Question 3 - A-Level Maths

OCR MEI M3 2009 June — Question 3 18 marks

Exam Board	OCR MEI
Module	M3 (Mechanics 3)
Year	2009
Session	June
Marks	18
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Dimensional Analysis
Type	Find exponents with all unknowns
Difficulty	Standard +0.3 Part (a) is a standard dimensional analysis exercise requiring students to equate dimensions and solve simultaneous equations for three exponents—routine for M3 students who have practiced this technique. Part (b) applies standard SHM formulas (ω from period, velocity from energy/SHM equations, and time from phase calculations) with straightforward substitution. While multi-part and requiring several techniques, all steps follow textbook methods without requiring novel insight or complex problem-solving.
Spec	4.10f Simple harmonic motion: x'' = -omega^2 x 6.01a Dimensions: M, L, T notation 6.01c Dimensional analysis: error checking 6.01d Unknown indices: using dimensions

1. Write down the dimensions of velocity, force and density (which is mass per unit volume). A vehicle moving with velocity $v$ experiences a force $F$, due to air resistance, given by $$F = \frac { 1 } { 2 } C \rho ^ { \alpha } v ^ { \beta } A ^ { \gamma }$$ where $\rho$ is the density of the air, $A$ is the cross-sectional area of the vehicle, and $C$ is a dimensionless quantity called the drag coefficient.
2. Use dimensional analysis to find $\alpha , \beta$ and $\gamma$.
A light rod is freely pivoted about a fixed point at one end and has a heavy weight attached to its other end. The rod with the weight attached is oscillating in a vertical plane as a simple pendulum with period 4.3 s . The maximum angle which the rod makes with the vertical is 0.08 radians. You may assume that the motion is simple harmonic.
1. Find the angular speed of the rod when it makes an angle of 0.05 radians with the vertical.
2. Find the time taken for the pendulum to swing directly from a position where the rod makes an angle of 0.05 radians on one side of the vertical to the position where the rod makes an angle of 0.05 radians on the other side of the vertical.

Show mark scheme Show mark scheme source

Question 3:

Part (a)(i):

Answer	Marks	Guidance
Working	Marks	Guidance
$[\text{Velocity}] = \text{LT}^{-1}$	B1	Deduct 1 mark for $\text{ms}^{-1}$ etc
$[\text{Force}] = \text{MLT}^{-2}$	B1
$[\text{Density}] = \text{ML}^{-3}$	B1
Total: 3

Part (a)(ii):

Answer	Marks	Guidance
Working	Marks	Guidance
$\text{MLT}^{-2} = (\text{ML}^{-3})^\alpha (\text{LT}^{-1})^\beta (\text{L}^2)^\gamma$
$\alpha = 1$	B1
$\beta = 2$	B1
$-3\alpha + \beta + 2\gamma = 1$	M1A1	(ft if equation involves $\alpha$, $\beta$ and $\gamma$)
$\gamma = 1$	A1
Total: 5

Part (b)(i):

Answer	Marks	Guidance
Working	Marks	Guidance
$\frac{2\pi}{\omega} = 4.3$	M1
$\omega = \frac{2\pi}{4.3}$ $(= 1.4612)$	A1
$\dot{\theta}^2 = 1.4612^2(0.08^2 - 0.05^2)$	M1	Using $\omega^2(A^2 - \theta^2)$
F1	For RHS
Angular speed is $0.0913\text{ rad s}^{-1}$	A1	(b.o.d. for $v = 0.0913\text{ ms}^{-1}$)
OR $\dot{\theta} = 0.08\omega\cos\omega t = 0.08 \times 1.4612\cos 0.6751 = 0.0913$	M1, F1, A1	Or $\dot{\theta} = (-)\,0.08\omega\sin\omega t = (-)\,0.08 \times 1.4612\sin 0.8957$
Total: 5

Part (b)(ii):

Answer	Marks	Guidance
Working	Marks	Guidance
$\theta = 0.08\sin\omega t$	B1	or $\theta = 0.08\cos\omega t$
When $\theta = 0.05$, $0.08\sin\omega t = 0.05$	M1	Using $\theta = (\pm)0.05$ to obtain equation for $t$; B1M1 above can be earned in (i)
$\omega t = 0.6751$		or $t = 0.613$ from $\theta = 0.08\cos\omega t$
$t = 0.462$	A1 cao	or $t = 1.537$ from $\theta = 0.08\cos\omega t$
Time taken is $2 \times 0.462$	M1	Strategy for finding the required time $(2\times 0.462$ or $\frac{1}{2}\times 4.3 - 2\times 0.613$)
$= 0.924$ s	A1 cao	or $(1.537 - 0.613)$ Dep on first M1
Total: 5	For $\theta = 0.05\sin\omega t$, max B0M1A0M0 (for $0.05 = 0.05\sin\omega t$)

