| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 14 |
| 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 question with connected particles on different surfaces. Part (a) requires routine application of Newton's second law to two particles with friction, using standard techniques (resolving forces, finding sin α from tan α). Part (b) involves finding limiting friction when the system is in equilibrium. While multi-step, all techniques are textbook-standard for M1 with no novel insight required, making it slightly easier than average. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03n Equilibrium in 2D: particle under forces3.03o Advanced connected particles: and pulleys3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
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| Q8 | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R = mg\) | B1 | Resolve vertically at \(Q\) |
| \(F = \frac{1}{2}R\) | B1 | Use of \(F = \mu R\) |
| \(T - F = ma\) | M1 | Equation of motion for \(Q\); no missing/additional terms; condone sign error(s) |
| A1 | ||
| \(2mg\sin\alpha - T = 2ma\) | M1 | Equation of motion for \(P\); no missing/additional terms; condone sign error(s) and sin/cos confusion |
| A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| DM1 | Solve for \(a\) or \(T\); dependent on 2 correct equations | |
| \(a = \frac{7g}{30} = 2.3\) or \(2.29\) ms\(^{-2}\) | A1 | \(a\) or \(T\) correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T = \frac{7mg}{30} + \frac{mg}{2}\) | DM1 | Solve for second unknown; dependent on 2 correct equations |
| \(= \frac{11mg}{15}\) | A1 | Both correct; accept \(T = 7.2m\) or better |
| (10) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 0 \Rightarrow 2mg\sin\alpha - T = 0\) | M1 | Use equation of motion of \(P\) to find \(T\) |
| \(\Rightarrow T = \frac{6mg}{5}\) | A1 | \((11.76m)\) |
| \(\mu mg \geq \frac{6mg}{5}\) | DM1 | For \(Q\), \(T \leq \mu R\); dependent on preceding M; condone use of \(T = \mu R\) |
| Least value is \(1.2\) | A1 | |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2mg\sin\alpha - \mu R = 0\) | M1A1 | Using the combined equation |
| \(\frac{6}{5}mg = \mu mg\) | M1 | Substitute for trig and \(R\) and solve |
| Least value is \(1.2\) | A1 | |
| (4) | ||
| [14] |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = mg$ | B1 | Resolve vertically at $Q$ |
| $F = \frac{1}{2}R$ | B1 | Use of $F = \mu R$ |
| $T - F = ma$ | M1 | Equation of motion for $Q$; no missing/additional terms; condone sign error(s) |
| | A1 | |
| $2mg\sin\alpha - T = 2ma$ | M1 | Equation of motion for $P$; no missing/additional terms; condone sign error(s) and sin/cos confusion |
| | A1 | |
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | DM1 | Solve for $a$ or $T$; dependent on 2 correct equations |
| $a = \frac{7g}{30} = 2.3$ or $2.29$ ms$^{-2}$ | A1 | $a$ or $T$ correct |
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = \frac{7mg}{30} + \frac{mg}{2}$ | DM1 | Solve for second unknown; dependent on 2 correct equations |
| $= \frac{11mg}{15}$ | A1 | Both correct; accept $T = 7.2m$ or better |
| | **(10)** | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 0 \Rightarrow 2mg\sin\alpha - T = 0$ | M1 | Use equation of motion of $P$ to find $T$ |
| $\Rightarrow T = \frac{6mg}{5}$ | A1 | $(11.76m)$ |
| $\mu mg \geq \frac{6mg}{5}$ | DM1 | For $Q$, $T \leq \mu R$; dependent on preceding M; condone use of $T = \mu R$ |
| Least value is $1.2$ | A1 | |
| | **(4)** | |
### Part (b) alt:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2mg\sin\alpha - \mu R = 0$ | M1A1 | Using the combined equation |
| $\frac{6}{5}mg = \mu mg$ | M1 | Substitute for trig and $R$ and solve |
| Least value is $1.2$ | A1 | |
| | **(4)** | |
| | **[14]** | |
8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5c3869c7-008f-4131-b68d-8ecdd4da3377-24_369_1200_248_370}
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\caption{Figure 4}
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\end{figure}
Two particles, $P$ and $Q$, with masses $2 m$ and $m$ respectively, are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed at the edge of a rough horizontal table. Particle $Q$ is held at rest on the table and particle $P$ is on the surface of a smooth inclined plane. The top of the plane coincides with the edge of the table. The plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac { 3 } { 4 }$, as shown in Figure 4. The string lies in a vertical plane containing the pulley and a line of greatest slope of the plane. The coefficient of friction between $Q$ and the table is $\frac { 1 } { 2 }$. Particle $Q$ is released from rest with the string taut and $P$ begins to slide down the plane.
\begin{enumerate}[label=(\alph*)]
\item By writing down an equation of motion for each particle,
\begin{enumerate}[label=(\roman*)]
\item find the initial acceleration of the system,
\item find the tension in the string.
Suppose now that the coefficient of friction between $Q$ and the table is $\mu$ and when $Q$ is released it remains at rest.
\end{enumerate}\item Find the smallest possible value of $\mu$.
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\hfill \mbox{\textit{Edexcel M1 2017 Q8 [14]}}