| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Year | 2018 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Dimensional analysis MI context |
| Difficulty | Standard +0.3 This is a standard dimensional analysis problem requiring students to equate dimensions and solve simultaneous equations. Part (a) is straightforward substitution, and part (b) involves setting up three equations from M, L, T dimensions—routine for Further Maths mechanics students with no novel insight required. |
| Spec | 6.01a Dimensions: M, L, T notation6.01d Unknown indices: using dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([E] = \text{ML}^2\text{T}^{-2}\), \([\omega] = \text{T}^{-1}\), \([I] = \frac{\text{ML}^2\text{T}^{-2}}{(\text{T}^{-1})^2} = \text{ML}^2\) | M1 | Recalls dimensions for energy and angular speed and forms equation for dimensional consistency |
| Verifies dimensions of \(I\) are \(\text{ML}^2\) | R1 | Completes rigorous argument using both dimensions to verify result |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[I^\alpha W^\beta h^\gamma\right] = \left(\text{ML}^2\right)^\alpha \left(\text{MLT}^{-2}\right)^\beta (\text{L})^\gamma = \text{M}^{\alpha+\beta}\text{L}^{2\alpha+\beta+\gamma}\text{T}^{-2\beta}\) | M1 | Uses dimensions to form correct expression for \(\left[I^\alpha W^\beta h^\gamma\right]\) |
| \(\alpha + \beta = 0\), \(-2\beta = 1\), \(2\alpha + \beta + \gamma = 0\) | M1 | Forms three simultaneous equations in three unknowns |
| \(\alpha = 0.5\), \(\beta = -0.5\), \(\gamma = -0.5\) | A1 | Obtains correct values for \(\alpha\), \(\beta\), \(\gamma\) — CAO |
| Total: 5 |
# Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[E] = \text{ML}^2\text{T}^{-2}$, $[\omega] = \text{T}^{-1}$, $[I] = \frac{\text{ML}^2\text{T}^{-2}}{(\text{T}^{-1})^2} = \text{ML}^2$ | M1 | Recalls dimensions for energy and angular speed and forms equation for dimensional consistency |
| Verifies dimensions of $I$ are $\text{ML}^2$ | R1 | Completes rigorous argument using both dimensions to verify result |
| **Total: 2** | | |
---
# Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[I^\alpha W^\beta h^\gamma\right] = \left(\text{ML}^2\right)^\alpha \left(\text{MLT}^{-2}\right)^\beta (\text{L})^\gamma = \text{M}^{\alpha+\beta}\text{L}^{2\alpha+\beta+\gamma}\text{T}^{-2\beta}$ | M1 | Uses dimensions to form correct expression for $\left[I^\alpha W^\beta h^\gamma\right]$ |
| $\alpha + \beta = 0$, $-2\beta = 1$, $2\alpha + \beta + \gamma = 0$ | M1 | Forms three simultaneous equations in three unknowns |
| $\alpha = 0.5$, $\beta = -0.5$, $\gamma = -0.5$ | A1 | Obtains correct values for $\alpha$, $\beta$, $\gamma$ — **CAO** |
| **Total: 5** | | |3 The kinetic energy, $E$, of a compound pendulum is given by
$$E = \frac { 1 } { 2 } I \omega ^ { 2 }$$
where $\omega$ is the angular speed and $I$ is a quantity called the moment of inertia.\\
3
\begin{enumerate}[label=(\alph*)]
\item Show that for this formula to be dimensionally consistent then $I$ must have dimensions $M L ^ { 2 }$, where $M$ represents mass and $L$ represents length.\\[0pt]
[2 marks]\\
3
\item The time, $T$, taken for one complete swing of a pendulum is thought to depend on its moment of inertia, $I$, its weight, $W$, and the distance, $h$, of the centre of mass of the pendulum from the point of suspension.
The formula being proposed is
$$T = k I ^ { \alpha } W ^ { \beta } h ^ { \gamma }$$
where $k$ is a dimensionless constant.
Determine the values of $\alpha , \beta$ and $\gamma$.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2018 Q3 [5]}}