| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion up then down slope |
| Difficulty | Standard +0.3 This is a standard M1 mechanics problem requiring application of Newton's second law with friction on an inclined plane in two phases (up and down). While it involves multiple parts and careful sign conventions, the techniques are routine for M1: resolving forces, using F=μR, applying suvat equations, and comparing forces to determine motion direction. The multi-part structure and need to handle two motion phases makes it slightly above average difficulty, but it follows a predictable template without requiring novel insight. |
| Spec | 3.03r Friction: concept and vector form3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0 = 6 + (\pm)1.5a\) | M1 | For using \(v = u + at\) with \(v = 0\) |
| \(a = (\mp)4\ \text{ms}^{-2}\) | A1 | |
| \(-mg\sin 15° - F = ma\) | M1 | For applying Newton's second law with 2 forces |
| \(-0.1 \times 9.8\sin 15° - F = 0.1 \times (-4)\) | A1 | |
| \(R = 0.1g\cos 15°\) | B1 | |
| \(0.146357\ldots = \mu \times 0.946607\) | M1 | For using \(F = \mu R\) |
| Coefficient is 0.155 | A1 | 7 Anything between 0.15 and 0.16 inclusive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(mg\sin 15° > \mu\, mg\cos 15°\) (or \(\tan 15° > \mu\)) | M1 | For comparing weight component with frictional force (or tan 'angle of friction' with \(\mu\)) |
| \(\rightarrow\) particle moves down | A1 | 2 Awarded if conclusion is correct even though values are wrong |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((6 + 0) \div 2 = s \div 1.5\) | M1 | For using \((u+v) \div 2 = s \div t\) |
| \(s = 4.5\) | A1 | |
| \(mg\sin 15° - F = ma\) | M1 | For using Newton's second law with 2 forces |
| \(0.25364\ldots - 0.146357\ldots = 0.1a\) | A1 | Values must be correct even if not explicitly stated. Note correct friction value may legitimately arise from wrong \(\mu\) and wrong \(R\) |
| \(v^2 = 2(1.07285\ldots) \times 4.5\) | M1 | For using \(v^2 = 2as\) with any value of \(a\) |
| Speed is \(3.11\ \text{ms}^{-1}\) | A1 | 6 Accept anything rounding to 3.1 from correct working |
# Question 7:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = 6 + (\pm)1.5a$ | M1 | For using $v = u + at$ with $v = 0$ |
| $a = (\mp)4\ \text{ms}^{-2}$ | A1 | |
| $-mg\sin 15° - F = ma$ | M1 | For applying Newton's second law with 2 forces |
| $-0.1 \times 9.8\sin 15° - F = 0.1 \times (-4)$ | A1 | |
| $R = 0.1g\cos 15°$ | B1 | |
| $0.146357\ldots = \mu \times 0.946607$ | M1 | For using $F = \mu R$ |
| Coefficient is 0.155 | A1 | **7** Anything between 0.15 and 0.16 inclusive |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $mg\sin 15° > \mu\, mg\cos 15°$ (or $\tan 15° > \mu$) | M1 | For comparing weight component with frictional force (or tan 'angle of friction' with $\mu$) |
| $\rightarrow$ particle moves down | A1 | **2** Awarded if conclusion is correct even though values are wrong |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(6 + 0) \div 2 = s \div 1.5$ | M1 | For using $(u+v) \div 2 = s \div t$ |
| $s = 4.5$ | A1 | |
| $mg\sin 15° - F = ma$ | M1 | For using Newton's second law with 2 forces |
| $0.25364\ldots - 0.146357\ldots = 0.1a$ | A1 | Values must be correct even if not explicitly stated. Note correct friction value may legitimately arise from wrong $\mu$ and wrong $R$ |
| $v^2 = 2(1.07285\ldots) \times 4.5$ | M1 | For using $v^2 = 2as$ with any value of $a$ |
| Speed is $3.11\ \text{ms}^{-1}$ | A1 | **6** Accept anything rounding to 3.1 from correct working |7 A particle of mass 0.1 kg is at rest at a point $A$ on a rough plane inclined at $15 ^ { \circ }$ to the horizontal. The particle is given an initial velocity of $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and starts to move up a line of greatest slope of the plane. The particle comes to instantaneous rest after 1.5 s .\\
(i) Find the coefficient of friction between the particle and the plane.\\
(ii) Show that, after coming to instantaneous rest, the particle moves down the plane.\\
(iii) Find the speed with which the particle passes through $A$ during its downward motion.
\hfill \mbox{\textit{OCR M1 2006 Q7 [15]}}