CAIE M1 2007 November — Question 1 4 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2007
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeAcceleration from power and speed
DifficultyModerate -0.3 This is a straightforward application of the power equation P = Fv, requiring students to find the driving force using F = ma + resistance, then substitute into the power formula. It's a standard M1 question with clear given values and a direct method, making it slightly easier than average but still requiring correct application of multiple mechanics principles.
Spec3.03d Newton's second law: 2D vectors6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product

A car of mass 900 kg travels along a horizontal straight road with its engine working at a constant rate of \(P\) kW. The resistance to motion of the car is 550 N. Given that the acceleration of the car is \(0.2 \text{ m s}^{-2}\) at an instant when its speed is \(30 \text{ m s}^{-1}\), find the value of \(P\). [4]

AnswerMarks Guidance
DF − 550 = 900x0.2M1 For using Newton's second law (3 terms)
[P = 730x30 ÷ 1000]M1 For using \(P = (DF)v\)
P = 21.9A1
A14
DF − 550 = 900x0.2 | M1 | For using Newton's second law (3 terms)
[P = 730x30 ÷ 1000] | M1 | For using $P = (DF)v$
P = 21.9 | A1 | 
| A1 | 4
A car of mass 900 kg travels along a horizontal straight road with its engine working at a constant rate of $P$ kW. The resistance to motion of the car is 550 N. Given that the acceleration of the car is $0.2 \text{ m s}^{-2}$ at an instant when its speed is $30 \text{ m s}^{-1}$, find the value of $P$. [4]

\hfill \mbox{\textit{CAIE M1 2007 Q1 [4]}}