| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion down rough slope |
| Difficulty | Moderate -0.3 This is a standard mechanics question on motion down a rough slope requiring resolution of forces perpendicular and parallel to the plane, application of F=ma, and use of the friction model. The trigonometry is straightforward (given tan α = 5/12, students can find sin α = 5/13 and cos α = 12/13). Part (b) is a 'show that' requiring algebraic manipulation but following a routine method. Part (c) tests conceptual understanding of limiting equilibrium. Slightly easier than average due to its structured nature and standard approach. |
| Spec | 3.03b Newton's first law: equilibrium3.03e Resolve forces: two dimensions3.03r Friction: concept and vector form3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((R =) mg\cos\alpha\) | M1 | 3.4 - Correct no. of terms, condone sin/cos confusion and sign errors, dimensionally correct. Allow use of different symbol for angle |
| \(= \frac{12}{13}mg\) | A1 | 1.1b - Accept \(0.92mg\) or better. Must be positive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of motion down the plane | M1 | 2.1 - Correct no. of terms, condone sin/cos confusion and sign errors. M0 if they use \(g\) for \(a\). Must be using \(mg\) or \(m\) not \(W\) for weight |
| \(mg\sin\alpha - F = ma\) or \(mg\sin\alpha - F = -ma\) | A1 | 1.1b - Any correct equation e.g. \(mg\sin\alpha = ma + F\). N.B. \(F\) does not need to be substituted |
| \((F=)\ \mu \times \text{their } R\) | M1 | 3.4 - \(\mu\times\) their \(R\) seen (possibly on diagram), any trig does not need to be replaced. M0 if they use \(\mu = \frac{5}{12}\). M0 for just \(\mu R\) with \(R\) not replaced |
| \(\frac{1}{13}g(5-12\mu)\) * | A1* | 2.2a - Given answer correctly obtained, with at least one further line of working with both trig ratios substituted as fractions. Allow \(\frac{g}{13}(5-12\mu)\) or \(\frac{g(5-12\mu)}{13}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P\) wouldn't move | B1 | 2.4 - e.g. \(P\) (or the particle/object) would stay at rest, would not slide down, would not roll down the plane, static equilibrium, equilibrium at rest. Allow 'it' for \(P\) or just 'stays at rest oe'. B0 for 'wouldn't accelerate', 'would be in equilibrium (only)', 'would stop'. N.B. Ignore reasons but not contradictions |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(R =) mg\cos\alpha$ | M1 | 3.4 - Correct no. of terms, condone sin/cos confusion and sign errors, dimensionally correct. Allow use of different symbol for angle |
| $= \frac{12}{13}mg$ | A1 | 1.1b - Accept $0.92mg$ or better. Must be positive |
**(2 marks)**
---
## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion down the plane | M1 | 2.1 - Correct no. of terms, condone sin/cos confusion and sign errors. M0 if they use $g$ for $a$. Must be using $mg$ or $m$ not $W$ for weight |
| $mg\sin\alpha - F = ma$ or $mg\sin\alpha - F = -ma$ | A1 | 1.1b - Any correct equation e.g. $mg\sin\alpha = ma + F$. N.B. $F$ does not need to be substituted |
| $(F=)\ \mu \times \text{their } R$ | M1 | 3.4 - $\mu\times$ **their** $R$ seen (possibly on diagram), any trig does not need to be replaced. M0 if they use $\mu = \frac{5}{12}$. M0 for just $\mu R$ with $R$ not replaced |
| $\frac{1}{13}g(5-12\mu)$ * | A1* | 2.2a - Given answer correctly obtained, with at least one further line of working with both trig ratios substituted as fractions. Allow $\frac{g}{13}(5-12\mu)$ or $\frac{g(5-12\mu)}{13}$ |
**(4 marks)**
---
## Question 3(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P$ wouldn't move | B1 | 2.4 - e.g. $P$ (or the particle/object) would stay at rest, would not slide down, would not roll down the plane, static equilibrium, equilibrium at rest. Allow 'it' for $P$ or just 'stays at rest oe'. B0 for 'wouldn't accelerate', 'would be in equilibrium (only)', 'would stop'. N.B. Ignore reasons but not contradictions |
**(1 mark)**
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{184043b7-1222-44fb-bc9f-3f484f72147b-08_408_606_246_731}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A particle $P$ of mass $m$ is held at rest at a point on a rough inclined plane, as shown in Figure 3.
It is given that
\begin{itemize}
\item the plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac { 5 } { 12 }$
\item the coefficient of friction between $P$ and the plane is $\mu$, where $\mu < \frac { 5 } { 12 }$
\end{itemize}
The particle $P$ is released from rest and slides down the plane.\\
Air resistance is modelled as being negligible.\\
Using the model,
\begin{enumerate}[label=(\alph*)]
\item find, in terms of $m$ and $g$, the magnitude of the normal reaction of the plane on $P$,
\item show that, as $P$ slides down the plane, the acceleration of $P$ down the plane is
$$\frac { 1 } { 13 } g ( 5 - 12 \mu )$$
\item State what would happen to $P$ if it is released from rest but $\mu \geqslant \frac { 5 } { 12 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 3 2024 Q3 [7]}}