| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: released from rest at natural length or above (string initially slack) |
| Difficulty | Standard +0.3 This is a standard M3 elastic string problem requiring energy conservation to find the extension, then Newton's second law for acceleration. Part (a) uses PE = EPE with straightforward algebra; part (b) applies F=ma at the turning point. Both are routine applications of well-practiced techniques with no novel insight required, making this slightly easier than average. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| EPE gained \(= \frac{14.7x^2}{2\times 1.5}\) | B1 | Correct EPE when extension is \(x\) |
| \(\frac{14.7x^2}{3} = 0.6g\times(x+1.5)\) | M1, A1ft | Equating EPE to GPE lost. EPE of form \(k\frac{\lambda x^2}{l}\); correct equation ft their EPE |
| \(5x^2-6x-9=0\) | — | |
| \(x = \frac{6\pm\sqrt{36+180}}{10} = 2.069\ldots\) (or \(-0.869\)) | DM1, A1 | Solve quadratic (formula must be correct); \(x=2.069\ldots\), neg value not needed |
| \(OA = 3.569\ldots \approx 3.6\) or \(3.57\) | A1 (6) | Add 1.5 to 2.069; answer to 2 or 3 sf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{14.7\times 2.069}{1.5} - 0.6\times 9.8 = 0.6a\) | M1, A1ft | Use NL2 at \(A\) inc use of Hooke's law. Formula for HL correct. Ext can be \((3.569-1.5)\); correct numbers ft their extension |
| \(a = 23.993\ldots \approx 24\) or \(24.0\) | A1cao (3) | 24 or 24.0 only. No negatives allowed |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| EPE gained $= \frac{14.7x^2}{2\times 1.5}$ | B1 | Correct EPE when extension is $x$ |
| $\frac{14.7x^2}{3} = 0.6g\times(x+1.5)$ | M1, A1ft | Equating EPE to GPE lost. EPE of form $k\frac{\lambda x^2}{l}$; correct equation ft their EPE |
| $5x^2-6x-9=0$ | — | |
| $x = \frac{6\pm\sqrt{36+180}}{10} = 2.069\ldots$ (or $-0.869$) | DM1, A1 | Solve quadratic (formula must be correct); $x=2.069\ldots$, neg value not needed |
| $OA = 3.569\ldots \approx 3.6$ or $3.57$ | A1 (6) | Add 1.5 to 2.069; answer to 2 or 3 sf |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{14.7\times 2.069}{1.5} - 0.6\times 9.8 = 0.6a$ | M1, A1ft | Use NL2 at $A$ inc use of Hooke's law. Formula for HL correct. Ext can be $(3.569-1.5)$; correct numbers ft their extension |
| $a = 23.993\ldots \approx 24$ or $24.0$ | A1cao (3) | 24 or 24.0 only. No negatives allowed |
---3. One end of a light elastic string, of natural length 1.5 m and modulus of elasticity 14.7 N ,\\
3. One end of a light elastic string, of natural length 1.5 m and modulus of elasticity 14.7 N , is attached to a fixed point $O$ on a ceiling. A particle $P$ of mass 0.6 kg is attached to the free end of the string. The particle is held at $O$ and released from rest. The particle comes to instantaneous rest for the first time at the point $A$.
Find
\begin{enumerate}[label=(\alph*)]
\item the distance $O A$,
\item the magnitude of the instantaneous acceleration of $P$ at $A$.\\
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4c1c51ff-6ae8-402d-b303-b656d26e4230-05_620_956_118_500}
\captionsetup{labelformat=empty}
\caption{igure 2}
\end{center}
\end{figure}
A uniform solid $S$ consists of two right circular cones of base radius $r$. The smaller cone has height $2 h$ and the centre of the plane face of this cone is $O$. The larger cone has height $k h$ where $k > 2$. The two cones are joined so that their plane faces coincide, as shown in Figure 2.\\
(a) Show that the distance of the centre of mass of $S$ from $O$ is
$$\frac { h } { 4 } ( k - 2 )$$
The point $A$ lies on the circumference of the base of one of the cones. The solid is suspended by a string attached at $A$ and hangs freely in equilibrium.
Given that $r = 3 h$ and $k = 6$\\
(b) find the size of the angle between $A O$ and the vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2016 Q3 [9]}}