| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Particle on slope with pulley |
| Difficulty | Standard +0.3 This is a standard M1 mechanics problem combining equilibrium on a slope with subsequent motion. Part (a) requires resolving forces in limiting equilibrium (routine technique), and part (b) applies energy methods or equations of motion after the string breaks. Both parts follow textbook procedures with no novel insight required, making it slightly easier than average for M1. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Perp. to plane for \(P\): \(R=mg\cos a\) | M1 A1 | Resolve perpendicular; condone sin/cos confusion and sign errors. Correct unsimplified equation |
| \(P\): \(T=mg\sin\alpha+F\) | M1 A1 | Form equilibrium equation for \(P\); correct no. of terms, dimensionally correct. If \(F=ma\) used then \(a\) must be zero |
| \(Q\): \(T=0.5mg\) | B1 | Correct equation |
| Use of \(F=\mu R\) | B1 | Seen or implied in an equation |
| \(0.5mg=\dfrac{5mg}{13}+\mu\dfrac{12mg}{13}\) | dM1 | Dependent on previous M mark; replace trig and form equation in \(\mu\) only |
| \(\mu=\dfrac{1}{8}\) | A1 | Correct answer. Accept 0.125, 0.13 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(mg\sin\alpha - F = ma\) | M1 A1 | Use of \(F=ma\) for \(P\); correct no. of terms, dimensionally correct, ignore sin/cos confusion. Correct equation, trig and \(F\) do not need to be substituted |
| \(\left(a=\dfrac{7g}{26}\ \text{ms}^{-2}\right)\) | ||
| \(V^2=0^2+2\left(\dfrac{7g}{26}\right)0.8\) | M1 | Use of calculated acceleration to form equation in \(V\). M0 if they use \(g\) |
| \(V=2.1\) or \(2.05\) | A1 | Correct answer 3 s.f. |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Perp. to plane for $P$: $R=mg\cos a$ | M1 A1 | Resolve perpendicular; condone sin/cos confusion and sign errors. Correct unsimplified equation |
| $P$: $T=mg\sin\alpha+F$ | M1 A1 | Form equilibrium equation for $P$; correct no. of terms, dimensionally correct. If $F=ma$ used then $a$ must be zero |
| $Q$: $T=0.5mg$ | B1 | Correct equation |
| Use of $F=\mu R$ | B1 | Seen or implied in an equation |
| $0.5mg=\dfrac{5mg}{13}+\mu\dfrac{12mg}{13}$ | dM1 | Dependent on previous M mark; replace trig and form equation in $\mu$ only |
| $\mu=\dfrac{1}{8}$ | A1 | Correct answer. Accept 0.125, 0.13 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $mg\sin\alpha - F = ma$ | M1 A1 | Use of $F=ma$ for $P$; correct no. of terms, dimensionally correct, ignore sin/cos confusion. Correct equation, trig and $F$ do not need to be substituted |
| $\left(a=\dfrac{7g}{26}\ \text{ms}^{-2}\right)$ | | |
| $V^2=0^2+2\left(\dfrac{7g}{26}\right)0.8$ | M1 | Use of calculated acceleration to form equation in $V$. M0 if they use $g$ |
| $V=2.1$ or $2.05$ | A1 | Correct answer 3 s.f. |8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e59a66b8-c2ad-41fd-9959-9d21e9455c37-24_346_961_246_543}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
A fixed rough plane is inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac { 5 } { 12 }$\\
A small smooth pulley is fixed at the top of the plane.\\
One end of a light inextensible string is attached to a particle $P$ which is at rest on the plane. The string passes over the pulley and the other end of the string is attached to a particle $Q$ which hangs vertically below the pulley, as shown in Figure 5.
Particle $P$ has mass $m$ and particle $Q$ has mass $0.5 m$\\
The string from $P$ to the pulley lies along a line of greatest slope of the plane.\\
The coefficient of friction between $P$ and the plane is $\mu$.\\
The system is in limiting equilibrium with the string taut and $P$ is on the point of slipping up the plane.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\mu$.
The string breaks and $P$ begins to move down the plane.\\
When particle $P$ has travelled a distance of 0.8 m down the plane, the speed of $P$ is $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$
\item Find the value of $V$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2024 Q8 [12]}}