| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Particle on rough incline, particle hanging |
| Difficulty | Standard +0.3 This is a standard M1 pulley system question with routine components: identifying string constraint (a), writing F=ma equations with friction (b-c), and using energy/kinematics for the final inequality (d). While multi-part, each step follows textbook methods with no novel insight required. The tan α = 3/4 setup and the final show-that calculation are typical M1 exam techniques, making this slightly easier than average. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| END |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Inextensible string | B1 | B0 if incorrect consequences stated; (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(4mg\sin\alpha - T - F = 4ma\) | M1 A2 | Equation of motion for \(P\); omission of 4 on RHS is M0 |
| \(T - mg = ma\) | M1 A1 | Equation of motion for \(Q\); (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(F = \frac{1}{4}R\) | B1 | Seen or implied |
| \(R = 4mg\cos\alpha\) | B1 | Seen or implied |
| \(\cos\alpha = \frac{4}{5}\) or \(\sin\alpha = \frac{3}{5}\) | B1 | Or appropriate correct trig ratio |
| Eliminating \(R\), \(F\) and \(T\) | M1 | Finding \(a\) value |
| \(a = \frac{3}{25}g = 1.2\) or \(1.18\ \text{ms}^{-2}\) | A1 | Must be positive; (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(v^2 = 2 \times \frac{3}{25}gh = \frac{6}{25}gh\) | M1 | |
| \(0^2 = \frac{6}{25}gh - 2gs\) | ||
| \(s = \frac{3}{25}h\) | M1 A1 | |
| \(d > \frac{3}{25}h + h = \frac{28}{25}h\) | DM1 A1 | Given answer; (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Find \(v\) or \(v^2\) for \(P\) using their \(a\) | M1 | M0 if \(g\) is used |
| Complete method to find \(s\), independent but must have found \(v\) or \(v^2\) | M1 | M0 if \(g\) not used |
| \(s = \frac{3}{25}h\) | A1 | oe |
| Adding \(h\) onto their \(s\) | DM1 | Dependent on previous two M marks; oe |
| Given answer | A1 | — |
# Question 7:
## Part 7(a):
| Working | Marks | Guidance |
|---------|-------|----------|
| Inextensible string | B1 | B0 if incorrect consequences stated; (1 mark) |
## Part 7(b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $4mg\sin\alpha - T - F = 4ma$ | M1 A2 | Equation of motion for $P$; omission of 4 on RHS is M0 |
| $T - mg = ma$ | M1 A1 | Equation of motion for $Q$; (5 marks) |
## Part 7(c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $F = \frac{1}{4}R$ | B1 | Seen or implied |
| $R = 4mg\cos\alpha$ | B1 | Seen or implied |
| $\cos\alpha = \frac{4}{5}$ or $\sin\alpha = \frac{3}{5}$ | B1 | Or appropriate correct trig ratio |
| Eliminating $R$, $F$ and $T$ | M1 | Finding $a$ value |
| $a = \frac{3}{25}g = 1.2$ or $1.18\ \text{ms}^{-2}$ | A1 | Must be positive; (5 marks) |
## Part 7(d):
| Working | Marks | Guidance |
|---------|-------|----------|
| $v^2 = 2 \times \frac{3}{25}gh = \frac{6}{25}gh$ | M1 | |
| $0^2 = \frac{6}{25}gh - 2gs$ | | |
| $s = \frac{3}{25}h$ | M1 A1 | |
| $d > \frac{3}{25}h + h = \frac{28}{25}h$ | DM1 A1 | Given answer; (5 marks) |
## Question 7(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Find $v$ or $v^2$ for $P$ using their $a$ | M1 | M0 if $g$ is used |
| Complete method to find $s$, independent but must have found $v$ or $v^2$ | M1 | M0 if $g$ not used |
| $s = \frac{3}{25}h$ | A1 | oe |
| Adding $h$ onto their $s$ | DM1 | Dependent on previous two M marks; oe |
| Given answer | A1 | — |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4fd21e83-0bdf-4bb1-8a3f-76beada511ae-24_391_917_251_516}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A particle $P$ of mass $4 m$ is held at rest at the point $X$ on the surface of a rough inclined plane which is fixed to horizontal ground. The point $X$ is a distance $h$ from the bottom of the inclined plane. The plane is inclined to the horizontal at an angle $\alpha$ where $\tan \alpha = \frac { 3 } { 4 }$. The coefficient of friction between $P$ and the plane is $\frac { 1 } { 4 }$. The particle $P$ is attached to one end of a light inextensible string. The string passes over a small smooth pulley which is fixed at the top of the plane. The other end of the string is attached to a particle $Q$ of mass $m$ which hangs freely at a distance $d$, where $d > h$, below the pulley, as shown in Figure 4.
The string lies in a vertical plane through a line of greatest slope of the inclined plane. The system is released from rest with the string taut and $P$ moves down the plane.
For the motion of the particles before $P$ hits the ground,
\begin{enumerate}[label=(\alph*)]
\item state which of the information given above implies that the magnitudes of the accelerations of the two particles are the same,
\item write down an equation of motion for each particle,
\item find the acceleration of each particle.
When $P$ hits the ground, it immediately comes to rest. Given that $Q$ comes to instantaneous rest before reaching the pulley,
\item show that $d > \frac { 28 h } { 25 }$.
\begin{center}
\end{center}
\begin{center}
\end{center}
\includegraphics[max width=\textwidth, alt={}, center]{4fd21e83-0bdf-4bb1-8a3f-76beada511ae-27_56_20_109_1950}\\
\begin{center}
\begin{tabular}{|l|l|}
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END & \\
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\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2018 Q7 [16]}}