CAIE M1 2014 June — Question 3

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2014
SessionJune
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeDistance from velocity-time graph
DifficultyModerate -0.5 This is a standard SUVAT kinematics problem involving a velocity-time graph with constant acceleration phases. Students need to find acceleration from the graph, calculate distance using area under the curve, and apply basic kinematic equations. These are routine M1 skills with straightforward multi-part structure, making it slightly easier than average but still requiring proper method application.
Spec3.03b Newton's first law: equilibrium3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces

3 \includegraphics[max width=\textwidth, alt={}, center]{77976dad-c055-45fd-93fe-e37fa8e9ae22-2_520_719_1137_712} \(A\) and \(B\) are fixed points of a vertical wall with \(A\) vertically above \(B\). A particle \(P\) of mass 0.7 kg is attached to \(A\) by a light inextensible string of length \(3 \mathrm {~m} . P\) is also attached to \(B\) by a light inextensible string of length \(2.5 \mathrm {~m} . P\) is maintained in equilibrium at a distance of 2.4 m from the wall by a horizontal force of magnitude 10 N acting on \(P\) (see diagram). Both strings are taut, and the 10 N force acts in the plane \(A P B\) which is perpendicular to the wall. Find the tensions in the strings. [6]

Question 3:
AnswerMarks Guidance
Working/AnswerMark Guidance
M1For resolving forces acting on P horizontally (3 terms)
\(0.8T_1 + 0.96T_2 = 10\) or \(T_1\cos 36.9 + T_2\cos 16.3 = 10\)A1
M1For resolving forces acting on P vertically (3 terms)
\(0.6T_1 - 0.28T_2 = 0.7g\) or \(T_1\sin 36.9 - T_2\sin 16.3 = 0.7g\)A1
M1For solving simultaneous equations and finding both \(T_1\) and \(T_2\)
\(T_1 = 11.9\) and \(T_2 = 0.5\)A1 Total: 6 marks
## Question 3:
| Working/Answer | Mark | Guidance |
|---|---|---|
| | M1 | For resolving forces acting on P horizontally (3 terms) |
| $0.8T_1 + 0.96T_2 = 10$ or $T_1\cos 36.9 + T_2\cos 16.3 = 10$ | A1 | |
| | M1 | For resolving forces acting on P vertically (3 terms) |
| $0.6T_1 - 0.28T_2 = 0.7g$ or $T_1\sin 36.9 - T_2\sin 16.3 = 0.7g$ | A1 | |
| | M1 | For solving simultaneous equations and finding both $T_1$ and $T_2$ |
| $T_1 = 11.9$ **and** $T_2 = 0.5$ | A1 | **Total: 6 marks** |

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\includegraphics[max width=\textwidth, alt={}, center]{77976dad-c055-45fd-93fe-e37fa8e9ae22-2_520_719_1137_712}\\
$A$ and $B$ are fixed points of a vertical wall with $A$ vertically above $B$. A particle $P$ of mass 0.7 kg is attached to $A$ by a light inextensible string of length $3 \mathrm {~m} . P$ is also attached to $B$ by a light inextensible string of length $2.5 \mathrm {~m} . P$ is maintained in equilibrium at a distance of 2.4 m from the wall by a horizontal force of magnitude 10 N acting on $P$ (see diagram). Both strings are taut, and the 10 N force acts in the plane $A P B$ which is perpendicular to the wall. Find the tensions in the strings. [6]

\hfill \mbox{\textit{CAIE M1 2014 Q3}}