12 \mathbf { i } - 5 \mathbf { j }$ and $\mathbf { q } = 16 \mathbf { i } + 1.5 \mathbf { j }$.\\
(i) Write down the force $\mathbf { p } + \mathbf { q }$ and show that it is parallel to $8 \mathbf { i } - \mathbf { j }$.\\
(ii) Show that the force $3 \mathbf { p } + 10 \mathbf { q }$ acts in the horizontal direction.\\
(iii) A particle is in equilibrium under forces $k \mathbf { p } , 3 \mathbf { q }$ and its weight $\mathbf { w }$.

Show that the value of $k$ must be - 4 and find the mass of the particle.

\hfill \mbox{\textit{OCR MEI M1 2014 Q2 [8]}}
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