CAIE M1 2022 November — Question 6 12 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeEnergy methods for pulley systems
DifficultyStandard +0.3 This is a standard two-part pulley problem requiring routine application of Newton's laws (part a: resolving forces with friction in limiting equilibrium) and energy conservation (part b: straightforward PE to KE conversion). While it involves multiple steps and careful bookkeeping of forces/energy, it follows predictable M1 patterns with no novel insight required, making it slightly easier than average.
Spec3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys6.02i Conservation of energy: mechanical energy principle
\includegraphics{figure_6_1} **Fig. 6.1** Fig. 6.1 shows particles \(A\) and \(B\), of masses 4 kg and 3 kg respectively, attached to the ends of a light inextensible string that passes over a small smooth pulley. The pulley is fixed at the top of a plane which is inclined at an angle of 30° to the horizontal. \(A\) hangs freely below the pulley and \(B\) is on the inclined plane. The string is taut and the section of the string between \(B\) and the pulley is parallel to a line of greatest slope of the plane.
  1. It is given that the plane is rough and the particles are in limiting equilibrium. Find the coefficient of friction between \(B\) and the plane. [6]
  2. \includegraphics{figure_6_2} **Fig. 6.2** It is given instead that the plane is smooth and the particles are released from rest when the difference in the vertical heights of the particles is 1 m (see Fig. 6.2). Use an energy method to find the speed of the particles at the instant when the particles are at the same horizontal level. [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 October/November 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 16
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
6(a)T =4g B1
R=3gcos30B1
Attempt to resolve parallel to the planeM1 3 terms, allow g missing.
Allow sign errors, sin/cos mix.
AnswerMarks Guidance
F =T −3gsin30*A1 May see F =25.
Eliminate T and use F =R to get an equation in  onlyDM1 Where R is a component of their weight.
Coefficient of friction = 0.962A1 5 3
allow .
9
allow 0.96.
If F negative must say why using positive for
this mark.
6
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
6(b)Find height gained by B relative to height lost by A M1
y
OR B gains ym in height and A loses .
sin30
2
EITHER x+xsin30=1 x=
3
y 1
OR y+ =1 y =
AnswerMarks
sin30 3A1
1 1  1 
Change in KE = 4v2 + 3v2 = 7v2
 
AnswerMarks
2 2  2 B1
 y 
Change in PE ( 4gx−3gxsin30 ) or  4g −3gy  OR ( 4gx−3gy)
AnswerMarks Guidance
 sin30 B1 x or y need not be substituted.
Conservation of energy
1 1
4gx−3gxsin30= 4v2 + 3v2
2 2
y 1 1
OR 4g −3gy= 4v2 + 3v2
sin30 2 2
1 1
OR 4gx−3gy = 4v2 + 3v2
AnswerMarks Guidance
2 2M1 4 terms.
x or y need not be substituted.
Must be same v for both particles.
100 10 21
Speed = = =2.18 ms-1
AnswerMarks Guidance
21 21A1 2.182178902
SC B1 B1 M1 3/6 max for using x= y=0.5
AnswerMarks Guidance
QuestionAnswer Marks
6(b)Alternative method 1 for final 4 marks of question 6(b)
T −3gsin30=3a
4g−T =4a
AnswerMarks Guidance
4g−3gsin30=( 4+3 )aM1 Attempt at 2 equations from N2L on either
particle or the system. Allow sign errors.
Allow sin/cos mix.
Correct number of terms.
18
Solve to get T = g 25.7
AnswerMarks Guidance
7A1 5 25
May see a= g = 3.57
14 7
y 1 1
T = 3v2 +3gy OR 4gx=Tx+ 4v2
AnswerMarks Guidance
sin30 2 2M1 Attempt at work energy using their
T(4g or 3gsin30 ) .
May be in terms of x and/or y.
100 10 21
Speed = = =2.18 ms-1
AnswerMarks Guidance
21 21A1
QuestionAnswer Marks
6(b)Alternative method 2 for final 4 marks of question 6(b): Special case where constant acceleration assumed. Score maximum 4/6
Find height gained by B relative to height lost by AM1 A loses xm in height, B gains xsin30
y
OR B gains ym in height and A loses .
sin30
2
EITHER x+xsin30=1 x=
3
y 1
OR y+ =1 y =
AnswerMarks
sin30 3A1
25
T −3gsin30=3a and 4g−T =4a a= =3.57
7
25
OR 4g−3gsin30=( 4+3 )a a= =3.57
AnswerMarks
7B1
100 10 21
Uses constant acceleration to get speed = = =2.18 m s–1
AnswerMarks
21 21B1
6
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 October/November 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 16
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | T =4g | B1 | soi
R=3gcos30 | B1
Attempt to resolve parallel to the plane | M1 | 3 terms, allow g missing.
Allow sign errors, sin/cos mix.
F =T −3gsin30 | *A1 | May see F =25.
Eliminate T and use F =R to get an equation in  only | DM1 | Where R is a component of their weight.
Coefficient of friction = 0.962 | A1 | 5 3
allow .
9
allow 0.96.
If F negative must say why using positive for
this mark.
6
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | Find height gained by B relative to height lost by A | M1 | A loses xm in height, B gains xsin30
y
OR B gains ym in height and A loses .
sin30
2
EITHER x+xsin30=1 x=
3
y 1
OR y+ =1 y =
sin30 3 | A1
1 1  1 
Change in KE = 4v2 + 3v2 = 7v2
 
