| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Equilibrium on slope with force at angle to slope |
| Difficulty | Standard +0.3 This is a standard M1 equilibrium problem on a slope with a force at an angle. It requires resolving forces in two directions (parallel and perpendicular to slope), applying F=μR for limiting friction, and checking equilibrium when the force is removed. The multi-part structure and calculation with specific angles makes it slightly above average difficulty, but it follows a well-practiced method with no novel insight required. |
| Spec | 3.03e Resolve forces: two dimensions3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Perpendicular to slope: \(R = 2.7g\cos 40° + 15\cos 40°\) | M1A2 | M1 for resolving perp. to slope with correct no. of terms, both \(2.7g\) and \(15\) terms resolved; A2 correct equation, \(-1\) each error |
| \(R = 31.8\) (N) or \(32\) (N) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Parallel to slope: \(F = 2.7g\sin 40° - 15\cos 50°\) \(\quad (F = 7.366\ldots)\) | M1A2 | M1 for resolving parallel to slope with correct no. of terms, both \(2.7g\) and \(15\) terms resolved; A2 correct equation, \(-1\) each error |
| Use of \(F = \mu R\) | M1 | |
| \(\mu = \dfrac{2.7g\sin 40° - 15\cos 50°}{R} = 0.23\) or \(0.232\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Component of weight parallel to slope \(= 2.7g\sin 40°\ (= 17.0)\) | B1 | \(17\) or better |
| \(F_{\max} = 0.232 \times 2.7 \times g \times \cos 40° = 4.7\ldots\) (N) | M1A1 | M0 if \(R\) from (a) used; A1 for \(4.7\) or better |
| \(17.0 > 4.70\) so the particle moves | A1 | Comparison and correct conclusion; N.B. if first A mark is 0, second A mark must also be 0 |
## Question 7:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Perpendicular to slope: $R = 2.7g\cos 40° + 15\cos 40°$ | M1A2 | M1 for resolving perp. to slope with correct no. of terms, both $2.7g$ and $15$ terms resolved; A2 correct equation, $-1$ each error |
| $R = 31.8$ (N) or $32$ (N) | A1 | |
---
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Parallel to slope: $F = 2.7g\sin 40° - 15\cos 50°$ $\quad (F = 7.366\ldots)$ | M1A2 | M1 for resolving parallel to slope with correct no. of terms, both $2.7g$ and $15$ terms resolved; A2 correct equation, $-1$ each error |
| Use of $F = \mu R$ | M1 | |
| $\mu = \dfrac{2.7g\sin 40° - 15\cos 50°}{R} = 0.23$ or $0.232$ | A1 | |
---
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Component of weight parallel to slope $= 2.7g\sin 40°\ (= 17.0)$ | B1 | $17$ or better |
| $F_{\max} = 0.232 \times 2.7 \times g \times \cos 40° = 4.7\ldots$ (N) | M1A1 | M0 if $R$ from (a) used; A1 for $4.7$ or better |
| $17.0 > 4.70$ so the particle moves | A1 | Comparison and correct conclusion; **N.B. if first A mark is 0, second A mark must also be 0** |
**N.B.** Only penalise over- or under-accuracy after using $g = 9.8$ (or $g = 9.81$) once in whole question.7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b896c631-00a0-46c5-bce9-16d65f6e3095-13_364_833_269_561}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A particle $P$ of mass 2.7 kg lies on a rough plane inclined at $40 ^ { \circ }$ to the horizontal. The particle is held in equilibrium by a force of magnitude 15 N acting at an angle of $50 ^ { \circ }$ to the plane, as shown in Figure 4. The force acts in a vertical plane containing a line of greatest slope of the plane. The particle is in equilibrium and is on the point of sliding down the plane.
Find
\begin{enumerate}[label=(\alph*)]
\item the magnitude of the normal reaction of the plane on $P$,
\item the coefficient of friction between $P$ and the plane.
The force of magnitude 15 N is removed.
\item Determine whether $P$ moves, justifying your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2014 Q7 [13]}}