| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion up rough slope |
| Difficulty | Standard +0.3 This is a standard mechanics problem requiring resolution of forces on an inclined plane, friction calculations, and work-energy principles. While it involves multiple parts and careful bookkeeping of forces, the techniques are routine for M1 level: resolving perpendicular/parallel to plane, applying F=ma, and calculating work done. The 'show that' in part (i) provides the answer to check against, reducing difficulty. Slightly above average due to the three-part structure and need for accuracy across multiple steps. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes6.02b Calculate work: constant force, resolved component |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R = 13\cos 67.4 = 13(5/13)\) \([R = 5]\) | B1 | Resolve forces perpendicular to plane. Allow 67.4 used |
| \(F + 13\sin 67.4 = F + 13(12/13) = 20\) \([F = 8]\) | B1 | Resolve forces parallel to plane. Allow 67.4 used |
| M1 | Use \(F = \mu R\) | |
| \(\mu = 8/5 = 1.6\) | A1 | AG Must be from exact working here |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(13\sin 67.4 - F = 1.3a\), \(F = \mu R = 8 \rightarrow [4 = 1.3a]\) | M1 | For applying Newton's second law along the plane and also using \(F = \mu R\) (3 terms) |
| \(a = 3.08\) ms\(^{-2}\) | A1 | Allow \(a = 40/13\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(s = 0 + 0.5 \times (40/13) \times 2^2\ [= 80/13 = 6.15]\) | M1 | Use \(s = ut + \frac{1}{2}at^2\) with \(u = 0\) and their \(a \neq \pm g\) to find the distance moved in the first 2 seconds |
| \(WD = 8 \times 6.15\) | M1 | \(WD = F \times d\) |
| \(WD = 49.2\) J | A1 | Allow \(WD = 640/13\) J |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(s = 0 + 0.5 \times (40/13) \times 2^2\ [= 80/13 = 6.15]\) | M1 | |
| \([v = (40/13) \times 2]\) and \([WD = 1.3g(80/13)(12/13) - \frac{1}{2} \times 1.3 \times (80/13)^2]\) | M1 | Finding \(v\) after 2 seconds and using \(WD =\) PE loss \(-\) KE gain |
| \(WD = 49.2\) J | A1 | Allow \(WD = 640/13\) J |
## Question 4(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = 13\cos 67.4 = 13(5/13)$ $[R = 5]$ | B1 | Resolve forces perpendicular to plane. Allow 67.4 used |
| $F + 13\sin 67.4 = F + 13(12/13) = 20$ $[F = 8]$ | B1 | Resolve forces parallel to plane. Allow 67.4 used |
| | M1 | Use $F = \mu R$ |
| $\mu = 8/5 = 1.6$ | A1 | AG Must be from exact working here |
## Question 4(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $13\sin 67.4 - F = 1.3a$, $F = \mu R = 8 \rightarrow [4 = 1.3a]$ | M1 | For applying Newton's second law along the plane and also using $F = \mu R$ (3 terms) |
| $a = 3.08$ ms$^{-2}$ | A1 | Allow $a = 40/13$ |
## Question 4(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $s = 0 + 0.5 \times (40/13) \times 2^2\ [= 80/13 = 6.15]$ | M1 | Use $s = ut + \frac{1}{2}at^2$ with $u = 0$ and their $a \neq \pm g$ to find the distance moved in the first 2 seconds |
| $WD = 8 \times 6.15$ | M1 | $WD = F \times d$ |
| $WD = 49.2$ J | A1 | Allow $WD = 640/13$ J |
**Alternative method for question 4(iii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $s = 0 + 0.5 \times (40/13) \times 2^2\ [= 80/13 = 6.15]$ | M1 | |
| $[v = (40/13) \times 2]$ and $[WD = 1.3g(80/13)(12/13) - \frac{1}{2} \times 1.3 \times (80/13)^2]$ | M1 | Finding $v$ after 2 seconds and using $WD =$ PE loss $-$ KE gain |
| $WD = 49.2$ J | A1 | Allow $WD = 640/13$ J |4 A particle of mass 1.3 kg rests on a rough plane inclined at an angle $\theta$ to the horizontal, where $\tan \theta = \frac { 12 } { 5 }$. The coefficient of friction between the particle and the plane is $\mu$.\\
(i) A force of magnitude 20 N parallel to a line of greatest slope of the plane is applied to the particle and the particle is on the point of moving up the plane. Show that $\mu = 1.6$.\\
The force of magnitude 20 N is now removed.\\
(ii) Find the acceleration of the particle.\\
(iii) Find the work done against friction during the first 2 s of motion.\\
\hfill \mbox{\textit{CAIE M1 2019 Q4 [9]}}