CAIE M1 2004 November — Question 3 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2004
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeConstant power on inclined plane
DifficultyStandard +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 resolve forces parallel to the slope, apply F=ma to find the driving force, then use P=Fv to find velocity. It requires multiple standard techniques but follows a predictable structure with no conceptual surprises, making it slightly easier than average.
Spec3.02d Constant acceleration: SUVAT formulae6.02l Power and velocity: P = Fv

3 A car of mass 1250 kg travels down a straight hill with the engine working at a power of 22 kW . The hill is inclined at \(3 ^ { \circ }\) to the horizontal and the resistance to motion of the car is 1130 N . Find the speed of the car at an instant when its acceleration is \(0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).

Question 3:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Equation \(F - 1130 + 1250g\sin3° = 1250 \times 0.2\), contains not more than one errorA1
Equation is correctA1
\(22000 = 725.8v\)M1 For using \(P = Fv\)
Speed is \(30.3 \text{ ms}^{-1}\)A1 5 marks total
# Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation $F - 1130 + 1250g\sin3° = 1250 \times 0.2$, contains not more than one error | A1 | |
| Equation is correct | A1 | |
| $22000 = 725.8v$ | M1 | For using $P = Fv$ |
| Speed is $30.3 \text{ ms}^{-1}$ | A1 | **5 marks total** |

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3 A car of mass 1250 kg travels down a straight hill with the engine working at a power of 22 kW . The hill is inclined at $3 ^ { \circ }$ to the horizontal and the resistance to motion of the car is 1130 N . Find the speed of the car at an instant when its acceleration is $0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.

\hfill \mbox{\textit{CAIE M1 2004 Q3 [5]}}