| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Challenging +1.2 This is a standard M3 circular motion problem requiring energy conservation and Newton's second law in the radial direction. Part (i) involves routine application of these principles to find when tension becomes zero, while part (ii) requires understanding of the critical speed condition. The multi-step nature and need to combine two key mechanics principles elevates it slightly above average, but the techniques are well-practiced in M3 and follow standard patterns. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| Height is \(\frac{2}{3}a\) | M1, A1, M1, A1, M1, A1 6 | Need correct 3 terms; allow wrong sign, missing/extra \(g\), missing \(m\) / \(a\); sin for cos correct. Need 3 terms and \(g\), allow sign slip, sin for cos. Ft if their \(T\) has right form |
| Answer | Marks | Guidance |
|---|---|---|
| \(U < \sqrt{(5ag)}\) | M1, A1, M1, M1, A1 5 | Allow '=' for all M marks. Allow ≥. Allow wrong sign. Allow wrong sign. Allow ≤ |
| Answer | Marks |
|---|---|
| \((θ = \frac{π}{2})\) gives \(U > \sqrt{(2ag)}\) | (M1), (A1), (M1), (A1), (A1) |
## (i)
By energy
$\frac{1}{2}m(4ag) = \frac{1}{2}mv^2 + mga(1 + \cos θ)$
Use of $F = ma$
$T + mg \cos θ = \frac{mv^2}{a}$
$T = 2mg - 3mg \cos θ$
Slack when $\cos θ = \frac{2}{3}$
Height is $\frac{2}{3}a$ | M1, A1, M1, A1, M1, A1 6 | Need correct 3 terms; allow wrong sign, missing/extra $g$, missing $m$ / $a$; sin for cos correct. Need 3 terms and $g$, allow sign slip, sin for cos. Ft if their $T$ has right form | $v^2 = 2ag - 2ag \cos θ$
## (ii)
If $0 > π/2$; $\frac{1}{3}mU^2 > mga$
$U > \sqrt{(2ag)}$
For no complete revolutions
$\frac{1}{2}mU^2 < \frac{1}{2}mu^2 + 2mga$
and $mg = m\frac{u^2}{a}$
$U < \sqrt{(5ag)}$ | M1, A1, M1, M1, A1 5 | Allow '=' for all M marks. Allow ≥. Allow wrong sign. Allow wrong sign. Allow ≤ | $u$ is vel at top. $\sqrt{2ag} < U < \sqrt{5ag}$
OR Use $\frac{1}{2}mU^2 = \frac{1}{2}mu^2 + mga(1 + \cos θ)$
and $T + mg \cos θ = \frac{mu^2}{a}$
To get $T = m\frac{u^2}{a} - 2mg - 3mg \cos θ$ oe
When T = 0, $U^2 = 2ag + 3ag \cos θ$
$(θ = 0)$ gives $U < \sqrt{(5ag)}$
$(θ = \frac{π}{2})$ gives $U > \sqrt{(2ag)}$ | (M1), (A1), (M1), (A1), (A1) |
---\includegraphics{figure_5}
One end of a light inextensible string of length $a$ is attached to a fixed point $O$. A particle $P$ of mass $m$ is attached to the other end of the string and hangs at rest. $P$ is then projected horizontally from this position with speed $2\sqrt{ag}$. When the string makes an angle $\theta$ with the upward vertical $P$ has speed $v$ (see diagram). The tension in the string is $T$.
\begin{enumerate}[label=(\roman*)]
\item Find an expression for $T$ in terms of $m$, $g$ and $\theta$, and hence find the height of $P$ above its initial level when the string becomes slack. [6]
\end{enumerate}
$P$ is now projected horizontally from the same initial position with speed $U$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the set of values of $U$ for which the string does not remain taut in the subsequent motion. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2016 Q5 [11]}}