| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | November |
| Marks | 5 |
| 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 work-energy principles with clearly stated values and standard formulas. Part (i) uses mgh directly, part (ii) applies energy conservation with constant speed (so KE change = 0), and part (iii) uses work-energy theorem with given resistance. All steps are routine M1 mechanics with no problem-solving insight required, making it slightly easier than average. |
| Spec | 6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Part (i) | ||
| Loss in PE is \(2.7 \times 10^6\) J | B1 | 1 |
| Part (ii) | ||
| WD is \(2.1 \times 10^6\) J | B1ft | 1 |
| Part (iii) | ||
| \(KE \text{ change} = \frac{1}{2} \times 15000(16^2 - 14^2)\) | B1 | |
| \([WD = \frac{1}{2} \times 15000(16^2 - 14^2) + 1600 \times 2500]\) | M1 | WD by DF = Gain in KE + WD by resistance |
| WD is \(4.45 \times 10^6\) J | A1 | 3 |
| SR for candidates who use Newton's Law method instead of energy (max 1/3) \(a = (16^2 - 14^2)(2 \times 2500) = 0.012\) \(DF = 1600 + 15000 \times 0.012 = 1780\) WD = \(1780 \times 2500 = 4.45 \times 10^6\) B1 |
| **Part (i)** | | |
|---|---|---|
| Loss in PE is $2.7 \times 10^6$ J | B1 | 1 |
| **Part (ii)** | | |
|---|---|---|
| WD is $2.1 \times 10^6$ J | B1ft | 1 | ft incorrect loss in PE |
| **Part (iii)** | | |
|---|---|---|
| $KE \text{ change} = \frac{1}{2} \times 15000(16^2 - 14^2)$ | B1 | |
| $[WD = \frac{1}{2} \times 15000(16^2 - 14^2) + 1600 \times 2500]$ | M1 | WD by DF = Gain in KE + WD by resistance |
| WD is $4.45 \times 10^6$ J | A1 | 3 |
| | | SR for candidates who use Newton's Law method instead of energy (max 1/3) $a = (16^2 - 14^2)(2 \times 2500) = 0.012$ $DF = 1600 + 15000 \times 0.012 = 1780$ WD = $1780 \times 2500 = 4.45 \times 10^6$ B1 |2 A lorry of mass 15000 kg moves with constant speed $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from the top to the bottom of a straight hill of length 900 m . The top of the hill is 18 m above the level of the bottom of the hill. The total work done by the resistive forces acting on the lorry, including the braking force, is $4.8 \times 10 ^ { 6 } \mathrm {~J}$. Find\\
(i) the loss in gravitational potential energy of the lorry,\\
(ii) the work done by the driving force.
On reaching the bottom of the hill the lorry continues along a straight horizontal road against a constant resistance of 1600 N . There is no braking force acting. The speed of the lorry increases from $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the bottom of the hill to $16 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the point $X$, where $X$ is 2500 m from the bottom of the hill.\\
(iii) By considering energy, find the work done by the driving force of the lorry while it travels from the bottom of the hill to $X$.
\hfill \mbox{\textit{CAIE M1 2009 Q2 [5]}}