| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - driving force up incline, find KE/PE changes as sub-parts |
| Difficulty | Moderate -0.3 This is a straightforward energy method problem requiring standard application of work-energy principles. Students must calculate kinetic energy gain (routine formula), express potential energy loss in terms of θ (basic trigonometry), then solve a linear equation using the work-energy theorem. All steps are standard textbook techniques with no novel insight required, making it slightly easier than average. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using KE gain \(= \frac{1}{2}mv_B^2\) or PE loss \(= mg \times AB\sin\theta\) | |
| KE gain \(= 4032 \times 10^3\) or PE loss \(= 42 \times 10^6\sin\theta\) | A1 | |
| PE loss \(= 42 \times 10^6\sin\theta\) or KE gain \(= 4032 \times 10^3\) | B1 | |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(WD\) by \(DF =\) KE gain \(-\) PE loss \(+\) WD by resistance | |
| \(5000 = 4032 - 42000\sin\theta + 3360\) | A1\(\checkmark\) | |
| \(\theta = 3.3°\) | A1 | |
| Total: 3 marks |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using KE gain $= \frac{1}{2}mv_B^2$ or PE loss $= mg \times AB\sin\theta$ |
| KE gain $= 4032 \times 10^3$ or PE loss $= 42 \times 10^6\sin\theta$ | A1 | |
| PE loss $= 42 \times 10^6\sin\theta$ or KE gain $= 4032 \times 10^3$ | B1 | |
| **Total: 3 marks** | | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $WD$ by $DF =$ KE gain $-$ PE loss $+$ WD by resistance |
| $5000 = 4032 - 42000\sin\theta + 3360$ | A1$\checkmark$ | |
| $\theta = 3.3°$ | A1 | |
| **Total: 3 marks** | | |
---4 A lorry of mass 14000 kg moves along a road starting from rest at a point $O$. It reaches a point $A$, and then continues to a point $B$ which it reaches with a speed of $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The part $O A$ of the road is straight and horizontal and has length 400 m . The part $A B$ of the road is straight and is inclined downwards at an angle of $\theta ^ { \circ }$ to the horizontal and has length 300 m .\\
(i) For the motion from $O$ to $B$, find the gain in kinetic energy of the lorry and express its loss in potential energy in terms of $\theta$.
The resistance to the motion of the lorry is 4800 N and the work done by the driving force of the lorry from $O$ to $B$ is 5000 kJ .\\
(ii) Find the value of $\theta$.
\hfill \mbox{\textit{CAIE M1 2015 Q4 [6]}}