CAIE M1 2010 June — Question 1 4 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2010
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeFind acceleration on incline given power
DifficultyModerate -0.3 This is a straightforward application of the power-force-velocity relationship (P = Fv) combined with Newton's second law on an incline. Students must find the driving force from power, resolve forces parallel to the slope (weight component and resistance), then apply F = ma. It's a standard M1 question requiring routine techniques with no novel insight, making it slightly easier than average.
Spec3.03c Newton's second law: F=ma one dimension6.02l Power and velocity: P = Fv

A car of mass \(1150 \text{ kg}\) travels up a straight hill inclined at \(1.2°\) to the horizontal. The resistance to motion of the car is \(975 \text{ N}\). Find the acceleration of the car at an instant when it is moving with speed \(16 \text{ m s}^{-1}\) and the engine is working at a power of \(35 \text{ kW}\). [4]

AnswerMarks Guidance
DF = 35000/16B1
DF − 1150g sin1.2° − 975 = 1150aM1 For using Newton's second law
Acceleration is 0.845 ms⁻²A1
[4]
DF = 35000/16 | B1 | 
DF − 1150g sin1.2° − 975 = 1150a | M1 | For using Newton's second law
Acceleration is 0.845 ms⁻² | A1 |
| | [4]
A car of mass $1150 \text{ kg}$ travels up a straight hill inclined at $1.2°$ to the horizontal. The resistance to motion of the car is $975 \text{ N}$. Find the acceleration of the car at an instant when it is moving with speed $16 \text{ m s}^{-1}$ and the engine is working at a power of $35 \text{ kW}$.
[4]

\hfill \mbox{\textit{CAIE M1 2010 Q1 [4]}}