| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Modelling assumptions and refinements |
| Difficulty | Moderate -0.3 This is a standard M1 equilibrium on a slope question with straightforward force resolution. Parts (i)-(ii) involve basic component resolution parallel to the plane, (iii) is a diagram, and (iv)-(v) require resolving forces in two directions with an angled string. All techniques are routine for M1 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03r Friction: concept and vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Up the plane: \(T - 4g\sin 25 = 0\) | M1 | Resolving parallel to the plane. If any other direction used, all forces must be present. Accept \(s \leftrightarrow c\). Allow use of \(m\). No extra forces |
| \(T = 16.5666...\) so 16.6 N (3 s.f.) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Down the plane: \((4+m)g\sin 25 - 50 = 0\) | M1 | No extra forces. Must attempt resolution in at least 1 term. Accept \(s \leftrightarrow c\). Accept \(Mg\sin 25\). Accept use of mass |
| \(m = 8.0724...\) so 8.07 (3 s.f.) | A1, A1 | Accept \(Mg\sin 25\) |
| Answer | Marks | Guidance |
|---|---|---|
| Diagram | B1 | Any 3 of weight, friction, normal reaction and \(P\) present |
| B1 | All forces present with suitable directions, labels and arrows. Accept \(W\), \(mg\), \(4g\) and 39.2 |
| Answer | Marks | Guidance |
|---|---|---|
| Resolving up the plane | M1 | Resolving parallel to the plane. All forces must be present. Accept \(s \leftrightarrow c\). Allow use of \(m\). At least one resolution attempted and accept wrong angles. Allow sign errors |
| \(P\cos 15\) term correct | B1 | Allow sign error |
| \(P\cos 15 - 20 - 4g\sin 25 = 0\) | B1 | Both resolutions correct. Weight used. Allow sign errors. FT use of \(P\sin 15\) |
| \(P = 37.8565...\) so 37.9 N (3 s.f.) | A1, A1 | All correct but FT use of \(P\sin 15\) |
| Answer | Marks | Guidance |
|---|---|---|
| Resolving perpendicular to the plane | M1 | May use other directions. All forces present. No extras. Allow \(s \leftrightarrow c\). Weight not mass used. Both resolutions attempted. Allow sign errors |
| \(R + P\sin 15 - 4g\cos 25 = 0\) | B1 | Both resolutions correct. Allow sign errors. Allow use of \(P\cos 15\) if \(P\sin 15\) used in (iv) |
| F1 | All correct. Only FT their \(P\) and their use of \(P\cos 15\) | |
| \(R = 25.729...\) so 25.7 N | A1 | cao |
# Question 6:
## Part (i)
| Up the plane: $T - 4g\sin 25 = 0$ | M1 | Resolving parallel to the plane. If any other direction used, all forces must be present. Accept $s \leftrightarrow c$. Allow use of $m$. No extra forces |
| $T = 16.5666...$ so 16.6 N (3 s.f.) | A1 | |
## Part (ii)
| Down the plane: $(4+m)g\sin 25 - 50 = 0$ | M1 | No extra forces. Must attempt resolution in at least 1 term. Accept $s \leftrightarrow c$. Accept $Mg\sin 25$. Accept use of mass |
| $m = 8.0724...$ so 8.07 (3 s.f.) | A1, A1 | Accept $Mg\sin 25$ |
## Part (iii)
| Diagram | B1 | Any 3 of weight, friction, normal reaction and $P$ present |
| | B1 | All forces present with suitable directions, labels and arrows. Accept $W$, $mg$, $4g$ and 39.2 |
## Part (iv)
| Resolving up the plane | M1 | Resolving parallel to the plane. All forces must be present. Accept $s \leftrightarrow c$. Allow use of $m$. At least one resolution attempted and accept wrong angles. Allow sign errors |
| $P\cos 15$ term correct | B1 | Allow sign error |
| $P\cos 15 - 20 - 4g\sin 25 = 0$ | B1 | Both resolutions correct. Weight used. Allow sign errors. FT use of $P\sin 15$ |
| $P = 37.8565...$ so 37.9 N (3 s.f.) | A1, A1 | All correct but FT use of $P\sin 15$ |
## Part (v)
| Resolving perpendicular to the plane | M1 | May use other directions. All forces present. No extras. Allow $s \leftrightarrow c$. Weight not mass used. Both resolutions attempted. Allow sign errors |
| $R + P\sin 15 - 4g\cos 25 = 0$ | B1 | Both resolutions correct. Allow sign errors. Allow use of $P\cos 15$ if $P\sin 15$ used in (iv) |
| | F1 | All correct. Only FT their $P$ and their use of $P\cos 15$ |
| $R = 25.729...$ so 25.7 N | A1 | cao |
---6 An empty open box of mass 4 kg is on a plane that is inclined at $25 ^ { \circ }$ to the horizontal.\\
In one model the plane is taken to be smooth.\\
The box is held in equilibrium by a string with tension $T \mathrm {~N}$ parallel to the plane, as shown in Fig. 6.1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d6e78f93-ac2c-4053-87e4-5e5537d6dc3d-5_314_575_621_785}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\end{center}
\end{figure}
(i) Calculate $T$.
A rock of mass $m \mathrm {~kg}$ is now put in the box. The system is in equilibrium when the tension in the string, still parallel to the plane, is 50 N .\\
(ii) Find $m$.
In a refined model the plane is rough.\\
The empty box, of mass 4 kg , is in equilibrium when a frictional force of 20 N acts down the plane and the string has a tension of $P \mathrm {~N}$ inclined at $15 ^ { \circ }$ to the plane, as shown in Fig. 6.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d6e78f93-ac2c-4053-87e4-5e5537d6dc3d-5_369_561_1653_790}
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
\end{center}
\end{figure}
(iii) Draw a diagram showing all the forces acting on the box.\\
(iv) Calculate $P$.\\
(v) Calculate the normal reaction of the plane on the box.
\hfill \mbox{\textit{OCR MEI M1 2009 Q6 [16]}}