CAIE M1 2024 June — Question 7 10 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeParticle on slope with pulley
DifficultyStandard +0.3 This is a standard two-particle connected system problem requiring resolution of forces, Newton's second law, and energy methods. Part (a) involves routine force resolution on an inclined plane with friction (a core M1 skill), while part (b) applies conservation of energy with clear geometric setup. The 10 marks reflect multiple steps rather than conceptual difficulty, and all techniques are standard textbook exercises with no novel insight required.
Spec3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle
\includegraphics{figure_7} Two particles \(P\) and \(Q\) of masses 2.5 kg and 0.5 kg respectively are connected by a light inextensible string that passes over a small smooth pulley fixed at the top of a plane inclined at an angle of \(30°\) to the horizontal. Particle \(P\) is on the plane and \(Q\) hangs below the pulley such that the level of \(Q\) is 2 m below the level of \(P\) (see diagram). Particle \(P\) is released from rest with the string taut and slides down the plane. The plane is rough with coefficient of friction 0.2 between the plane and \(P\).
  1. Find the acceleration of \(P\). [5]
  2. Use an energy method to find the speed of the particles at the instant when they are at the same vertical height. [5]
Question 7:
AnswerMarks
7(a) 25 3 
R2.5gcos30 21.65063509
2
AnswerMarks Guidance
 B1 5 3
Note: F 0.5gcos30 4.330127019.
2
AnswerMarks Guidance
Attempt at Newton’s second law*M1 Correct number of dimensionally correct/relevant terms;
allow sign errors; allow sin/cos mix.
Using this twice to get equations for P and Q; allow
different T’s (equations with 0.5 and 2.5).
Using once to get a system equation (equation with
0.52.5).
EITHER: T 0.5g 0.5a AND 2.5gsin30F T 2.5a
AnswerMarks Guidance
OR: 2.5gsin30F0.5g2.50.5aA1 EITHER: Both correct; allow their F; must be the same T.
OR: correct system equation; allow their F.
Use F 0.2R to get an equation in a only
2.5gsin300.22.5gcos300.5g 2.50.5a
AnswerMarks Guidance
 DM1 Where R is a component of weight of P only; from
equation(s) with the correct number of dimensionally
correct/relevant terms.
AnswerMarks Guidance
a1.06ms–2A1 155 3
Allow .
6
1.05662433.
AWRT 1.06 from correct work.
5
AnswerMarks Guidance
QuestionAnswer Marks
4
xsin302x   x
3
2
OR 2 ysin30 y   y
 
3
1 1 
OR 1.056t2  1.056t2sin302 t1.5886
 
2 2 
 1 1.0561.5886..2  2
x  sin30 
 
 2  3
 1 1.0561.5886..2 4
OR  y  
 
AnswerMarks Guidance
 2  3*B1 Where x is the distance P moves down the plane or Q
moves vertically upwards, or where y is the vertical
distance of Q below P’s starting point.
Allow x1.3or better; allow y0.67 or better.
 3 
Change in PE 2.5gtheir x sin300.5gtheir x   gtheir x 
 
 4 
OR 2.5g2their y sin300.5g2their y 
 3their y 1g
AnswerMarks Guidance
 B1 Using their x 2 or 1 or 0 , 0 x2;
or theiry 2 or 1 or 0 , 0 y2.
WD against friction 0.22.5gcos30their x  4.33their x
 
AnswerMarks Guidance
OR WD against friction 0.22.5gcos302their y B1 Using their x 2 or 1 or 0 , 0 x2;
or theiry 2 or 1 or 0 , 0 y2.
1 1
2.5v2  0.5v2 
2 2
4 4  4 
2.5g sin300.5g 0.22.5gcos30
     
3 3  3 
1 1
OR 2.5v2  0.5v2 
2 2
 2   2   2 
2.5g 2 sin300.5g 2 0.22.5gcos30 2
     
AnswerMarks Guidance
 3   3   3 DM1 Attempt at work energy equation; dimensionally correct; 5
relevant terms; allow sign errors; allow sin/cos mix.
Must be using correct values of x or y.
AnswerMarks Guidance
QuestionAnswer Marks
v1.68A1 1.67859014
AWRT 1.68 from correct work.
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(b)Alternative Method for Question 7(b): Considering energy on Q only
Must be using tension and mass 0.5kg only to be awarded the last 4 marks
4
xsin302x   x
3
2
OR 2 ysin30 y   y
 
3
1 1 
OR 1.056t2  1.056t2sin302 t1.5886.
 
