OCR M2 2009 January — Question 5 12 marks

Exam BoardOCR
ModuleM2 (Mechanics 2)
Year2009
SessionJanuary
Marks12
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Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeTwo strings, two fixed points
DifficultyStandard +0.3 This is a standard M2 circular motion problem with two strings. It requires resolving forces vertically and horizontally, using Pythagoras to find geometry, and applying F=mrω². The question guides students through part (i) and the method is straightforward textbook application. Slightly above average difficulty due to the two-string geometry, but well within typical M2 scope.
Spec3.03d Newton's second law: 2D vectors6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks
5 \includegraphics[max width=\textwidth, alt={}, center]{dd23f4a8-f7e7-4f80-bad6-7e9ec21565fc-3_729_739_868_703} A particle \(P\) of mass 0.2 kg is attached to one end of each of two light inextensible strings, one of length 0.4 m and one of length 0.3 m . The other end of the longer string is attached to a fixed point \(A\), and the other end of the shorter string is attached to a fixed point \(B\), which is vertically below \(A\). The particle moves in a horizontal circle of radius 0.24 m at a constant angular speed of \(8 \mathrm { rad } \mathrm { s } ^ { - 1 }\) (see diagram). Both strings are taut, the tension in \(A P\) is \(S \mathrm {~N}\) and the tension in \(B P\) is \(T \mathrm {~N}\).
  1. By resolving vertically, show that \(4 S = 3 T + 9.8\).
  2. Find another equation connecting \(S\) and \(T\) and hence calculate the tensions, correct to 1 decimal place. \section*{[Questions 6 and 7 are printed overleaf.]}
5\\
\includegraphics[max width=\textwidth, alt={}, center]{dd23f4a8-f7e7-4f80-bad6-7e9ec21565fc-3_729_739_868_703}

A particle $P$ of mass 0.2 kg is attached to one end of each of two light inextensible strings, one of length 0.4 m and one of length 0.3 m . The other end of the longer string is attached to a fixed point $A$, and the other end of the shorter string is attached to a fixed point $B$, which is vertically below $A$. The particle moves in a horizontal circle of radius 0.24 m at a constant angular speed of $8 \mathrm { rad } \mathrm { s } ^ { - 1 }$ (see diagram). Both strings are taut, the tension in $A P$ is $S \mathrm {~N}$ and the tension in $B P$ is $T \mathrm {~N}$.\\
(i) By resolving vertically, show that $4 S = 3 T + 9.8$.\\
(ii) Find another equation connecting $S$ and $T$ and hence calculate the tensions, correct to 1 decimal place.

\section*{[Questions 6 and 7 are printed overleaf.]}

\hfill \mbox{\textit{OCR M2 2009 Q5 [12]}}
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