| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Practical friction scenarios |
| Difficulty | Moderate -0.3 This is a standard M1 friction question with multiple straightforward parts: resolving forces in equilibrium, describing motion when forces are unbalanced, calculating acceleration using F=ma, and finding friction on a slope. All parts use routine mechanics techniques with no novel problem-solving required, making it slightly easier than average but not trivial due to the multiple contexts and careful resolution needed. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Diagram | B1, B1 | Weight, friction and 121 N present with arrows. All forces with suitable labels. Accept \(W\), \(mg\), \(100g\), 980. No extra forces |
| Resolve \(\rightarrow\): \(121\cos34 - F = 0\), \(F = 100.313\ldots\) so \(100\) N (3 s.f.) | M1, E1 | Resolving horiz. Accept \(s\leftrightarrow c\). Some evidence required for the show, e.g. at least 4 figures. Accept \(\pm\) |
| Resolve \(\uparrow\): \(R + 121\sin34 - 980 = 0\) | M1, B1 | Resolve vert. Accept \(s\leftrightarrow c\) and sign errors. All correct |
| \(R = 912.337\ldots\) so \(912\) N (3 s.f.) | A1 | |
| 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| It will continue to move at constant speed of \(0.5\ \text{m s}^{-1}\) | E1, E1 | Accept no reference to direction (×2). [Do not isw: conflicting statements get zero] |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using N2L horizontally: \(155\cos34 - 95 = 100a\) | M1 | Use of N2L. Allow \(F=mga\), \(F\) omitted and 155 not resolved |
| A1 | Use of \(F=ma\) with resistance and \(T\) resolved. Allow \(s\leftrightarrow c\) and signs as the only errors | |
| \(a = 0.335008\ldots\) so \(0.335\ \text{m s}^{-2}\) (3 s.f.) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 5\div2 = 2.5\) | M1, A1 | Attempt to find \(a\) from information |
| N2L down the slope: \(100g\sin26 - F = 100\times2.5\) | M1 | \(F=ma\) using their "new" \(a\). All forces present. No extras. Require attempt at wt component. Allow \(s\leftrightarrow c\) and sign errors |
| B1 | Weight term resolved correctly, seen in equation or on diagram | |
| \(F = 179.603\ldots\) so \(180\) N (3 s.f.) | A1 | cao. Accept \(-180\) N if consistent with direction of \(F\) on their diagram |
| 5 |
# Question 7:
**(i)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Diagram | B1, B1 | Weight, friction and 121 N present with arrows. All forces with suitable labels. Accept $W$, $mg$, $100g$, 980. No extra forces |
| Resolve $\rightarrow$: $121\cos34 - F = 0$, $F = 100.313\ldots$ so $100$ N (3 s.f.) | M1, E1 | Resolving horiz. Accept $s\leftrightarrow c$. Some evidence required for the show, e.g. at least 4 figures. Accept $\pm$ |
| Resolve $\uparrow$: $R + 121\sin34 - 980 = 0$ | M1, B1 | Resolve vert. Accept $s\leftrightarrow c$ and sign errors. All correct |
| $R = 912.337\ldots$ so $912$ N (3 s.f.) | A1 | |
| | **7** | |
**(ii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| It will continue to move at constant speed of $0.5\ \text{m s}^{-1}$ | E1, E1 | Accept no reference to direction (×2). [Do not isw: conflicting statements get zero] |
| | **2** | |
**(iii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using N2L horizontally: $155\cos34 - 95 = 100a$ | M1 | Use of N2L. Allow $F=mga$, $F$ omitted and 155 not resolved |
| | A1 | Use of $F=ma$ with resistance and $T$ resolved. Allow $s\leftrightarrow c$ and signs as the **only** errors |
| $a = 0.335008\ldots$ so $0.335\ \text{m s}^{-2}$ (3 s.f.) | A1 | |
| | **3** | |
**(iv)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 5\div2 = 2.5$ | M1, A1 | Attempt to find $a$ from information |
| N2L down the slope: $100g\sin26 - F = 100\times2.5$ | M1 | $F=ma$ using their "new" $a$. All forces present. No extras. Require attempt at wt component. Allow $s\leftrightarrow c$ and sign errors |
| | B1 | Weight term resolved correctly, seen in equation or on diagram |
| $F = 179.603\ldots$ so $180$ N (3 s.f.) | A1 | cao. Accept $-180$ N if consistent with direction of $F$ on their diagram |
| | **5** | |
---7 An explorer is trying to pull a loaded sledge of total mass 100 kg along horizontal ground using a light rope. The only resistance to motion of the sledge is from friction between it and the ground.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{93a5d409-ade4-418b-9c09-620d97df97de-4_327_1013_482_566}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
Initially she pulls with a force of 121 N on the rope inclined at $34 ^ { \circ }$ to the horizontal, as shown in Fig. 7, but the sledge does not move.\\
(i) Draw a diagram showing all the forces acting on the sledge.
Show that the frictional force between the ground and the sledge is 100 N , correct to 3 significant figures.
Calculate the normal reaction of the ground on the sledge.
The sledge is given a small push to set it moving at $0.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The explorer continues to pull on the rope with the same force and the same angle as before. The frictional force is also unchanged.\\
(ii) Describe the subsequent motion of the sledge.
The explorer now pulls the rope, still at an angle of $34 ^ { \circ }$ to the horizontal, so that the tension in it is 155 N . The frictional force is now 95 N .\\
(iii) Calculate the acceleration of the sledge.
In a new situation, there is no rope and the sledge slides down a uniformly rough slope inclined at $26 ^ { \circ }$ to the horizontal. The sledge starts from rest and reaches a speed of $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in 2 seconds.\\
(iv) Calculate the frictional force between the slope and the sledge.
\hfill \mbox{\textit{OCR MEI M1 2009 Q7 [17]}}