| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle on inclined plane with friction |
| Difficulty | Standard +0.3 This is a standard M2 inclined plane problem requiring work-energy methods. Part (i) involves calculating friction force (R = mg cos θ, F = μR) and multiplying by distance—straightforward application of formulas. Part (ii) uses energy conservation with given values. The 55° angle and coefficient 0.15 require careful calculation but no conceptual insight beyond textbook methods. Slightly above average due to the two-part structure and numerical complexity, but well within typical M2 scope. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| \(Fr = 0.15(2.5g \cos 55)\) | M1 | Resolving perpendicular and use of \(Fr = \mu R\) |
| M1 | Use of work done = friction × distance \(AB\), using their friction; if combined with work done by other forces | |
| Work done = 8.43 N m | A1 [3] | Work done = 8.4315736... |
| Answer | Marks | Guidance |
|---|---|---|
| \(2.5g(4 \sin 55)\) | B1 | PE term; 80.27690034 |
| \(\pm\left(\frac{1}{2}(2.5)u^2 - \frac{1}{2}(2.5)(6.7)^2\right)\) | B1 | Change in KE |
| \(\frac{1}{2}(2.5)u^2 - 2.5g(\sin 55) - \frac{1}{2}(2.5)(6.7)^2 = cv(8.43)\) | M1 | Use of work–energy principle (4 terms); dimensionally correct |
| \(u = 10.8 \text{ m s}^{-1}\) | A1 [4] | \(u = 10.763678...\) |
| OR | ||
| \(+/-2.5a = -0.15(2.5g\cos 55) - 2.5g\sin 55\) | M1* A1 | Use of N2L with 3 terms (\(a = -8.87074739...\); allow \(a\) positive) |
| \(6.7^2 = u^2 + 2 \times 4 \times a\) | M1 dep* | Method to find \(u\) using their deceleration, leading to \(u > 6.7\) \(u = 10.763678....\) |
| \(u = 10.8 \text{ m s}^{-1}\) | A1 |
## (i)
$Fr = 0.15(2.5g \cos 55)$ | M1 | Resolving perpendicular and use of $Fr = \mu R$
| M1 | Use of work done = friction × distance $AB$, using their friction; if combined with work done by other forces
Work done = 8.43 N m | A1 [3] | Work done = 8.4315736...
## (ii)
$2.5g(4 \sin 55)$ | B1 | PE term; 80.27690034
$\pm\left(\frac{1}{2}(2.5)u^2 - \frac{1}{2}(2.5)(6.7)^2\right)$ | B1 | Change in KE
$\frac{1}{2}(2.5)u^2 - 2.5g(\sin 55) - \frac{1}{2}(2.5)(6.7)^2 = cv(8.43)$ | M1 | Use of work–energy principle (4 terms); dimensionally correct
$u = 10.8 \text{ m s}^{-1}$ | A1 [4] | $u = 10.763678...$
OR | |
$+/-2.5a = -0.15(2.5g\cos 55) - 2.5g\sin 55$ | M1* A1 | Use of N2L with 3 terms ($a = -8.87074739...$; allow $a$ positive)
$6.7^2 = u^2 + 2 \times 4 \times a$ | M1 dep* | Method to find $u$ using their deceleration, leading to $u > 6.7$ $u = 10.763678....$
$u = 10.8 \text{ m s}^{-1}$ | A1 |
---$A$ and $B$ are two points on a line of greatest slope of a plane inclined at $55°$ to the horizontal. $A$ is below the level of $B$ and $AB = 4$ m. A particle $P$ of mass 2.5 kg is projected up the plane from $A$ towards $B$ and the speed of $P$ at $B$ is $6.7 \text{ m s}^{-1}$. The coefficient of friction between the plane and $P$ is 0.15. Find
\begin{enumerate}[label=(\roman*)]
\item the work done against the frictional force as $P$ moves from $A$ to $B$, [3]
\item the initial speed of $P$ at $A$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR M2 2016 Q2 [7]}}