CAIE M1 2019 June — Question 6 7 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2019
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeForce from power and speed
DifficultyStandard +0.8 This is a multi-step mechanics problem requiring students to set up two simultaneous equations from power-force-velocity relationships (P=Fv) in different scenarios, then solve for two unknowns. It involves understanding steady speed (zero acceleration, so driving force equals resistance plus component of weight on incline), applying P=Fv correctly in both cases, and solving the resulting system. While the individual concepts are standard M1 material, the problem requires careful setup, algebraic manipulation, and synthesis of multiple ideas, making it moderately challenging but not requiring novel insight.
Spec6.02l Power and velocity: P = Fv

A car has mass 1000 kg. When the car is travelling at a steady speed of \(v\) m s\(^{-1}\), where \(v > 2\), the resistance to motion of the car is \((Av + B)\) N, where \(A\) and \(B\) are constants. The car can travel along a horizontal road at a steady speed of 18 m s\(^{-1}\) when its engine is working at 36 kW. The car can travel up a hill inclined at an angle of \(\theta\) to the horizontal, where \(\sin \theta = 0.05\), at a steady speed of 12 m s\(^{-1}\) when its engine is working at 21 kW. Find \(A\) and \(B\). [7]

Question 6:
AnswerMarks
6Case 1: DF = 36000/18
or
AnswerMarks Guidance
Case 2: DF = 21000/12B1 DF = P/v in either case
18A + B = DF
AnswerMarks Guidance
[36000/18 = 18A + B = 2000]M1 Use DF = resistance (case 1)
18A + B = 2000 oeA1 Correct equation, unsimplified
12A + B = DF + weight component
AnswerMarks Guidance
[21000/12 = 12A + B + 1000 g × 1/20]M1 Use DF = resistance + weight component (case 2)
12A + B = 1250 oeA1 Correct equation, unsimplified
DM1Solve two simultaneous equations in A and B only for A or B
Dependent on both previous M1’s
AnswerMarks Guidance
A = 125, B = –250A1 Both correct
7
AnswerMarks Guidance
QuestionAnswer Marks
Question 6:
6 | Case 1: DF = 36000/18
or
Case 2: DF = 21000/12 | B1 | DF = P/v in either case
18A + B = DF
[36000/18 = 18A + B = 2000] | M1 | Use DF = resistance (case 1)
18A + B = 2000 oe | A1 | Correct equation, unsimplified
12A + B = DF + weight component
[21000/12 = 12A + B + 1000 g × 1/20] | M1 | Use DF = resistance + weight component (case 2)
12A + B = 1250 oe | A1 | Correct equation, unsimplified
DM1 | Solve two simultaneous equations in A and B only for A or B
Dependent on both previous M1’s
A = 125, B = –250 | A1 | Both correct
7
Question | Answer | Marks | Guidance
A car has mass 1000 kg. When the car is travelling at a steady speed of $v$ m s$^{-1}$, where $v > 2$, the resistance to motion of the car is $(Av + B)$ N, where $A$ and $B$ are constants. The car can travel along a horizontal road at a steady speed of 18 m s$^{-1}$ when its engine is working at 36 kW. The car can travel up a hill inclined at an angle of $\theta$ to the horizontal, where $\sin \theta = 0.05$, at a steady speed of 12 m s$^{-1}$ when its engine is working at 21 kW. Find $A$ and $B$. [7]

\hfill \mbox{\textit{CAIE M1 2019 Q6 [7]}}