2 2 
 1 1.0561.5886..2  2
x  sin30 
 
 2  3
 1 1.0561.5886..2 4
OR  y  
 
AnswerMarks Guidance
 2  3(*B1) Where x is the distance P moves down the plane or Q
moves vertically upwards, or where y is the vertical
distance of Q below P’s starting point.
Allow x1.3or better; allow y0.67 or better.
Change in PE 0.5gtheir x 
AnswerMarks Guidance
OR Change in PE 0.5g2their y (B1) Using their x 2 or 1 or 0 , 0 x2;
or theiry 2 or 1 or 0 , 0 y2.
WD by tension their 5.528312164their x 
AnswerMarks Guidance
OR WD by tension their 5.5283121642their y (B1) Using their tension from 7(a) from equation(s) with the
correct number of dimensionally correct/relevant terms.
Using their x 2 or 1 or 0 , 0 x2;
or theiry 2 or 1 or 0 , 0 y2.
4  1  4 
0.5g  0.5v2 their 5.528312164
   
3  2  3 
 2  1  2 
OR 0.5g  2   0.5v2 their 5.528312164  2 
AnswerMarks Guidance
 3  2  3 (DM1) Using their tension from 7(a) from equation(s) with the
correct number of dimensionally correct/relevant terms.
Attempt at work energy equation; dimensionally correct; 3
relevant terms; allow sign errors.
Must be using correct values of x or y.
AnswerMarks Guidance
v1.68(A1) 1.67859014
AWRT 1.68 from correct work.
AnswerMarks Guidance
QuestionAnswer Marks
7(b)Alternative Method 2 for Question 7(b): Considering energy on P only
Note: must be using tension and mass 2.5kg only to be awarded the last 4 marks
4 2
xsin302x   x OR 2 ysin30 y   y
 
3 3
1 1 
OR 1.056t2  1.056t2sin302 t1.5886.
 
2 2 
 1 1.0561.5886..2  2
x  sin30 
 
 2  3
 1 1.0561.5886..2 4
OR  y  
 
AnswerMarks Guidance
 2  3(*B1) Where x is the distance P moves down the plane or Q
moves vertically upwards, or where y is the vertical
distance of Q below P’s starting point.
Allow x1.3or better; allow y0.67 or better.
AnswerMarks Guidance
Change in PE 2.5gtheir x sin30 OR 2.5gtheir y(B1) Using their x 2 or 1 or 0 , 0 x2;
or theiry 2 or 1 or 0 , 0 y2.
WD by tension their 5.528312164their x 
AnswerMarks Guidance
OR WD against friction 0.22.5gcos30their x (B1) Using their tension from 7(a) from equation(s) with the
correct number of dimensionally correct/relevant terms.
Using their x 2 or 1 or 0 , 0 x2;
or theiry 2 or 1 or 0 , 0 y2.
4  1
2.5g sin30 2.5v2 
 
3  2
4  4 
0.22.5gcos30 their 5.528312164
   
AnswerMarks Guidance
3  3 (DM1) Using their tension from 7(a) from equation(s) with the
correct number of dimensionally correct/relevant terms.
Attempt at work energy equation; dimensionally correct; 4
relevant terms; allow sign errors; allow sin/cos mix.
Must be using correct values of x or y.
AnswerMarks Guidance
v1.68(A1) 1.67859014
AWRT 1.68 from correct work.
AnswerMarks Guidance
QuestionAnswer Marks
7(b)Special Case for using constant acceleration: Maximum 2 marks
4
xsin302x   x
3
2
OR 2 ysin30 y   y
 
3
1 1 
OR 1.056t2  1.056t2sin302 t1.5886.
 
