CAIE M1 2018 June — Question 2 4 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2018
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeFind acceleration on incline given power
DifficultyStandard +0.3 This is a straightforward mechanics problem requiring application of Newton's second law combined with the power equation P=Fv. Students must resolve forces parallel to the slope (weight component, resistance, driving force), use F=ma to find the driving force, then calculate power. It's slightly above average difficulty due to multiple steps and the inclined plane context, but follows a standard template with no novel insight required.
Spec6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product

2 A train of mass 240000 kg travels up a slope inclined at an angle of \(4 ^ { \circ }\) to the horizontal. There is a constant resistance of magnitude 18000 N acting on the train. At an instant when the speed of the train is \(15 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) its deceleration is \(0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). Find the power of the engine of the train.

Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
Driving force \(DF = \frac{P}{15}\)B1 Correct use of \(P = Fv\)
\([DF - 240\,000g\sin 4 - 18\,000 = 240\,000 \times (-0.2)]\)M1 A four-term Newton's 2nd law equation
A1Correct equation
Power is \(2\,060\,000\) (W)A1 Allow 2060 kW or 2.06 MW
Total: 4
**Question 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Driving force $DF = \frac{P}{15}$ | B1 | Correct use of $P = Fv$ |
| $[DF - 240\,000g\sin 4 - 18\,000 = 240\,000 \times (-0.2)]$ | M1 | A four-term Newton's 2nd law equation |
| | A1 | Correct equation |
| Power is $2\,060\,000$ (W) | A1 | Allow 2060 kW or 2.06 MW |
| **Total: 4** | | |
2 A train of mass 240000 kg travels up a slope inclined at an angle of $4 ^ { \circ }$ to the horizontal. There is a constant resistance of magnitude 18000 N acting on the train. At an instant when the speed of the train is $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ its deceleration is $0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Find the power of the engine of the train.\\

\hfill \mbox{\textit{CAIE M1 2018 Q2 [4]}}