CAIE M1 2022 June — Question 5 9 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionJune
Marks9
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Mark schemeDownload PDF ↗
TopicPower and driving force
TypeFind steady/maximum speed given power
DifficultyStandard +0.3 This is a straightforward work-energy problem requiring application of standard formulas across three parts. Part (a) uses work-energy theorem with given values to show d=100. Part (b) applies the same principle to car B. Part (c) uses P=Fv at steady speed where driving force equals resistance. All parts involve direct application of memorized formulas with minimal problem-solving insight, making it slightly easier than average.
Spec6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02l Power and velocity: P = Fv
Two racing cars \(A\) and \(B\) are at rest alongside each other at a point \(O\) on a straight horizontal test track. The mass of \(A\) is 1200 kg. The engine of \(A\) produces a constant driving force of 4500 N. When \(A\) arrives at a point \(P\) its speed is 25 m s\(^{-1}\). The distance \(OP\) is \(d\) m. The work done against the resistance force experienced by \(A\) between \(O\) and \(P\) is 75 000 J.
  1. Show that \(d = 100\). [3]
Car \(B\) starts off at the same instant as car \(A\). The two cars arrive at \(P\) simultaneously and with the same speed. The engine of \(B\) produces a driving force of 3200 N and the car experiences a constant resistance to motion of 1200 N.
  1. Find the mass of \(B\). [3]
  2. Find the steady speed which \(B\) can maintain when its engine is working at the same rate as it is at \(P\). [3]
Question 5:
AnswerMarks
5Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
AnswerMarks Guidance
5(a)For attempt at work energy equation M1
M0 for (constant) acceleration method
1
4500d – 75 000 = 1200252 [= 375 000]
AnswerMarks Guidance
2A1 Correct equation
d = 100A1 AG
Accept verification with d substituted in above line to show LHS =
375 000 or LHS −RHS = 0
1
If no marks scored allow SCB1 for 1200252
2
3
AnswerMarks Guidance
5(b)252 02a100 [leading to a = 3.125] B1
stated here
AnswerMarks Guidance
3200 – 1200 = m  3.125M1 Newton’s second law with 3 terms. Allow sign errors and their a.
Mass of car B = 640 kgA1
Alternative mark scheme for question 5(b)
AnswerMarks Guidance
For attempt at work energy equationM1 3 terms. Allow sign errors.
1
(3200 – 1200)  100 = m252
AnswerMarks Guidance
2A1 Correct equation
Mass of car B = 640 kgA1
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
5(c)At P power = 3200  25 [= 80 000] B1
80000
12000
AnswerMarks Guidance
vM1 Attempt Newton’s second law for car B with a = 0
Allow their 80 000 (dimensionally correct)
AnswerMarks Guidance
Steady speed = 66.7 m s−1A1 200 2
Allow 66
3 3
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | For attempt at work energy equation | M1 | 3 terms. Allow sign errors.
M0 for (constant) acceleration method
1
4500d – 75 000 = 1200252 [= 375 000]
2 | A1 | Correct equation
d = 100 | A1 | AG
Accept verification with d substituted in above line to show LHS =
375 000 or LHS −RHS = 0
1
If no marks scored allow SCB1 for 1200252
2
3
--- 5(b) ---
5(b) | 252 02a100 [leading to a = 3.125] | B1 | Allow B1 if acceleration found in part (a) as 3.125 and used or
stated here
3200 – 1200 = m  3.125 | M1 | Newton’s second law with 3 terms. Allow sign errors and their a.
Mass of car B = 640 kg | A1
Alternative mark scheme for question 5(b)
For attempt at work energy equation | M1 | 3 terms. Allow sign errors.
1
(3200 – 1200)  100 = m252
2 | A1 | Correct equation
Mass of car B = 640 kg | A1
3
Question | Answer | Marks | Guidance
--- 5(c) ---
5(c) | At P power = 3200  25 [= 80 000] | B1 | For use of power = Fv
80000
12000
v | M1 | Attempt Newton’s second law for car B with a = 0
Allow their 80 000 (dimensionally correct)
Steady speed = 66.7 m s−1 | A1 | 200 2
Allow 66
3 3
3
Question | Answer | Marks | Guidance
Two racing cars $A$ and $B$ are at rest alongside each other at a point $O$ on a straight horizontal test track. The mass of $A$ is 1200 kg. The engine of $A$ produces a constant driving force of 4500 N. When $A$ arrives at a point $P$ its speed is 25 m s$^{-1}$. The distance $OP$ is $d$ m. The work done against the resistance force experienced by $A$ between $O$ and $P$ is 75 000 J.

\begin{enumerate}[label=(\alph*)]
\item Show that $d = 100$. [3]
\end{enumerate}

Car $B$ starts off at the same instant as car $A$. The two cars arrive at $P$ simultaneously and with the same speed. The engine of $B$ produces a driving force of 3200 N and the car experiences a constant resistance to motion of 1200 N.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the mass of $B$. [3]

\item Find the steady speed which $B$ can maintain when its engine is working at the same rate as it is at $P$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2022 Q5 [9]}}
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