| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | January |
| Marks | 16 |
| 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 +0.8 This is a multi-part pulley problem requiring resolution of forces on two inclined planes with friction, application of Newton's second law to a connected system, and vector addition to find the resultant force on the pulley. While it involves several standard M1 techniques (friction, connected particles, trigonometry with tan α = 4/3), the combination of multiple friction coefficients, inclined geometry, and the final vector calculation makes it moderately challenging but still within typical M1 scope. |
| Spec | 3.03o Advanced connected particles: and pulleys3.03r Friction: concept and vector form3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For \(B\): \(S = 3mg\cos\alpha\) | M1 A1 | Resolve perpendicular to plane |
| For \(B\): \(3mg\sin\alpha - T - F_1 = 3ma\) | M1 A2 | Equation of motion parallel to inclined plane; -1 each error |
| For \(A\): \(R = mg\) | B1 | cao |
| For \(A\): \(T - F_2 = ma\) | M1 A1 | Equation of motion horizontally |
| \(F_1 = \frac{1}{3}S\); \(F_2 = \frac{1}{5}R\) | M1 | Using \(F = \mu R\) correctly twice |
| Solving for \(T\) | DM1 | Dependent on all M marks; must be in terms of \(m\) only |
| \(T = \frac{3mg}{5}\) or \(5.88m\) | A1 | cao. N.B. Use of \(\sin(4/5)\) treated as A error but allow recovery. Whole system equation \(3mg\sin\alpha - F_1 - F_2 = 4ma\) also acceptable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Constant tension throughout the string | B1 | Penalise extra wrong answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R = 2T\cos\frac{(180°-\alpha)}{2}\) \(\left(= 2T\sin\frac{1}{2}\alpha\right)\) | M1 A1 | Correct expression for \(R\) in terms of \(T\) and \(\alpha\); condone cos/sin confusion but must use correct angle |
| \(= 2 \times \frac{3mg}{5} \times \frac{\sqrt{5}}{5}\) | DM1 | Substituting \(T\) (must be in terms of \(m\)) and correct value of \(\alpha\) |
| \(= \frac{6mg\sqrt{5}}{25}\) (\(5.3m\) or \(5.26m\)) | A1 | Any equivalent surd form. Special case: allow M1A1DM0A0 if \(m\) lost from \(T\) but expression for \(R\) otherwise correct |
| OR: \(R = \sqrt{(T-T\cos\alpha)^2+(T\sin\alpha)^2}\) or \(R = \sqrt{T^2+T^2-2T^2\cos\alpha}\) | M1 A1 | |
| Substitute \(T\) in terms of \(m\) and correct \(\alpha\) | DM1 | |
| \(= \frac{6mg\sqrt{5}}{25}\) | A1 |
## Question 7:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For $B$: $S = 3mg\cos\alpha$ | M1 A1 | Resolve perpendicular to plane |
| For $B$: $3mg\sin\alpha - T - F_1 = 3ma$ | M1 A2 | Equation of motion parallel to inclined plane; -1 each error |
| For $A$: $R = mg$ | B1 | cao |
| For $A$: $T - F_2 = ma$ | M1 A1 | Equation of motion horizontally |
| $F_1 = \frac{1}{3}S$; $F_2 = \frac{1}{5}R$ | M1 | Using $F = \mu R$ correctly twice |
| Solving for $T$ | DM1 | Dependent on all M marks; must be in terms of $m$ only |
| $T = \frac{3mg}{5}$ or $5.88m$ | A1 | cao. N.B. Use of $\sin(4/5)$ treated as A error but allow recovery. Whole system equation $3mg\sin\alpha - F_1 - F_2 = 4ma$ also acceptable |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Constant tension throughout the string | B1 | Penalise extra wrong answers |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = 2T\cos\frac{(180°-\alpha)}{2}$ $\left(= 2T\sin\frac{1}{2}\alpha\right)$ | M1 A1 | Correct expression for $R$ in terms of $T$ and $\alpha$; condone cos/sin confusion but must use correct angle |
| $= 2 \times \frac{3mg}{5} \times \frac{\sqrt{5}}{5}$ | DM1 | Substituting $T$ (must be in terms of $m$) and correct value of $\alpha$ |
| $= \frac{6mg\sqrt{5}}{25}$ ($5.3m$ or $5.26m$) | A1 | Any equivalent surd form. Special case: allow M1A1DM0A0 if $m$ lost from $T$ but expression for $R$ otherwise correct |
| OR: $R = \sqrt{(T-T\cos\alpha)^2+(T\sin\alpha)^2}$ or $R = \sqrt{T^2+T^2-2T^2\cos\alpha}$ | M1 A1 | |
| Substitute $T$ in terms of $m$ and correct $\alpha$ | DM1 | |
| $= \frac{6mg\sqrt{5}}{25}$ | A1 | |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0d5a56ba-6a33-4dc8-b612-d2957211124f-20_410_1091_258_440}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Two particles $A$ and $B$ have masses $m$ and $3 m$ respectively. The particles are attached to the ends of a light inextensible string. Particle $A$ is held at rest on a rough horizontal table. The coefficient of friction between particle $A$ and the table is $\frac { 1 } { 5 }$. The string lies along the table and passes over a small smooth light pulley that is fixed at the edge of the table. Particle $B$ is at rest on a rough plane that is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac { 4 } { 3 }$, as shown in Figure 4. The coefficient of friction between particle $B$ and the inclined plane is $\frac { 1 } { 3 }$. The string lies 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 and $B$ slides down the inclined plane. Given that $A$ does not reach the pulley,
\begin{enumerate}[label=(\alph*)]
\item find the tension in the string,
\item state where in your working you have used the fact that the string is modelled as being light,
\item find the magnitude of the force exerted on the pulley by the string.\\
\begin{center}
\end{center}
\includegraphics[max width=\textwidth, alt={}, center]{0d5a56ba-6a33-4dc8-b612-d2957211124f-24_172_1824_2581_123}\\
\includegraphics[max width=\textwidth, alt={}, center]{0d5a56ba-6a33-4dc8-b612-d2957211124f-24_157_85_2595_1966}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2019 Q7 [16]}}