CAIE M2 2002 November — Question 4 6 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2002
SessionNovember
Marks6
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Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeTwo strings, two fixed points
DifficultyStandard +0.3 This is a standard circular motion problem with two strings requiring resolution of forces in horizontal (centripetal) and vertical (equilibrium) directions. The setup is straightforward with given values, requiring application of F=mrω² and vertical force balance. Slightly above average difficulty due to the two-string configuration, but follows a well-practiced method with no conceptual surprises.
Spec6.05c Horizontal circles: conical pendulum, banked tracks
4 \includegraphics[max width=\textwidth, alt={}, center]{fcf239a6-6558-43ec-b404-70aa349af6a9-3_604_490_258_831} A small ball \(B\) of mass 0.5 kg is attached to points \(P\) and \(Q\) on a fixed vertical axis by two light inextensible strings of equal length. Both of the strings are taut and each is inclined at \(60 ^ { \circ }\) to the vertical, as shown in the diagram. The ball moves with constant speed \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a horizontal circle of radius 0.8 m . Find the tension in the string \(P B\).
AnswerMarks
\(a = 4^2/0.8 = [20]\)B1
Uses Newton's 2nd law horizontally to obtain a 3 term equationM1
Obtains \((T_P + T_Q)\cos 30° = 0.5 \times 20\) \([T_P + T_Q = \frac{20}{\sqrt{3}}]\)A1 ft
Resolves forces vertically to obtain a 3 term equationM1
Obtains \(T_P\cos 60° = T_Q\cos 60° + 5\) \([T_P - T_Q = 10]\)A1
Alternatively for the above 4 marks:
AnswerMarks
Uses Newton's 2nd law perpendicular to \(BQ\) to obtain a 3 term equationM2
Obtains \(T_P\cos 30° - 0.5g\cos 30° = 0.5 \times 20\cos 60°\) \([T_P = 5 + \frac{10}{\sqrt{3}}]\)A2 ft
[SR Allow A1 with 1 sign or trigonometric error]
AnswerMarks Guidance
Obtains tension in \(PB\) as 10.8 N (10.7735)A1 6 marks
*NB: Use of equal tensions (and using B1, M1, A1, M1, A1 in order).* 
$a = 4^2/0.8 = [20]$ | B1 |
Uses Newton's 2nd law horizontally to obtain a 3 term equation | M1 |
Obtains $(T_P + T_Q)\cos 30° = 0.5 \times 20$ $[T_P + T_Q = \frac{20}{\sqrt{3}}]$ | A1 ft |
Resolves forces vertically to obtain a 3 term equation | M1 |
Obtains $T_P\cos 60° = T_Q\cos 60° + 5$ $[T_P - T_Q = 10]$ | A1 |

**Alternatively for the above 4 marks:**
Uses Newton's 2nd law perpendicular to $BQ$ to obtain a 3 term equation | M2 |
Obtains $T_P\cos 30° - 0.5g\cos 30° = 0.5 \times 20\cos 60°$ $[T_P = 5 + \frac{10}{\sqrt{3}}]$ | A2 ft |

**[SR Allow A1 with 1 sign or trigonometric error]**
Obtains tension in $PB$ as 10.8 N (10.7735) | A1 | 6 marks |

*NB: Use of equal tensions (and using B1, M1, A1, M1, A1 in order).* | |
4\\
\includegraphics[max width=\textwidth, alt={}, center]{fcf239a6-6558-43ec-b404-70aa349af6a9-3_604_490_258_831}

A small ball $B$ of mass 0.5 kg is attached to points $P$ and $Q$ on a fixed vertical axis by two light inextensible strings of equal length. Both of the strings are taut and each is inclined at $60 ^ { \circ }$ to the vertical, as shown in the diagram. The ball moves with constant speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a horizontal circle of radius 0.8 m . Find the tension in the string $P B$.

\hfill \mbox{\textit{CAIE M2 2002 Q4 [6]}}
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