| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Coefficient of friction from motion |
| Difficulty | Standard +0.3 This is a standard M1 inclined plane question with straightforward application of Newton's second law, friction, and kinematics. Parts (a)-(c) involve routine resolution of forces and use of F=ma, while part (d) requires basic suvat equations. The multi-part structure and friction calculation add some complexity, but all techniques are standard textbook exercises with no novel insight required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\text{resolve for } P \text{ down slope } \quad mgcos30 = ma\) | M1 A1 | |
| \(a = \frac{g\sqrt{3}}{2} = 8.49 \text{ m s}^{-2}\) | A1 | |
| (b) \(s = \frac{1}{\cos 30} = 2\sqrt{3}, u = 0, a = \frac{g\sqrt{3}}{2}, \text{ use } s = ut + \frac{1}{2}at^2\) | B1 M1 | |
| \(2\sqrt{3} = 0 + \frac{1}{2}(\frac{g\sqrt{3}}{2})t^2 \therefore t^2 = \frac{8}{g} \text{ and so } t = 0.904 \text{ seconds (3sf)}\) | M2 A1 | |
| (c) \(\text{resolving perp. to plane: } R - mg\sin60 = 0 \text{ so } R = \frac{s\sqrt{3}}{2}mg\) | M1 | |
| \(F = \mu R = \mu \frac{s\sqrt{3}}{2}mg\) | A1 | |
| \(\text{resolving down the plane: } mgcos60 - F = ma\) | M1 | |
| \(\frac{1}{2}mg - \mu \frac{s\sqrt{3}}{2}mg = 3m \therefore \mu\sqrt{3} = g - 6\) | M1 | |
| \(\text{giving } \mu = \frac{(g-6)\sqrt{3}}{3g} = 0.224 \text{ (3sf)}\) | A1 | |
| (d) \(s = \frac{6}{\sin 30} = 6, u = 0, a = 3, \text{ use } s = ut + \frac{1}{2}at^2\) | M1 | |
| \(6 = 0 + \frac{1}{2}(3)t^2 \therefore t^2 = 4 \text{ and so } t = 2 \text{ seconds}\) | M1 A1 | |
| \(\text{for } P \text{ and } Q \text{ to arrive at the same time, "} t^{''} = 2 - 0.904 = 1.10 \text{ (2dp)}\) | A1 | (17 marks) |
(a) $\text{resolve for } P \text{ down slope } \quad mgcos30 = ma$ | M1 A1 |
$a = \frac{g\sqrt{3}}{2} = 8.49 \text{ m s}^{-2}$ | A1 |
(b) $s = \frac{1}{\cos 30} = 2\sqrt{3}, u = 0, a = \frac{g\sqrt{3}}{2}, \text{ use } s = ut + \frac{1}{2}at^2$ | B1 M1 |
$2\sqrt{3} = 0 + \frac{1}{2}(\frac{g\sqrt{3}}{2})t^2 \therefore t^2 = \frac{8}{g} \text{ and so } t = 0.904 \text{ seconds (3sf)}$ | M2 A1 |
(c) $\text{resolving perp. to plane: } R - mg\sin60 = 0 \text{ so } R = \frac{s\sqrt{3}}{2}mg$ | M1 |
$F = \mu R = \mu \frac{s\sqrt{3}}{2}mg$ | A1 |
$\text{resolving down the plane: } mgcos60 - F = ma$ | M1 |
$\frac{1}{2}mg - \mu \frac{s\sqrt{3}}{2}mg = 3m \therefore \mu\sqrt{3} = g - 6$ | M1 |
$\text{giving } \mu = \frac{(g-6)\sqrt{3}}{3g} = 0.224 \text{ (3sf)}$ | A1 |
(d) $s = \frac{6}{\sin 30} = 6, u = 0, a = 3, \text{ use } s = ut + \frac{1}{2}at^2$ | M1 |
$6 = 0 + \frac{1}{2}(3)t^2 \therefore t^2 = 4 \text{ and so } t = 2 \text{ seconds}$ | M1 A1 |
$\text{for } P \text{ and } Q \text{ to arrive at the same time, "} t^{''} = 2 - 0.904 = 1.10 \text{ (2dp)}$ | A1 | (17 marks)\includegraphics{figure_4}
Figure 4 shows two golf balls $P$ and $Q$ being held at the top of planes inclined at $30°$ and $60°$ to the vertical respectively. Both planes slope down to a common hole at $H$, which is 3 m vertically below $P$ and $Q$.
$P$ is released from rest and travels down the line of greatest slope of the plane it is on which is assumed to be smooth.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of $P$ down the slope. [3 marks]
\item Show that the time taken for $P$ to reach the hole is 0.904 seconds, correct to 3 significant figures. [5 marks]
$Q$ travels down the line of greatest slope of the plane it is on which is rough. The coefficient of friction between $Q$ and the plane is $\mu$.
Given that the acceleration of $Q$ down the slope is $3 \text{ m s}^{-2}$,
\item find, correct to 3 significant figures, the value of $\mu$. [5 marks]
In order for the two balls to arrive at the hole at the same time, $Q$ must be released $t$ seconds before $P$.
\item Find the value of $t$ correct to 2 decimal places. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q7 [17]}}