| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | Specimen |
| Marks | 15 |
| 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 problem with particles on inclined planes. While it involves multiple steps (resolving forces on two planes, using friction at limiting equilibrium, and finding resultant force on pulley), all techniques are routine for M1. The given tan values lead to convenient 3-4-5 triangles, simplifying calculations. This is slightly easier than average as it's a textbook application of standard methods with no novel insight required. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case3.03b Newton's first law: equilibrium3.03e Resolve forces: two dimensions3.03k Connected particles: pulleys and equilibrium3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| VIAV SIHI NI BIIIM ION OC | VGHV SIHI NI GHIYM ION OC | VJ4V SIHI NI JIIYM ION OC |
| VIAV SIHI NI JIIYM IONOO | VI4V SIHI NI IIIIMM ION OO | VEYV SIHI NI JLIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = 0.25R\) | B1 | |
| \(\sin\alpha = \frac{3}{5}\) or \(\cos\alpha = \frac{4}{5}\); \(\sin\beta = \frac{4}{5}\) or \(\cos\beta = \frac{3}{5}\) | B1 | Use of correct trig ratios for \(\alpha\) or \(\beta\) |
| \(R = 4g\cos\alpha \quad (= 31.36)\) | M1 | Normal reaction on \(P\). Condone trig confusion (using \(\alpha\)) |
| Correct equation | A1 | |
| \(T + F = 4g\sin\alpha\) | M1 | Equation of motion for \(P\). Requires all 3 terms. Condone consistent trig confusion. Condone acceleration \(\neq 0\): \(T + F - 4g\sin\alpha = 4a\) |
| \(T + 7.84 = 23.52 \Rightarrow T = 15.68\) | A1 | Correct equation |
| \(T = mg\sin\beta\) | M1 | Equation of motion for \(Q\). Condone trig confusion. Condone acceleration \(\neq 0\): \(T - mg\sin\beta = -ma\) |
| \(T = 7.84m\) | A1 | Correct equation |
| Solve for \(m\) | DM1 | Dependent on 3 preceding M marks. Not available if equations used \(a \neq 0\) |
| \(m = 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = \sqrt{T^2 + T^2}\) or \(2T\cos 45°\) or \(\dfrac{T}{\cos 45}\) | M1 | Complete method for finding \(F\) in terms of \(T\). Accept \(\sqrt{(R_h)^2 + (R_v)^2}\) |
| Correct expression in \(T\) | A1 | |
| Substitute their \(T\) into correct expression | DM1 | Dependent on previous M mark |
| \(F = \sqrt{2} \cdot \dfrac{8g}{5} = 22\) or 22.2 (N) | A1 | Watch out: resolving vertically is not a correct method and gives 21.9 N |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Along the angle bisector at the pulley | B1 | Or equivalent - accept angle + arrow shown on diagram. (\(8.1°\) to downward vertical). Do not accept a bearing |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = 0.25R$ | B1 | |
| $\sin\alpha = \frac{3}{5}$ or $\cos\alpha = \frac{4}{5}$; $\sin\beta = \frac{4}{5}$ or $\cos\beta = \frac{3}{5}$ | B1 | Use of correct trig ratios for $\alpha$ or $\beta$ |
| $R = 4g\cos\alpha \quad (= 31.36)$ | M1 | Normal reaction on $P$. Condone trig confusion (using $\alpha$) |
| Correct equation | A1 | |
| $T + F = 4g\sin\alpha$ | M1 | Equation of motion for $P$. Requires all 3 terms. Condone consistent trig confusion. Condone acceleration $\neq 0$: $T + F - 4g\sin\alpha = 4a$ |
| $T + 7.84 = 23.52 \Rightarrow T = 15.68$ | A1 | Correct equation |
| $T = mg\sin\beta$ | M1 | Equation of motion for $Q$. Condone trig confusion. Condone acceleration $\neq 0$: $T - mg\sin\beta = -ma$ |
| $T = 7.84m$ | A1 | Correct equation |
| Solve for $m$ | DM1 | Dependent on 3 preceding M marks. Not available if equations used $a \neq 0$ |
| $m = 2$ | A1 | |
**NB: Condone whole system equation $4g\sin\alpha - F = mg\sin\beta$ followed by $m=2$ for 6/6**
**Total: (10)**
---
## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = \sqrt{T^2 + T^2}$ or $2T\cos 45°$ or $\dfrac{T}{\cos 45}$ | M1 | Complete method for finding $F$ in terms of $T$. Accept $\sqrt{(R_h)^2 + (R_v)^2}$ |
| Correct expression in $T$ | A1 | |
| Substitute their $T$ into correct expression | DM1 | Dependent on previous M mark |
| $F = \sqrt{2} \cdot \dfrac{8g}{5} = 22$ or 22.2 (N) | A1 | Watch out: resolving vertically is not a correct method and gives 21.9 N |
**Total: (4)**
---
## Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Along the angle bisector at the pulley | B1 | Or equivalent - accept angle + arrow shown on diagram. ($8.1°$ to downward vertical). Do not accept a bearing |
**Total: (1)**
**Overall Total: (15 marks)**7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6ab8838f-d6f8-4761-8def-1022d97d4e82-20_568_1045_264_461}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A particle $P$ of mass 4 kg is attached to one end of a light inextensible string. A particle $Q$ of mass $m \mathrm {~kg}$ is attached to the other end of the string. The string passes over a small smooth pulley which is fixed at a point on the intersection of two fixed inclined planes. The string lies in a vertical plane that contains a line of greatest slope of each of the two inclined planes. The first plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac { 3 } { 4 }$ and the second plane is inclined to the horizontal at an angle $\beta$, where $\tan \beta = \frac { 4 } { 3 }$. Particle $P$ is on the first plane and particle $Q$ is on the second plane with the string taut, as shown in Figure 3.
The first plane is rough and the coefficient of friction between $P$ and the plane is $\frac { 1 } { 4 }$. The second plane is smooth. The system is in limiting equilibrium.
Given that $P$ is on the point of slipping down the first plane,
\begin{enumerate}[label=(\alph*)]
\item find the value of $m$,
\item find the magnitude of the force exerted on the pulley by the string,
\item find the direction of the force exerted on the pulley by the string.
\includegraphics[max width=\textwidth, alt={}, center]{6ab8838f-d6f8-4761-8def-1022d97d4e82-21_2258_50_314_37}\\
\begin{center}
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VIAV SIHI NI BIIIM ION OC & VGHV SIHI NI GHIYM ION OC & VJ4V SIHI NI JIIYM ION OC \\
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\end{center}
\includegraphics[max width=\textwidth, alt={}, center]{6ab8838f-d6f8-4761-8def-1022d97d4e82-23_2258_50_314_37}\\
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{6ab8838f-d6f8-4761-8def-1022d97d4e82-24_2655_1830_105_121}
\end{center}
\begin{center}
\begin{tabular}{|l|l|l|}
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VIAV SIHI NI JIIYM IONOO & VI4V SIHI NI IIIIMM ION OO & VEYV SIHI NI JLIYM ION OC \\
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\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2018 Q7 [15]}}