CAIE M1 2018 November — Question 4 8 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2018
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeApplied force in addition to weights
DifficultyStandard +0.3 This is a standard A-level mechanics pulley problem with friction on an inclined plane. Part (i) requires resolving forces, finding limiting friction, and applying equilibrium conditions—routine techniques. Part (ii) involves using F=ma with the constraint that the string is inextensible. While it requires careful bookkeeping of forces and signs, it follows standard methods taught in M1 with no novel insight required. Slightly easier than average due to being a textbook-style question with clear structure.
Spec3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes

\includegraphics{figure_4} Two particles \(P\) and \(Q\), of masses \(0.4\) kg and \(0.7\) kg respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley which is attached to the edge of a rough plane. The coefficient of friction between \(P\) and the plane is \(0.5\). The plane is inclined at an angle \(\alpha\) to the horizontal, where \(\tan \alpha = \frac{3}{4}\). Particle \(P\) lies on the plane and particle \(Q\) hangs vertically. The string between \(P\) and the pulley is parallel to a line of greatest slope of the plane (see diagram). A force of magnitude \(X\) N, acting directly down the plane, is applied to \(P\).
  1. Show that the greatest value of \(X\) for which \(P\) remains stationary is \(6.2\). [4]
  2. Given instead that \(X = 0.8\), find the acceleration of \(P\). [4]

Question 4:

AnswerMarks Guidance
4(i)T = 0.7g B1
R = 0.4g × 4/ [ = 16/ = 3.2]
AnswerMarks Guidance
5 5B1 Normal reaction on particle P
[X + 0.4g × 3/ – F – T = 0]
AnswerMarks Guidance
5M1 Attempt to resolve forces along the plane
X = 6.2A1 AG
4

AnswerMarks
4(ii)[0.7g – T = 0.7a]
[T – 0.8 – 0.4g × 3/ – F= 0.4a]
5
[0.7g – 0.8 – 0.4g × 3/ – F = (0.7 + 0.4)a] System
AnswerMarks Guidance
5M1 For using Newton’s 2nd law for both particle P and particle Q or the
system equation
AnswerMarks
A1Both equations correct or system equation correct
M1Solve either the system equation or solve two simultaneous equations to
find a
AnswerMarks
a = 2 m s–2A1
4
AnswerMarks Guidance
QuestionAnswer Marks
Question 4:
--- 4(i) ---
4(i) | T = 0.7g | B1
R = 0.4g × 4/ [ = 16/ = 3.2]
5 5 | B1 | Normal reaction on particle P
[X + 0.4g × 3/ – F – T = 0]
5 | M1 | Attempt to resolve forces along the plane
X = 6.2 | A1 | AG
4
--- 4(ii) ---
4(ii) | [0.7g – T = 0.7a]
[T – 0.8 – 0.4g × 3/ – F= 0.4a]
5
[0.7g – 0.8 – 0.4g × 3/ – F = (0.7 + 0.4)a] System
5 | M1 | For using Newton’s 2nd law for both particle P and particle Q or the
system equation
A1 | Both equations correct or system equation correct
M1 | Solve either the system equation or solve two simultaneous equations to
find a
a = 2 m s–2 | A1
4
Question | Answer | Marks | Guidance
\includegraphics{figure_4}

Two particles $P$ and $Q$, of masses $0.4$ kg and $0.7$ kg respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley which is attached to the edge of a rough plane. The coefficient of friction between $P$ and the plane is $0.5$. The plane is inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac{3}{4}$. Particle $P$ lies on the plane and particle $Q$ hangs vertically. The string between $P$ and the pulley is parallel to a line of greatest slope of the plane (see diagram). A force of magnitude $X$ N, acting directly down the plane, is applied to $P$.

\begin{enumerate}[label=(\roman*)]
\item Show that the greatest value of $X$ for which $P$ remains stationary is $6.2$. [4]

\item Given instead that $X = 0.8$, find the acceleration of $P$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2018 Q4 [8]}}