| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Horizontal force on slope |
| Difficulty | Standard +0.3 This is a standard M1 inclined plane problem with friction requiring resolution of forces and Newton's second law. Part (i) involves finding acceleration from kinematics then applying F=ma with friction opposing motion. Part (ii) requires resolving a horizontal force and finding limiting friction. Both parts are routine applications of core mechanics techniques with no novel insight required, making it slightly easier than average. |
| Spec | 3.03m 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 |
|---|---|---|
| (i) Acceleration is \(\frac{0.8}{10} = 0.08\) m s\(^{-2}\) | B1 | For 0.8 ÷ 10 stated or implied |
| \(R = 25g \cos 30°\) | B1 | For correct resolving \(\perp\) plane |
| \(T - 25g \sin 30° - 0.2 \times 25g \cos 30° = 25 \times 0.08\) | M1 | For attempting Newton II |
| B1 | For upwards force \(T - 25g \sin 30° - F\) | |
| For \(F = 0.2 \times 25g \cos 30°\) | B1 | |
| Hence the tension is 167 N | A1 | For correct value 167 |
| Subtotal: 6 marks | ||
| (ii) \(R' = P \sin 30° + 175 g \cos 30°\) | M1 | For resolving \(\perp\) plane, with 3 forces |
| A1 | For correct equation | |
| \(P \cos 30° + 0.2R' = 175 g \sin 30°\) | M1 | For resolving |
| A1 | For correct equation | |
| \(P(\cos 30° + 0.2 \sin 30°) = 175 g(\sin 30° - 0.2 \cos 30°)\) | M1 | For attempting elimination of \(R'\) |
| Hence \(P = \frac{175 g(\sin 30° - 0.2 \cos 30°)}{\cos 30° + 0.2 \sin 30°} = 580\) | M1, A1 | For solving a relevant equation for \(P\); For correct value 580 |
| Subtotal: 7 marks | ||
| Total: 13 marks |
**(i)** Acceleration is $\frac{0.8}{10} = 0.08$ m s$^{-2}$ | B1 | For 0.8 ÷ 10 stated or implied
$R = 25g \cos 30°$ | B1 | For correct resolving $\perp$ plane
$T - 25g \sin 30° - 0.2 \times 25g \cos 30° = 25 \times 0.08$ | M1 | For attempting Newton II || plane
| B1 | For upwards force $T - 25g \sin 30° - F$
For $F = 0.2 \times 25g \cos 30°$ | B1 |
Hence the tension is 167 N | A1 | For correct value 167
| | **Subtotal: 6 marks**
**(ii)** $R' = P \sin 30° + 175 g \cos 30°$ | M1 | For resolving $\perp$ plane, with 3 forces
| A1 | For correct equation
$P \cos 30° + 0.2R' = 175 g \sin 30°$ | M1 | For resolving || plane, with 3 forces
| A1 | For correct equation
$P(\cos 30° + 0.2 \sin 30°) = 175 g(\sin 30° - 0.2 \cos 30°)$ | M1 | For attempting elimination of $R'$
Hence $P = \frac{175 g(\sin 30° - 0.2 \cos 30°)}{\cos 30° + 0.2 \sin 30°} = 580$ | M1, A1 | For solving a relevant equation for $P$; For correct value 580
| | **Subtotal: 7 marks**
| | **Total: 13 marks**A sledge of mass 25 kg is on a plane inclined at $30°$ to the horizontal. The coefficient of friction between the sledge and the plane is 0.2.
\begin{enumerate}[label=(\roman*)]
\item
\includegraphics{figure_6}
The sledge is pulled up the plane, with constant acceleration, by means of a light cable which is parallel to a line of greatest slope (see Fig. 1). The sledge starts from rest and acquires a speed of $0.8 \text{ m s}^{-1}$ after being pulled for 10 s. Ignoring air resistance, find the tension in the cable. [6]
\item
\includegraphics{figure_7}
On a subsequent occasion the cable is not in use and two people of total mass 150 kg are seated in the sledge. The sledge is held at rest by a horizontal force of magnitude $P$ newtons, as shown in Fig. 2. Find the least value of $P$ which will prevent the sledge from sliding down the plane. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR M1 Q7 [13]}}