| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2019 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Coplanar forces in equilibrium |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem involving resolving forces and using symmetry. Part (a) requires recognizing symmetry, (b) is basic trigonometry and force resolution, (c) uses simple geometry with the constraint that paths are 6m apart, (d) repeats the force resolution with new angles, and (e)-(f) are conceptual. All steps are standard textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | Symmetry |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | sinπΌ = 0.6 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2πcosπΌ = 1200 | M1 | 3.3 |
| 750 (N) | A1 | 1.1 |
| Answer | Marks |
|---|---|
| (c) | 1 52 +62 β72 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2Γ5Γ6 5 | B1 | 3.1a |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2Γ6Γ7 7 | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (d) | π cosβ +π cosπ = 1200 | |
| 1 2 | M1 | 3.3 |
| Answer | Marks |
|---|---|
| 1 5 2 7 | T is tension in Jackβs |
| Answer | Marks | Guidance |
|---|---|---|
| 1 2 | M1 | 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| variable | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (e) | Tensions will both be the same | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| part b) | B1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| (f) | Resistance to the motion of the boat may well | |
| be different when it is closer to the bank in (d) | B1 | 3.5b |
Question 5:
5 | (a) | Symmetry | B1 | 1.2 | Oe: tension comp perp to canal in
each rope is the same, and ropes
make same angle with canal.
[1]
(b) | sinπΌ = 0.6 | B1 | 1.1a | Oe: where Ξ± is angle between
rope and canal
2πcosπΌ = 1200 | M1 | 3.3
750 (N) | A1 | 1.1
[3]
(c) | 1 52 +62 β72 1
sinβ
= cos( πββ
) = =
2 2Γ5Γ6 5 | B1 | 3.1a | By cos rule
1 62 +72 β52 5
sinπ = cos( πβπ) = =
2 2Γ6Γ7 7 | B1 | 1.1
[2]
(d) | π cosβ
+π cosπ = 1200
1 2 | M1 | 3.3 | β24 β24
π +π = 1200
1 5 2 7 | T is tension in Jackβs
1
rope, T in Jemimaβs
2
π sinβ
= π sinπ
1 2 | M1 | 1.1a | 1 5
π = π
15 27
Attempt to solve by finding equation in 1
variable | M1 | 1.1 | Can be impled by 1 right answer
T = 1020(.62) (N)
1 | A1 | 1.1 | 1250
β6 cao
3
T = 286 (285.77) (N)
2 | A1 | 1.1 | 350
β6 cao
3
[5]
(e) | Tensions will both be the same | B1 | 3.4
Tensions will (both) be less than 750 N(their
part b) | B1 | 3.4
[2]
(f) | Resistance to the motion of the boat may well
be different when it is closer to the bank in (d) | B1 | 3.5b | Or other valid comment
[1]5 Jack and Jemima are pulling a boat along a straight level canal.\\
The resistance to the motion of the boat is modelled as constant and equal to 1200 N .\\
Jack and Jemima walk in the same direction on paths on opposite sides of the canal. They each walk forwards at the same steady speed, keeping level with each other so that the distance between them is always 6 m . Jack and Jemima each pull a long light inextensible rope attached to the boat; initially they hold their ropes so the distance from each of them to the boat is 5 m , as shown in Fig. 5.1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b808042-95b8-4862-8355-3979c1981089-4_417_1109_605_246}
\captionsetup{labelformat=empty}
\caption{Fig. 5.1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Explain why the tension will be the same in each rope.
\item Find the tension in each rope.
Jemima then gradually releases more rope, so that the distance between her and the boat is 7 m . Jack and Jemima continue to walk at the same steady speed along the paths, but the position of the boat changes so that Jemima's rope makes an angle of $\theta$ with the path and Jack's rope makes an angle of $\phi$ with the path, as shown in Fig. 5.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b808042-95b8-4862-8355-3979c1981089-4_513_1109_1610_246}
\captionsetup{labelformat=empty}
\caption{Fig. 5.2}
\end{center}
\end{figure}
\item - Show that $\sin \phi = \frac { 1 } { 5 }$.
\begin{itemize}
\item Show that $\sin \theta = \frac { 5 } { 7 }$.
\item Find the tension in each rope in this new equilibrium position.
\item Without further calculation, state the effect on the tensions in the ropes if Jack now lengthens his rope to 7 m , the same length as Jemima's rope.
\item Suggest how the modelling assumption used in this question could be improved.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2019 Q5 [14]}}