| Exam Board | OCR |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2018 |
| Session | March |
| Marks | 12 |
| Topic | Dimensional Analysis |
| Type | Find exponents with all unknowns |
| Difficulty | Standard +0.3 This is a standard dimensional analysis question with routine steps: deriving dimensions of pressure from first principles, solving a system of three linear equations for exponents, and applying the formula. Part (iv) adds a simple dimensional consistency check. While it requires careful algebraic manipulation, it follows a well-established template with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship6.01d Unknown indices: using dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([F] = \text{MLT}^{-2}\) | B1 | |
| \([p] = \text{MLT}^{-2}L^{-2} = \text{ML}^{-1}\text{T}^{-2}\) | B1 [2] | AG Must see intermediate step |
| \([\rho] = \text{ML}^{-3}\) | B1 | |
| \([g] = \text{LT}^{-2}\) and \([h] = \text{L}\) | B1 | |
| \(\text{ML}^{-1}\text{T}^{-2} = \text{M}^\alpha\text{L}^{-3\alpha}\text{L}^\gamma\text{T}^{-2\beta}\text{L}^\gamma\) | M1 | |
| \(\alpha = 1, \beta = 1\) | A1 [6] | |
| \(-1 = -3\alpha + \beta + \gamma\) | M1 | |
| \(\gamma = 1\) | A1 | |
| \(S = 4.39 \times 10^{-4}\) | M1 | Accept 4.38 or in exact or uncalculated form |
| \(h = \text{awrt } 2.0 \text{ m}\) | A1 [2] | |
| \([hv^2] = \text{L}^3\text{T}^{-2}\) | M1 | |
| This is not consistent with the other terms and so the equation cannot be correct | A1 [2] |
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[F] = \text{MLT}^{-2}$ | B1 | |
| $[p] = \text{MLT}^{-2}L^{-2} = \text{ML}^{-1}\text{T}^{-2}$ | B1 [2] | AG Must see intermediate step |
| $[\rho] = \text{ML}^{-3}$ | B1 | |
| $[g] = \text{LT}^{-2}$ and $[h] = \text{L}$ | B1 | |
| $\text{ML}^{-1}\text{T}^{-2} = \text{M}^\alpha\text{L}^{-3\alpha}\text{L}^\gamma\text{T}^{-2\beta}\text{L}^\gamma$ | M1 | |
| $\alpha = 1, \beta = 1$ | A1 [6] | |
| $-1 = -3\alpha + \beta + \gamma$ | M1 | |
| $\gamma = 1$ | A1 | |
| $S = 4.39 \times 10^{-4}$ | M1 | Accept 4.38 or in exact or uncalculated form |
| $h = \text{awrt } 2.0 \text{ m}$ | A1 [2] | |
| $[hv^2] = \text{L}^3\text{T}^{-2}$ | M1 | |
| This is not consistent with the other terms and so the equation cannot be correct | A1 [2] | |3 A student is investigating fluid flowing through a pipe.\\
In her first model she assumes a relationship of the form $P = S \rho ^ { \alpha } g ^ { \beta } h ^ { \gamma }$ where $\rho$ is the density of the fluid, $h$ is the length of the pipe, $P$ is the pressure difference between the ends of the pipe, $g$ is the acceleration due to gravity and $S$ is a dimensionless constant. You are given that $\rho$ is measured in $\mathrm { kg } \mathrm { m } ^ { - 3 }$.\\
(i) Use the fact that pressure is force per unit area to show that $[ P ] = \mathrm { ML } ^ { - 1 } \mathrm {~T} ^ { - 2 }$.\\
(ii) Find the values of $\alpha , \beta$ and $\gamma$.
The density of the fluid the student is using is $540 \mathrm {~kg} \mathrm {~m} ^ { - 3 }$. In her experiment she finds that when the length of the pipe is 1.40 m the pressure difference between the ends of the pipe is $3.25 \mathrm { Nm } ^ { - 2 }$.\\
(iii) Find the length of the pipe for which her first model would predict a pressure difference between the ends of the pipe of $4.65 \mathrm { Nm } ^ { - 2 }$.
In an alternative model the student suggests a modified relationship of the form $P = S \rho ^ { \alpha } g ^ { \beta } h ^ { \gamma } + \frac { 1 } { 2 } h v ^ { 2 }$, where $v$ is the average velocity of the fluid in the pipe.\\
(iv) Use dimensional analysis to assess the validity of her alternative model.
\hfill \mbox{\textit{OCR FM1 AS 2018 Q3 [12]}}