| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | November |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: complete revolution conditions |
| Difficulty | Standard +0.8 This is a classic vertical circular motion problem requiring energy conservation and circular motion dynamics across multiple parts. Part (i) requires deriving the critical condition using energy methods and the minimum tension condition at the top. Parts (ii) and (iii) involve applying Newton's second law for circular motion at different positions. While the techniques are standard for M2, the multi-step reasoning, algebraic manipulation, and need to correctly apply both energy conservation and centripetal force equations make this moderately challenging—harder than routine exercises but not requiring novel insight. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05f Vertical circle: motion including free fall |
| Answer | Marks |
|---|---|
| 4 (i) | 0.2a = 0.024 t – 0.2 g × 0.3 |
| Answer | Marks |
|---|---|
| v = 0.06(t2 – 50t + 15) AG | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | [4] | Uses N2L |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) | t2 – 50t + 15 = 0 | |
| t = 0.302 | M1 | |
| A1 | [2] | Solves 3 term quadratic |
| Answer | Marks | Guidance |
|---|---|---|
| Page 5 | Mark Scheme | Syllabus |
| GCE A LEVEL – October/November 2013 | 9709 | 53 |
| (iii) | 0.024t = 0.3 × 0.2 g | |
| t = 25 | M1 | |
| A1 | [2] | Equates tractive force and friction |
| force | 8 |
Question 4:
--- 4 (i) ---
4 (i) | 0.2a = 0.024 t – 0.2 g × 0.3
a = 0.12t – 3
∫ dv = ∫ (0.12t – 3)dt,
v = 0.12t 2 /2 – 3t + c, t = 0, v = 0.9
hence c = 0.9
v = 0.06(t2 – 50t + 15) AG | M1
A1
M1
A1 | [4] | Uses N2L
Integrates and finds c
(ii) | t2 – 50t + 15 = 0
t = 0.302 | M1
A1 | [2] | Solves 3 term quadratic
Smaller +ve root only
Page 5 | Mark Scheme | Syllabus | Paper
GCE A LEVEL – October/November 2013 | 9709 | 53
(iii) | 0.024t = 0.3 × 0.2 g
t = 25 | M1
A1 | [2] | Equates tractive force and friction
force | 8A particle of mass $m$ is attached to one end of a light inextensible string of length $l$. The other end of the string is attached to a fixed point. The particle moves in a vertical circle.
\begin{enumerate}[label=(\roman*)]
\item Show that the speed $v$ at the lowest point of the circle must satisfy $v^2 \geq 5gl$ for the particle to complete the circle.
\item Given that the particle just completes the circle, find the tensions in the string at the highest and lowest points of the circle.
\item Given that $v^2 = 6gl$ at the lowest point, find the tension in the string when the particle has risen through an angle $\theta$ from the lowest point.
\end{enumerate}
[14]
\hfill \mbox{\textit{CAIE M2 2013 Q4 [14]}}