| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | String through hole – lower particle also moves in horizontal circle (conical pendulum below) |
| Difficulty | Standard +0.3 This is a standard circular motion problem with coupled particles requiring application of Newton's second law and tension equilibrium. Part (i) involves straightforward force balance (tension = weight for Q, tension provides centripetal force for P). Part (ii) adds geometric complexity with the inclined string but follows the same principles with trigonometric resolution. The calculations are routine for M2 level with no novel insight required, making it slightly easier than average. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = 0.5g\) | B1 | \(T = 5\) |
| \(T = 0.8 \times 6.25^2 \times r\) | M1 | |
| \(r = 0.16\,\mathrm{m}\) | A1 | |
| \(v = 1\,\mathrm{ms}^{-1}\) | B1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(T\cos 60 = 0.5g\) | B1 | \(T = 10\) |
| \(r = 0.32\,\mathrm{m}\) | B1ft | \(\unicode{x2713}\) (2 × candidate's value of \(r\)) |
| \(T\sin 60 = 0.5 \times 6.25^2 \times R\) | M1 | Newton's Second Law with component of \(T\) |
| \(R = 0.443(40)\,\mathrm{m}\) | A1 | |
| \(L = 0.32 + 0.443(4)/\sin 60\) | M1 | |
| \(L = 0.832\,\mathrm{m}\) | A1 | [6] |
**(i)**
$T = 0.5g$ | B1 | $T = 5$
$T = 0.8 \times 6.25^2 \times r$ | M1 |
$r = 0.16\,\mathrm{m}$ | A1 |
$v = 1\,\mathrm{ms}^{-1}$ | B1 | [4]
**(ii)**
$T\cos 60 = 0.5g$ | B1 | $T = 10$
$r = 0.32\,\mathrm{m}$ | B1ft | $\unicode{x2713}$ (2 × candidate's value of $r$)
$T\sin 60 = 0.5 \times 6.25^2 \times R$ | M1 | Newton's Second Law with component of $T$
$R = 0.443(40)\,\mathrm{m}$ | A1 |
$L = 0.32 + 0.443(4)/\sin 60$ | M1 |
$L = 0.832\,\mathrm{m}$ | A1 | [6] | [10]7 Particles $P$ and $Q$, of masses 0.8 kg and 0.5 kg respectively, are attached to the ends of a light inextensible string which passes through a small hole in a smooth horizontal table of negligible thickness. $P$ moves with constant angular speed $6.25 \mathrm { rad } \mathrm { s } ^ { - 1 }$ in a circular path on the surface of the table.\\
(i) It is given that $Q$ is stationary and that the part of string attached to $Q$ is vertical. Calculate the radius of the path of $P$, and find the speed of $P$.\\
(ii) It is given instead that the part of string attached to $Q$ is inclined at $60 ^ { \circ }$ to the vertical, and that $Q$ moves in a horizontal circular path below the table, also with constant angular speed $6.25 \mathrm { rad } \mathrm { s } ^ { - 1 }$. Calculate the total length of the string.\\[0pt]
[6]
\hfill \mbox{\textit{CAIE M2 2012 Q7 [10]}}