| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - driving force on horizontal road |
| Difficulty | Standard +0.3 This is a straightforward multi-part work-energy question requiring standard applications of formulas (work = force × distance, change in KE and PE, power = force × velocity). Part (i) uses energy conservation on an incline, part (ii) applies work-energy theorem with given speeds, and part (iii) is a simple verification using P = Fv. All steps are routine M1 mechanics with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using WD by driving force \(=\) Gain in PE \(+\) WD against resistance | |
| WD against resistance \(= 4500 \times 1200 - 16000g \times 18\) | A1 | |
| WD against resistance \(= 2.52 \times 10^6\) J | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([R + 16000g \times 18/1200 = 4500]\) | M1 | For resolving along the plane |
| \([WD = (4500 - 16000g \times 18/1200) \times 1200]\) | M1 | For using WD against resistance \(= Rs\) |
| WD against resistance \(= 2.52 \times 10^6\) J | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(KE\ \text{gain} = \frac{1}{2}\ 16000(21^2 - 9^2)\) J | B1 | |
| M1 | For using \(F = (\text{KE gain} + 2000 \times 2400) \div 2400\) | |
| \(F = 7680000 \div 2400 = 3200\) | A1 [3] | |
| SR (max 1/3) for using \(v^2=u^2+2as\) and Newton's 2nd law: \(21^2 - 9^2 = 2a \times 2400,\ a = 0.075\); \(F - 2000 = 16000 \times 0.075\); \(F = 3200\) B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([P_A = (3200 + 1280) \times 9\) and \(P_B = (3200 - 1280) \times 21]\) | M1 | For using \(P = Fv\) to find \(P_A\) and \(P_B\) |
| \(P_A = P_B = 40320\) W | A1 [2] |
## Question 5:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using WD by driving force $=$ Gain in PE $+$ WD against resistance |
| WD against resistance $= 4500 \times 1200 - 16000g \times 18$ | A1 | |
| WD against resistance $= 2.52 \times 10^6$ J | A1 [3] | |
**Alternative Method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[R + 16000g \times 18/1200 = 4500]$ | M1 | For resolving along the plane |
| $[WD = (4500 - 16000g \times 18/1200) \times 1200]$ | M1 | For using WD against resistance $= Rs$ |
| WD against resistance $= 2.52 \times 10^6$ J | A1 [3] | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $KE\ \text{gain} = \frac{1}{2}\ 16000(21^2 - 9^2)$ J | B1 | |
| | M1 | For using $F = (\text{KE gain} + 2000 \times 2400) \div 2400$ |
| $F = 7680000 \div 2400 = 3200$ | A1 [3] | |
| | | **SR** (max 1/3) for using $v^2=u^2+2as$ and Newton's 2nd law: $21^2 - 9^2 = 2a \times 2400,\ a = 0.075$; $F - 2000 = 16000 \times 0.075$; $F = 3200$ B1 |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[P_A = (3200 + 1280) \times 9$ and $P_B = (3200 - 1280) \times 21]$ | M1 | For using $P = Fv$ to find $P_A$ and $P_B$ |
| $P_A = P_B = 40320$ W | A1 [2] | |
---5 A lorry of mass 16000 kg travels at constant speed from the bottom, $O$, to the top, $A$, of a straight hill. The distance $O A$ is 1200 m and $A$ is 18 m above the level of $O$. The driving force of the lorry is constant and equal to 4500 N .\\
(i) Find the work done against the resistance to the motion of the lorry.
On reaching $A$ the lorry continues along a straight horizontal road against a constant resistance of 2000 N . The driving force of the lorry is not now constant, and the speed of the lorry increases from $9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $A$ to $21 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the point $B$ on the road. The distance $A B$ is 2400 m .\\
(ii) Use an energy method to find $F$, where $F \mathrm {~N}$ is the average value of the driving force of the lorry while moving from $A$ to $B$.\\
(iii) Given that the driving force at $A$ is 1280 N greater than $F \mathrm {~N}$ and that the driving force at $B$ is 1280 N less than $F \mathrm {~N}$, show that the power developed by the lorry's engine is the same at $B$ as it is at $A$.
\hfill \mbox{\textit{CAIE M1 2014 Q5 [8]}}