CAIE M1 2018 June — Question 7 12 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2018
SessionJune
Marks12
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 on inclined planes with straightforward resolution of forces and energy methods. Part (i) uses either Newton's second law or energy conservation with given angles and masses. Part (ii) requires setting up limiting equilibrium with friction forces, but the symmetry of having the same coefficient μ on both faces simplifies the algebra. The angles (45° and 30°) yield clean trigonometric values, and the problem follows a well-established template for M1 mechanics questions.
Spec3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution3.03r Friction: concept and vector form3.03u Static equilibrium: on rough surfaces6.02i Conservation of energy: mechanical energy principle
\includegraphics{figure_7} The diagram shows a triangular block with sloping faces inclined to the horizontal at \(45°\) and \(30°\). Particle \(A\) of mass \(0.8 \text{ kg}\) lies on the face inclined at \(45°\) and particle \(B\) of mass \(1.2 \text{ kg}\) lies on the face inclined at \(30°\). The particles are connected by a light inextensible string which passes over a small smooth pulley \(P\) fixed at the top of the faces. The parts \(AP\) and \(BP\) of the string are parallel to lines of greatest slope of the respective faces. The particles are released from rest with both parts of the string taut. In the subsequent motion neither particle reaches the pulley and neither particle reaches the bottom of a face.
  1. Given that both faces are smooth, find the speed of \(A\) after each particle has travelled a distance of \(0.4 \text{ m}\). [6]
  2. It is given instead that both faces are rough. The coefficient of friction between each particle and a face of the block is \(\mu\). Find the value of \(\mu\) for which the system is in limiting equilibrium. [6]
Question 7:
AnswerMarks
7(i)A T – 0.8 g sin 45 = 0.8a
B 1.2g sin 30 – T = 1.2a
AnswerMarks Guidance
System 1.2 g sin 30 – 0.8 g sin 45 = 2aM1 Apply Newton 2nd law to either A or to B
or to the system
AnswerMarks Guidance
A1One correct equation
A1A second correct equation
a = 0.171M1 Solve for a
v2 = 2 × a × 0.4M1 Use v2 = u2 + 2as with u = 0
v = 0.370 so speed of A is 0.370 ms–1A1
6
Alternative scheme for Question 7(i)
AnswerMarks
M1Attempt KE gain or PE loss
1 1
KE gain = × 0.8 × v2 + × 1.2 × v2
AnswerMarks Guidance
2 2A1 v is the required speed of A
PE loss =
AnswerMarks
1.2 g × 0.4 sin 30 – 0.8 g × 0.4 sin 45A1
1 1
× 0.8 × v2 + × 1.2 × v2 =
2 2
AnswerMarks Guidance
1.2 g × 0.4 sin 30 – 0.8 g × 0.4 sin 45M1 4 term energy equation
M1Solving for v
v = 0.370 so speed of A is 0.370 ms–1A1
QuestionAnswer Marks
AnswerMarks
7(ii)R = 0.8gcos45=4 2
A
R = 1.2gcos30=6 3
AnswerMarks Guidance
BB1 For either R or R
A B
F = 4 2 µ and F = 6 3µ
AnswerMarks Guidance
A BM1 Either F or F used
A B
A 0.8 g sin 45 + F = T
A
B 1.2 g sin 30 – F = T
B
or system equation:
12 sin 30 – 8 sin 45 = F + F
AnswerMarks Guidance
A BM1 Resolve parallel to the plane either for
both particles A and B or for the system
equation
AnswerMarks
Correct equation(s)A1
M1Eliminate T and solve for µ
( )
6−4 2
µ = ( )
6 3+4√2
AnswerMarks
= 0.0214A1
6
Question 7:
--- 7(i) ---
7(i) | A T – 0.8 g sin 45 = 0.8a
B 1.2g sin 30 – T = 1.2a
System 1.2 g sin 30 – 0.8 g sin 45 = 2a | M1 | Apply Newton 2nd law to either A or to B
or to the system
A1 | One correct equation
A1 | A second correct equation
a = 0.171 | M1 | Solve for a
v2 = 2 × a × 0.4 | M1 | Use v2 = u2 + 2as with u = 0
v = 0.370 so speed of A is 0.370 ms–1 | A1
6
Alternative scheme for Question 7(i)
M1 | Attempt KE gain or PE loss
1 1
KE gain = × 0.8 × v2 + × 1.2 × v2
2 2 | A1 | v is the required speed of A
PE loss =
1.2 g × 0.4 sin 30 – 0.8 g × 0.4 sin 45 | A1
1 1
× 0.8 × v2 + × 1.2 × v2 =
2 2
1.2 g × 0.4 sin 30 – 0.8 g × 0.4 sin 45 | M1 | 4 term energy equation
M1 | Solving for v
v = 0.370 so speed of A is 0.370 ms–1 | A1
Question | Answer | Marks | Guidance
--- 7(ii) ---
7(ii) | R = 0.8gcos45=4 2
A
R = 1.2gcos30=6 3
B | B1 | For either R or R
A B
F = 4 2 µ and F = 6 3µ
A B | M1 | Either F or F used
A B
A 0.8 g sin 45 + F = T
A
B 1.2 g sin 30 – F = T
B
or system equation:
12 sin 30 – 8 sin 45 = F + F
A B | M1 | Resolve parallel to the plane either for
both particles A and B or for the system
equation
Correct equation(s) | A1
M1 | Eliminate T and solve for µ
( )
6−4 2
µ = ( )
6 3+4√2
= 0.0214 | A1
6
\includegraphics{figure_7}

The diagram shows a triangular block with sloping faces inclined to the horizontal at $45°$ and $30°$. Particle $A$ of mass $0.8 \text{ kg}$ lies on the face inclined at $45°$ and particle $B$ of mass $1.2 \text{ kg}$ lies on the face inclined at $30°$. The particles are connected by a light inextensible string which passes over a small smooth pulley $P$ fixed at the top of the faces. The parts $AP$ and $BP$ of the string are parallel to lines of greatest slope of the respective faces. The particles are released from rest with both parts of the string taut. In the subsequent motion neither particle reaches the pulley and neither particle reaches the bottom of a face.

\begin{enumerate}[label=(\roman*)]
\item Given that both faces are smooth, find the speed of $A$ after each particle has travelled a distance of $0.4 \text{ m}$. [6]
\item It is given instead that both faces are rough. The coefficient of friction between each particle and a face of the block is $\mu$. Find the value of $\mu$ for which the system is in limiting equilibrium. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2018 Q7 [12]}}
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