CAIE M1 2024 November — Question 7 8 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeVariable resistance: find constant speed
DifficultyStandard +0.3 This is a standard power-force-velocity mechanics problem requiring straightforward application of P=Fv and F=ma. Part (a)(i) is a simple one-step 'show that' calculation. Parts (a)(ii) and (b) involve routine manipulation of the power equation with resistance and weight components. The multi-part structure and 8 total marks indicate moderate length, but all steps follow textbook methods with no novel insight required, making it slightly easier than average.
Spec6.02l Power and velocity: P = Fv
A car has mass \(1200\) kg. When the car is travelling at a speed of \(v \text{ ms}^{-1}\), there is a resistive force of magnitude \(kv\) N. The maximum power of the car's engine is \(92.16\) kW.
  1. The car travels along a straight level road.
    1. The car has a greatest possible constant speed of \(48 \text{ ms}^{-1}\). Show that \(k = 40\). [1]
    2. At an instant when its speed is \(45 \text{ ms}^{-1}\), find the greatest possible acceleration of the car. [3]
  2. The car now travels at a constant speed up a hill inclined at an angle of \(\sin^{-1} 0.15\) to the horizontal. Find the greatest possible speed of the car going up the hill. [4]
Question 7:
AnswerMarks Guidance
7(a)(i)Power = k482 =92160  k =40 B1
1
AnswerMarks
7(a)(ii)92160
[DF =] [=2048]
AnswerMarks Guidance
45B1 For any use of power = Fv e.g. 45DF=92160.
2048−4045=1200aM1 Apply N2L using their DF92160,92.16,1920.
3 terms; allow sign errors. Dimensionally correct.
248 31
a= = ms−2
AnswerMarks Guidance
1200 150A1 Allow 0.207 or better.
3
AnswerMarks Guidance
7(b)DF =40v+1200g0.15 *M1
errors and sin/cos mix – dimensionally correct.
92160
=40v+1200g0.15
AnswerMarks Guidance
vDM1 Set up an equation in v only – must be usingDFv=92160.
40v2 +1800v−92160 [=0]DM1 Attempt to solve their 3TQ in v – dependent on both previous M
marks.
AnswerMarks Guidance
v=30.5ms−1A1 30.51179…
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 7:
--- 7(a)(i) ---
7(a)(i) | Power = k482 =92160  k =40 | B1 | AG
1
--- 7(a)(ii) ---
7(a)(ii) | 92160
[DF =] [=2048]
45 | B1 | For any use of power = Fv e.g. 45DF=92160.
2048−4045=1200a | M1 | Apply N2L using their DF92160,92.16,1920.
3 terms; allow sign errors. Dimensionally correct.
248 31
a= = ms−2
1200 150 | A1 | Allow 0.207 or better.
3
--- 7(b) ---
7(b) | DF =40v+1200g0.15 | *M1 | Two term expression for the driving force up the hill, allow sign
errors and sin/cos mix – dimensionally correct.
92160
=40v+1200g0.15
v | DM1 | Set up an equation in v only – must be usingDFv=92160.
40v2 +1800v−92160 [=0] | DM1 | Attempt to solve their 3TQ in v – dependent on both previous M
marks.
v=30.5ms−1 | A1 | 30.51179…
4
Question | Answer | Marks | Guidance
A car has mass $1200$ kg. When the car is travelling at a speed of $v \text{ ms}^{-1}$, there is a resistive force of magnitude $kv$ N. The maximum power of the car's engine is $92.16$ kW.

\begin{enumerate}[label=(\alph*)]
\item The car travels along a straight level road.
\begin{enumerate}[label=(\roman*)]
\item The car has a greatest possible constant speed of $48 \text{ ms}^{-1}$.

Show that $k = 40$. [1]

\item At an instant when its speed is $45 \text{ ms}^{-1}$, find the greatest possible acceleration of the car. [3]
\end{enumerate}

\item The car now travels at a constant speed up a hill inclined at an angle of $\sin^{-1} 0.15$ to the horizontal.

Find the greatest possible speed of the car going up the hill. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2024 Q7 [8]}}
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