| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: general symbolic/proof questions |
| Difficulty | Standard +0.3 This is a standard M3 elastic string energy problem with straightforward application of conservation of energy (GPE + KE + EPE), followed by routine calculus to find maxima. The multi-part structure and need to work with the given energy equation adds slight complexity, but all steps follow predictable patterns for this topic with no novel insight required. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gain in EE \(= 20x^2/(2\times2)\) | B1 | Accept \(0.8gx\) if gain in KE is \(\frac{1}{2}0.8(v^2 - 19.6)\) |
| Loss in GPE \(= 0.8g(2+x)\) | B1 | |
| \([\frac{1}{2}\cdot0.8v^2 = (15.68 + 7.84x) - 5x^2]\) | M1 | For using p.c. energy |
| \(v^2 = 39.2 + 19.6x - 12.5x^2\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Maximum extension is \(2.72\)m | M1 | For attempting to solve \(v^2 = 0\) |
| A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([19.6 - 25x = 0\), \(v^2 = 46.8832 - 12.5(x - 0.784)^2]\) | M1 | For solving \(20x/2 = 0.8g\) or differentiating and attempting to solve \(d(v^2)/dx = 0\) or \(dv/dx = 0\), or expressing \(v^2\) in form \(c - a(x-b)^2\) |
| \(x = 0.784\) or \(c = 46.9\) | A1 | |
| \([v_{\max}^2 = 39.2 + 15.3664 - 7.6832]\) | M1 | For substituting \(x = 0.784\) in expression for \(v^2\) or evaluating \(\sqrt{c}\) |
| Maximum speed is \(6.85\ \text{ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\pm(0.8g - 20x/2) = 0.8a\) | M1 | For using Newton's second law (3 terms required) or \(a = v\ dv/dx\) |
| or \(2v\ dv/dx = 19.6 - 25x\) | A1 | |
| \(a = \pm(9.8 - 12.5x)\) | ||
| or \(\ddot{y} = -12.5y\) where \(y = x - 0.784\) | A1 | |
| \([\lvert a\rvert_{\max} = \lvert 9.8 - 12.5\times2.72\rvert]\) | M1 | For substituting \(x = \text{ans(ii)(a)}\) into \(a(x)\) or \(y = \text{ans(ii)(a)} - 0.784\) into \(\ddot{y}(y)\) |
| or \(\lvert\ddot{y}_{\max}\rvert = \lvert{-12.5(2.72 - 0.784)}\rvert\) | A1 | |
| Maximum magnitude is \(24.2\ \text{ms}^{-2}\) |
## Question 7:
**Part (i)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gain in EE $= 20x^2/(2\times2)$ | B1 | Accept $0.8gx$ if gain in KE is $\frac{1}{2}0.8(v^2 - 19.6)$ |
| Loss in GPE $= 0.8g(2+x)$ | B1 | |
| $[\frac{1}{2}\cdot0.8v^2 = (15.68 + 7.84x) - 5x^2]$ | M1 | For using p.c. energy |
| $v^2 = 39.2 + 19.6x - 12.5x^2$ | A1 | AG |
**Part (ii)(a)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Maximum extension is $2.72$m | M1 | For attempting to solve $v^2 = 0$ |
| | A1 | |
**Part (ii)(b)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[19.6 - 25x = 0$, $v^2 = 46.8832 - 12.5(x - 0.784)^2]$ | M1 | For solving $20x/2 = 0.8g$ or differentiating and attempting to solve $d(v^2)/dx = 0$ or $dv/dx = 0$, or expressing $v^2$ in form $c - a(x-b)^2$ |
| $x = 0.784$ or $c = 46.9$ | A1 | |
| $[v_{\max}^2 = 39.2 + 15.3664 - 7.6832]$ | M1 | For substituting $x = 0.784$ in expression for $v^2$ or evaluating $\sqrt{c}$ |
| Maximum speed is $6.85\ \text{ms}^{-1}$ | A1 | |
**Part (ii)(c)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm(0.8g - 20x/2) = 0.8a$ | M1 | For using Newton's second law (3 terms required) or $a = v\ dv/dx$ |
| or $2v\ dv/dx = 19.6 - 25x$ | A1 | |
| $a = \pm(9.8 - 12.5x)$ | | |
| or $\ddot{y} = -12.5y$ where $y = x - 0.784$ | A1 | |
| $[\lvert a\rvert_{\max} = \lvert 9.8 - 12.5\times2.72\rvert]$ | M1 | For substituting $x = \text{ans(ii)(a)}$ into $a(x)$ or $y = \text{ans(ii)(a)} - 0.784$ into $\ddot{y}(y)$ |
| or $\lvert\ddot{y}_{\max}\rvert = \lvert{-12.5(2.72 - 0.784)}\rvert$ | A1 | |
| Maximum magnitude is $24.2\ \text{ms}^{-2}$ | | |7 A particle of mass 0.8 kg is attached to one end of a light elastic string of natural length 2 m and modulus of elasticity 20 N . The other end of the string is attached to a fixed point $O$. The particle is held at rest at $O$ and then released. When the extension of the string is $x \mathrm {~m}$, the particle is moving with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) By considering energy show that $v ^ { 2 } = 39.2 + 19.6 x - 12.5 x ^ { 2 }$.\\
(ii) Hence find
\begin{enumerate}[label=(\alph*)]
\item the maximum extension of the string,
\item the maximum speed of the particle,
\item the maximum magnitude of the acceleration of the particle.
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2009 Q7 [15]}}