| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | December |
| Marks | 11 |
| Topic | Motion on a slope |
| Type | Energy methods on slope |
| Difficulty | Standard +0.8 This is a multi-stage mechanics problem requiring energy methods or kinematics across two different surfaces (smooth then rough), calculation of friction forces on an incline, and geometric understanding of contact forces. It involves more steps and conceptual integration than a typical M1 question, but uses standard techniques throughout. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Acceleration component \(= g\sin 30°\) | B1 | Correct acceleration component seen |
| \(v_M^2 = 4.2^2 + 2(g\sin 30°)x\) | M1 | Use of \(v^2 = u^2 + 2as\) for the motion from A to M |
| \(R = mg\cos 30°\) | B1 | Resolving perpendicular to the plane |
| \(F = \frac{\sqrt{3}}{6}mg\cos 30°\) | M1 | Use of \(F = \mu R\) for the motion of \(P\) between \(M\) and \(B\) |
| \(mg\sin 30° - F = ma\) | M1* | Use of Newton's 2nd Law for the motion of \(P\) between \(M\) and \(B\) |
| \(12.6^2 = v_M^2 + 2g(\sin 30° - \frac{\sqrt{3}}{6}\cos 30°)(20 - x)\) | M1dep* | Correct use of \(v^2 = u^2 + 2as\) for the motion from \(M\) to \(B\) with their \(a\) and correct \(x\) |
| \(12.6^2 = 4.2^2 + 2(g\sin 30°)x + 2g(20-x)(\sin 30° - \frac{\sqrt{3}}{6}\cos 30°)\) | M1 | Substitute their expression for \(v_M\) to obtain an equation in \(x\) only |
| \(x = 8.8 \text{ so the distance } AM \text{ is } 8.8 \text{ m}\) | A1 | BC |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\alpha = \frac{R}{F} = \frac{mg\cos 30°}{\frac{\sqrt{3}}{6}mg\cos 30°}\) | M1* | Equates ratio of contact forces to \(\tan\alpha\) |
| \(\text{angle} = 180° - \alpha = 106.1°\) | M1dep* | |
| \(= 106.1°\) | A1 | Correct answer (to at least 3 sf) |
## Part (a)
Acceleration component $= g\sin 30°$ | B1 | Correct acceleration component seen
$v_M^2 = 4.2^2 + 2(g\sin 30°)x$ | M1 | Use of $v^2 = u^2 + 2as$ for the motion from A to M | $x$ is the distance $AM$ and $v_M$ is the speed of $P$ at $M$
$R = mg\cos 30°$ | B1 | Resolving perpendicular to the plane | $R$ is the normal contact force between $P$ and the plane, $m$ is the mass of $P$
$F = \frac{\sqrt{3}}{6}mg\cos 30°$ | M1 | Use of $F = \mu R$ for the motion of $P$ between $M$ and $B$
$mg\sin 30° - F = ma$ | M1* | Use of Newton's 2nd Law for the motion of $P$ between $M$ and $B$
$12.6^2 = v_M^2 + 2g(\sin 30° - \frac{\sqrt{3}}{6}\cos 30°)(20 - x)$ | M1dep* | Correct use of $v^2 = u^2 + 2as$ for the motion from $M$ to $B$ with their $a$ and correct $x$
$12.6^2 = 4.2^2 + 2(g\sin 30°)x + 2g(20-x)(\sin 30° - \frac{\sqrt{3}}{6}\cos 30°)$ | M1 | Substitute their expression for $v_M$ to obtain an equation in $x$ only
$x = 8.8 \text{ so the distance } AM \text{ is } 8.8 \text{ m}$ | A1 | BC
## Part (b)
$\tan\alpha = \frac{R}{F} = \frac{mg\cos 30°}{\frac{\sqrt{3}}{6}mg\cos 30°}$ | M1* | Equates ratio of contact forces to $\tan\alpha$
$\text{angle} = 180° - \alpha = 106.1°$ | M1dep* |
$= 106.1°$ | A1 | Correct answer (to at least 3 sf) | $106.102 113\ldots$\includegraphics{figure_10}
$A$ and $B$ are points at the upper and lower ends, respectively, of a line of greatest slope on a plane inclined at $30°$ to the horizontal. The distance $AB$ is $20\text{m}$. $M$ is a point on the plane between $A$ and $B$. The surface of the plane is smooth between $A$ and $M$, and rough between $M$ and $B$.
A particle $P$ is projected with speed $4.2\text{m s}^{-1}$ from $A$ down the line of greatest slope (see diagram). $P$ moves down the plane and reaches $B$ with speed $12.6\text{m s}^{-1}$. The coefficient of friction between $P$ and the rough part of the plane is $\frac{\sqrt{3}}{6}$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance $AM$. [8]
\item Find the angle between the contact force and the downward direction of the line of greatest slope when $P$ is in motion between $M$ and $B$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q10 [11]}}