| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2001 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Vehicle on slope with resistance |
| Difficulty | Standard +0.3 This is a standard M1 mechanics problem requiring resolution of forces on a slope with friction in two scenarios. While it involves multiple steps (finding normal reaction, friction force, applying F=ma twice), the techniques are routine and well-practiced. The trig is given, and both parts follow textbook methods without requiring novel insight. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03e Resolve forces: two dimensions3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\alpha = \arctan\frac{5}{12}\) | M1 A1 | \(\alpha \approx 22.6°\); \(\cos\alpha = \frac{12}{13}\), \(\sin\alpha = \frac{5}{13}\) |
| \(R = 78g\cos\alpha\) | B1 | |
| \(F = 78g\cos\alpha(0.25)\) | M1 A1 f.t. | |
| \(G = 78g\sin\alpha\) | B1 | |
| Newton II along slope attempted with \(T\), \(F\), \(G\) included | M1 | |
| \(T - F - G = 78(0.5)\) | DM1 | |
| Solving for \(T\) (dependent on M1) | ||
| \(T = 509.4\) (accept 510 to 2 s.f. or 509 to 3 s.f. — must only be these) | A1 | (9) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Accelerating force down slope is \(G - F\) (friction reversed and \(T\) no longer included) | M1, M1 | |
| Newton II: \(G - F = 78a\) | ||
| \(a = g\sin\alpha - \mu g\cos\alpha\) | A1 | |
| \(= 9.8\left(\frac{5}{13} - \frac{3}{13}\right)\) or \(\frac{2}{13}g\) | DM1 | |
| \(= 1.5,\ 1.50,\ 1.51\) (Score A2); other answers which round to 1.5 score A1 | \(A_{2,1,0}\) | (6) |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha = \arctan\frac{5}{12}$ | M1 A1 | $\alpha \approx 22.6°$; $\cos\alpha = \frac{12}{13}$, $\sin\alpha = \frac{5}{13}$ |
| $R = 78g\cos\alpha$ | B1 | |
| $F = 78g\cos\alpha(0.25)$ | M1 A1 f.t. | |
| $G = 78g\sin\alpha$ | B1 | |
| Newton II along slope attempted with $T$, $F$, $G$ included | M1 | |
| $T - F - G = 78(0.5)$ | DM1 | |
| Solving for $T$ (dependent on M1) | | |
| $T = 509.4$ (accept 510 to 2 s.f. or 509 to 3 s.f. — must only be these) | A1 | (9) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Accelerating force down slope is $G - F$ (friction reversed and $T$ no longer included) | M1, M1 | |
| Newton II: $G - F = 78a$ | | |
| $a = g\sin\alpha - \mu g\cos\alpha$ | A1 | |
| $= 9.8\left(\frac{5}{13} - \frac{3}{13}\right)$ or $\frac{2}{13}g$ | DM1 | |
| $= 1.5,\ 1.50,\ 1.51$ (Score A2); other answers which round to 1.5 score A1 | $A_{2,1,0}$ | (6) |7. A sledge of mass 78 kg is pulled up a slope by means of a rope. The slope is modelled as a rough plane inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac { 5 } { 12 }$. The rope is modelled as light and inextensible and is in a line of greatest slope of the plane. The coefficient of friction between the sledge and the slope is 0.25 . Given that the sledge is accelerating up the slope with acceleration $0.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$,
\begin{enumerate}[label=(\alph*)]
\item find the tension in the rope.
The rope suddenly breaks. Subsequently the sledge comes to instantaneous rest and then starts sliding down the slope.
\item Find the acceleration of the sledge down the slope after it has come to instantaneous rest.\\
(6 marks)\\
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2001 Q7 [15]}}