| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Verify dimensional consistency |
| Difficulty | Standard +0.8 This is a substantial multi-part question requiring elastic energy calculations, energy conservation, equilibrium analysis, and dimensional analysis to find unknown exponents in a formula. While parts (i)-(iii) are standard M3 mechanics, parts (iv)-(v) require systematic dimensional analysis—a technique that's conceptually straightforward but requires careful bookkeeping. The question is longer and more involved than typical A-level questions but doesn't require exceptional insight, placing it moderately above average difficulty. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 06.02g 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 | Marks | Guidance |
| Elastic energy is \(\frac{1}{2}\times\frac{573.3}{3.9}\times 0.9^2\) | M1 | Allow one error |
| \(= 59.535\) J | A1 | (Allow 60, A0 for 59) |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Length of string at bottom is \(2\sqrt{1.8^2 + 2.4^2}\ (= 6)\) | M1 | Finding length of string (or half-string) |
| \(\frac{1}{2}\times\frac{573.3}{3.9}\times(2.1^2 - 0.9^2) = m\times 9.8\times 1.8\) | M1 B1B1 | Equation involving EE and PE; For change in EE and change in PE |
| \(324.135 - 59.535 = 17.64m\) | ||
| Mass is \(15\) kg | E1 | |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Length of string is \(2\sqrt{1.0^2 + 2.4^2} = 5.2\) | ||
| Tension \(T = \frac{573.3}{3.9}\times 1.3\ (= 191.1)\) | M1 A1 | Finding tension (via Hooke's law) |
| \(2T\sin\alpha - mg = 2\times 191.1\times\frac{1.0}{2.6} - 15\times 9.8\) | M1 | Finding vertical component of tension; Give A1 for \(T = 191.1\) obtained from resolving vertically |
| \(= 147 - 147 = 0\), hence it is in equilibrium | E1 | *SC* If 573.3 is used as stiffness: (i) M1A0 (ii) M1M1B0B1E0 (iii) M1A1 (745.29) M1E0 |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([8\pi^2 h^3] = L^3\), \([8h^3 - ad^2] = L^3\) | *Condone* '\(L^3/L^3 = 0\), dimensionless'; But E0 for \(\frac{L^3}{L^3 - L^3} = \frac{L^3}{0}\) | |
| So \(\frac{8\pi^2 h^3}{8h^3 - ad^2}\) is dimensionless | E1 | |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T = M^\alpha L^\beta (MLT^{-2})^\gamma\) | B1 | For \([\lambda] = MLT^{-2}\) |
| \(\gamma = -\frac{1}{2}\) | B1 | If \(\gamma\) is wrong but non-zero, give B1 ft for \(\alpha = \beta = -\gamma\) |
| \(\alpha + \gamma = 0\), so \(\alpha = \frac{1}{2}\) | B1 | |
| \(\beta + \gamma = 0\), so \(\beta = \frac{1}{2}\) | B1 | |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a = 3.9,\ \lambda = 573.3,\ d = 4.8,\ h = 2.6,\ m = 15\) | ||
| Period is \(\sqrt{\frac{8\pi^2 h^3}{8h^3 - ad^2}}\ m^{1/2}\ a^{1/2}\ \lambda^{-1/2} = 1.67\) s (3 sf) | M1 A1 cao | Using formula with numerical \(\alpha, \beta, \gamma\) (must use the complete formula) |
| Total | 2 |
# Question 3:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Elastic energy is $\frac{1}{2}\times\frac{573.3}{3.9}\times 0.9^2$ | M1 | Allow one error |
| $= 59.535$ J | A1 | (Allow 60, A0 for 59) |
| **Total** | **2** | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Length of string at bottom is $2\sqrt{1.8^2 + 2.4^2}\ (= 6)$ | M1 | Finding length of string (or half-string) |
| $\frac{1}{2}\times\frac{573.3}{3.9}\times(2.1^2 - 0.9^2) = m\times 9.8\times 1.8$ | M1 B1B1 | Equation involving EE and PE; For change in EE and change in PE |
