CAIE M1 2023 June — Question 7 11 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2023
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeFind acceleration given power
DifficultyStandard +0.3 This is a standard M1 power-driving force question with straightforward application of P=Fv, F=ma, and energy methods. Parts (a) and (b) are routine textbook exercises. Part (c) requires energy conservation with multiple terms but follows a clear template with all values given explicitly, requiring no novel insight—slightly above average due to the multi-step energy calculation.
Spec6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02j Conservation with elastics: springs and strings6.02l Power and velocity: P = Fv
A car of mass \(1200\) kg is travelling along a straight horizontal road. The power of the car's engine is constant and is equal to \(16\) kW. There is a constant resistance to motion of magnitude \(500\) N.
  1. Find the acceleration of the car at an instant when its speed is \(20\) m s\(^{-1}\). [3]
  2. Assuming that the power and the resistance forces remain unchanged, find the steady speed at which the car can travel. [2]
The car comes to the bottom of a straight hill of length \(316\) m, inclined at an angle to the horizontal of \(\sin^{-1}(\frac{4}{65})\). The power remains constant at \(16\) kW, but the magnitude of the resistance force is no longer constant and changes such that the work done against the resistance force in ascending the hill is \(128400\) J. The time taken to ascend the hill is \(15\) s.
  1. Given that the car is travelling at a speed of \(20\) m s\(^{-1}\) at the bottom of the hill, find its speed at the top of the hill. [6]
Question 7:
AnswerMarks
7(a)16000
Driving force F  [800]
AnswerMarks Guidance
20B1 OE e.g. 1600020F
F 5001200aM1 Use of Newton’s second law; allow sign errors but
must be 3 terms. Allow F or any non-zero value for the
driving force (allow 0.8 from using 16 rather than
16000) but not 16000, 16, 20 or 500 for F.
AnswerMarks
a = 0.25 (ms–2)A1
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(b)16000
5000
AnswerMarks Guidance
vM1 Allow sign errors but must be 2 terms. Condone
16
5000 for M1.
v
AnswerMarks
v 32 (ms–1)A1
2
AnswerMarks Guidance
7(c)Work done by engine = 1600015240000 B1
Or 16000 .
15
1 1 
KE change =   1200v2  1200202 
AnswerMarks Guidance
2 2 B1 (600v2 240000)
 1 
PE change =   1200g316  63200
AnswerMarks Guidance
 60B1 Allow 1200g316sin0.955 or
79
1200g316sin0.95 or 1200g or
15
1200g5.266....
AnswerMarks Guidance
Attempt at work-energy equation.M1 Use of work-energy principle with 5 terms;
dimensionally correct. Allow sign errors and sin/cos
mix on PE term
1 1 1
1600015128400 1200v2  1200202 1200g316
2 2 60
AnswerMarks Guidance
(240000128400600v2 24000063200)A1 Allow a value in the interval [62870,63600] for the PE
term from using non-exact values for the given angle
(but not if from incorrect working).
AnswerMarks Guidance
v = 21.9 (ms–1)A1 21.924111…
6
Question 7:
--- 7(a) ---
7(a) | 16000
Driving force F  [800]
20 | B1 | OE e.g. 1600020F
F 5001200a | M1 | Use of Newton’s second law; allow sign errors but
must be 3 terms. Allow F or any non-zero value for the
driving force (allow 0.8 from using 16 rather than
16000) but not 16000, 16, 20 or 500 for F.
a = 0.25 (ms–2) | A1
3
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | 16000
5000
v | M1 | Allow sign errors but must be 2 terms. Condone
16
5000 for M1.
v
v 32 (ms–1) | A1
2
--- 7(c) ---
7(c) | Work done by engine = 1600015240000 | B1 | WD
Or 16000 .
15
1 1 
KE change =   1200v2  1200202 
2 2  | B1 | (600v2 240000)
 1 
PE change =   1200g316  63200
 60 | B1 | Allow 1200g316sin0.955 or
79
1200g316sin0.95 or 1200g or
15
1200g5.266....
Attempt at work-energy equation. | M1 | Use of work-energy principle with 5 terms;
dimensionally correct. Allow sign errors and sin/cos
mix on PE term
1 1 1
1600015128400 1200v2  1200202 1200g316
2 2 60
(240000128400600v2 24000063200) | A1 | Allow a value in the interval [62870,63600] for the PE
term from using non-exact values for the given angle
(but not if from incorrect working).
v = 21.9 (ms–1) | A1 | 21.924111…
6
A car of mass $1200$ kg is travelling along a straight horizontal road. The power of the car's engine is constant and is equal to $16$ kW. There is a constant resistance to motion of magnitude $500$ N.

\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the car at an instant when its speed is $20$ m s$^{-1}$. [3]
\item Assuming that the power and the resistance forces remain unchanged, find the steady speed at which the car can travel. [2]
\end{enumerate}

The car comes to the bottom of a straight hill of length $316$ m, inclined at an angle to the horizontal of $\sin^{-1}(\frac{4}{65})$. The power remains constant at $16$ kW, but the magnitude of the resistance force is no longer constant and changes such that the work done against the resistance force in ascending the hill is $128400$ J. The time taken to ascend the hill is $15$ s.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Given that the car is travelling at a speed of $20$ m s$^{-1}$ at the bottom of the hill, find its speed at the top of the hill. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2023 Q7 [11]}}
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