CAIE M1 2012 June — Question 1 4 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2012
SessionJune
Marks4
PaperDownload PDF ↗
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) combined with Newton's second law. Students must find the driving force using F = ma + resistance, then calculate P = Fv. It requires two standard formulas with clear given values and minimal problem-solving insight, making it slightly easier than average.
Spec6.02l Power and velocity: P = Fv

1 A car of mass 880 kg travels along a straight horizontal road with its engine working at a constant rate of \(P \mathrm {~W}\). The resistance to motion is 700 N . At an instant when the car's speed is \(16 \mathrm {~ms} ^ { - 1 }\) its acceleration is \(0.625 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). Find the value of \(P\).

Question 1:
AnswerMarks Guidance
Working/AnswerMark Guidance
\(DF - 700 = 880 \times 0.625\)A1
\([P = 1250 \times 16]\)M1 For using \(P = (DF)v\)
\(P = 20\,000\)A1 [4]
## Question 1:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $DF - 700 = 880 \times 0.625$ | A1 | |
| $[P = 1250 \times 16]$ | M1 | For using $P = (DF)v$ |
| $P = 20\,000$ | A1 [4] | |

---
1 A car of mass 880 kg travels along a straight horizontal road with its engine working at a constant rate of $P \mathrm {~W}$. The resistance to motion is 700 N . At an instant when the car's speed is $16 \mathrm {~ms} ^ { - 1 }$ its acceleration is $0.625 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Find the value of $P$.

\hfill \mbox{\textit{CAIE M1 2012 Q1 [4]}}