CAIE M1 2017 November — Question 6 10 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2017
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeParticle on slope with pulley
DifficultyStandard +0.3 This is a standard M1 mechanics problem involving forces on a slope with friction and a pulley system. Part (i) requires resolving forces in limiting equilibrium (routine technique), and part (ii) applies Newton's second law with the given acceleration. Both parts follow standard textbook methods with no novel insight required, making it slightly easier than average.
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_6} Two particles \(P\) and \(Q\), each of mass \(m\) kg, 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 plane is inclined at an angle \(α\) to the horizontal, where \(\tan α = \frac{4}{3}\). Particle \(P\) rests on the plane and particle \(Q\) hangs vertically, as shown in the diagram. The string between \(P\) and the pulley is parallel to a line of greatest slope of the plane. The system is in limiting equilibrium.
  1. Show that the coefficient of friction between \(P\) and the plane is \(\frac{4}{3}\). [5]
A force of magnitude 10 N is applied to \(P\), acting up a line of greatest slope of the plane, and \(P\) accelerates at 2.5 m s\(^{-2}\).
  1. Find the value of \(m\). [5]

Question 6:

AnswerMarks Guidance
6(i)R = mg cos α (R = 9.6m) B1
[T = mg
AnswerMarks Guidance
F = mg sin α + T ]M1 For resolving forces on P and Q and
eliminating T or for considering the
equilibrium of the system
AnswerMarks Guidance
F = mg sin α + mgA1 (F = 12.8m)
M1For use of F = µR
4
Coefficient of friction = 1⅓ =
AnswerMarks Guidance
3A1 AG so must be from exact working
5
AnswerMarks Guidance
QuestionAnswer Marks

AnswerMarks
6(ii)EITHER:
P equation is
10 – mg sin α – F – T = 2.5 m
Q equation is
AnswerMarks Guidance
T – mg = 2.5m(*M1 For applying Newton’s 2nd law to
P (5 terms) or Q (3 terms)
AnswerMarks
*M1For applying Newton’s 2nd law to the
other particle and eliminate T
10 – mg sin α – µmg cos α
AnswerMarks Guidance
– mg = 2m (2.5)A1 If evaluated then this is
10 – 2.8m – 12.8m – 10m = 5m
AnswerMarks
DM1For solving this equation for m as far as
m = Dependent on one or other of the
previous M marks having been scored
AnswerMarks Guidance
m = 0.327A1) 50
Allow m =
153
OR:
AnswerMarks Guidance
[10 – mg sin α –F – mg = m(2.5 + 2.5)](*M1 For applying Newton’s 2nd law to the
system. Allow with 5 terms
AnswerMarks
*M1System equation with all 6 terms
10 – mg sin α – µmg cos α
AnswerMarks
– mg = 2m (2.5)A1
DM1For solving this equation for m as far as
m = Dependent on one or other of the
previous M marks having been scored
AnswerMarks Guidance
m = 0.327A1) 50
Allow m =
153
5
AnswerMarks Guidance
QuestionAnswer Marks
Question 6:
--- 6(i) ---
6(i) | R = mg cos α (R = 9.6m) | B1 | Allow use of α = 16.3º throughout
[T = mg
F = mg sin α + T ] | M1 | For resolving forces on P and Q and
eliminating T or for considering the
equilibrium of the system
F = mg sin α + mg | A1 | (F = 12.8m)
M1 | For use of F = µR
4
Coefficient of friction = 1⅓ =
3 | A1 | AG so must be from exact working
5
Question | Answer | Marks | Guidance
--- 6(ii) ---
6(ii) | EITHER:
P equation is
10 – mg sin α – F – T = 2.5 m
Q equation is
T – mg = 2.5m | (*M1 | For applying Newton’s 2nd law to
P (5 terms) or Q (3 terms)
*M1 | For applying Newton’s 2nd law to the
other particle and eliminate T
10 – mg sin α – µmg cos α
– mg = 2m (2.5) | A1 | If evaluated then this is
10 – 2.8m – 12.8m – 10m = 5m
DM1 | For solving this equation for m as far as
m = Dependent on one or other of the
previous M marks having been scored
m = 0.327 | A1) | 50
Allow m =
153
OR:
[10 – mg sin α –F – mg = m(2.5 + 2.5)] | (*M1 | For applying Newton’s 2nd law to the
system. Allow with 5 terms
*M1 | System equation with all 6 terms
10 – mg sin α – µmg cos α
– mg = 2m (2.5) | A1
DM1 | For solving this equation for m as far as
m = Dependent on one or other of the
previous M marks having been scored
m = 0.327 | A1) | 50
Allow m =
153
5
Question | Answer | Marks | Guidance
\includegraphics{figure_6}

Two particles $P$ and $Q$, each of mass $m$ kg, 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 plane is inclined at an angle $α$ to the horizontal, where $\tan α = \frac{4}{3}$. Particle $P$ rests on the plane and particle $Q$ hangs vertically, as shown in the diagram. The string between $P$ and the pulley is parallel to a line of greatest slope of the plane. The system is in limiting equilibrium.

\begin{enumerate}[label=(\roman*)]
\item Show that the coefficient of friction between $P$ and the plane is $\frac{4}{3}$. [5]
\end{enumerate}

A force of magnitude 10 N is applied to $P$, acting up a line of greatest slope of the plane, and $P$ accelerates at 2.5 m s$^{-2}$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the value of $m$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2017 Q6 [10]}}