# Question 3:

## Part (a)(i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $[\text{Velocity}] = \text{LT}^{-1}$ | B1 | Deduct 1 mark for $\text{ms}^{-1}$ etc |
| $[\text{Force}] = \text{MLT}^{-2}$ | B1 | |
| $[\text{Density}] = \text{ML}^{-3}$ | B1 | |
| | **Total: 3** | |

## Part (a)(ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\text{MLT}^{-2} = (\text{ML}^{-3})^\alpha (\text{LT}^{-1})^\beta (\text{L}^2)^\gamma$ | | |
| $\alpha = 1$ | B1 | |
| $\beta = 2$ | B1 | |
| $-3\alpha + \beta + 2\gamma = 1$ | M1A1 | (ft if equation involves $\alpha$, $\beta$ and $\gamma$) |
| $\gamma = 1$ | A1 | |
| | **Total: 5** | |

## Part (b)(i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{2\pi}{\omega} = 4.3$ | M1 | |
| $\omega = \frac{2\pi}{4.3}$ $(= 1.4612)$ | A1 | |
| $\dot{\theta}^2 = 1.4612^2(0.08^2 - 0.05^2)$ | M1 | Using $\omega^2(A^2 - \theta^2)$ |
| | F1 | For RHS |
| Angular speed is $0.0913\text{ rad s}^{-1}$ | A1 | (b.o.d. for $v = 0.0913\text{ ms}^{-1}$) |
| OR $\dot{\theta} = 0.08\omega\cos\omega t = 0.08 \times 1.4612\cos 0.6751 = 0.0913$ | M1, F1, A1 | Or $\dot{\theta} = (-)\,0.08\omega\sin\omega t = (-)\,0.08 \times 1.4612\sin 0.8957$ |
| | **Total: 5** | |

## Part (b)(ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\theta = 0.08\sin\omega t$ | B1 | or $\theta = 0.08\cos\omega t$ |
| When $\theta = 0.05$, $0.08\sin\omega t = 0.05$ | M1 | Using $\theta = (\pm)0.05$ to obtain equation for $t$; B1M1 above can be earned in (i) |
| $\omega t = 0.6751$ | | or $t = 0.613$ from $\theta = 0.08\cos\omega t$ |
| $t = 0.462$ | A1 cao | or $t = 1.537$ from $\theta = 0.08\cos\omega t$ |
| Time taken is $2 \times 0.462$ | M1 | Strategy for finding the required time $(2\times 0.462$ or $\frac{1}{2}\times 4.3 - 2\times 0.613$) |
| $= 0.924$ s | A1 cao | or $(1.537 - 0.613)$ Dep on first M1 |
| | **Total: 5** | For $\theta = 0.05\sin\omega t$, max B0M1A0M0 (for $0.05 = 0.05\sin\omega t$) |

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Show LaTeX source

3
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down the dimensions of velocity, force and density (which is mass per unit volume).

A vehicle moving with velocity $v$ experiences a force $F$, due to air resistance, given by

$$F = \frac { 1 } { 2 } C \rho ^ { \alpha } v ^ { \beta } A ^ { \gamma }$$

where $\rho$ is the density of the air, $A$ is the cross-sectional area of the vehicle, and $C$ is a dimensionless quantity called the drag coefficient.
\item Use dimensional analysis to find $\alpha , \beta$ and $\gamma$.
\end{enumerate}\item A light rod is freely pivoted about a fixed point at one end and has a heavy weight attached to its other end. The rod with the weight attached is oscillating in a vertical plane as a simple pendulum with period 4.3 s . The maximum angle which the rod makes with the vertical is 0.08 radians. You may assume that the motion is simple harmonic.
\begin{enumerate}[label=(\roman*)]
\item Find the angular speed of the rod when it makes an angle of 0.05 radians with the vertical.
\item Find the time taken for the pendulum to swing directly from a position where the rod makes an angle of 0.05 radians on one side of the vertical to the position where the rod makes an angle of 0.05 radians on the other side of the vertical.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI M3 2009 Q3 [18]}}

This paper (4 questions)

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Q1 19 Q2 17 Q3 18 Q4 18

Answer	Marks	Guidance
Working	Marks	Guidance
\(\theta = 0.08\sin\omega t\)	B1	or \(\theta = 0.08\cos\omega t\)
When \(\theta = 0.05\), \(0.08\sin\omega t = 0.05\)	M1	Using \(\theta = (\pm)0.05\) to obtain equation for \(t\); B1M1 above can be earned in (i)
\(\omega t = 0.6751\)		or \(t = 0.613\) from \(\theta = 0.08\cos\omega t\)
\(t = 0.462\)	A1 cao	or \(t = 1.537\) from \(\theta = 0.08\cos\omega t\)
Time taken is \(2 \times 0.462\)	M1	Strategy for finding the required time \((2\times 0.462\) or \(\frac{1}{2}\times 4.3 - 2\times 0.613\))
\(= 0.924\) s	A1 cao	or \((1.537 - 0.613)\) Dep on first M1
Total: 5	For \(\theta = 0.05\sin\omega t\), max B0M1A0M0 (for \(0.05 = 0.05\sin\omega t\))