CAIE M1 2011 November — Question 3 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2011
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeHeavier particle hits ground, lighter continues upward - vertical strings
DifficultyChallenging +1.2 This is a multi-stage pulley problem requiring students to: (1) find Q's mass using tension equation, (2) calculate Q's velocity when P hits ground using energy or kinematics, (3) determine additional height Q rises after string goes slack, and (4) calculate total distance (up then down). While it involves several connected steps and careful tracking of motion phases, each individual calculation uses standard M1 techniques (Newton's laws, SUVAT, energy). The conceptual demand is moderate—understanding that Q continues upward after P stops is a key insight but fairly standard for pulley problems.
Spec3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys

3 Particles \(P\) and \(Q\) are attached to opposite ends of a light inextensible string which passes over a fixed smooth pulley. The system is released from rest with the string taut, with its straight parts vertical, and with both particles at a height of 2 m above horizontal ground. \(P\) moves vertically downwards and does not rebound when it hits the ground. At the instant that \(P\) hits the ground, \(Q\) is at the point \(X\), from where it continues to move vertically upwards without reaching the pulley. Given that \(P\) has mass 0.9 kg and that the tension in the string is 7.2 N while \(P\) is moving, find the total distance travelled by \(Q\) from the instant it first reaches \(X\) until it returns to \(X\).

AnswerMarks Guidance
M1For using Newton's second law
\(0.9g - 7.2 = 0.9a\)A1 \((a = 2)\)
\([v^2 = 2 \times (0.9g - 7.2)/0.9 \times 2]\)M1 For using \(v^2 = (0)^2 + 2ah\)
\(u_{\text{slack}} = v_{\text{taut}} = 2\sqrt{g - 8}\)B1ft
\([\text{distance} = 4 - 32/g]\)M1 For using \((0)^2 = u^2 - 2gh\) and distance = 2h
Distance is 0.8 mA1 6
| M1 | For using Newton's second law
$0.9g - 7.2 = 0.9a$ | A1 | $(a = 2)$
$[v^2 = 2 \times (0.9g - 7.2)/0.9 \times 2]$ | M1 | For using $v^2 = (0)^2 + 2ah$
$u_{\text{slack}} = v_{\text{taut}} = 2\sqrt{g - 8}$ | B1ft | | It incorrect equation for a
$[\text{distance} = 4 - 32/g]$ | M1 | For using $(0)^2 = u^2 - 2gh$ and distance = 2h
Distance is 0.8 m | A1 | 6
3 Particles $P$ and $Q$ are attached to opposite ends of a light inextensible string which passes over a fixed smooth pulley. The system is released from rest with the string taut, with its straight parts vertical, and with both particles at a height of 2 m above horizontal ground. $P$ moves vertically downwards and does not rebound when it hits the ground. At the instant that $P$ hits the ground, $Q$ is at the point $X$, from where it continues to move vertically upwards without reaching the pulley. Given that $P$ has mass 0.9 kg and that the tension in the string is 7.2 N while $P$ is moving, find the total distance travelled by $Q$ from the instant it first reaches $X$ until it returns to $X$.

\hfill \mbox{\textit{CAIE M1 2011 Q3 [6]}}