| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Derive dimensions from formula |
| Difficulty | Moderate -0.8 This is a straightforward dimensional analysis question requiring only systematic application of standard techniques. Parts (i), (ii), and (iv) involve basic recall and substitution of dimensions, while parts (iii) and (v) require routine unit conversion and solving simultaneous equations from dimensional consistency—all standard M3 exercises with no novel problem-solving required. |
| Spec | 6.01a Dimensions: M, L, T notation6.01c Dimensional analysis: error checking6.01d Unknown indices: using dimensions6.01e Formulate models: dimensional arguments |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Velocity: \(LT^{-1}\) | B1 | |
| Acceleration: \(LT^{-2}\) | B1 | |
| Force: \(MLT^{-2}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([F] = [G][m_1][m_2]/[r^2]\), so \([G] = [F][r^2]/[m_1][m_2]\) | M1 | Rearranging for G |
| \(= MLT^{-2} \cdot L^2 / M^2 = M^{-1}L^3T^{-2}\) | A1 | Correct result shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Need to convert: 1 kg = \(\frac{1}{0.4536}\) lb, 1 m = \(\frac{1}{0.3048}\) ft | M1 | Setting up conversion |
| \(G = 6.67 \times 10^{-11} \times \frac{(0.3048)^3}{0.4536}\) | M1 | Correct powers of conversion factors |
| \(G \approx 1.07 \times 10^{-9}\) (imperial units) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([v] = LT^{-1}\) | B1 | |
| \(\left[\sqrt{\frac{2Gm}{r}}\right] = \sqrt{M^{-1}L^3T^{-2} \cdot M \cdot L^{-1}}\) | M1 | Substituting dimensions of G |
| \(= \sqrt{L^2T^{-2}} = LT^{-1}\) ✓ | A1 | Correct conclusion stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([t] = T\), \([G^\alpha M^\beta R^\gamma] = (M^{-1}L^3T^{-2})^\alpha \cdot M^\beta \cdot L^\gamma\) | M1 | Setting up dimensional equation |
| M: \(0 = -\alpha + \beta\) | A1 | Correct M equation |
| L: \(0 = 3\alpha + \gamma\) | A1 | Correct L equation |
| T: \(1 = -2\alpha\) | A1 | Correct T equation |
| \(\alpha = -\frac{1}{2}\), \(\beta = -\frac{1}{2}\), \(\gamma = \frac{3}{2}\) | A1 | All three correct |
# Question 1:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Velocity: $LT^{-1}$ | B1 | |
| Acceleration: $LT^{-2}$ | B1 | |
| Force: $MLT^{-2}$ | B1 | |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[F] = [G][m_1][m_2]/[r^2]$, so $[G] = [F][r^2]/[m_1][m_2]$ | M1 | Rearranging for G |
| $= MLT^{-2} \cdot L^2 / M^2 = M^{-1}L^3T^{-2}$ | A1 | Correct result shown |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Need to convert: 1 kg = $\frac{1}{0.4536}$ lb, 1 m = $\frac{1}{0.3048}$ ft | M1 | Setting up conversion |
| $G = 6.67 \times 10^{-11} \times \frac{(0.3048)^3}{0.4536}$ | M1 | Correct powers of conversion factors |
| $G \approx 1.07 \times 10^{-9}$ (imperial units) | A1 | Correct answer |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[v] = LT^{-1}$ | B1 | |
| $\left[\sqrt{\frac{2Gm}{r}}\right] = \sqrt{M^{-1}L^3T^{-2} \cdot M \cdot L^{-1}}$ | M1 | Substituting dimensions of G |
| $= \sqrt{L^2T^{-2}} = LT^{-1}$ ✓ | A1 | Correct conclusion stated |
## Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[t] = T$, $[G^\alpha M^\beta R^\gamma] = (M^{-1}L^3T^{-2})^\alpha \cdot M^\beta \cdot L^\gamma$ | M1 | Setting up dimensional equation |
| M: $0 = -\alpha + \beta$ | A1 | Correct M equation |
| L: $0 = 3\alpha + \gamma$ | A1 | Correct L equation |
| T: $1 = -2\alpha$ | A1 | Correct T equation |
| $\alpha = -\frac{1}{2}$, $\beta = -\frac{1}{2}$, $\gamma = \frac{3}{2}$ | A1 | All three correct |
---1 (i) Write down the dimensions of velocity, acceleration and force.
The force $F$ of gravitational attraction between two objects with masses $m _ { 1 }$ and $m _ { 2 }$, at a distance $r$ apart, is given by
$$F = \frac { G m _ { 1 } m _ { 2 } } { r ^ { 2 } }$$
where $G$ is the universal constant of gravitation.\\
(ii) Show that the dimensions of $G$ are $\mathrm { M } ^ { - 1 } \mathrm {~L} ^ { 3 } \mathrm {~T} ^ { - 2 }$.\\
(iii) In SI units (based on the kilogram, metre and second) the value of $G$ is $6.67 \times 10 ^ { - 11 }$.
Find the value of $G$ in imperial units based on the pound $( 0.4536 \mathrm {~kg} )$, foot $( 0.3048 \mathrm {~m} )$ and second.\\
(iv) For a planet of mass $m$ and radius $r$, the escape velocity $v$ from the planet's surface is given by
$$v = \sqrt { \frac { 2 G m } { r } }$$
Show that this formula is dimensionally consistent.\\
(v) For a planet in circular orbit of radius $R$ round a star of mass $M$, the time $t$ taken to complete one orbit is given by
$$t = k G ^ { \alpha } M ^ { \beta } R ^ { \gamma }$$
where $k$ is a dimensionless constant.\\
Use dimensional analysis to find $\alpha , \beta$ and $\gamma$.
\hfill \mbox{\textit{OCR MEI M3 2007 Q1 [16]}}