| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | October |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Variable mass or unknown mass |
| Difficulty | Standard +0.3 This is a standard M1 pulley system question requiring resolution of forces on an incline with friction, Newton's second law for connected particles, and vector addition for pulley forces. While multi-part with several steps, it follows a routine template with given values that simplify calculations (e.g., 7-24-25 triangle). The techniques are all standard M1 content with no novel insight required, making it slightly easier than average. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution3.03o Advanced connected particles: and pulleys3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For \(A\): \(\frac{4mg}{3} - mg\sin\alpha - F = ma\) | M1 A1 | Use \(F=ma\) parallel to plane, dimensionally correct, correct no. of terms, condone sin/cos confusion. If \(T\) used and never replaced, allow M1 |
| \(R = mg\cos\alpha\) | M1 A1 | Resolve perpendicular to plane, dimensionally correct, correct no. of terms |
| Use of \(F = \frac{1}{3}R\) in an equation | M1 | — |
| \(a = \frac{11g}{15}\) or \(0.73g\) or better | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For \(B\): \(kmg - \frac{4mg}{3} = kma\) | M1 A1 | Use \(F=ma\) vertically, dimensionally correct, correct no. of terms. Must have \(km\) on both sides. If \(T\) used and never replaced, allow M1 |
| \(k = 5\) | A1 | N.B. Either equation of motion could be replaced by whole system equation: \(kmg - mg\sin\alpha - F = (k+1)ma\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2T\cos\left(\frac{90°-\alpha}{2}\right)\) | M1 A1 | Complete method to find resultant force on pulley |
| Substitute \(T = \frac{4mg}{3}\) and trig | dM1 | Dependent on previous M mark |
| \(\frac{32mg}{15}\) or \(2.1mg\) or better | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sqrt{T^2+T^2-2(T)(T)\cos(90°+\alpha)}\) | M1 A1 | Use of cosine rule |
| Substitute \(T=\frac{4mg}{3}\) and trig | dM1 | — |
| \(\frac{32mg}{15}\) or allow \(\sqrt{\frac{1024m^2g^2}{225}}\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sqrt{(T\cos\alpha)^2+(T+T\sin\alpha)^2}\) | M1 A1 | Use of vertical and horizontal components |
| Substitute \(T=\frac{4mg}{3}\) and trig | dM1 | — |
| \(\frac{32mg}{15}\) or allow \(\sqrt{\frac{1024m^2g^2}{225}}\) | A1 | — |
## Question 7:
### Part 7(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For $A$: $\frac{4mg}{3} - mg\sin\alpha - F = ma$ | M1 A1 | Use $F=ma$ parallel to plane, dimensionally correct, correct no. of terms, condone sin/cos confusion. If $T$ used and never replaced, allow M1 |
| $R = mg\cos\alpha$ | M1 A1 | Resolve perpendicular to plane, dimensionally correct, correct no. of terms |
| Use of $F = \frac{1}{3}R$ in an equation | M1 | — |
| $a = \frac{11g}{15}$ or $0.73g$ or better | A1 | — |
### Part 7(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For $B$: $kmg - \frac{4mg}{3} = kma$ | M1 A1 | Use $F=ma$ vertically, dimensionally correct, correct no. of terms. Must have $km$ on both sides. If $T$ used and never replaced, allow M1 |
| $k = 5$ | A1 | N.B. Either equation of motion could be replaced by whole system equation: $kmg - mg\sin\alpha - F = (k+1)ma$ |
### Part 7(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2T\cos\left(\frac{90°-\alpha}{2}\right)$ | M1 A1 | Complete method to find resultant force on pulley |
| Substitute $T = \frac{4mg}{3}$ and trig | dM1 | Dependent on previous M mark |
| $\frac{32mg}{15}$ or $2.1mg$ or better | A1 | — |
**ALT1:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sqrt{T^2+T^2-2(T)(T)\cos(90°+\alpha)}$ | M1 A1 | Use of cosine rule |
| Substitute $T=\frac{4mg}{3}$ and trig | dM1 | — |
| $\frac{32mg}{15}$ or allow $\sqrt{\frac{1024m^2g^2}{225}}$ | A1 | — |
**ALT2:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sqrt{(T\cos\alpha)^2+(T+T\sin\alpha)^2}$ | M1 A1 | Use of vertical and horizontal components |
| Substitute $T=\frac{4mg}{3}$ and trig | dM1 | — |
| $\frac{32mg}{15}$ or allow $\sqrt{\frac{1024m^2g^2}{225}}$ | A1 | — |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{017cc2b0-9ec3-45ff-94c0-9d989badfd5d-24_339_942_244_635}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a block $A$ of mass $m$ held at rest on a rough plane.\\
The plane is inclined at an angle $\alpha$ to the horizontal and the coefficient of friction between the block and the plane is $\mu$.
One end of a light inextensible string is now attached to $A$. 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 block $B$ of mass $k m$.\\
Block $B$ hangs vertically below the pulley, with the string taut.\\
The string from $A$ to the pulley lies along a line of greatest slope of the plane.\\
Both $A$ and $B$ are modelled as particles.\\
When the system is released from rest, $A$ moves up the plane and the tension in the string is $\frac { 4 m g } { 3 }$
Given that $\mu = \frac { 1 } { 3 }$ and $\tan \alpha = \frac { 7 } { 24 }$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item find the magnitude of the acceleration of $A$, giving your answer in terms of $g$,
\item find the value of $k$.
\end{enumerate}\item Find the magnitude of the resultant force exerted on the pulley by the string, giving your answer in terms of $m$ and $g$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2023 Q7 [13]}}