| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Verify dimensional consistency |
| Difficulty | Moderate -0.3 This is a straightforward dimensional analysis question requiring routine application of standard techniques. Part (a) involves basic dimensional checking and simple algebraic correction of a formula, while part (b) uses standard elastic energy and conservation of energy methods. All steps are textbook-standard with no novel problem-solving 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 | Mark | Guidance |
| Power = Work/time = Force × velocity | M1 | Or energy/time |
| \(= \text{MLT}^{-2} \times \text{LT}^{-1}\) | A1 | |
| \([\text{P}] = \text{ML}^2\text{T}^{-3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([x] = \text{L}\) | B1 | |
| \([r^3] = \text{L}^3\), \([u^2] = \text{L}^2\text{T}^{-2}\), \([\rho] = \text{ML}^{-3}\) | B1 | At least two correct |
| \(\left[\frac{r^3 u^2 \rho}{P}\right] = \frac{\text{L}^3 \cdot \text{L}^2\text{T}^{-2} \cdot \text{ML}^{-3}}{\text{ML}^2\text{T}^{-3}}\) | M1 | Substituting dimensions |
| \(= \frac{\text{ML}^2\text{T}^{-2}}{\text{ML}^2\text{T}^{-3}} = \text{T}\) | A1 | |
| This gives \([\text{T}] \neq \text{L}\), so not consistent | A1 | Clear conclusion required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Need extra factor of \(\text{LT}^{-1}\), i.e. one more power of \(u\) | M1 | Identifying \(u\) needs different power |
| \(x = \dfrac{28\pi r^3 u^3 \rho}{9P}\) | A1 | |
| Check: dimensions are now L ✓ | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Extension \(= 2.4 - 1.6 = 0.8\) m | B1 | |
| \(E = \frac{1}{2} \times 150 \times 0.8^2 = 48\) J | M1 A1 | Using \(E = \frac{1}{2}ke^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| New length of each half: \(\sqrt{1.2^2 + 0.9^2} = 1.5\) m | M1 | |
| Natural length of each half \(= 0.8\) m, extension \(= 0.7\) m | A1 | |
| Energy stored in each half \(= \frac{1}{2} \times 150 \times 0.7^2 = 36.75\) J | M1 | |
| Total energy stored \(= 2 \times 36.75 = 73.5\) J | A1 | |
| Energy conservation: \(\frac{1}{2}mv^2 + 48 = 73.5\) | M1 | |
| \(\frac{1}{2}m \times 100 = 25.5\) | A1 | |
| \(m = 0.51\) kg | A1 |
# Question 1:
## Part (a)(i) - Dimensions of Power
| Answer | Mark | Guidance |
|--------|------|----------|
| Power = Work/time = Force × velocity | M1 | Or energy/time |
| $= \text{MLT}^{-2} \times \text{LT}^{-1}$ | A1 | |
| $[\text{P}] = \text{ML}^2\text{T}^{-3}$ | A1 | |
## Part (a)(ii) - Show formula not dimensionally consistent
| Answer | Mark | Guidance |
|--------|------|----------|
| $[x] = \text{L}$ | B1 | |
| $[r^3] = \text{L}^3$, $[u^2] = \text{L}^2\text{T}^{-2}$, $[\rho] = \text{ML}^{-3}$ | B1 | At least two correct |
| $\left[\frac{r^3 u^2 \rho}{P}\right] = \frac{\text{L}^3 \cdot \text{L}^2\text{T}^{-2} \cdot \text{ML}^{-3}}{\text{ML}^2\text{T}^{-3}}$ | M1 | Substituting dimensions |
| $= \frac{\text{ML}^2\text{T}^{-2}}{\text{ML}^2\text{T}^{-3}} = \text{T}$ | A1 | |
| This gives $[\text{T}] \neq \text{L}$, so not consistent | A1 | Clear conclusion required |
## Part (a)(iii) - Correct formula
| Answer | Mark | Guidance |
|--------|------|----------|
| Need extra factor of $\text{LT}^{-1}$, i.e. one more power of $u$ | M1 | Identifying $u$ needs different power |
| $x = \dfrac{28\pi r^3 u^3 \rho}{9P}$ | A1 | |
| Check: dimensions are now L ✓ | A1 | |
## Part (b)(i) - Energy stored in string
| Answer | Mark | Guidance |
|--------|------|----------|
| Extension $= 2.4 - 1.6 = 0.8$ m | B1 | |
| $E = \frac{1}{2} \times 150 \times 0.8^2 = 48$ J | M1 A1 | Using $E = \frac{1}{2}ke^2$ |
## Part (b)(ii) - Mass of particle
| Answer | Mark | Guidance |
|--------|------|----------|
| New length of each half: $\sqrt{1.2^2 + 0.9^2} = 1.5$ m | M1 | |
| Natural length of each half $= 0.8$ m, extension $= 0.7$ m | A1 | |
| Energy stored in each half $= \frac{1}{2} \times 150 \times 0.7^2 = 36.75$ J | M1 | |
| Total energy stored $= 2 \times 36.75 = 73.5$ J | A1 | |
| Energy conservation: $\frac{1}{2}mv^2 + 48 = 73.5$ | M1 | |
| $\frac{1}{2}m \times 100 = 25.5$ | A1 | |
| $m = 0.51$ kg | A1 | |
---1
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the dimensions of power.
In a particle accelerator operating at power $P$, a charged sphere of radius $r$ and density $\rho$ has its speed increased from $u$ to $2 u$ over a distance $x$. A student derives the formula
$$x = \frac { 28 \pi r ^ { 3 } u ^ { 2 } \rho } { 9 P }$$
\item Show that this formula is not dimensionally consistent.
\item Given that there is only one error in this formula for $x$, obtain the correct formula.
\end{enumerate}\item A light elastic string, with natural length 1.6 m and stiffness $150 \mathrm { Nm } ^ { - 1 }$, is stretched between fixed points A and B which are 2.4 m apart on a smooth horizontal surface.
\begin{enumerate}[label=(\roman*)]
\item Find the energy stored in the string.
A particle is attached to the mid-point of the string. The particle is given a horizontal velocity of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ perpendicular to AB (see Fig. 1.1), and it comes instantaneously to rest after travelling a distance of 0.9 m (see Fig. 1.2).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5bb02383-91c0-4454-aaea-0bd6af6ba325-2_524_305_1274_639}
\captionsetup{labelformat=empty}
\caption{Fig. 1.1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5bb02383-91c0-4454-aaea-0bd6af6ba325-2_524_305_1274_1128}
\captionsetup{labelformat=empty}
\caption{Fig. 1.2}
\end{center}
\end{figure}
\item Find the mass of the particle.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2006 Q1 [18]}}