| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Block on horizontal plane motion |
| Difficulty | Moderate -0.3 This is a standard M1 friction problem requiring repeated application of Newton's second law and friction formulae across four cases with varying angles. While it involves multiple parts and careful resolution of forces, the method is routine and well-practiced, making it slightly easier than average for A-level mechanics questions. |
| Spec | 3.03d Newton's second law: 2D vectors3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(Fr = 0.2 \times 0.4g\) | B1 | |
| \(1.2 - 0.2 \times 0.4g = 0.4a\) | M1 | N2L, 2 forces |
| \(a = 1.04\) m s\(^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R = 0.4g - 1.2\sin20\) | B1 | SC \(R = 0.4g + 1.2\sin20\) |
| \(1.2\cos20 - 0.2(0.4g - 1.2\sin20) = 0.4a\) | M1 | N2L, 2 forces, cmpt of 1.2 and \(Fr\) not \(Fr\)(i); \(1.2\cos20 - 0.2(0.4g + 1.2\sin20) = 0.4a\) |
| \(a = 1.06\) m s\(^{-2}\) | A1 | \(a = 0.654\) m s\(^{-2}\); Give B1M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1.2\cos70 - 0.2(0.4g - 1.2\sin70)\) | M1 | Difference of two relevant forces, neither used earlier (or find and compare); SC \(1.2\cos70 - 0.2(0.4g + 1.2\sin70)\); Mark as correct case |
| (Total is negative,) friction not overcome by (tractive) force | A1 | Only finding a negative acceleration scores maximum M1 in both cases |
| \(a = 0\) m s\(^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1.2 < 0.4g\) (oe, soi) | M1 | Comparison of weight and 1.2 without involving \(R\); SC Sum of weight and 1.2 |
| P cannot rise from table *or* \(a = 0\) m s\(^{-2}\) | A1 | Only finding a negative acceleration scores M0; P can't go through the table *or* \(a=0\): B1 only |
## Question 5:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Fr = 0.2 \times 0.4g$ | B1 | |
| $1.2 - 0.2 \times 0.4g = 0.4a$ | M1 | N2L, 2 forces |
| $a = 1.04$ m s$^{-2}$ | A1 | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = 0.4g - 1.2\sin20$ | B1 | SC $R = 0.4g + 1.2\sin20$ |
| $1.2\cos20 - 0.2(0.4g - 1.2\sin20) = 0.4a$ | M1 | N2L, 2 forces, cmpt of 1.2 and $Fr$ not $Fr$(i); $1.2\cos20 - 0.2(0.4g + 1.2\sin20) = 0.4a$ |
| $a = 1.06$ m s$^{-2}$ | A1 | $a = 0.654$ m s$^{-2}$; Give B1M1A1 |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.2\cos70 - 0.2(0.4g - 1.2\sin70)$ | M1 | Difference of two relevant forces, neither used earlier (or find and compare); SC $1.2\cos70 - 0.2(0.4g + 1.2\sin70)$; Mark as correct case |
| (Total is negative,) friction not overcome by (tractive) force | A1 | Only finding a negative acceleration scores maximum M1 in both cases |
| $a = 0$ m s$^{-2}$ | A1 | |
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.2 < 0.4g$ (oe, soi) | M1 | Comparison of weight and 1.2 without involving $R$; SC Sum of weight and 1.2 |
| P cannot rise from table *or* $a = 0$ m s$^{-2}$ | A1 | Only finding a negative acceleration scores M0; P can't go through the table *or* $a=0$: B1 only |
---5 A particle $P$ of mass 0.4 kg is at rest on a horizontal surface. The coefficient of friction between $P$ and the surface is 0.2 . A force of magnitude 1.2 N acting at an angle of $\theta ^ { \circ }$ above the horizontal is then applied to $P$. Find the acceleration of $P$ in each of the following cases:\\
(i) $\theta = 0$;\\
(ii) $\theta = 20$;\\
(iii) $\theta = 70$;\\
(iv) $\theta = 90$.
\hfill \mbox{\textit{OCR M1 2015 Q5 [11]}}