| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Force on pulley from string |
| Difficulty | Standard +0.8 This is a substantial multi-part mechanics problem requiring resolution of forces on two inclined planes, connected particle dynamics, and energy/kinematics calculations. While the techniques are standard M1 content (resolving forces, Newton's second law, pulley systems), the problem demands careful setup across multiple parts, geometric reasoning about the system configuration, and vector resolution for the pulley force. The algebraic manipulation to reach the given acceleration expression and the need to find when Y hits the ground (requiring geometric analysis of the 2.5m string constraint) elevate this above routine exercises, though it remains within the scope of well-prepared M1 students. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02e Two-dimensional constant acceleration: with vectors3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| (a) for \(X\): \(T - 3g\sin30° = 3a \therefore T - \frac{3}{2}g = 3a\) (1) for \(Y\): \(2g\cos30° - T = 2a \therefore g\sqrt{3} - T = 2a\) (2) (1) + (2) gives \(g\sqrt{3} - \frac{3}{2}g = 5a\) \(a = \frac{g\sqrt{3} - 3g}{5} \therefore a = \frac{g}{10}(2\sqrt{3} - 3)\) | M1 A1 M1 A1 A1 | |
| (b) sub. \(a\) into (1) to get \(T = 3a + \frac{3g}{2} = \frac{3}{10}(2\sqrt{3} - 3) + \frac{3g}{2}\) \(T = 16.0645\) for on pulley \(= \sqrt{(T^2 + T^2)} = T\sqrt{2}\) force on pulley \(= 22.7 \text{ N}\) force acts at an angle 45° to each plane i.e. 15° to vertical | M1 A1 M1 A1 M1 A1 | |
| (c) initially, \(Y\) is at \(C\) and \(CB = 4\sin30° = 2 \text{ m}\) use \(v^2 = u^2 + 2as\) with \(u = 0\), \(s = 2\), \(a = \frac{g}{10}(2\sqrt{3} - 3)\) \(v^2 = 0 + \frac{4g}{10}(2\sqrt{3} - 3)\) so \(v = 1.35 \text{ ms}^{-1}\) (2dp) | M1 M1 M1 A1 | (18) |
**(a)** for $X$: $T - 3g\sin30° = 3a \therefore T - \frac{3}{2}g = 3a$ (1) for $Y$: $2g\cos30° - T = 2a \therefore g\sqrt{3} - T = 2a$ (2) (1) + (2) gives $g\sqrt{3} - \frac{3}{2}g = 5a$ $a = \frac{g\sqrt{3} - 3g}{5} \therefore a = \frac{g}{10}(2\sqrt{3} - 3)$ | M1 A1 M1 A1 A1 |
**(b)** sub. $a$ into (1) to get $T = 3a + \frac{3g}{2} = \frac{3}{10}(2\sqrt{3} - 3) + \frac{3g}{2}$ $T = 16.0645$ for on pulley $= \sqrt{(T^2 + T^2)} = T\sqrt{2}$ force on pulley $= 22.7 \text{ N}$ force acts at an angle 45° to each plane i.e. 15° to vertical | M1 A1 M1 A1 M1 A1 |
**(c)** initially, $Y$ is at $C$ and $CB = 4\sin30° = 2 \text{ m}$ use $v^2 = u^2 + 2as$ with $u = 0$, $s = 2$, $a = \frac{g}{10}(2\sqrt{3} - 3)$ $v^2 = 0 + \frac{4g}{10}(2\sqrt{3} - 3)$ so $v = 1.35 \text{ ms}^{-1}$ (2dp) | M1 M1 M1 A1 | (18)
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**Total: (75)**\includegraphics{figure_3}
Figure 3 shows a particle $X$ of mass 3 kg on a smooth plane inclined at an angle 30° to the horizontal, and a particle $Y$ of mass 2 kg on a smooth plane inclined at an angle 60° to the horizontal. The two particles are connected by a light, inextensible string of length 2.5 metres passing over a smooth pulley at $C$ which is the highest point of the two planes.
Initially, $Y$ is at a point just below $C$ touching the pulley with the string taut. When the particles are released from rest they travel along the lines of greatest slope, $AC$ in the case of $X$ and $BC$ in the case of $Y$, of their respective planes. $A$ and $B$ are the points where the planes meet the horizontal ground and $AB = 4$ metres.
\begin{enumerate}[label=(\alph*)]
\item Show that the initial acceleration of the system is given by $\frac{g}{10}\left(2\sqrt{3} - 3\right)$ ms$^{-2}$. [7 marks]
\item By finding the tension in the string, or otherwise, find the magnitude of the force exerted on the pulley and the angle that this force makes with the vertical. [7 marks]
\item Find, correct to 2 decimal places, the speed with which $Y$ hits the ground. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q7 [18]}}