Question 6 - A-Level Maths

CAIE M1 2023 March — Question 6 9 marks

Exam Board	CAIE
Module	M1 (Mechanics 1)
Year	2023
Session	March
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Motion on a slope
Type	String at angle to slope
Difficulty	Standard +0.3 This is a standard mechanics problem involving forces on an inclined plane with a rope at an angle. It requires resolving forces in two directions and applying F=ma, which are routine M1 techniques. The multi-part structure and need to find the normal reaction force add some complexity, but the problem follows a predictable template with no novel insight required, making it slightly easier than average.
Spec	3.03e Resolve forces: two dimensions 3.03m Equilibrium: sum of resolved forces = 0 3.03v Motion on rough surface: including inclined planes

\includegraphics{figure_6} A block \(B\), of mass 2 kg, lies on a rough inclined plane sloping at \(30°\) to the horizontal. A light rope, inclined at an angle of \(20°\) above a line of greatest slope, is attached to \(B\). The tension in the rope is \(T\) N. There is a friction force of \(F\) N acting on \(B\) (see diagram). The coefficient of friction between \(B\) and the plane is \(\mu\).

It is given that \(F = 5\) and that the acceleration of \(B\) up the plane is \(1.2\,\text{m}\,\text{s}^{-2}\).
1. Find the value of \(T\). [3]
2. Find the value of \(\mu\). [3]
It is given instead that \(\mu = 0.8\) and \(T = 15\). Determine whether \(B\) will move up the plane. [3]

Show mark scheme Show mark scheme source

Question 6:

Answer	Marks
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

9709/42 Cambridge International AS & A Level – Mark Scheme February/March 2023

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 2023 Page 5 of 19

Answer	Marks	Guidance
Question	Answer	Marks

Answer	Marks	Guidance
6(a)(i)	Attempt to resolve parallel to the plane	M1
Tcos20−5−2gsin30=21.2	A1	Correct equation.
T =18.5	A1	awrt 18.5 .

3

Answer	Marks	Guidance
Question	Answer	Marks

Answer	Marks	Guidance
6(a)(ii)	Attempt resolve perpendicular to the plane
R=2g cos30−Tsin20	M1	3 terms; allow sin/cos mix: allow sign errors; allow with T or

their T ; allow g missing.

Use of 5 =R to get an equation in  only

 5=(2g cos30−Tsin20)

Answer	Marks	Guidance
	M1	Where R is a two term expression with a component of 2g and

a component of their T ; allow g missing.

Answer	Marks	Guidance
=0.455	A1	5

awrt 0.455; allow 0.46 or 0.45; do not allow .

11

3

Answer	Marks	Guidance
6(b)	Max F =0.8(2g cos30−15sin20)
=0.812.1902=9.7521	*B1
Net force up the plane =15cos20−2gsin30 =4.0953	*B1	15cos20−2gsin30−0.8(2g cos30−15sin20)

OR

=2a   −5.6567=2a  a=−2.8283..

If max F incorrect and use F = ma then allow B1 for

15cos20−2gsin30−theirmaxF.

[State 4.09539.7521 ,] hence the block does not move

Answer	Marks	Guidance
[up the plane]	DB1	Must have correct values (to at least 1 sf) to compare for this

mark. No incorrect statement seen.

Alternative Method 1 for Question 6(b)

Max force down plane=0.8(2g cos30−15sin20)+2gsin30

Answer	Marks	Guidance
=0.812.1902+10=19.7521	*B1
Force up plane=15cos20 =14.0953	*B1	i.e. using it to compare with their max force down the plane.

[State 14.095319.7521 ,] hence the block does not move

Answer	Marks	Guidance
[up the plane]	DB1	Must have correct values (to at least 1 sf) to compare for this

mark. No incorrect statement seen.

