| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particles at coordinate positions |
| Difficulty | Moderate -0.3 This is a straightforward circular motion problem requiring standard application of T = mrω². Part (i) is direct substitution, part (ii) requires recognizing that AP must provide tension for both particles, and part (iii) is routine KE calculation. All steps follow standard M2 procedures with no novel insight required, making it slightly easier than average. |
| Spec | 6.02e Calculate KE and PE: using formulae6.05b Circular motion: v=r*omega and a=v^2/r |
4 A light inextensible string of length 0.6 m has one end fixed to a point $A$ on a smooth horizontal plane. The other end of the string is attached to a particle $B$, of mass 0.4 kg , which rotates about $A$ with constant angular speed $2 \mathrm { rad } \mathrm { s } ^ { - 1 }$ on the surface of the plane.\\
(i) Calculate the tension in the string.
A particle $P$ of mass 0.1 kg is attached to the mid-point of the string. The line $A P B$ is straight and rotation continues at $2 \mathrm { rad } \mathrm { s } ^ { - 1 }$.\\
(ii) Calculate the tension in the section of the string $A P$.\\
(iii) Calculate the total kinetic energy of the system.\\
\hfill \mbox{\textit{OCR M2 2009 Q4 [11]}}