| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Find exponents with partial constraints |
| Difficulty | Standard +0.3 This is a straightforward dimensional analysis question requiring students to equate dimensions on both sides and solve simultaneous equations, followed by a simple substitution case. The method is standard and well-practiced, though it requires careful algebraic manipulation. Slightly easier than average due to the systematic nature of the approach. |
| Spec | 6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship6.01c Dimensional analysis: error checking6.01d Unknown indices: using dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha = \beta\) | B1 | soi – does not need justification |
| \([u] = LT^{-1}\) or \([v] = LT^{-1}\) | B1 | Seen |
| \(L = L^\alpha T^{-\alpha} T^\gamma\) or \(L^\alpha T^{\gamma - \alpha}\) | M1 | No \(k\). Could be \(\beta\). Allow \(L = L^\alpha T^{-\alpha} T^\gamma + L^\beta T^{-\beta} T^\gamma\) with consistent indices, must be expanded, use BOD |
| \(\alpha = 1\) | A1 | or \(\beta = 1\) |
| \(\gamma - \alpha = 0\) | M1 | |
| \(\gamma = 1\) and \(\beta = 1\) | A1 | or \(\alpha = 1\) if \(\beta\) found |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If \(a = 0\) then \(u = v\) and \(s = 2kut...\) | M1 | |
| ...but "dist = speed \(\times\) time" so \(k = \frac{1}{2}\) | A1 | Must include justification. Do not accept use of prior knowledge of uvast |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha = \beta$ | B1 | soi – does not need justification |
| $[u] = LT^{-1}$ or $[v] = LT^{-1}$ | B1 | Seen |
| $L = L^\alpha T^{-\alpha} T^\gamma$ or $L^\alpha T^{\gamma - \alpha}$ | M1 | No $k$. Could be $\beta$. Allow $L = L^\alpha T^{-\alpha} T^\gamma + L^\beta T^{-\beta} T^\gamma$ with consistent indices, must be expanded, use BOD |
| $\alpha = 1$ | A1 | or $\beta = 1$ |
| $\gamma - \alpha = 0$ | M1 | |
| $\gamma = 1$ and $\beta = 1$ | A1 | or $\alpha = 1$ if $\beta$ found |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| If $a = 0$ then $u = v$ **and** $s = 2kut...$ | M1 | |
| ...but "dist = speed $\times$ time" so $k = \frac{1}{2}$ | A1 | Must include justification. Do not accept use of prior knowledge of uvast |
---3 A particle moves in a straight line with constant acceleration. Its initial and final velocities are $u$ and $v$ respectively and at time $t$ its displacement from its starting position is $s$. An equation connecting these quantities is $s = k \left( u ^ { \alpha } + v ^ { \beta } \right) t ^ { \gamma }$, where $k$ is a dimensionless constant.\\
(i) Use dimensional analysis to find the values of $\alpha , \beta$ and $\gamma$.\\
(ii) By considering the case where the acceleration is zero, determine the value of $k$.
\hfill \mbox{\textit{OCR Further Mechanics AS 2018 Q3 [8]}}