| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Force on pulley from string |
| Difficulty | Standard +0.8 This is a multi-part M1 pulley question requiring resolution of forces on an inclined plane, friction calculations, and vector addition to find the resultant force on the pulley. While it involves several standard techniques (Newton's second law, friction, connected particles), part (d) requires finding the vector sum of two tensions at an angle, which is less routine and demands careful geometric reasoning beyond typical M1 exercises. |
| Spec | 3.03e Resolve forces: two dimensions3.03f Weight: W=mg3.03o Advanced connected particles: and pulleys3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Eqn. of motion for 5 kg mass: \(5g - T = 5a\) (1) | M1 | |
| Eqn. of motion for 4 kg mass: \(T - \mu R - 4g\sin30 = 4a\) (2) | M1 | |
| Resolving perp. to plane: \(R - 4g\cos30 = 0 \therefore R = 2g\sqrt{3}\) | M1 A1 | |
| Sub. for \(R\) in (2): \(T - 2\mu g\sqrt{3} - 2g = 4a\) (3) | A1 | |
| \((1)+(3)\): \(3g - 2\mu g\sqrt{3} = 9a \therefore a = \frac{1}{9}(3-2\mu\sqrt{3})g\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Since motion takes place, \(a > 0\) | B1 | |
| i.e. \(3 - 2\mu\sqrt{3} > 0 \therefore \mu < \frac{\sqrt{3}}{2}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mu = \frac{1}{2}\) means \(a = \frac{3-\sqrt{3}}{9}g\) | B1 | |
| \(T = 5g - 5a = 5g - 5\left(\frac{3-\sqrt{3}}{9}\right)g\) | M1 | |
| \(T = \frac{5}{9}(6+\sqrt{3})g\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Force on pulley \(= 2T\cos30\) | M1 A1 | |
| \(= \frac{10}{9}(6+\sqrt{3})g \cdot \frac{\sqrt{3}}{2} = \frac{5}{9}(6\sqrt{3}+3)g\) | M2 | |
| \(= \frac{5}{3}(2\sqrt{3}+1)g\) N | A1 | (19) |
## Question 7:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Eqn. of motion for 5 kg mass: $5g - T = 5a$ (1) | M1 | |
| Eqn. of motion for 4 kg mass: $T - \mu R - 4g\sin30 = 4a$ (2) | M1 | |
| Resolving perp. to plane: $R - 4g\cos30 = 0 \therefore R = 2g\sqrt{3}$ | M1 A1 | |
| Sub. for $R$ in (2): $T - 2\mu g\sqrt{3} - 2g = 4a$ (3) | A1 | |
| $(1)+(3)$: $3g - 2\mu g\sqrt{3} = 9a \therefore a = \frac{1}{9}(3-2\mu\sqrt{3})g$ | M1 A1 | |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Since motion takes place, $a > 0$ | B1 | |
| i.e. $3 - 2\mu\sqrt{3} > 0 \therefore \mu < \frac{\sqrt{3}}{2}$ | M1 A1 | |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mu = \frac{1}{2}$ means $a = \frac{3-\sqrt{3}}{9}g$ | B1 | |
| $T = 5g - 5a = 5g - 5\left(\frac{3-\sqrt{3}}{9}\right)g$ | M1 | |
| $T = \frac{5}{9}(6+\sqrt{3})g$ | M1 A1 | |
### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Force on pulley $= 2T\cos30$ | M1 A1 | |
| $= \frac{10}{9}(6+\sqrt{3})g \cdot \frac{\sqrt{3}}{2} = \frac{5}{9}(6\sqrt{3}+3)g$ | M2 | |
| $= \frac{5}{3}(2\sqrt{3}+1)g$ N | A1 | **(19)** |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6fb27fe5-055a-4701-bd80-e66ebd57292a-5_417_1016_237_440}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
Figure 3 shows a particle of mass 4 kg resting on the surface of a rough plane inclined at an angle of $30 ^ { \circ }$ to the horizontal. It is connected by a light inextensible string passing over a smooth pulley at the top of the plane, to a particle of mass 5 kg which hangs freely.
The coefficient of friction between the 4 kg mass and the plane is $\mu$ and when the system is released from rest the 4 kg mass starts to move up the slope.
\begin{enumerate}[label=(\alph*)]
\item Show that the acceleration of the system is $\frac { 1 } { 9 } ( 3 - 2 \mu \sqrt { 3 } ) \mathrm { g } \mathrm { ms } ^ { - 2 }$.
\item Hence, find the maximum value of $\mu$.
Given that $\mu = \frac { 1 } { 2 }$,
\item find the tension in the string in terms of $g$,
\item show that the magnitude of the force on the pulley is given by $\frac { 5 } { 3 } ( 2 \sqrt { 3 } + 1 ) \mathrm { g }$.
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q7 [19]}}