2 2 
 1 1.0561.5886..2  2
x  sin30 
 
 2  3
 1 1.0561.5886..2 4
OR  y  
 
AnswerMarks Guidance
 2  3(B1) Where x is the distance P moves down the plane or Q
moves vertically upwards, or where y is the vertical
distance of Q below P’s starting point.
Allow x1.3or better; allow y0.67 or better.
 4 
v2 21.06  v1.68
 
AnswerMarks Guidance
 3 (B1) 1.67859014
AWRT 1.68 from correct work.
5
Question 7:
--- 7(a) ---
7(a) |  25 3 
R2.5gcos30 21.65063509
2
  | B1 | 5 3
Note: F 0.5gcos30 4.330127019.
2
Attempt at Newton’s second law | *M1 | Correct number of dimensionally correct/relevant terms;
allow sign errors; allow sin/cos mix.
Using this twice to get equations for P and Q; allow
different T’s (equations with 0.5 and 2.5).
Using once to get a system equation (equation with
0.52.5).
EITHER: T 0.5g 0.5a AND 2.5gsin30F T 2.5a
OR: 2.5gsin30F0.5g2.50.5a | A1 | EITHER: Both correct; allow their F; must be the same T.
OR: correct system equation; allow their F.
Use F 0.2R to get an equation in a only
2.5gsin300.22.5gcos300.5g 2.50.5a
  | DM1 | Where R is a component of weight of P only; from
equation(s) with the correct number of dimensionally
correct/relevant terms.
a1.06ms–2 | A1 | 155 3
Allow .
6
1.05662433.
AWRT 1.06 from correct work.
5
Question | Answer | Marks | Guidance
4
xsin302x   x
3
2
OR 2 ysin30 y   y
 
3
1 1 
OR 1.056t2  1.056t2sin302 t1.5886
 
2 2 
 1 1.0561.5886..2  2
x  sin30 
 
 2  3
 1 1.0561.5886..2 4
OR  y  
 
 2  3 | *B1 | Where x is the distance P moves down the plane or Q
moves vertically upwards, or where y is the vertical
distance of Q below P’s starting point.
Allow x1.3or better; allow y0.67 or better.
 3 
Change in PE 2.5gtheir x sin300.5gtheir x   gtheir x 
 
 4 
OR 2.5g2their y sin300.5g2their y 
 3their y 1g
  | B1 | Using their x 2 or 1 or 0 , 0 x2;
or theiry 2 or 1 or 0 , 0 y2.
WD against friction 0.22.5gcos30their x  4.33their x
 
OR WD against friction 0.22.5gcos302their y  | B1 | Using their x 2 or 1 or 0 , 0 x2;
or theiry 2 or 1 or 0 , 0 y2.
1 1
2.5v2  0.5v2 
2 2
4 4  4 
2.5g sin300.5g 0.22.5gcos30
     
3 3  3 
1 1
OR 2.5v2  0.5v2 
2 2
 2   2   2 
2.5g 2 sin300.5g 2 0.22.5gcos30 2
     
 3   3   3  | DM1 | Attempt at work energy equation; dimensionally correct; 5
relevant terms; allow sign errors; allow sin/cos mix.
Must be using correct values of x or y.
Question | Answer | Marks | Guidance
v1.68 | A1 | 1.67859014
AWRT 1.68 from correct work.
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | Alternative Method for Question 7(b): Considering energy on Q only
Must be using tension and mass 0.5kg only to be awarded the last 4 marks
4
xsin302x   x
3
2
OR 2 ysin30 y   y
 
3
1 1 
OR 1.056t2  1.056t2sin302 t1.5886.
 
