| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Verify dimensional consistency |
| Difficulty | Moderate -0.5 Part (a) is a standard M3 elastic strings problem requiring routine equilibrium analysis, energy calculations, and conservation of energy. Part (b) is a straightforward dimensional analysis exercise requiring only basic manipulation of dimensions [M], [L], [T]. While multi-part with several marks, all components are textbook-standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 6.01a Dimensions: M, L, T notation6.01c Dimensional analysis: error checking6.01d Unknown indices: using dimensions6.02g 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 principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AP = \sqrt{2.4^2 + 0.7^2} = 2.5\) | M1 | |
| Tension \(T = 70 \times 0.35 = 24.5\) | A1, M1 | Attempting to resolve vertically |
| Resultant vertical force on P is \(2T\cos\theta - mg\) | ||
| \(= 2 \times 24.5 \times \frac{2.4}{2.5} - 4.8 \times 9.8\) | B1 | For \(T \times \frac{2.4}{2.5}\) (or \(T\cos 16.3°\) etc) |
| \(= 47.04 - 47.04 = 0\) | B1 | For \(4.8 \times 9.8\) |
| Hence P is in equilibrium | E1 | Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(EE = \frac{1}{2} \times 70 \times 0.35^2\) | M1 | (M0 for \(\frac{1}{2} \times 70 \times 0.35\)) |
| Elastic energy is \(4.2875\) J | A1 | *Note* If 70 is used as modulus instead of stiffness: (i) M1A0M1B1B1E0 (ii) M1 A1 for 1.99 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Initial \(KE = \frac{1}{2} \times 4.8 \times 3.5^2\) | B1 | |
| By conservation of energy: \(4.8 \times 9.8h = 2 \times 4.2875 + \frac{1}{2} \times 4.8 \times 3.5^2\) | M1, F1 | Equation involving EE, KE and PE |
| \(47.04h = 8.575 + 29.4\) | ||
| Height is \(0.807\) m (3 sf) | A1 | (A0 for 0.8) ft is \(\frac{2\times(\text{ii})+29.4}{47.04}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([\text{Force}] = \text{MLT}^{-2}\) | B1 | *Deduct 1 mark if units are used* |
| \([\text{Stiffness}] = \text{MT}^{-2}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{LT}^{-1} = \text{M}^\alpha(\text{MT}^{-2})^\beta \text{L}^\gamma\) | ||
| \(\gamma = 1\) | B1 | |
| \(\beta = \frac{1}{2}\) | B1 | |
| \(0 = \alpha + \beta\) | M1 | Considering powers of M |
| \(\alpha = -\frac{1}{2}\) | A1 | *When [Stiffness] is wrong in (i), allow all marks ft provided the work is comparable and answers are non-zero* |
# Question 1:
## Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AP = \sqrt{2.4^2 + 0.7^2} = 2.5$ | M1 | |
| Tension $T = 70 \times 0.35 = 24.5$ | A1, M1 | Attempting to resolve vertically |
| Resultant vertical force on P is $2T\cos\theta - mg$ | | |
| $= 2 \times 24.5 \times \frac{2.4}{2.5} - 4.8 \times 9.8$ | B1 | For $T \times \frac{2.4}{2.5}$ (or $T\cos 16.3°$ etc) |
| $= 47.04 - 47.04 = 0$ | B1 | For $4.8 \times 9.8$ |
| Hence P is in equilibrium | E1 | Correctly shown |
## Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $EE = \frac{1}{2} \times 70 \times 0.35^2$ | M1 | (M0 for $\frac{1}{2} \times 70 \times 0.35$) |
| Elastic energy is $4.2875$ J | A1 | *Note* If 70 is used as modulus instead of stiffness: (i) M1A0M1B1B1E0 (ii) M1 A1 for 1.99 |
## Part (a)(iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Initial $KE = \frac{1}{2} \times 4.8 \times 3.5^2$ | B1 | |
| By conservation of energy: $4.8 \times 9.8h = 2 \times 4.2875 + \frac{1}{2} \times 4.8 \times 3.5^2$ | M1, F1 | Equation involving EE, KE and PE |
| $47.04h = 8.575 + 29.4$ | | |
| Height is $0.807$ m (3 sf) | A1 | (A0 for 0.8) ft is $\frac{2\times(\text{ii})+29.4}{47.04}$ |
## Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\text{Force}] = \text{MLT}^{-2}$ | B1 | *Deduct 1 mark if units are used* |
| $[\text{Stiffness}] = \text{MT}^{-2}$ | B1 | |
## Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{LT}^{-1} = \text{M}^\alpha(\text{MT}^{-2})^\beta \text{L}^\gamma$ | | |
| $\gamma = 1$ | B1 | |
| $\beta = \frac{1}{2}$ | B1 | |
| $0 = \alpha + \beta$ | M1 | Considering powers of M |
| $\alpha = -\frac{1}{2}$ | A1 | *When [Stiffness] is wrong in (i), allow all marks ft provided the work is comparable and answers are non-zero* |
---1
\begin{enumerate}[label=(\alph*)]
\item Two light elastic strings, each having natural length 2.15 m and stiffness $70 \mathrm {~N} \mathrm {~m} ^ { - 1 }$, are attached to a particle P of mass 4.8 kg . The other ends of the strings are attached to fixed points A and B , which are 1.4 m apart at the same horizontal level. The particle P is placed 2.4 m vertically below the midpoint of AB , as shown in Fig. 1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c93aed95-f655-45cb-805f-7114a15acccf-2_677_474_482_877}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Show that P is in equilibrium in this position.
\item Find the energy stored in the string AP .
Starting in this equilibrium position, P is set in motion with initial velocity $3.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ vertically upwards. You are given that P first comes to instantaneous rest at a point C where the strings are slack.
\item Find the vertical height of C above the initial position of P .
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Write down the dimensions of force and stiffness (of a spring).
A particle of mass $m$ is performing oscillations with amplitude $a$ on the end of a spring with stiffness $k$. The maximum speed $v$ of the particle is given by $v = c m ^ { \alpha } k ^ { \beta } a ^ { \gamma }$, where $c$ is a dimensionless constant.
\item Use dimensional analysis to find $\alpha , \beta$ and $\gamma$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2010 Q1 [18]}}