CAIE M1 2022 June — Question 6 10 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeParticle on smooth incline, particle hanging
DifficultyStandard +0.3 This is a standard M1 pulley system problem with two particles on inclines. Part (a) involves routine application of Newton's second law to connected particles on a smooth plane—a textbook exercise requiring F=ma for both particles and solving simultaneous equations. Part (b) adds friction and equilibrium conditions but remains methodical: resolve forces, apply limiting friction, and solve for μ. While it requires careful bookkeeping of forces and angles, it follows standard procedures without requiring novel insight or particularly challenging problem-solving, making it slightly easier than the average A-level question.
Spec3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes

  1. It is given that the plane \(B C\) is smooth and that the particles are released from rest. Find the tension in the string and the magnitude of the acceleration of the particles.
  2. It is given instead that the plane \(B C\) is rough. A force of magnitude 3 N is applied to \(Q\) directly up the plane along a line of greatest slope of the plane. Find the least value of the coefficient of friction between \(Q\) and the plane \(B C\) for which the particles remain at rest.

Question 6:
Part (a):
AnswerMarks Guidance
For \(P\): \(0.3g\sin 60 - T = 0.3a\)A1 For any one equation
For \(Q\): \(T + 0.2g\sin 30 = 0.2a\); System: \(0.3g\sin 60 + 0.2g\sin 30 = 0.5a\)A1 For any second equation
\(\dfrac{0.3g\sin 60 - T}{0.3} = \dfrac{0.2g\sin 30 + T}{0.2}\)
\(0.3g\sin 60 + 0.2g\sin 30 = 0.5a \quad a = \ldots\)M1 For solving for \(a\) or \(T\)
Magnitude of acceleration \(= 7.20 \text{ ms}^{-2}\), Tension \(= 0.439 \text{ N}\)A1 Use of Newton's Second Law (attempt) M1
Part (b):
AnswerMarks Guidance
\(R = 0.2g\cos 30\)B1
\([0.3g\sin 60 - T = 0] \quad T = \dfrac{3\sqrt{3}}{2}\) or \(T = 2.598\ldots\)B1 Equilibrium for \(P\)
\(T + 0.2g\sin 30 - F - 3 = 0\)M1 Equilibrium for \(Q\) on the point of moving down
\(\dfrac{3\sqrt{3}}{2} + 0.2g\sin 30 - \mu(0.2g\cos 30) - 3 = 0\)M1 Use of \(F = \mu R\)
\(\mu = 0.345\)A1
## Question 6:

### Part (a):
For $P$: $0.3g\sin 60 - T = 0.3a$ | A1 | For any one equation

For $Q$: $T + 0.2g\sin 30 = 0.2a$; System: $0.3g\sin 60 + 0.2g\sin 30 = 0.5a$ | A1 | For any second equation

$\dfrac{0.3g\sin 60 - T}{0.3} = \dfrac{0.2g\sin 30 + T}{0.2}$ | | |

$0.3g\sin 60 + 0.2g\sin 30 = 0.5a \quad a = \ldots$ | M1 | For solving for $a$ or $T$

Magnitude of acceleration $= 7.20 \text{ ms}^{-2}$, Tension $= 0.439 \text{ N}$ | A1 | Use of Newton's Second Law (attempt) M1

### Part (b):
$R = 0.2g\cos 30$ | B1 |

$[0.3g\sin 60 - T = 0] \quad T = \dfrac{3\sqrt{3}}{2}$ or $T = 2.598\ldots$ | B1 | Equilibrium for $P$

$T + 0.2g\sin 30 - F - 3 = 0$ | M1 | Equilibrium for $Q$ on the point of moving down

$\dfrac{3\sqrt{3}}{2} + 0.2g\sin 30 - \mu(0.2g\cos 30) - 3 = 0$ | M1 | Use of $F = \mu R$

$\mu = 0.345$ | A1 |

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\begin{enumerate}[label=(\alph*)]
\item It is given that the plane $B C$ is smooth and that the particles are released from rest.

Find the tension in the string and the magnitude of the acceleration of the particles.
\item It is given instead that the plane $B C$ is rough. A force of magnitude 3 N is applied to $Q$ directly up the plane along a line of greatest slope of the plane.

Find the least value of the coefficient of friction between $Q$ and the plane $B C$ for which the particles remain at rest.
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2022 Q6 [10]}}