| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Motion on rough inclined plane |
| Difficulty | Standard +0.3 This is a standard two-part mechanics question requiring energy methods on an inclined plane. Part (i) is straightforward application of conservation of energy (KE = PE). Part (ii) requires calculating work done against friction in both directions, which is routine for M1 students who have covered energy methods with friction. The calculations are methodical rather than conceptually challenging, making this slightly easier than average. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (PE gain \(=\)) \(18gd\sin30°\) or (KE loss \(=\)) \(\frac{1}{2} \times 18 \times 20^2\) | B1 | |
| (PE gain \(=\)) \(18gd\sin30°\) and (KE loss \(=\)) \(\frac{1}{2} \times 18 \times 20^2\) | B1 | |
| \([18gd\sin30° = \frac{1}{2} \times 18 \times 20^2]\) or \([18gh = \frac{1}{2} \times 18 \times 20^2]\) | M1 | Energy equation (PE gain = KE loss) |
| Distance up plane \(= 40\text{ m}\) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R = 18g\cos30°\) \((90\sqrt{3}\) or \(155.884...)\) | B1 | |
| \([F = 0.25(18g\cos30°)]\) \((45\sqrt{3}/2\) or \(38.971...)\) | M1 | Use of \(F = \mu R\) |
| \([18g\sin30° + 0.25(18g\cos30°) = -18a \rightarrow a = ...]\) \((a = -7.165..)\) | M1 | Newton's Second Law (3 term equation) |
| \([0^2 = 20^2 + 2 \times -7.165.. \times s \rightarrow s = ...]\) | M1 | Use of suvat to find \(s\) |
| \(s = 27.913...\) | A1 | |
| \([18g\sin30° - 0.25(18g\cos30°) = 18a \rightarrow a = ...]\) \((a = 2.835..)\) | M1 | \((a = 2.835..)\) Newton's Second Law (3 term equation) |
| \([v^2 = 0^2 + 2 \times 2.835.. \times 27.913.. \rightarrow v = ...]\) | M1 | Use of suvat to find \(s\) |
| \(v = 12.6 \text{ ms}^{-1}\) | A1 | \((12.580...)\) |
| 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R = 18g\cos30°\) \((90\sqrt{3}\) or \(155.884...)\) | B1 | |
| \([F = 0.25(18g\cos30°)]\) \((45\sqrt{3}/2\) or \(38.971...)\) | M1 | Use of \(F = \mu R\) |
| [KE gain \(= \frac{1}{2} \times 18 \times 20^2\) and PE loss \(= 18gh\) or \(18gs(\sin30°)\)] | M1 | Use of \(KE = \frac{1}{2}mv^2\) and \(PE = mgh\) |
| \([\frac{1}{2} \times 18 \times 20^2 = 18gs(\sin30°) + 45\cos30° \times s]\) | M1 | Work/Energy equation (up plane) |
| \(s = 27.913...\) | A1 | |
| \([WD = 45\cos30° \times 27.91...]\) | M1 | Work done against friction |
| \([\frac{1}{2} \times 18v^2 = (18g\sin30°) \times 27.91.. - 45\cos30° \times 27.91...]\) | M1 | Work/Energy equation (down plane) |
| \(v = 12.6 \text{ ms}^{-1}\) | A1 | \((12.580...)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([WD = 2 \times 45\cos30° \times 27.91...]\) | M1 | WD against friction (up and down) |
| \([\frac{1}{2} \times 18 \times 20^2 - \frac{1}{2} \times 18v^2 = 2 \times 45\cos30° \times 27.91...]\) | M1 | Uses KE loss = total WD against friction |
| \(v = 12.6 \text{ ms}^{-1}\) | A1 | \((12.580...)\) |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| (PE gain $=$) $18gd\sin30°$ **or** (KE loss $=$) $\frac{1}{2} \times 18 \times 20^2$ | B1 | |
| (PE gain $=$) $18gd\sin30°$ **and** (KE loss $=$) $\frac{1}{2} \times 18 \times 20^2$ | B1 | |
| $[18gd\sin30° = \frac{1}{2} \times 18 \times 20^2]$ **or** $[18gh = \frac{1}{2} \times 18 \times 20^2]$ | M1 | Energy equation (PE gain = KE loss) |
| Distance up plane $= 40\text{ m}$ | A1 | |
| | **4** | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = 18g\cos30°$ $(90\sqrt{3}$ or $155.884...)$ | B1 | |
| $[F = 0.25(18g\cos30°)]$ $(45\sqrt{3}/2$ or $38.971...)$ | M1 | Use of $F = \mu R$ |
| $[18g\sin30° + 0.25(18g\cos30°) = -18a \rightarrow a = ...]$ $(a = -7.165..)$ | M1 | Newton's Second Law (3 term equation) |
| $[0^2 = 20^2 + 2 \times -7.165.. \times s \rightarrow s = ...]$ | M1 | Use of suvat to find $s$ |
| $s = 27.913...$ | A1 | |
| $[18g\sin30° - 0.25(18g\cos30°) = 18a \rightarrow a = ...]$ $(a = 2.835..)$ | M1 | $(a = 2.835..)$ Newton's Second Law (3 term equation) |
| $[v^2 = 0^2 + 2 \times 2.835.. \times 27.913.. \rightarrow v = ...]$ | M1 | Use of suvat to find $s$ |
| $v = 12.6 \text{ ms}^{-1}$ | A1 | $(12.580...)$ |
| | **8** | |
**Alternative Method 1 for 5(ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = 18g\cos30°$ $(90\sqrt{3}$ or $155.884...)$ | B1 | |
| $[F = 0.25(18g\cos30°)]$ $(45\sqrt{3}/2$ or $38.971...)$ | M1 | Use of $F = \mu R$ |
| [KE gain $= \frac{1}{2} \times 18 \times 20^2$ and PE loss $= 18gh$ or $18gs(\sin30°)$] | M1 | Use of $KE = \frac{1}{2}mv^2$ and $PE = mgh$ |
| $[\frac{1}{2} \times 18 \times 20^2 = 18gs(\sin30°) + 45\cos30° \times s]$ | M1 | Work/Energy equation (up plane) |
| $s = 27.913...$ | A1 | |
| $[WD = 45\cos30° \times 27.91...]$ | M1 | Work done against friction |
| $[\frac{1}{2} \times 18v^2 = (18g\sin30°) \times 27.91.. - 45\cos30° \times 27.91...]$ | M1 | Work/Energy equation (down plane) |
| $v = 12.6 \text{ ms}^{-1}$ | A1 | $(12.580...)$ |
**Alternative Method 2 for 5(ii) (last 3 marks):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[WD = 2 \times 45\cos30° \times 27.91...]$ | M1 | WD against friction (up and down) |
| $[\frac{1}{2} \times 18 \times 20^2 - \frac{1}{2} \times 18v^2 = 2 \times 45\cos30° \times 27.91...]$ | M1 | Uses KE loss = total WD against friction |
| $v = 12.6 \text{ ms}^{-1}$ | A1 | $(12.580...)$ |5 A particle of mass 18 kg is on a plane inclined at an angle of $30 ^ { \circ }$ to the horizontal. The particle is projected up a line of greatest slope of the plane with a speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Given that the plane is smooth, use an energy method to find the distance the particle moves up the plane before coming to instantaneous rest.\\
(ii) Given instead that the plane is rough and the coefficient of friction between the particle and the plane is 0.25 , find the speed of the particle as it returns to its starting point.\\
\hfill \mbox{\textit{CAIE M1 2019 Q5 [12]}}