CAIE M1 2013 June — Question 3 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2013
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeVariable power or two power scenarios
DifficultyStandard +0.3 This is a standard two-equation mechanics problem requiring application of F=ma and P=Fv at two different speeds. While it involves simultaneous equations and careful algebraic manipulation, it follows a well-established template for constant power problems with no novel insight required, making it slightly easier than average.
Spec3.03c Newton's second law: F=ma one dimension6.02l Power and velocity: P = Fv

3 A car has mass 800 kg . The engine of the car generates constant power \(P \mathrm {~kW}\) as the car moves along a straight horizontal road. The resistance to motion is constant and equal to \(R \mathrm {~N}\). When the car's speed is \(14 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) its acceleration is \(1.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }\), and when the car's speed is \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) its acceleration is \(0.33 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). Find the values of \(P\) and \(R\).

Question 3:
AnswerMarks Guidance
Answer/WorkingMark Guidance
M1For using \(DF = P/v\)
M1For using Newton's 2nd law for both speeds/accelerations
\(1000P/14 - R = 800 \times 1.4\) and \(1000P/25 - R = 800 \times 0.33\)A1
M1For solving for \(P\)
\(P = 27.2\)A1
\(R = 825\)B1 [6] Accept 825.5
## Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $DF = P/v$ |
| | M1 | For using Newton's 2nd law for both speeds/accelerations |
| $1000P/14 - R = 800 \times 1.4$ and $1000P/25 - R = 800 \times 0.33$ | A1 | |
| | M1 | For solving for $P$ |
| $P = 27.2$ | A1 | |
| $R = 825$ | B1 [6] | Accept 825.5 |

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3 A car has mass 800 kg . The engine of the car generates constant power $P \mathrm {~kW}$ as the car moves along a straight horizontal road. The resistance to motion is constant and equal to $R \mathrm {~N}$. When the car's speed is $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ its acceleration is $1.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, and when the car's speed is $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ its acceleration is $0.33 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Find the values of $P$ and $R$.

\hfill \mbox{\textit{CAIE M1 2013 Q3 [6]}}