| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Difficulty | Standard +0.3 This is a straightforward energy/work problem requiring application of standard formulas (work-energy theorem, work = force × distance, change in KE and PE). Students must use Newton's second law to find driving forces at two points, then apply energy principles systematically. While multi-step, it involves only routine M1 techniques with no novel insight required, making it slightly above average difficulty. |
| Spec | 6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([30000/v - 1000 - 1250g \times 30/500 = 1250a]\) | M1 | For using Newton's 2nd law |
| \(v_{\text{bottom}} = 30000/(1250 \times 4 + 1000 + 750)\) | M1 | |
| \(v_{\text{top}} = 30000/(1250 \times 0.2 + 1000 + 750)\) | A1 | |
| \([\frac{1}{2} \times 1250(15^2 - 4.44...^2)]\) | M1 | For using KE gain \(= \frac{1}{2}m(v_{\text{top}}^2 - v_{\text{bottom}}^2)\) |
| Increase in KE is 128000 J (128 kJ) | A1 | Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([F - 1000 - 1250g \times 30/500 = 1250a]\) | M1 | For using Newton's second law to find the driving force at the bottom and the top |
| \(F_{\text{bottom}} = 1250 \times 4 + 1000 + 750 = 6750\) and \(F_{\text{top}} = 1250 \times 0.2 + 1000 + 750 = 2000\) | A1 | |
| \([v_{\text{bottom}} = 30000/6750\) and \(v_{\text{top}} = 30000/2000]\) | M1 | For using \(DF = 30000/v\) to find \(v_{\text{bottom}}\) and \(v_{\text{top}}\) |
| \([\frac{1}{2} \times 1250(15^2 - 4.44...^2)]\) | M1 | For using KE gain \(= \frac{1}{2}m(v_{\text{top}}^2 - v_{\text{bottom}}^2)\) |
| Increase in KE is 128000 J (128 kJ) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v_{\text{bottom}} = 30000/(1250 \times 4 + 1000)\) and \(v_{\text{top}} = 30000/(1250 \times 0.2 + 1000)\) | B1 | |
| \([\frac{1}{2} \times 1250(24^2 - 5^2)]\) | M1 | For using KE gain \(= \frac{1}{2}m(v_{\text{top}}^2 - v_{\text{bottom}}^2)\) |
| Increase in KE is 344000 J (344 kJ) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| PE gain \(= 1250g \times 30\) | ||
| WD against resistance \(= 1000 \times 500\) | B1 | |
| \([WD_{\text{car}} = 128000 + 375000 + 500000]\) | M1 | For using WD by car's engine \(=\) KE gain \(+\) PE gain \(+\) WD against resistance |
| Work done is 1000 000 J (1000 kJ) | A1ft | Total: 3 |
# Question 6:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[30000/v - 1000 - 1250g \times 30/500 = 1250a]$ | M1 | For using Newton's 2nd law |
| $v_{\text{bottom}} = 30000/(1250 \times 4 + 1000 + 750)$ | M1 | |
| $v_{\text{top}} = 30000/(1250 \times 0.2 + 1000 + 750)$ | A1 | |
| $[\frac{1}{2} \times 1250(15^2 - 4.44...^2)]$ | M1 | For using KE gain $= \frac{1}{2}m(v_{\text{top}}^2 - v_{\text{bottom}}^2)$ |
| Increase in KE is 128000 J (128 kJ) | A1 | **Total: 5** |
**Alternative for part (i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[F - 1000 - 1250g \times 30/500 = 1250a]$ | M1 | For using Newton's second law to find the driving force at the bottom and the top |
| $F_{\text{bottom}} = 1250 \times 4 + 1000 + 750 = 6750$ and $F_{\text{top}} = 1250 \times 0.2 + 1000 + 750 = 2000$ | A1 | |
| $[v_{\text{bottom}} = 30000/6750$ and $v_{\text{top}} = 30000/2000]$ | M1 | For using $DF = 30000/v$ to find $v_{\text{bottom}}$ and $v_{\text{top}}$ |
| $[\frac{1}{2} \times 1250(15^2 - 4.44...^2)]$ | M1 | For using KE gain $= \frac{1}{2}m(v_{\text{top}}^2 - v_{\text{bottom}}^2)$ |
| Increase in KE is 128000 J (128 kJ) | A1 | |
**Special Ruling** applying to part (i) for candidates who omit the weight component in applying Newton's second law. (Max 3 out of 5)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v_{\text{bottom}} = 30000/(1250 \times 4 + 1000)$ and $v_{\text{top}} = 30000/(1250 \times 0.2 + 1000)$ | B1 | |
| $[\frac{1}{2} \times 1250(24^2 - 5^2)]$ | M1 | For using KE gain $= \frac{1}{2}m(v_{\text{top}}^2 - v_{\text{bottom}}^2)$ |
| Increase in KE is 344000 J (344 kJ) | A1 | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| PE gain $= 1250g \times 30$ | | |
| WD against resistance $= 1000 \times 500$ | B1 | |
| $[WD_{\text{car}} = 128000 + 375000 + 500000]$ | M1 | For using WD by car's engine $=$ KE gain $+$ PE gain $+$ WD against resistance |
| Work done is 1000 000 J (1000 kJ) | A1ft | **Total: 3** | ft incorrect answer in (i) |
---6 A car of mass 1250 kg moves from the bottom to the top of a straight hill of length 500 m . The top of the hill is 30 m above the level of the bottom. The power of the car's engine is constant and equal to 30000 W . The car's acceleration is $4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ at the bottom of the hill and is $0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ at the top. The resistance to the car's motion is 1000 N . Find\\
(i) the car's gain in kinetic energy,\\
(ii) the work done by the car's engine.
\hfill \mbox{\textit{CAIE M1 2012 Q6 [8]}}