2 2 
 1 1.0561.5886..2  2
x  sin30 
 
 2  3
 1 1.0561.5886..2 4
OR  y  
 
 2  3 | (*B1) | Where x is the distance P moves down the plane or Q
moves vertically upwards, or where y is the vertical
distance of Q below P’s starting point.
Allow x1.3or better; allow y0.67 or better.
Change in PE 0.5gtheir x 
OR Change in PE 0.5g2their y  | (B1) | Using their x 2 or 1 or 0 , 0 x2;
or theiry 2 or 1 or 0 , 0 y2.
WD by tension their 5.528312164their x 
OR WD by tension their 5.5283121642their y  | (B1) | Using their tension from 7(a) from equation(s) with the
correct number of dimensionally correct/relevant terms.
Using their x 2 or 1 or 0 , 0 x2;
or theiry 2 or 1 or 0 , 0 y2.
4  1  4 
0.5g  0.5v2 their 5.528312164
   
3  2  3 
 2  1  2 
OR 0.5g  2   0.5v2 their 5.528312164  2 
 3  2  3  | (DM1) | Using their tension from 7(a) from equation(s) with the
correct number of dimensionally correct/relevant terms.
Attempt at work energy equation; dimensionally correct; 3
relevant terms; allow sign errors.
Must be using correct values of x or y.
v1.68 | (A1) | 1.67859014
AWRT 1.68 from correct work.
Question | Answer | Marks | Guidance
7(b) | Alternative Method 2 for Question 7(b): Considering energy on P only
Note: must be using tension and mass 2.5kg only to be awarded the last 4 marks
4 2
xsin302x   x OR 2 ysin30 y   y
 
3 3
1 1 
OR 1.056t2  1.056t2sin302 t1.5886.
 
2 2 
 1 1.0561.5886..2  2
x  sin30 
 
 2  3
 1 1.0561.5886..2 4
OR  y  
 
 2  3 | (*B1) | Where x is the distance P moves down the plane or Q
moves vertically upwards, or where y is the vertical
distance of Q below P’s starting point.
Allow x1.3or better; allow y0.67 or better.
Change in PE 2.5gtheir x sin30 OR 2.5gtheir y | (B1) | Using their x 2 or 1 or 0 , 0 x2;
or theiry 2 or 1 or 0 , 0 y2.
WD by tension their 5.528312164their x 
OR WD against friction 0.22.5gcos30their x  | (B1) | Using their tension from 7(a) from equation(s) with the
correct number of dimensionally correct/relevant terms.
Using their x 2 or 1 or 0 , 0 x2;
or theiry 2 or 1 or 0 , 0 y2.
4  1
2.5g sin30 2.5v2 
 
3  2
4  4 
0.22.5gcos30 their 5.528312164
   
3  3  | (DM1) | Using their tension from 7(a) from equation(s) with the
correct number of dimensionally correct/relevant terms.
Attempt at work energy equation; dimensionally correct; 4
relevant terms; allow sign errors; allow sin/cos mix.
Must be using correct values of x or y.
v1.68 | (A1) | 1.67859014
AWRT 1.68 from correct work.
Question | Answer | Marks | Guidance
7(b) | Special Case for using constant acceleration: Maximum 2 marks
4
xsin302x   x
3
2
OR 2 ysin30 y   y
 
3
1 1 
OR 1.056t2  1.056t2sin302 t1.5886.
 
2 2 
 1 1.0561.5886..2  2
x  sin30 
 
 2  3
 1 1.0561.5886..2 4
OR  y  
 
 2  3 | (B1) | Where x is the distance P moves down the plane or Q
moves vertically upwards, or where y is the vertical
distance of Q below P’s starting point.
Allow x1.3or better; allow y0.67 or better.
 4 
v2 21.06  v1.68
 
 3  | (B1) | 1.67859014
AWRT 1.68 from correct work.
5
\includegraphics{figure_7}

Two particles $P$ and $Q$ of masses 2.5 kg and 0.5 kg respectively are connected by a light inextensible string that passes over a small smooth pulley fixed at the top of a plane inclined at an angle of $30°$ to the horizontal. Particle $P$ is on the plane and $Q$ hangs below the pulley such that the level of $Q$ is 2 m below the level of $P$ (see diagram).

Particle $P$ is released from rest with the string taut and slides down the plane. The plane is rough with coefficient of friction 0.2 between the plane and $P$.

\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of $P$. [5]

\item Use an energy method to find the speed of the particles at the instant when they are at the same vertical height. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2024 Q7 [10]}}
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