| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| 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 vertical circle problem with multiple parts requiring energy conservation, circular motion equations, and projectile motion. While it involves several steps and careful application of mechanics principles, the techniques are all standard for this topic with no novel insights required. The algebraic manipulation is moderate, making it slightly above average difficulty for A-level. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02h Motion under gravity: vector form6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(900\frac{dv}{dt} = \frac{63000}{kt^2}\) | M1 | |
| \(\int_0^{10.5} dv = \int_1^4 \frac{70}{kt^2}\, dt\) | ||
| \([v]_0^{10.5} = \left[-\frac{70}{kt}\right]_1^4\) | DM1, A1 | Integration, limits not needed |
| \(10.5(-0) = -\frac{70}{4k} + \frac{70}{k}\) | M1 | Substitute limits |
| \(k = 5\) | A1 | Correct value |
| \(\int_0^v dv = \int_1^t \frac{14}{t^2}\, dt\) | A1 | Integrate again with limits as shown |
| \(v = 14 - \frac{14}{t}\) ✱ | A1 | Obtain GIVEN answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(900\frac{dv}{dt} = \frac{63000}{kt^2}\) | M1 | |
| \(\int_0^v dv = \int_1^t \frac{70}{kt^2}\, dt\) | ||
| \([v]_0^v = \left[-\frac{70}{kt}\right]_1^t\) | DM1, A1 | Integration, limits not needed |
| \(v = \frac{70}{k}\left[-\frac{1}{t}\right]_1^t = \frac{70}{k}\left(1 - \frac{1}{t}\right)\) | M1 | Substitute limits and \(v = 10.5\), \(t = 4\) |
| \(k = 5\) | A1 | Correct value |
| \(v = \frac{70}{5}\left(1 - \frac{1}{t}\right)\) | A1 | Substitute |
| \(v = 14 - \frac{14}{t}\) ✱ | A1 | Obtain GIVEN answer |
## Question 4(a): (by definite integration)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $900\frac{dv}{dt} = \frac{63000}{kt^2}$ | M1 | |
| $\int_0^{10.5} dv = \int_1^4 \frac{70}{kt^2}\, dt$ | | |
| $[v]_0^{10.5} = \left[-\frac{70}{kt}\right]_1^4$ | DM1, A1 | Integration, limits not needed |
| $10.5(-0) = -\frac{70}{4k} + \frac{70}{k}$ | M1 | Substitute limits |
| $k = 5$ | A1 | Correct value |
| $\int_0^v dv = \int_1^t \frac{14}{t^2}\, dt$ | A1 | Integrate again with limits as shown |
| $v = 14 - \frac{14}{t}$ ✱ | A1 | Obtain GIVEN answer |
**OR:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $900\frac{dv}{dt} = \frac{63000}{kt^2}$ | M1 | |
| $\int_0^v dv = \int_1^t \frac{70}{kt^2}\, dt$ | | |
| $[v]_0^v = \left[-\frac{70}{kt}\right]_1^t$ | DM1, A1 | Integration, limits not needed |
| $v = \frac{70}{k}\left[-\frac{1}{t}\right]_1^t = \frac{70}{k}\left(1 - \frac{1}{t}\right)$ | M1 | Substitute limits and $v = 10.5$, $t = 4$ |
| $k = 5$ | A1 | Correct value |
| $v = \frac{70}{5}\left(1 - \frac{1}{t}\right)$ | A1 | Substitute |
| $v = 14 - \frac{14}{t}$ ✱ | A1 | Obtain GIVEN answer |
---4.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{ab85ec29-b1fc-45a9-9343-09feb33ab6c5-006_574_510_324_726}
\end{center}
\end{figure}
A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a point $O$. The point $A$ is vertically below $O$, and $O A = a$. The particle is projected horizontally from $A$ with speed $\sqrt { } ( 3 a g )$. When $O P$ makes an angle $\theta$ with the upward vertical through $O$ and the string is still taut, the tension in the string is $T$ and the speed of $P$ is $v$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $a , g$ and $\theta$, an expression for $v ^ { 2 }$.
\item Show that $T = ( 1 - 3 \cos \theta ) m g$.
The string becomes slack when $P$ is at the point $B$.
\item Find, in terms of $a$, the vertical height of $B$ above $A$.
After the string becomes slack, the highest point reached by $P$ is $C$.
\item Find, in terms of $a$, the vertical height of $C$ above $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q4}}