CAIE M1 2014 November — Question 2

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2014
SessionNovember
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeDistance from velocity-time graph
DifficultyEasy -1.2 This appears to be a standard CAIE M1 kinematics question involving SUVAT equations and possibly a velocity-time graph interpretation. Such questions typically require straightforward application of 2-3 kinematic equations with minimal problem-solving insight. The multi-part structure is typical of routine M1 questions that test basic competency rather than deep understanding.
Spec3.02a Kinematics language: position, displacement, velocity, acceleration3.02h Motion under gravity: vector form

2 \includegraphics[max width=\textwidth, alt={}, center]{9c7e8624-c4cd-4a8e-83d9-f92d0bd6f95b-2_262_1004_760_575} The tops of each of two smooth inclined planes \(A\) and \(B\) meet at a right angle. Plane \(A\) is inclined at angle \(\alpha\) to the horizontal and plane \(B\) is inclined at angle \(\beta\) to the horizontal, where \(\sin \alpha = \frac { 63 } { 65 }\) and \(\sin \beta = \frac { 16 } { 65 }\). A small smooth pulley is fixed at the top of the planes and a light inextensible string passes over the pulley. Two particles \(P\) and \(Q\), each of mass 0.65 kg , are attached to the string, one at each end. Particle \(Q\) is held at rest at a point of the same line of greatest slope of the plane \(B\) as the pulley. Particle \(P\) rests freely below the pulley in contact with plane \(A\) (see diagram). Particle \(Q\) is released and the particles start to move with the string taut. Find the tension in the string.

Question 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
M1For applying Newton's 2nd law to \(P\) or to \(Q\)
\(0.65 \times 10 \times (63/65) - T = 0.65a\) or \(T - 0.65 \times 10 \times (16/65) = 0.65a\)A1
\(T - 0.65 \times 10 \times (16/65) = 0.65a\) or \(0.65 \times 10 \times (63/65) - T = 0.65a\) or \(0.65 \times 10 \times (63-16)/65 = 2 \times 0.65a\)B1
\([T - 1.6 = 6.3 - T]\) or \([T = 6.3 - 0.65 \times (47/13)]\) or \([T = 1.6 + 0.65 \times (47/13)]\)M1 For eliminating \(a\)
Tension is \(3.95\) NA1 [5 marks]
## Question 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For applying Newton's 2nd law to $P$ or to $Q$ |
| $0.65 \times 10 \times (63/65) - T = 0.65a$ **or** $T - 0.65 \times 10 \times (16/65) = 0.65a$ | A1 | |
| $T - 0.65 \times 10 \times (16/65) = 0.65a$ **or** $0.65 \times 10 \times (63/65) - T = 0.65a$ **or** $0.65 \times 10 \times (63-16)/65 = 2 \times 0.65a$ | B1 | |
| $[T - 1.6 = 6.3 - T]$ **or** $[T = 6.3 - 0.65 \times (47/13)]$ **or** $[T = 1.6 + 0.65 \times (47/13)]$ | M1 | For eliminating $a$ |
| Tension is $3.95$ N | A1 | **[5 marks]** |

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\includegraphics[max width=\textwidth, alt={}, center]{9c7e8624-c4cd-4a8e-83d9-f92d0bd6f95b-2_262_1004_760_575}

The tops of each of two smooth inclined planes $A$ and $B$ meet at a right angle. Plane $A$ is inclined at angle $\alpha$ to the horizontal and plane $B$ is inclined at angle $\beta$ to the horizontal, where $\sin \alpha = \frac { 63 } { 65 }$ and $\sin \beta = \frac { 16 } { 65 }$. A small smooth pulley is fixed at the top of the planes and a light inextensible string passes over the pulley. Two particles $P$ and $Q$, each of mass 0.65 kg , are attached to the string, one at each end. Particle $Q$ is held at rest at a point of the same line of greatest slope of the plane $B$ as the pulley. Particle $P$ rests freely below the pulley in contact with plane $A$ (see diagram). Particle $Q$ is released and the particles start to move with the string taut. Find the tension in the string.

\hfill \mbox{\textit{CAIE M1 2014 Q2}}