| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - driving force on inclined plane (down hill) |
| Difficulty | Moderate -0.3 This is a straightforward application of the work-energy principle with clearly stated forces and motion. Part (i) requires resolving forces on an incline and calculating work done (standard M1 content). Part (ii) applies conservation of energy with given forces, requiring calculation of work done by driving force, resistance, and gravity, then using the work-energy theorem. While multi-step, each step follows standard procedures without requiring problem-solving insight or novel approaches. |
| Spec | 6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gain in \(PE = 15000g \times 500\sin 2.5°\ \text{J}\) | B1 | |
| WD against the resistance \(= 800 \times 500\ \text{J}\) | B1 | |
| \([3271454 + 400000]\) | M1 | For using WD by driving force = Gain in PE + WD against resistance |
| Work done is \(3670000\ \text{J}\) or \(3670\ \text{kJ}\) | A1 [4] | |
| *Alternatively:* For resolving forces up the plane | M1 | |
| Driving Force \(= 800 + 15000g\sin 2.5°\) | A1 | |
| For using \(WD = \text{Driving Force} \times 500\) | M1 | |
| Work done is \(3670000\ \text{J}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Work done by \(DF = 2000 \times 500\ \text{J}\) | B1 | |
| Gain in \(KE = \frac{1}{2} \times 15000(v^2 - 20^2)\) | B1 | |
| M1 | For using Gain in KE = Loss in PE − WD against resistance + WD by driving force | |
| \(\frac{1}{2} \times 15000(v^2 - 20^2) = 3271454 - 400000 + 1000000\) | A1 | |
| Speed of the lorry is \(30.3\ \text{ms}^{-1}\) | A1 [5] | |
| *Alternatively:* For applying Newton's second law | M1 | |
| \(2000 + 15000g\sin 2.5° - 800 = 15000a\) | A1 | |
| For using \(v^2 = u^2 + 2as\) | M1 | |
| \(v^2 = 20^2 + 2 \times 0.5162 \times 500\) | A1 | |
| Speed is \(30.3\ \text{ms}^{-1}\) | A1 |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gain in $PE = 15000g \times 500\sin 2.5°\ \text{J}$ | B1 | |
| WD against the resistance $= 800 \times 500\ \text{J}$ | B1 | |
| $[3271454 + 400000]$ | M1 | For using WD by driving force = Gain in PE + WD against resistance |
| Work done is $3670000\ \text{J}$ or $3670\ \text{kJ}$ | A1 **[4]** | |
| *Alternatively:* For resolving forces up the plane | M1 | |
| Driving Force $= 800 + 15000g\sin 2.5°$ | A1 | |
| For using $WD = \text{Driving Force} \times 500$ | M1 | |
| Work done is $3670000\ \text{J}$ | A1 | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Work done by $DF = 2000 \times 500\ \text{J}$ | B1 | |
| Gain in $KE = \frac{1}{2} \times 15000(v^2 - 20^2)$ | B1 | |
| | M1 | For using Gain in KE = Loss in PE − WD against resistance + WD by driving force |
| $\frac{1}{2} \times 15000(v^2 - 20^2) = 3271454 - 400000 + 1000000$ | A1 | |
| Speed of the lorry is $30.3\ \text{ms}^{-1}$ | A1 **[5]** | |
| *Alternatively:* For applying Newton's second law | M1 | |
| $2000 + 15000g\sin 2.5° - 800 = 15000a$ | A1 | |
| For using $v^2 = u^2 + 2as$ | M1 | |
| $v^2 = 20^2 + 2 \times 0.5162 \times 500$ | A1 | |
| Speed is $30.3\ \text{ms}^{-1}$ | A1 | |
---6 A lorry of mass 15000 kg climbs a hill of length 500 m at a constant speed. The hill is inclined at $2.5 ^ { \circ }$ to the horizontal. The resistance to the lorry's motion is constant and equal to 800 N .\\
(i) Find the work done by the lorry's driving force.
On its return journey the lorry reaches the top of the hill with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and continues down the hill with a constant driving force of 2000 N . The resistance to the lorry's motion is again constant and equal to 800 N .\\
(ii) Find the speed of the lorry when it reaches the bottom of the hill.
\hfill \mbox{\textit{CAIE M1 2011 Q6 [9]}}