2 2  2  | B1
 y 
Change in PE ( 4gx−3gxsin30 ) or  4g −3gy  OR ( 4gx−3gy)
 sin30  | B1 | x or y need not be substituted.
Conservation of energy
1 1
4gx−3gxsin30= 4v2 + 3v2
2 2
y 1 1
OR 4g −3gy= 4v2 + 3v2
sin30 2 2
1 1
OR 4gx−3gy = 4v2 + 3v2
2 2 | M1 | 4 terms.
x or y need not be substituted.
Must be same v for both particles.
100 10 21
Speed = = =2.18 ms-1
21 21 | A1 | 2.182178902
SC B1 B1 M1 3/6 max for using x= y=0.5
Question | Answer | Marks | Guidance
6(b) | Alternative method 1 for final 4 marks of question 6(b)
T −3gsin30=3a
4g−T =4a
4g−3gsin30=( 4+3 )a | M1 | Attempt at 2 equations from N2L on either
particle or the system. Allow sign errors.
Allow sin/cos mix.
Correct number of terms.
18
Solve to get T = g 25.7
7 | A1 | 5 25
May see a= g = 3.57
14 7
y 1 1
T = 3v2 +3gy OR 4gx=Tx+ 4v2
sin30 2 2 | M1 | Attempt at work energy using their
T(4g or 3gsin30 ) .
May be in terms of x and/or y.
100 10 21
Speed = = =2.18 ms-1
21 21 | A1
Question | Answer | Marks | Guidance
6(b) | Alternative method 2 for final 4 marks of question 6(b): Special case where constant acceleration assumed. Score maximum 4/6
Find height gained by B relative to height lost by A | M1 | A loses xm in height, B gains xsin30
y
OR B gains ym in height and A loses .
sin30
2
EITHER x+xsin30=1 x=
3
y 1
OR y+ =1 y =
sin30 3 | A1
25
T −3gsin30=3a and 4g−T =4a a= =3.57
7
25
OR 4g−3gsin30=( 4+3 )a a= =3.57
7 | B1
100 10 21
Uses constant acceleration to get speed = = =2.18 m s–1
21 21 | B1
6
\includegraphics{figure_6_1}

**Fig. 6.1**

Fig. 6.1 shows particles $A$ and $B$, of masses 4 kg and 3 kg respectively, attached to the ends of a light inextensible string that passes over a small smooth pulley. The pulley is fixed at the top of a plane which is inclined at an angle of 30° to the horizontal. $A$ hangs freely below the pulley and $B$ is on the inclined plane. The string is taut and the section of the string between $B$ and the pulley is parallel to a line of greatest slope of the plane.

\begin{enumerate}[label=(\alph*)]
\item It is given that the plane is rough and the particles are in limiting equilibrium.

Find the coefficient of friction between $B$ and the plane. [6]

\item \includegraphics{figure_6_2}

**Fig. 6.2**

It is given instead that the plane is smooth and the particles are released from rest when the difference in the vertical heights of the particles is 1 m (see Fig. 6.2).

Use an energy method to find the speed of the particles at the instant when the particles are at the same horizontal level. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2022 Q6 [12]}}
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