| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2004 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - driving force up incline, find work done by engine/force |
| Difficulty | Standard +0.3 This is a straightforward multi-part work-energy question requiring standard formula application (mgh, Fs, work-energy theorem). All parts follow directly from definitions with no problem-solving insight needed. The constant speed in part (i)-(iii) simplifies the analysis. Slightly above average difficulty only due to the multi-step nature and careful bookkeeping required in part (iv). |
| Spec | 6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gain in GPE \(= 3.2 \times 10^6 \text{ J}\) | B1 | From \(16000 \times 10 \times 20\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| WD by driving force \(= 5 \times 10^6 \text{ J}\) | B1 | From \(5000 \times 1000\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Work done is \(1.8 \times 10^6 \text{ J}\) | B1ft | From ans (ii) \(-\) ans (i) or from \((5000 - 160000 \times 20/1000)1000\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(KE = \frac{1}{2}mv^2\) | |
| Increase in KE \(= \frac{1}{2} \times 16000(25^2 - 10^2)\) | A1 | |
| WD by resistance \(= 1500 \times 2000\) | B1 | |
| WD by driving force \(= 4.2 \times 10^6 + 3 \times 10^6\) | M1 | WD by driving force \(=\) increase in KE \(+\) WD by resistance |
| WD by driving force \(= 7.2 \times 10^6 \text{ J}\) | A1 | 5 marks total |
| Answer | Marks |
|---|---|
| - \(WD = (2100 + 1500) \times 2000 = 7.2 \times 10^6 \text{ J}\) | B2 |
# Question 4:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gain in GPE $= 3.2 \times 10^6 \text{ J}$ | B1 | From $16000 \times 10 \times 20$ |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| WD by driving force $= 5 \times 10^6 \text{ J}$ | B1 | From $5000 \times 1000$ |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Work done is $1.8 \times 10^6 \text{ J}$ | B1ft | From ans **(ii)** $-$ ans **(i)** or from $(5000 - 160000 \times 20/1000)1000$ |
## Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $KE = \frac{1}{2}mv^2$ |
| Increase in KE $= \frac{1}{2} \times 16000(25^2 - 10^2)$ | A1 | |
| WD by resistance $= 1500 \times 2000$ | B1 | |
| WD by driving force $= 4.2 \times 10^6 + 3 \times 10^6$ | M1 | WD by driving force $=$ increase in KE $+$ WD by resistance |
| WD by driving force $= 7.2 \times 10^6 \text{ J}$ | A1 | **5 marks total** |
*SR for candidates who assume implicitly that the driving force is constant: max 2/5*
- $a = (625 - 100)/(2 \times 2000)$
- $DF = 16000 \times 0.13125 = 2100$
- $WD = (2100 + 1500) \times 2000 = 7.2 \times 10^6 \text{ J}$ | B2 |
---4 A lorry of mass 16000 kg climbs from the bottom to the top of a straight hill of length 1000 m at a constant speed of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The top of the hill is 20 m above the level of the bottom of the hill. The driving force of the lorry is constant and equal to 5000 N . Find\\
(i) the gain in gravitational potential energy of the lorry,\\
(ii) the work done by the driving force,\\
(iii) the work done against the force resisting the motion of the lorry.
On reaching the top of the hill the lorry continues along a straight horizontal road against a constant resistance of 1500 N . The driving force of the lorry is not now constant, and the speed of the lorry increases from $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the top of the hill to $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the point $P$. The distance of $P$ from the top of the hill is 2000 m .\\
(iv) Find the work done by the driving force of the lorry while the lorry travels from the top of the hill to $P$.
\hfill \mbox{\textit{CAIE M1 2004 Q4 [8]}}