| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Motion on rough inclined plane |
| Difficulty | Moderate -0.3 This is a straightforward application of the work-energy principle on an inclined plane. Part (i) requires direct substitution into KE and PE formulas with clearly given values. Part (ii) uses work done by resistance equals the difference between PE loss and KE gain—a standard M1 technique with no conceptual challenges beyond routine problem-solving. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| KE gain \(= \frac{1}{2} \times 8 \times 4.5^2 = 81\) J | B1 | |
| Decrease \(= 8g \times 12 \times \frac{1}{8}\) | M1 | For using \(PE = mgh\) and \(h = d\sin\alpha\) |
| PE loss \(= 120\) J | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(81 = 120 - 12R\) | M1 | For using KE gain \(=\) PE loss \(-\) WD by resistance |
| Resisting force is \(3.25\) N | A1 | [2] Allow \(R = \frac{13}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(4.5^2 = 2 \times a \times 12 \Rightarrow a = \frac{27}{32} = 0.84375\) | M1 | For using \(v^2 = u^2 + 2as\) to find \(a\) and using Newton's 2nd law to find \(R\) |
| \(8g\sin\alpha - R = 8 \times \frac{27}{32}\) | ||
| Resisting force is \(3.25\) N | A1 | [2] |
## Question 3:
### Part (i)
| Working/Answer | Mark | Guidance |
|---|---|---|
| KE gain $= \frac{1}{2} \times 8 \times 4.5^2 = 81$ J | B1 | |
| Decrease $= 8g \times 12 \times \frac{1}{8}$ | M1 | For using $PE = mgh$ and $h = d\sin\alpha$ |
| PE loss $= 120$ J | A1 | **[3]** |
### Part (ii)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $81 = 120 - 12R$ | M1 | For using KE gain $=$ PE loss $-$ WD by resistance |
| Resisting force is $3.25$ N | A1 | **[2]** Allow $R = \frac{13}{4}$ |
**Alternative method for (ii):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $4.5^2 = 2 \times a \times 12 \Rightarrow a = \frac{27}{32} = 0.84375$ | M1 | For using $v^2 = u^2 + 2as$ to find $a$ **and** using Newton's 2nd law to find $R$ |
| $8g\sin\alpha - R = 8 \times \frac{27}{32}$ | | |
| Resisting force is $3.25$ N | A1 | **[2]** |
---3 A plane is inclined at an angle of $\sin ^ { - 1 } \left( \frac { 1 } { 8 } \right)$ to the horizontal. $A$ and $B$ are two points on the same line of greatest slope with $A$ higher than $B$. The distance $A B$ is 12 m . A small object $P$ of mass 8 kg is released from rest at $A$ and slides down the plane, passing through $B$ with speed $4.5 \mathrm {~ms} ^ { - 1 }$. For the motion of $P$ from $A$ to $B$, find\\
(i) the increase in kinetic energy of $P$ and the decrease in potential energy of $P$,\\
(ii) the magnitude of the constant resisting force that opposes the motion of $P$.
\hfill \mbox{\textit{CAIE M1 2015 Q3 [5]}}