Answer	Marks	Guidance
Question	Answer	Marks
6(b)	Alternative Method 2 for Question 6(b)

F =15cos20−2gsin30 =4.9053

Answer	Marks
Or R=2g cos30−15sin20 =12.1902	*B1

15cos20−2gsin30

Get = =0.3359

Answer	Marks
2g cos30−15sin20	*B1

[State 0.33590.8,] hence the block does not move [up the

Answer	Marks	Guidance
plane]	DB1	Must have correct value of  (to at least 1 sf) to compare for

this mark. No incorrect statement seen.

3

Answer	Marks	Guidance
Question	Answer	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
9709/42 Cambridge International AS & A Level – Mark Scheme February/March 2023
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 2023 Page 5 of 19
Question | Answer | Marks | Guidance
--- 6(a)(i) ---
6(a)(i) | Attempt to resolve parallel to the plane | M1 | 4 terms; allow sin/cos mix; allow sign errors; allow g missing.
Tcos20−5−2gsin30=21.2 | A1 | Correct equation.
T =18.5 | A1 | awrt 18.5 .
3
Question | Answer | Marks | Guidance
--- 6(a)(ii) ---
6(a)(ii) | Attempt resolve perpendicular to the plane
R=2g cos30−Tsin20 | M1 | 3 terms; allow sin/cos mix: allow sign errors; allow with T or
their T ; allow g missing.
Use of 5 =R to get an equation in  only
 5=(2g cos30−Tsin20)
 | M1 | Where R is a two term expression with a component of 2g and
a component of their T ; allow g missing.
=0.455 | A1 | 5
awrt 0.455; allow 0.46 or 0.45; do not allow .
11
3
--- 6(b) ---
6(b) | Max F =0.8(2g cos30−15sin20)
=0.812.1902=9.7521 | *B1
Net force up the plane =15cos20−2gsin30 =4.0953 | *B1 | 15cos20−2gsin30−0.8(2g cos30−15sin20)
OR
=2a   −5.6567=2a  a=−2.8283..
If max F incorrect and use F = ma then allow B1 for
15cos20−2gsin30−theirmaxF.
[State 4.09539.7521 ,] hence the block does not move
[up the plane] | DB1 | Must have correct values (to at least 1 sf) to compare for this
mark. No incorrect statement seen.
Alternative Method 1 for Question 6(b)
Max force down plane=0.8(2g cos30−15sin20)+2gsin30
=0.812.1902+10=19.7521 | *B1
Force up plane=15cos20 =14.0953 | *B1 | i.e. using it to compare with their max force down the plane.
[State 14.095319.7521 ,] hence the block does not move
[up the plane] | DB1 | Must have correct values (to at least 1 sf) to compare for this
mark. No incorrect statement seen.
Question | Answer | Marks | Guidance
6(b) | Alternative Method 2 for Question 6(b)
F =15cos20−2gsin30 =4.9053
Or R=2g cos30−15sin20 =12.1902 | *B1
15cos20−2gsin30
Get = =0.3359
2g cos30−15sin20 | *B1
[State 0.33590.8,] hence the block does not move [up the
plane] | DB1 | Must have correct value of  (to at least 1 sf) to compare for
this mark. No incorrect statement seen.
3
Question | Answer | Marks | Guidance

Show LaTeX source

\includegraphics{figure_6}

A block $B$, of mass 2 kg, lies on a rough inclined plane sloping at $30°$ to the horizontal. A light rope, inclined at an angle of $20°$ above a line of greatest slope, is attached to $B$. The tension in the rope is $T$ N. There is a friction force of $F$ N acting on $B$ (see diagram). The coefficient of friction between $B$ and the plane is $\mu$.

\begin{enumerate}[label=(\alph*)]
\item It is given that $F = 5$ and that the acceleration of $B$ up the plane is $1.2\,\text{m}\,\text{s}^{-2}$.
\begin{enumerate}[label=(\roman*)]
\item Find the value of $T$. [3]
\item Find the value of $\mu$. [3]
\end{enumerate}

\item It is given instead that $\mu = 0.8$ and $T = 15$.

Determine whether $B$ will move up the plane. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2023 Q6 [9]}}

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