CAIE M1 2022 June — Question 6 10 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeVerifying given motion properties
DifficultyStandard +0.3 This is a straightforward variable acceleration question requiring standard calculus techniques: integrating velocity to find displacement (part a), differentiating to find acceleration and solving simultaneous equations (part b). While it involves multiple steps across 10 marks, each step uses routine A-level mechanics methods with no novel problem-solving insight required, making it slightly easier than average.
Spec3.02f Non-uniform acceleration: using differentiation and integration
A particle starts from a point \(O\) and moves in a straight line. The velocity \(v\) m s\(^{-1}\) of the particle at time \(t\) s after leaving \(O\) is given by $$v = k(3t^2 - 2t^3),$$ where \(k\) is a constant.
  1. Verify that the particle returns to \(O\) when \(t = 2\). [4]
  2. It is given that the acceleration of the particle is \(-13.5\) m s\(^{-2}\) for the positive value of \(t\) at which \(v = 0\). Find \(k\) and hence find the total distance travelled in the first two seconds of motion. [6]
Question 6:
AnswerMarks
6Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9709/41 Cambridge International AS & A Level – Mark Scheme May/June 2022
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2022 Page 5 of 13
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
6(a)For attempt at integration *M1
least 1 term.
Allow if k is omitted.
 1 
k t3 t4 C
 
AnswerMarks Guidance
 2 A1 Allow unsimplified
 1 
k 8 16 0
 
AnswerMarks Guidance
 2 DM1 For use of limits 0 and 2 or substituting t = 2
OR equate to 0, then solve a quartic equation in t.
Allow if k is omitted.
AnswerMarks Guidance
0A1 AG, CWO including stating C = 0 if not using limits
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
6(b)v = 0 when k  3t2 2t3 0 M1
Leading to t = 1.5 [or t = 0]A1 A0 for other solutions not discarded
 6t6t2
AnswerMarks Guidance
a  kM1 For differentiation, the power of t must decrease by 1 with a
change of coefficient in at least 1 term.
Allow if k is omitted
AnswerMarks
[13.5 k  61.561.52 ⇒ 13.5 k913.5⇒] k = 3A1
Distance from t = 0 to t = 1.5 is
1.5
  1   1 
3 t3  t4 3 1.53  1.54 0
    
  2   2 
AnswerMarks Guidance
0DM1 1.5 2
For use of limits. 2  or 2 or both integrals. Dependent
0 15
on first M in part (a), unless they restart
AnswerMarks Guidance
[= 2.53125] So total distance = 2  2.53125 = 5.06 mA1 81 1
Allow distance  5
16 16
If DM0 then SCB1 for 5.06 without working
6
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9709/41 Cambridge International AS & A Level – Mark Scheme May/June 2022
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2022 Page 5 of 13
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | For attempt at integration | *M1 | The power of t must increase by 1 with a change of coefficient in at
least 1 term.
Allow if k is omitted.
 1 
k t3 t4 C
 
 2  | A1 | Allow unsimplified
 1 
k 8 16 0
 
 2  | DM1 | For use of limits 0 and 2 or substituting t = 2
OR equate to 0, then solve a quartic equation in t.
Allow if k is omitted.
0 | A1 | AG, CWO including stating C = 0 if not using limits
4
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | v = 0 when k  3t2 2t3 0 | M1 | For solving for t
Leading to t = 1.5 [or t = 0] | A1 | A0 for other solutions not discarded
 6t6t2
a  k | M1 | For differentiation, the power of t must decrease by 1 with a
change of coefficient in at least 1 term.
Allow if k is omitted
[13.5 k  61.561.52 ⇒ 13.5 k913.5⇒] k = 3 | A1
Distance from t = 0 to t = 1.5 is
1.5
  1   1 
3 t3  t4 3 1.53  1.54 0
    
  2   2 
0 | DM1 | 1.5 2
For use of limits. 2  or 2 or both integrals. Dependent
0 15
on first M in part (a), unless they restart
[= 2.53125] So total distance = 2  2.53125 = 5.06 m | A1 | 81 1
Allow distance  5
16 16
If DM0 then SCB1 for 5.06 without working
6
Question | Answer | Marks | Guidance
A particle starts from a point $O$ and moves in a straight line. The velocity $v$ m s$^{-1}$ of the particle at time $t$ s after leaving $O$ is given by
$$v = k(3t^2 - 2t^3),$$
where $k$ is a constant.

\begin{enumerate}[label=(\alph*)]
\item Verify that the particle returns to $O$ when $t = 2$. [4]

\item It is given that the acceleration of the particle is $-13.5$ m s$^{-2}$ for the positive value of $t$ at which $v = 0$.

Find $k$ and hence find the total distance travelled in the first two seconds of motion. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2022 Q6 [10]}}
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