OCR M3 2011 January — Question 2 6 marks

Exam BoardOCR
ModuleM3 (Mechanics 3)
Year2011
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 2
TypeVertical circle: tension at specific point
DifficultyStandard +0.3 This is a standard vertical circle problem requiring conservation of energy to find speeds at different positions, then applying Newton's second law for circular motion. The setup is straightforward (projected from lowest point) and both parts follow routine procedures taught in M3, though it requires careful application of multiple concepts across two parts.
Spec6.02e Calculate KE and PE: using formulae6.05e Radial/tangential acceleration

2 A particle of mass 0.4 kg is attached to a fixed point \(O\) by a light inextensible string of length 0.5 m . The particle is projected horizontally with speed \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from the point 0.5 m vertically below \(O\). The particle moves in a complete circle. Find the tension in the string when
  1. the string is horizontal,
  2. the particle is vertically above \(O\).

AnswerMarks Guidance
\(\frac{1}{2}(m)(v^2 - 6^2) = -(m)g \times 0.5\) in (i) or \(\frac{1}{2}(m)(v^2 - 6^2) = -(mg \times 1\) in (ii)M1 For using the principle of conservation of energy in (i) or (ii)
\(v^2 = 26.2\) in (i) and \(16.4\) in (ii)A1
\(T = 0.4v^2/0.5\) in (i) or \(T + 0.4g = 0.4v^2/0.5\)M1 For using Newton's second law with \(a = v^2/L\), M1 for either attempt, A1 for both right
\(T\)A1
Tension is \(21.0N\) in (i) (\(20.96\)), \(9.2N\) in (ii)A1
[6]
$\frac{1}{2}(m)(v^2 - 6^2) = -(m)g \times 0.5$ in (i) or $\frac{1}{2}(m)(v^2 - 6^2) = -(mg \times 1$ in (ii) | M1 | For using the principle of conservation of energy in (i) or (ii)
$v^2 = 26.2$ in (i) and $16.4$ in (ii) | A1 |

$T = 0.4v^2/0.5$ in (i) or $T + 0.4g = 0.4v^2/0.5$ | M1 | For using Newton's second law with $a = v^2/L$, M1 for either attempt, A1 for both right
$T$ | A1 |
Tension is $21.0N$ in (i) ($20.96$), $9.2N$ in (ii) | A1 |
| [6] |
2 A particle of mass 0.4 kg is attached to a fixed point $O$ by a light inextensible string of length 0.5 m . The particle is projected horizontally with speed $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from the point 0.5 m vertically below $O$. The particle moves in a complete circle. Find the tension in the string when\\
(i) the string is horizontal,\\
(ii) the particle is vertically above $O$.

\hfill \mbox{\textit{OCR M3 2011 Q2 [6]}}