OCR M2 2015 June — Question 2 6 marks

Exam BoardOCR
ModuleM2 (Mechanics 2)
Year2015
SessionJune
Marks6
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TopicCircular Motion 1
TypeConical pendulum – horizontal circle in free space (no surface)
DifficultyModerate -0.8 This is a standard conical pendulum problem requiring straightforward application of circular motion equations (T sin θ = mrω², T cos θ = mg) with given numerical values. It involves routine two-equation resolution with no conceptual challenges beyond basic setup, making it easier than average for A-level.
Spec6.05c Horizontal circles: conical pendulum, banked tracks
2 A particle of mass 0.3 kg is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point \(A\). The particle moves in a horizontal circle of radius 0.343 m , with centre vertically below \(A\), at a constant angular speed of \(6 \mathrm { rad } \mathrm { s } ^ { - 1 }\). Find the tension in the string and the angle at which the string is inclined to the vertical.
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(T\cos\theta = 0.3g\ (= 2.94)\)B1 Resolve vertically or \(T\sin a = 0.3g\); component of \(T\) required
M1Attempt at use of N2L with \(a = r\omega^2\) or \(a = v^2/r\)
\(T\sin\theta = 0.3(0.343)(6^2)\ (= 3.7044)\)A1 Resolve horizontally or \(T\cos a = 0.3(0.343)(6^2)\); component of \(T\) required
M1Attempt to solve for \(T\) or \(\theta\); component of \(T\) required in both equations
\(T = 4.73 \text{ N}\)A1
\(\theta = 51.6°\)A1 SC if 38.4° then B1M1A1M1A1A0 5/6 Max 
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $T\cos\theta = 0.3g\ (= 2.94)$ | B1 | Resolve vertically or $T\sin a = 0.3g$; component of $T$ required |
| | M1 | Attempt at use of N2L with $a = r\omega^2$ or $a = v^2/r$ |
| $T\sin\theta = 0.3(0.343)(6^2)\ (= 3.7044)$ | A1 | Resolve horizontally or $T\cos a = 0.3(0.343)(6^2)$; component of $T$ required |
| | M1 | Attempt to solve for $T$ or $\theta$; component of $T$ required in both equations |
| $T = 4.73 \text{ N}$ | A1 | |
| $\theta = 51.6°$ | A1 | SC if 38.4° then B1M1A1M1A1A0 5/6 Max |

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2 A particle of mass 0.3 kg is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point $A$. The particle moves in a horizontal circle of radius 0.343 m , with centre vertically below $A$, at a constant angular speed of $6 \mathrm { rad } \mathrm { s } ^ { - 1 }$. Find the tension in the string and the angle at which the string is inclined to the vertical.

\hfill \mbox{\textit{OCR M2 2015 Q2 [6]}}
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