| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| 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 mechanics problem requiring resolution of forces on a slope with friction. Students must resolve the horizontal force into components, apply F=ma, and use kinematic equations. While it involves multiple steps and careful component resolution, it follows a well-practiced procedure with no novel insights required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(5.5 = \frac{1}{2}a \cdot 2^2\) | M1 | Complete method using suvat equations to form an equation in \(a\) only |
| \(\Rightarrow a = 2.75\) | A1 | |
| Total: (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(R = 30\sin\alpha + 2g\cos\alpha\) | M1 | Resolve perpendicular to plane to find expression for \(R\). Must have all terms. Condone sign errors and sin/cos confusion |
| A2 | \(-1\) each error. All correct A1A1, one error A1A0, two or more errors A0A0 \((R = 33.68)\) | |
| \(-F + 30\cos\alpha - 2g\sin\alpha = 2a\) | M1 | Equation of motion parallel to plane with \(a\) or their \(a\). Must have all terms. Condone sign errors and sin/cos confusion |
| A2 | \(-1\) each error \((F = 6.74)\) | |
| \(\mu = \dfrac{30\cos\alpha - 2g\sin\alpha - 5.5}{30\sin\alpha + 2g\cos\alpha}\) | DM1 | Use \(F = \mu R\). Dependent on the 2 previous M marks |
| \(= 0.200\) or \(0.20\) | A1 | Do not accept \(0.2\) |
| Total: (8) |
# Question 3(a):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $5.5 = \frac{1}{2}a \cdot 2^2$ | M1 | Complete method using suvat equations to form an equation in $a$ only |
| $\Rightarrow a = 2.75$ | A1 | |
| **Total: (2)** | | |
---
# Question 3(b):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $R = 30\sin\alpha + 2g\cos\alpha$ | M1 | Resolve perpendicular to plane to find expression for $R$. Must have all terms. Condone sign errors and sin/cos confusion |
| | A2 | $-1$ each error. All correct A1A1, one error A1A0, two or more errors A0A0 $(R = 33.68)$ |
| $-F + 30\cos\alpha - 2g\sin\alpha = 2a$ | M1 | Equation of motion parallel to plane with $a$ or their $a$. Must have all terms. Condone sign errors and sin/cos confusion |
| | A2 | $-1$ each error $(F = 6.74)$ |
| $\mu = \dfrac{30\cos\alpha - 2g\sin\alpha - 5.5}{30\sin\alpha + 2g\cos\alpha}$ | DM1 | Use $F = \mu R$. Dependent on the 2 previous M marks |
| $= 0.200$ or $0.20$ | A1 | Do not accept $0.2$ |
| **Total: (8)** | | |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3280fdf1-d81a-4729-b065-e84dece6a220-05_325_947_267_493}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A particle $P$ of mass 2 kg is pushed by a constant horizontal force of magnitude 30 N up a line of greatest slope of a rough plane. The plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac { 3 } { 4 }$, as shown in Figure 1. The line of action of the force lies in the vertical plane containing $P$ and the line of greatest slope of the plane. The particle $P$ starts from rest. The coefficient of friction between $P$ and the plane is $\mu$. After 2 seconds, $P$ has travelled a distance of 5.5 m up the plane.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of $P$ up the plane.
\item Find the value of $\mu$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2015 Q3 [10]}}