| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - driving force on inclined plane (down hill) |
| Difficulty | Standard +0.3 This is a straightforward work-energy problem requiring standard application of formulas. Part (i) uses work-energy balance at constant speed (driving work = resistance work + PE gain). Part (ii) applies work-energy theorem with given values. Both parts involve routine substitution into well-practiced formulas with no conceptual subtlety or novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| WD against resistance \(= 800 \times 500\) | B1 | |
| \([2\,800\,000 = \text{PE gain} + 400\,000]\) | M1 | For using WD by the driving force \(=\) PE gain \(+\) WD against resistance |
| \([2\,400\,000 = 16000g \times 500\sin\alpha]\) | M1 | For using PE gain \(= mgL\sin\alpha\) |
| \(\alpha = 1.7\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([\text{KE gain} = 2\,400\,000 + 2\,400\,000 - 800\,000]\) | M1 | For using KE gain \(=\) WD by the driving force \(+\) PE loss \(-\) WD against resistance |
| \(4\,000\,000\) J | A1ft | ft PE gain |
| \([\frac{1}{2} \times 16000(v^2 - 20^2) = 4\,000\,000]\) | M1 | For KE gain \(= \frac{1}{2}m(v^2 - 20^2)\) and attempting to solve for \(v\) |
| Speed is \(30\ \text{ms}^{-1}\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Driving force \(= 2\,800\,000 \div 500\) | B1 | |
| \([DF - mg\sin\alpha - R = m \times 0]\) | M1 | For using Newton's second law |
| \([16000 \times 10\sin\alpha = 5600 - 800]\) | DM1 | |
| For solving the resultant equation for \(\alpha\); \(\alpha = 1.7\) | A1 [4] |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| WD against resistance $= 800 \times 500$ | B1 | |
| $[2\,800\,000 = \text{PE gain} + 400\,000]$ | M1 | For using WD by the driving force $=$ PE gain $+$ WD against resistance |
| $[2\,400\,000 = 16000g \times 500\sin\alpha]$ | M1 | For using PE gain $= mgL\sin\alpha$ |
| $\alpha = 1.7$ | A1 [4] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\text{KE gain} = 2\,400\,000 + 2\,400\,000 - 800\,000]$ | M1 | For using KE gain $=$ WD by the driving force $+$ PE loss $-$ WD against resistance |
| $4\,000\,000$ J | A1ft | ft PE gain |
| $[\frac{1}{2} \times 16000(v^2 - 20^2) = 4\,000\,000]$ | M1 | For KE gain $= \frac{1}{2}m(v^2 - 20^2)$ and attempting to solve for $v$ |
| Speed is $30\ \text{ms}^{-1}$ | A1 [4] | |
#### Special Ruling (max 2/4):
Uses Newton's Second Law and $v^2 = u^2 + 2as$: $[4800 + 16000g\sin\alpha - 1600 = 16000a,\ v^2 = 20^2 + 2a \times 500]$, Speed is $30\ \text{ms}^{-1}$ — M1, A1
#### Alternative Method for Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Driving force $= 2\,800\,000 \div 500$ | B1 | |
| $[DF - mg\sin\alpha - R = m \times 0]$ | M1 | For using Newton's second law |
| $[16000 \times 10\sin\alpha = 5600 - 800]$ | DM1 | |
| For solving the resultant equation for $\alpha$; $\alpha = 1.7$ | A1 [4] | |
---5 A lorry of mass 16000 kg moves on a straight hill inclined at angle $\alpha ^ { \circ }$ to the horizontal. The length of the hill is 500 m .\\
(i) While the lorry moves from the bottom to the top of the hill at constant speed, the resisting force acting on the lorry is 800 N and the work done by the driving force is 2800 kJ . Find the value of $\alpha$.\\
(ii) On the return journey the speed of the lorry is $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the top of the hill. While the lorry travels down the hill, the work done by the driving force is 2400 kJ and the work done against the resistance to motion is 800 kJ . Find the speed of the lorry at the bottom of the hill.\\[0pt]
[4]
\hfill \mbox{\textit{CAIE M1 2012 Q5 [8]}}