| $324.135 - 59.535 = 17.64m$ | | |
| Mass is $15$ kg | E1 | |
| **Total** | **5** | |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Length of string is $2\sqrt{1.0^2 + 2.4^2} = 5.2$ | | |
| Tension $T = \frac{573.3}{3.9}\times 1.3\ (= 191.1)$ | M1 A1 | Finding tension (via Hooke's law) |
| $2T\sin\alpha - mg = 2\times 191.1\times\frac{1.0}{2.6} - 15\times 9.8$ | M1 | Finding vertical component of tension; Give A1 for $T = 191.1$ obtained from resolving vertically |
| $= 147 - 147 = 0$, hence it is in equilibrium | E1 | *SC* If 573.3 is used as stiffness: (i) M1A0 (ii) M1M1B0B1E0 (iii) M1A1 (745.29) M1E0 |
| **Total** | **4** | |
## Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[8\pi^2 h^3] = L^3$, $[8h^3 - ad^2] = L^3$ | | *Condone* '$L^3/L^3 = 0$, dimensionless'; But E0 for $\frac{L^3}{L^3 - L^3} = \frac{L^3}{0}$ |
| So $\frac{8\pi^2 h^3}{8h^3 - ad^2}$ is dimensionless | E1 | |
| **Total** | **1** | |
## Part (v)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = M^\alpha L^\beta (MLT^{-2})^\gamma$ | B1 | For $[\lambda] = MLT^{-2}$ |
| $\gamma = -\frac{1}{2}$ | B1 | If $\gamma$ is wrong but non-zero, give B1 ft for $\alpha = \beta = -\gamma$ |
| $\alpha + \gamma = 0$, so $\alpha = \frac{1}{2}$ | B1 | |
| $\beta + \gamma = 0$, so $\beta = \frac{1}{2}$ | B1 | |
| **Total** | **4** | |
## Part (vi)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a = 3.9,\ \lambda = 573.3,\ d = 4.8,\ h = 2.6,\ m = 15$ | | |
| Period is $\sqrt{\frac{8\pi^2 h^3}{8h^3 - ad^2}}\ m^{1/2}\ a^{1/2}\ \lambda^{-1/2} = 1.67$ s (3 sf) | M1 A1 cao | Using formula with numerical $\alpha, \beta, \gamma$ (must use the complete formula) |
| **Total** | **2** | |
---3 Fixed points A and B are 4.8 m apart on the same horizontal level. The midpoint of AB is M . A light elastic string, with natural length 3.9 m and modulus of elasticity 573.3 N , has one end attached to A and the other end attached to $\mathbf { B }$.\\
(i) Find the elastic energy stored in the string.
A particle P is attached to the midpoint of the string, and is released from rest at M . It comes instantaneously to rest when P is 1.8 m vertically below M .\\
(ii) Show that the mass of P is 15 kg .\\
(iii) Verify that P can rest in equilibrium when it is 1.0 m vertically below M .
In general, a light elastic string, with natural length $a$ and modulus of elasticity $\lambda$, has its ends attached to fixed points which are a distance $d$ apart on the same horizontal level. A particle of mass $m$ is attached to the midpoint of the string, and in the equilibrium position each half of the string has length $h$, as shown in Fig. 3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5ecb198d-7863-4fc2-81b6-c8b6c37b1859-4_280_755_1064_696}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
When the particle makes small oscillations in a vertical line, the period of oscillation is given by the formula
$$\sqrt { \frac { 8 \pi ^ { 2 } h ^ { 3 } } { 8 h ^ { 3 } - a d ^ { 2 } } } m ^ { \alpha } a ^ { \beta } \lambda ^ { \gamma }$$
(iv) Show that $\frac { 8 \pi ^ { 2 } h ^ { 3 } } { 8 h ^ { 3 } - a d ^ { 2 } }$ is dimensionless.\\
(v) Use dimensional analysis to find $\alpha , \beta$ and $\gamma$.\\
(vi) Hence find the period when the particle P makes small oscillations in a vertical line centred on the position of equilibrium given in part (iii).
\hfill \mbox{\textit{OCR MEI M3 2011 Q3 [18]}}