CAIE M1 2002 November — Question 1 3 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2002
SessionNovember
Marks3
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Mark schemeDownload PDF ↗
TopicPower and driving force
TypeFind acceleration given power
DifficultyModerate -0.3 This is a straightforward application of the power-force-velocity relationship (P=Fv) to find driving force, then using F=ma with given resistance. It requires only direct substitution into standard formulas with no problem-solving insight, making it slightly easier than average but not trivial since it involves multiple connected steps.
Spec6.02l Power and velocity: P = Fv
1 A car of mass 1000 kg travels along a horizontal straight road with its engine working at a constant rate of 20 kW . The resistance to motion of the car is 600 N . Find the acceleration of the car at an instant when its speed is \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Driving force \(= \frac{20000}{25}\)B1
For using Newton's 2nd law (3 terms needed): \(\frac{20000}{25} - 600 = 1000a\)M1 \(\frac{20000-600}{25} = 1000a\) scores B0 M1; \(\frac{20000}{25} - \frac{600}{25} = 1000a\) scores B1 M0
Acceleration is \(0.2 \text{ ms}^{-2}\)A1 Total: 3 
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Driving force $= \frac{20000}{25}$ | B1 | |
| For using Newton's 2nd law (3 terms needed): $\frac{20000}{25} - 600 = 1000a$ | M1 | $\frac{20000-600}{25} = 1000a$ scores B0 M1; $\frac{20000}{25} - \frac{600}{25} = 1000a$ scores B1 M0 |
| Acceleration is $0.2 \text{ ms}^{-2}$ | A1 | **Total: 3** |

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1 A car of mass 1000 kg travels along a horizontal straight road with its engine working at a constant rate of 20 kW . The resistance to motion of the car is 600 N . Find the acceleration of the car at an instant when its speed is $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.

\hfill \mbox{\textit{CAIE M1 2002 Q1 [3]}}
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