CAIE M1 2022 June — Question 3 7 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeHeavier particle hits ground, lighter continues upward - vertical strings
DifficultyStandard +0.3 This is a standard two-stage pulley problem requiring Newton's second law for connected particles, then energy/kinematics after one particle lands. The tension calculation is guided ('show that'), and the methods are routine M1 techniques with straightforward arithmetic. Slightly above average due to the two-stage nature, but well within typical M1 scope.
Spec3.03k Connected particles: pulleys and equilibrium6.02i Conservation of energy: mechanical energy principle
Two particles \(A\) and \(B\), of masses \(2.4\text{kg}\) and \(1.2\text{kg}\) respectively, are connected by a light inextensible string which passes over a fixed smooth pulley. \(A\) is held at a distance of \(2.1\text{m}\) above a horizontal plane and \(B\) is \(1.5\text{m}\) above the plane. The particles hang vertically and are released from rest. In the subsequent motion \(A\) reaches the plane and does not rebound and \(B\) does not reach the pulley.
  1. Show that the tension in the string before \(A\) reaches the plane is \(16\text{N}\) and find the magnitude of the acceleration of the particles before \(A\) reaches the plane. [4]
  2. Find the greatest height of \(B\) above the plane. [3]
Question 3:
AnswerMarks
3(a)2.4gT 2.4a
T 1.2g 1.2a
AnswerMarks Guidance
2.4g1.2g 2.41.2aM1 Attempt at Newton’s second law on either particle or the
system with correct number of terms; allow sign errors.
AnswerMarks
A1Any 2 consistent and correct
May have an a in opposite direction to our a
AnswerMarks Guidance
Attempt to solve for a or TM1 From equation(s) with correct number of relevant terms.
If g missing then M0A0M1A0, maximum1/4.
Must get a  or T 
Must not assume T 16. May attempt to verify a value of a
using T 16 in 2 equations
10
T 16 N and a ms-2
AnswerMarks Guidance
3A1 Both correct; allow a3.33. AG for T 16.
Assuming T 16 and only one equation is M1A0M0A0
maximum 1/4. Withhold A mark if T 15.9...16, but
condone T 1.23.331216 or
T 242.43.3316
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
3(b)10
v2 2 2.1 14 v 14 3.741
 
3
1
OR 2.4v2 2.4g2.1162.1
2
1 1
OR 2.4v2  1.2v2 2.4g2.11.2g2.1
AnswerMarks Guidance
2 2M1 Use of suvat or use energy to find v or v2, using their
a  g (unless 10 comes from their attempt at a) from (a),
s2.1
0142gs s
1  2
or 1.2 14 1.2gh  h
AnswerMarks Guidance
2M1 Attempt to use v2 u2 2as (or other complete method),
using a g , to find additional height after string slack,
using their v or v2.
AnswerMarks Guidance
s 1.52.10.7 4.3 mA1 AWRT 4.3(0);
Allow use of a3.33 to give s 4.29934.30
Allow use of v3.74 to give s 4.299384.30
Alternative for question 3(b) - using energy on particle B
AnswerMarks Guidance
162.11.2gHM1 Apply energy to B, 2 terms
H 2.8A1
s 1.52.8 4.3 mA1
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
--- 3(a) ---
3(a) | 2.4gT 2.4a
T 1.2g 1.2a
2.4g1.2g 2.41.2a | M1 | Attempt at Newton’s second law on either particle or the
system with correct number of terms; allow sign errors.
A1 | Any 2 consistent and correct
May have an a in opposite direction to our a
Attempt to solve for a or T | M1 | From equation(s) with correct number of relevant terms.
If g missing then M0A0M1A0, maximum1/4.
Must get a  or T 
Must not assume T 16. May attempt to verify a value of a
using T 16 in 2 equations
10
T 16 N and a ms-2
3 | A1 | Both correct; allow a3.33. AG for T 16.
Assuming T 16 and only one equation is M1A0M0A0
maximum 1/4. Withhold A mark if T 15.9...16, but
condone T 1.23.331216 or
T 242.43.3316
4
Question | Answer | Marks | Guidance
--- 3(b) ---
3(b) | 10
v2 2 2.1 14 v 14 3.741
 
3
1
OR 2.4v2 2.4g2.1162.1
2
1 1
OR 2.4v2  1.2v2 2.4g2.11.2g2.1
2 2 | M1 | Use of suvat or use energy to find v or v2, using their
a  g (unless 10 comes from their attempt at a) from (a),
s2.1
0142gs s
1  2
or 1.2 14 1.2gh  h
2 | M1 | Attempt to use v2 u2 2as (or other complete method),
using a g , to find additional height after string slack,
using their v or v2.
s 1.52.10.7 4.3 m | A1 | AWRT 4.3(0);
Allow use of a3.33 to give s 4.29934.30
Allow use of v3.74 to give s 4.299384.30
Alternative for question 3(b) - using energy on particle B
162.11.2gH | M1 | Apply energy to B, 2 terms
H 2.8 | A1
s 1.52.8 4.3 m | A1
3
Question | Answer | Marks | Guidance
Two particles $A$ and $B$, of masses $2.4\text{kg}$ and $1.2\text{kg}$ respectively, are connected by a light inextensible string which passes over a fixed smooth pulley. $A$ is held at a distance of $2.1\text{m}$ above a horizontal plane and $B$ is $1.5\text{m}$ above the plane. The particles hang vertically and are released from rest. In the subsequent motion $A$ reaches the plane and does not rebound and $B$ does not reach the pulley.

\begin{enumerate}[label=(\alph*)]
\item Show that the tension in the string before $A$ reaches the plane is $16\text{N}$ and find the magnitude of the acceleration of the particles before $A$ reaches the plane. [4]

\item Find the greatest height of $B$ above the plane. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2022 Q3 [7]}}
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