CAIE M1 2018 November — Question 3 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2018
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeEnergy method - driving force on inclined plane (down hill)
DifficultyModerate -0.3 This is a straightforward work-energy problem requiring students to apply conservation of energy with multiple forces. It involves standard bookwork (calculating gravitational PE loss, KE change, work against resistance) and combining these using the work-energy principle. The calculation is multi-step but follows a predictable template with no conceptual surprises, making it slightly easier than average.
Spec6.02a Work done: concept and definition6.02i Conservation of energy: mechanical energy principle
3 A van of mass 2500 kg descends a hill of length 0.4 km inclined at \(4 ^ { \circ }\) to the horizontal. There is a constant resistance to motion of 600 N and the speed of the van increases from \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to \(30 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) as it descends the hill. Find the work done by the van's engine as it descends the hill.
Question 3:
AnswerMarks Guidance
\(\text{KE gained} = \frac{1}{2} \times 2500 \times (30^2 - 20^2) = 625000\text{ J}\)M1 KE gained or PE lost attempted
\(\text{PE lost} = 2500g \times 400\sin 4 = 697564.7\text{ J}\)A1 Both KE and PE correct
WD by engine \(+2500g \times 400\sin 4 + \frac{1}{2} \times 2500 \times 20^2 = 600 \times 400 + \frac{1}{2} \times 2500 \times 30^2\)M1 Using work-energy equation: WD by engine + PE lost = WD against F + KE gain
Work done by engine + PE lost \(= 600 \times 400 + 625000\)A1 Work-energy equation all correct
Work done \(= 167000\text{ J}\ (167435.2\ldots)\)A1 
## Question 3:

| $\text{KE gained} = \frac{1}{2} \times 2500 \times (30^2 - 20^2) = 625000\text{ J}$ | M1 | KE gained or PE lost attempted |
|---|---|---|
| $\text{PE lost} = 2500g \times 400\sin 4 = 697564.7\text{ J}$ | A1 | Both KE and PE correct |
| WD by engine $+2500g \times 400\sin 4 + \frac{1}{2} \times 2500 \times 20^2 = 600 \times 400 + \frac{1}{2} \times 2500 \times 30^2$ | M1 | Using work-energy equation: WD by engine + PE lost = WD against F + KE gain |
| Work done by engine + PE lost $= 600 \times 400 + 625000$ | A1 | Work-energy equation all correct |
| Work done $= 167000\text{ J}\ (167435.2\ldots)$ | A1 | |

---
3 A van of mass 2500 kg descends a hill of length 0.4 km inclined at $4 ^ { \circ }$ to the horizontal. There is a constant resistance to motion of 600 N and the speed of the van increases from $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ as it descends the hill. Find the work done by the van's engine as it descends the hill.\\

\hfill \mbox{\textit{CAIE M1 2018 Q3 [5]}}
This paper (7 questions)
View full paper
Q1 4 Q2 4 Q3 5 Q4 6 Q5 9 Q6 10 Q7 12