| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Particle on rough incline connected to particle on horizontal surface or other incline |
| Difficulty | Standard +0.3 This is a standard M1 pulley system with connected particles on different surfaces. While it involves multiple components (inclined plane with friction, horizontal plane, pulley forces), the solution follows a routine procedure: resolve forces, apply F=ma to each particle, solve simultaneous equations. The trig is given (tan α = 3/4 → sin α = 3/5, cos α = 4/5), and part (d) tests understanding rather than calculation. Slightly easier than average due to its methodical, textbook nature. |
| Spec | 3.03e Resolve forces: two dimensions3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T = ma\) (allow \(-a\)) | B1 | Equation must appear in (a) to earn B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4mg\sin\alpha - T - F = 4ma\) OR \(4mg\sin\alpha - F = 5ma\) | M1A1 | Equation of motion for \(P\) parallel to plane, correct no. of terms; condone sign errors and sin/cos confusion |
| \(F = \frac{1}{4}R\) | B1 | Could just be on the diagram |
| \(R = 4mg\cos\alpha\) | M1A1 | Resolve perpendicular to plane for \(P\); condone sign errors and sin/cos confusion |
| Solve for \(T\) in terms of \(mg\) only | DM1 | Dependent on both M marks; must be in terms of \(mg\) only (must be of form \(kmg\)) |
| \(T = \frac{8mg}{25}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2T\sin\frac{1}{2}\alpha\) oe, e.g. \(\sqrt{T^2 + T^2 - 2T^2\cos\alpha}\) using cosine rule; or \(\frac{T\sin\alpha}{\sin(90° - \frac{1}{2}\alpha)}\) using sine rule; or \(\sqrt{(T - T\cos\alpha)^2 + (T\sin\alpha)^2}\) using components and Pythagoras | M1A1 | If resolving, condone cos/sin confusion and sign errors but must have correct angle; A1 any correct unsimplified expression in terms of \(T\) and \(\alpha\) |
| Substitute for \(T\) and trig | M1 | Must substitute their \(T\) (must be of form \(kmg\)) and correct values for trig |
| \(\frac{8mg\sqrt{10}}{125}\) oe, \(2m\) or \(2.0m\) or \(1.98m\) or \(0.2mg\) or better | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Tension will be the same throughout a section of the string | B1 | B0 for 'tension is the same throughout the string'; B0 if incorrect extras |
## Question 7:
### Part 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = ma$ (allow $-a$) | B1 | Equation must appear in (a) to earn B1 |
### Part 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4mg\sin\alpha - T - F = 4ma$ **OR** $4mg\sin\alpha - F = 5ma$ | M1A1 | Equation of motion for $P$ parallel to plane, correct no. of terms; condone sign errors and sin/cos confusion |
| $F = \frac{1}{4}R$ | B1 | Could just be on the diagram |
| $R = 4mg\cos\alpha$ | M1A1 | Resolve perpendicular to plane for $P$; condone sign errors and sin/cos confusion |
| Solve for $T$ in terms of $mg$ only | DM1 | Dependent on both M marks; must be in terms of $mg$ only (must be of form $kmg$) |
| $T = \frac{8mg}{25}$ | A1 | cao |
### Part 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2T\sin\frac{1}{2}\alpha$ oe, e.g. $\sqrt{T^2 + T^2 - 2T^2\cos\alpha}$ using cosine rule; or $\frac{T\sin\alpha}{\sin(90° - \frac{1}{2}\alpha)}$ using sine rule; or $\sqrt{(T - T\cos\alpha)^2 + (T\sin\alpha)^2}$ using components and Pythagoras | M1A1 | If resolving, condone cos/sin confusion and sign errors but must have correct angle; A1 any correct unsimplified expression in terms of $T$ and $\alpha$ |
| Substitute for $T$ and trig | M1 | Must substitute their $T$ (must be of form $kmg$) and correct values for trig |
| $\frac{8mg\sqrt{10}}{125}$ oe, $2m$ or $2.0m$ or $1.98m$ or $0.2mg$ or better | A1 | cao |
### Part 7(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Tension will be the same throughout a **section** of the string | B1 | B0 for 'tension is the same throughout the string'; B0 if incorrect extras |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f1bdc84b-c8a1-4e7c-a2ba-48b40c6a6d36-22_342_1203_246_374}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A particle $P$ of mass $4 m$ lies on the surface of a fixed rough inclined plane.\\
The plane is inclined to the horizontal at an angle $\alpha$ where $\tan \alpha = \frac { 3 } { 4 }$\\
The particle $P$ is attached to one end of a light inextensible string.\\
The string passes over a small smooth pulley that is fixed at the top of the plane. The other end of the string is attached to a particle $Q$ of mass $m$ which lies on a smooth horizontal plane.
The string lies along the horizontal plane and in the vertical plane that contains the pulley and a line of greatest slope of the inclined plane.
The system is released from rest with the string taut, as shown in Figure 4, and $P$ moves down the plane.
The coefficient of friction between $P$ and the plane is $\frac { 1 } { 4 }$\\
For the motion before $Q$ reaches the pulley
\begin{enumerate}[label=(\alph*)]
\item write down an equation of motion for $Q$,
\item find, in terms of $m$ and $g$, the tension in the string,
\item find the magnitude of the force exerted on the pulley by the string.
\item State where in your working you have used the information that the string is light.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2022 Q7 [13]}}