| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion up rough slope |
| Difficulty | Standard +0.3 This is a standard M1 mechanics problem involving forces on a slope with friction. Part (a) requires resolving forces in two directions and using limiting friction (F=μR), which is routine for M1. Part (b) applies constant acceleration equations after finding net force. The 45° angle simplifies calculations, and the problem follows a predictable structure with no novel insights required—slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Resolve horizontally: \(T + \frac{2R}{\sqrt{2}} = \frac{R}{\sqrt{2}}\); \(T = -\frac{R}{7\sqrt{2}}\) | M1 A1 A1 | |
| Resolve vertically: \(mg = \frac{R}{\sqrt{2}} + \frac{2R}{\sqrt{2}} = \frac{9R}{\sqrt{2}}\); \(R = \frac{\sqrt{2}}{5}mg\) | M1 A1 A1 | |
| \(T = \frac{5}{7\sqrt{2}} \times \frac{\sqrt{2}}{9}mg = \frac{5mg}{9}\) | M1 A1 | |
| (b) Down wire: \(\frac{mg}{\sqrt{2}} - \frac{2mg}{\sqrt{2}} = ma\); \(a = \frac{5g}{7\sqrt{2}} = 4.95 \text{ ms}^{-2}\) | M1 A1 M1 A1 | |
| \(s = \frac{1}{2}at^2: 0.1 = 2.475t^2\); \(t^2 = 0.0404\); \(t = 0.201 \text{ s}\) | M1 A1 A1 | 15 marks |
(a) Resolve horizontally: $T + \frac{2R}{\sqrt{2}} = \frac{R}{\sqrt{2}}$; $T = -\frac{R}{7\sqrt{2}}$ | M1 A1 A1 |
Resolve vertically: $mg = \frac{R}{\sqrt{2}} + \frac{2R}{\sqrt{2}} = \frac{9R}{\sqrt{2}}$; $R = \frac{\sqrt{2}}{5}mg$ | M1 A1 A1 |
$T = \frac{5}{7\sqrt{2}} \times \frac{\sqrt{2}}{9}mg = \frac{5mg}{9}$ | M1 A1 |
(b) Down wire: $\frac{mg}{\sqrt{2}} - \frac{2mg}{\sqrt{2}} = ma$; $a = \frac{5g}{7\sqrt{2}} = 4.95 \text{ ms}^{-2}$ | M1 A1 M1 A1 |
$s = \frac{1}{2}at^2: 0.1 = 2.475t^2$; $t^2 = 0.0404$; $t = 0.201 \text{ s}$ | M1 A1 A1 | 15 marks6. A small ring, of mass $m \mathrm {~kg}$, can slide along a straight wire which is fixed at an angle of $45 ^ { \circ }$ to the horizontal as shown. The coefficient of friction between the ring and the wire is $\frac { 2 } { 7 }$.\\
The ring rests in equilibrium on the wire and is just prevented from\\
\includegraphics[max width=\textwidth, alt={}, center]{cc75a4a5-1c3a-4e36-acfd-21f6246f2a38-2_273_296_1192_1617}\\
sliding down the wire when a horizontal string is attached to it, as shown
\begin{enumerate}[label=(\alph*)]
\item Show that the tension in the string has magnitude $\frac { 5 m g } { 9 } \mathrm {~N}$.
The string is now removed and the ring starts to slide down the wire.
\item Find the time that elapses before the ring has moved 10 cm along the wire.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q6 [15]}}