| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Year | 2020 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Verify dimensional consistency |
| Difficulty | Standard +0.3 This is a straightforward dimensional analysis question requiring basic manipulation of dimensions (M, L, T) and physical reasoning. Part (a) involves routine equating of dimensions to find exponents, parts (b) and (d) require simple physical insight (mass independence, sign error), and part (c) is mechanical verification. Easier than average as it's mostly procedural with minimal problem-solving. |
| Spec | 6.01b Units vs dimensions: relationship6.01c Dimensional analysis: error checking6.01d Unknown indices: using dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | [ RHS ]=M α( LT −1 )β L γ ( =M α L β+γ T −β) |
| Answer | Marks | Guidance |
|---|---|---|
| So α=0, β+γ=0, −β=1 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| u | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | For a fixed projection speed, the model suggests it is possible | |
| for the particle to reach any height given enough time. | B1 | 3.5b |
| Answer | Marks |
|---|---|
| (c) | u2= ( LT −1 )2 =L2T −2 |
| Answer | Marks | Guidance |
|---|---|---|
| gh | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| u | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| | A1 | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (d) | Since u2 +gh >u, this model predicts a negative value for t | B1 |
Question 2:
2 | (a) | [ RHS ]=M α( LT −1 )β L γ ( =M α L β+γ T −β) | M1 | 1.1 | Setting up the model in terms of M, L and T with
at least two terms correct
So α=0, β+γ=0, −β=1 | A1 | 1.1 | Setting up all three correct equations in α,β and
γ
⇒β=−1, γ=1
kh
So t =
u | A1 | 1.1 | AG (as answer given sufficient working must be
shown)
SC1 for verifying that the given formula is
dimensionally consistent
[3]
(b) | For a fixed projection speed, the model suggests it is possible
for the particle to reach any height given enough time. | B1 | 3.5b | or, (e.g.) ‘Constant speed’
‘Unrealistic to ignore gravity’
[1]
(c) | u2= ( LT −1 )2 =L2T −2
[ ]=LT −2⋅L=L2T −2
gh | M1 | 1.1 | Attempt to check consistency beneath the square root.
Implied by correct terms
u2 +gh = ( L2T −2 )1 2 =LT −1
[ ]=LT −1
u | M1 | 1.1 | Attempt to check consistency in the numerator.
u− u2 +gh = LT −1 =T and [ t ]=T
g LT −2
| A1 | 2.1 | Correctly showing that both sides have the same
dimensions
A0 if working includes
L2T −2 +L2T −2 =LT −1+LT −1 etc
[3]
(d) | Since u2 +gh >u, this model predicts a negative value for t | B1 | 2.3 | oe
[1]2 George is investigating the time it takes for a ball to reach a certain height when projected vertically upwards. George believes that the time, $t$, for the ball to reach a certain height, $h$, depends on
\begin{itemize}
\item the ball's mass $m$,
\item the projection speed $u$, and
\item the height $h$.
\end{itemize}
George suggests the following formula to model this situation\\
$t = k m ^ { \alpha } u ^ { \beta } h ^ { \gamma }$,\\
where $k$ is a dimensionless constant.
\begin{enumerate}[label=(\alph*)]
\item Use dimensional analysis to show that $t = \frac { k h } { u }$.
\item Hence explain why George's formula is unrealistic.
Mandy argues that any model of this situation must consider the acceleration due to gravity, $g$. She suggests the alternative formula\\
$t = \frac { u - \sqrt { u ^ { 2 } + g h } } { g }$.
\item Show that Mandy's formula is dimensionally consistent.
\item Explain why Mandy's formula is incorrect.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2020 Q2 [8]}}