| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 15 |
| 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 problem with connected particles. It requires resolving forces on an incline (with given tan α to find sin/cos), applying friction in the direction of motion, using F=ma for both particles with the constraint that accelerations are equal, then finding resultant force on the pulley using vector addition. While multi-step, all techniques are routine M1 procedures with no novel insight required, making it slightly easier than average. |
| Spec | 3.03o Advanced connected particles: and pulleys3.03p Resultant forces: using vectors3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R = 4g\cos\alpha\) | M1 A1 | First M1 for resolving perpendicular to plane; First A1 for correct equation |
| \(T - 0.5g = 0.5a\) | M1 A1 | Second M1 for resolving vertically; Second A1 for correct equation in terms of \(a\) and \(T\) |
| \(4g\sin\alpha - T - F = 4a\) (OR: \(4g\sin\alpha - F - 0.5g = 4.5a\)) | M1 A1 | Third M1 for resolving parallel to slope; Third A1 for correct equation in terms of \(a\), \(F\) and \(T\) |
| \(F = \frac{1}{2}R\) | B1 | First B1 for \(F = \frac{1}{2}R\) seen or implied |
| \(\sin\alpha = \frac{4}{5}\) or \(\cos\alpha = \frac{3}{5}\) | B1 | Second B1 for \(\sin\alpha = 0.8\) or \(\cos\alpha = 0.6\) seen or implied |
| Eliminating \(a\) or finding \(a\) | M1 | Fourth M1 independent for eliminating or finding \(a\) |
| Solving for \(T\) (must have had an \(a\)) | M1 | Fifth M1 for solving for \(T\); must have had an \(a\) |
| \(T = \frac{2g}{3}\) N or 6.5N or 6.53N | A1 | Fourth A1 for \(\frac{2g}{3}\), 6.5 or 6.53 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Magnitude \(= 2T\cos\!\left(\dfrac{90°-\alpha}{2}\right)\) | M1 A1 | M1 for complete method for finding magnitude of resultant (M0 if same tensions used); A1 for correct expression in terms of \(T\) and \(\alpha\) |
| \(= 2 \times \dfrac{2g}{3} \times \dfrac{3}{\sqrt{10}}\ (0.94868\ldots)\) | A1 ft | A1 ft on their \(T\) for a 'correct' single numerical answer |
| \(= 12\text{N}\) or \(12.4\text{N}\ \left(\dfrac{4g}{\sqrt{10}}\right)\) | A1 | cao for 12 (N) or 12.4 (N) |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = 4g\cos\alpha$ | M1 A1 | First M1 for resolving perpendicular to plane; First A1 for correct equation |
| $T - 0.5g = 0.5a$ | M1 A1 | Second M1 for resolving vertically; Second A1 for correct equation in terms of $a$ and $T$ |
| $4g\sin\alpha - T - F = 4a$ (OR: $4g\sin\alpha - F - 0.5g = 4.5a$) | M1 A1 | Third M1 for resolving parallel to slope; Third A1 for correct equation in terms of $a$, $F$ and $T$ |
| $F = \frac{1}{2}R$ | B1 | First B1 for $F = \frac{1}{2}R$ seen or implied |
| $\sin\alpha = \frac{4}{5}$ or $\cos\alpha = \frac{3}{5}$ | B1 | Second B1 for $\sin\alpha = 0.8$ or $\cos\alpha = 0.6$ seen or implied |
| Eliminating $a$ or finding $a$ | M1 | Fourth M1 independent for eliminating or finding $a$ |
| Solving for $T$ (must have had an $a$) | M1 | Fifth M1 for solving for $T$; must have had an $a$ |
| $T = \frac{2g}{3}$ N or 6.5N or 6.53N | A1 | Fourth A1 for $\frac{2g}{3}$, 6.5 or 6.53 | **(11)** |
**N.B.** If $a = 0$ used in any of the above 3 equations and used to find $T$, they lose both marks for that equation and the two M marks for eliminating and solving. Their $a$ could be UP the slope — all 4 marks available with $-a$ replacing $a$, provided consistent.
---
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Magnitude $= 2T\cos\!\left(\dfrac{90°-\alpha}{2}\right)$ | M1 A1 | M1 for complete method for finding magnitude of resultant (M0 if same tensions used); A1 for correct expression in terms of $T$ and $\alpha$ |
| $= 2 \times \dfrac{2g}{3} \times \dfrac{3}{\sqrt{10}}\ (0.94868\ldots)$ | A1 ft | A1 ft on their $T$ for a 'correct' single numerical answer |
| $= 12\text{N}$ or $12.4\text{N}\ \left(\dfrac{4g}{\sqrt{10}}\right)$ | A1 | cao for 12 (N) or 12.4 (N) | **(4)** |
**OR methods accepted:**
- $\sqrt{(T + T\sin\alpha)^2 + (T\cos\alpha)^2}$ — allow sin/cos confusion; allow omission of $\sqrt{}$ only if $R^2 =\ldots$ included
- $\sqrt{T^2 + T^2 - 2T^2\cos(90°+\alpha)}$ — allow $(90°-\alpha)$ but must be cos
- $\dfrac{T\sin(90°+\alpha)}{\sin\!\left(\dfrac{90°-\alpha}{2}\right)}$ (Sine Rule) — allow sign errors in angles but must be sin
**Total: 15 marks**8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{de3245a7-cf6e-423e-8689-9a074bdbc23b-14_643_931_118_534}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Two particles $P$ and $Q$ have mass 4 kg and 0.5 kg respectively. The particles are attached to the ends of a light inextensible string. Particle $P$ is held at rest on a fixed rough plane, which is inclined to the horizontal at an angle $\alpha$ where $\tan \alpha = \frac { 4 } { 3 }$. The coefficient of friction between $P$ and the plane is 0.5 . The string lies along the plane and passes over a small smooth light pulley which is fixed at the top of the plane. Particle $Q$ hangs freely at rest vertically below the pulley. The string lies in the vertical plane which contains the pulley and a line of greatest slope of the inclined plane, as shown in Figure 4. Particle $P$ is released from rest with the string taut and slides down the plane.
Given that $Q$ has not hit the pulley, find
\begin{enumerate}[label=(\alph*)]
\item the tension in the string during the motion,
\item the magnitude of the resultant force exerted by the string on the pulley.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2015 Q8 [15]}}