OCR MEI M3 2007 June — Question 1 18 marks

Exam BoardOCR MEI
ModuleM3 (Mechanics 3)
Year2007
SessionJune
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDimensional Analysis
TypeVerify dimensional consistency
DifficultyEasy -1.2 Part (a) is pure recall of standard dimensional formulas and routine verification that terms in Bernoulli's equation have matching dimensions—a textbook exercise requiring no problem-solving. Part (b) involves standard SHM with given amplitude and period, requiring only direct application of formulas. The entire question tests basic recall and straightforward application of well-practiced techniques with no novel insight required.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship

1
    1. Write down the dimensions of the following quantities. \begin{displayquote} Velocity
      Acceleration
      Force
      Density (which is mass per unit volume)
      Pressure (which is force per unit area) \end{displayquote} For a fluid with constant density \(\rho\), the velocity \(v\), pressure \(P\) and height \(h\) at points on a streamline are related by Bernoulli's equation $$P + \frac { 1 } { 2 } \rho v ^ { 2 } + \rho g h = \mathrm { constant } ,$$ where \(g\) is the acceleration due to gravity.
    2. Show that the left-hand side of Bernoulli's equation is dimensionally consistent.
  1. In a wave tank, a float is performing simple harmonic motion with period 3.49 s in a vertical line. The height of the float above the bottom of the tank is \(h \mathrm {~m}\) at a time \(t \mathrm {~s}\). When \(t = 0\), the height has its maximum value. The value of \(h\) varies between 1.6 and 2.2.
    1. Sketch a graph showing how \(h\) varies with \(t\).
    2. Express \(h\) in terms of \(t\).
    3. Find the magnitude and direction of the acceleration of the float when \(h = 1.7\).

Question 1:
Part (a)(i)
AnswerMarks Guidance
AnswerMarks Guidance
Velocity: \(LT^{-1}\)B1
Acceleration: \(LT^{-2}\)B1
Force: \(MLT^{-2}\)B1
Density: \(ML^{-3}\)B1
Pressure: \(ML^{-1}T^{-2}\)B1
Part (a)(ii)
AnswerMarks Guidance
AnswerMarks Guidance
\([P] = ML^{-1}T^{-2}\)B1 May be implied
\([\frac{1}{2}\rho v^2] = ML^{-3} \cdot L^2T^{-2} = ML^{-1}T^{-2}\)M1 A1 Correct working shown
\([\rho g h] = ML^{-3} \cdot LT^{-2} \cdot L = ML^{-1}T^{-2}\)A1 All three terms shown equal
Part (b)(i)
AnswerMarks Guidance
AnswerMarks Guidance
Cosine curve starting at maximum \(h = 2.2\) when \(t=0\)B1 Correct shape
Period 3.49, amplitude 0.3, oscillating between 1.6 and 2.2B1 Correct features labelled
Part (b)(ii)
AnswerMarks Guidance
AnswerMarks Guidance
Centre: \(\frac{1.6+2.2}{2} = 1.9\), amplitude \(= 0.3\)B1
\(\omega = \frac{2\pi}{3.49}\)M1
\(h = 1.9 + 0.3\cos\!\left(\frac{2\pi t}{3.49}\right)\)A1 A1 A1 for 1.9 + 0.3cos, A1 for correct \(\omega\)
Part (b)(iii)
AnswerMarks Guidance
AnswerMarks Guidance
\(h - 1.9 = 0.3\cos(\omega t)\), so displacement from centre \(x = h - 1.9 = -0.2\)M1
\(a = -\omega^2 x = -\left(\frac{2\pi}{3.49}\right)^2(-0.2)\)M1
\(a = 0.647\ \text{m s}^{-2}\) (upwards, towards centre)A1
# Question 1:

## Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Velocity: $LT^{-1}$ | B1 | |
| Acceleration: $LT^{-2}$ | B1 | |
| Force: $MLT^{-2}$ | B1 | |
| Density: $ML^{-3}$ | B1 | |
| Pressure: $ML^{-1}T^{-2}$ | B1 | |

## Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P] = ML^{-1}T^{-2}$ | B1 | May be implied |
| $[\frac{1}{2}\rho v^2] = ML^{-3} \cdot L^2T^{-2} = ML^{-1}T^{-2}$ | M1 A1 | Correct working shown |
| $[\rho g h] = ML^{-3} \cdot LT^{-2} \cdot L = ML^{-1}T^{-2}$ | A1 | All three terms shown equal |

## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Cosine curve starting at maximum $h = 2.2$ when $t=0$ | B1 | Correct shape |
| Period 3.49, amplitude 0.3, oscillating between 1.6 and 2.2 | B1 | Correct features labelled |

## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Centre: $\frac{1.6+2.2}{2} = 1.9$, amplitude $= 0.3$ | B1 | |
| $\omega = \frac{2\pi}{3.49}$ | M1 | |
| $h = 1.9 + 0.3\cos\!\left(\frac{2\pi t}{3.49}\right)$ | A1 A1 | A1 for 1.9 + 0.3cos, A1 for correct $\omega$ |

## Part (b)(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $h - 1.9 = 0.3\cos(\omega t)$, so displacement from centre $x = h - 1.9 = -0.2$ | M1 | |
| $a = -\omega^2 x = -\left(\frac{2\pi}{3.49}\right)^2(-0.2)$ | M1 | |
| $a = 0.647\ \text{m s}^{-2}$ (upwards, towards centre) | A1 | |

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1
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down the dimensions of the following quantities.

\begin{displayquote}
Velocity\\
Acceleration\\
Force\\
Density (which is mass per unit volume)\\
Pressure (which is force per unit area)
\end{displayquote}

For a fluid with constant density $\rho$, the velocity $v$, pressure $P$ and height $h$ at points on a streamline are related by Bernoulli's equation

$$P + \frac { 1 } { 2 } \rho v ^ { 2 } + \rho g h = \mathrm { constant } ,$$

where $g$ is the acceleration due to gravity.
\item Show that the left-hand side of Bernoulli's equation is dimensionally consistent.
\end{enumerate}\item In a wave tank, a float is performing simple harmonic motion with period 3.49 s in a vertical line. The height of the float above the bottom of the tank is $h \mathrm {~m}$ at a time $t \mathrm {~s}$. When $t = 0$, the height has its maximum value. The value of $h$ varies between 1.6 and 2.2.
\begin{enumerate}[label=(\roman*)]
\item Sketch a graph showing how $h$ varies with $t$.
\item Express $h$ in terms of $t$.
\item Find the magnitude and direction of the acceleration of the float when $h = 1.7$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI M3 2007 Q1 [18]}}