| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Find exponents with all unknowns |
| Difficulty | Standard +0.3 This is a straightforward dimensional analysis problem requiring students to set up and solve three simultaneous equations from matching dimensions. Part (a) is routine mechanics; part (b) is simple ratio calculation; part (c) tests conceptual understanding that dimensionless quantities cannot be determined by dimensional analysis. Easier than average as it's a standard textbook application with clear structure. |
| Spec | 6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship6.01c Dimensional analysis: error checking6.01d Unknown indices: using dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | [a] = LT2 |
| Answer | Marks |
|---|---|
| = 1 and = 1 | B1 |
| Answer | Marks |
|---|---|
| [6] | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 1.1 | www | |
| 2 | (b) | (9.11×10−31)−1 or (1.67×10−27)−1 oe |
| Answer | Marks |
|---|---|
| or awrt 1830 (:1) | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.4 |
| Answer | Marks |
|---|---|
| 1.1 | 1 |
| Answer | Marks |
|---|---|
| times the acceleration of the proton) | Could be either way round |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (c) | Because N is a dimensionless quantity |
| [1] | 3.5b | Must see ‘dimensionless’ or ‘no |
Question 2:
2 | (a) | [a] = LT2
[h] = L and [m] = M and [v] = LT1
M: 0 = 1 + so = 1
L: 1 = 1 + +
T: 2 = 1
= 1 and = 1 | B1
B1
B1
M1
M1
A1
[6] | 3.3
1.2
3.3
1.1
3.4
1.1 | www
2 | (b) | (9.11×10−31)−1 or (1.67×10−27)−1 oe
(9.11×10−31)−1: (1.67×10−27)−1
1.671027:9.111031
or awrt 1830 (:1) | M1
M1
A1
[3] | 3.4
2.2a
1.1 | 1
Using a
m
Division of values or forming a ratio
Accept answers in words (eg the
acceleration of the electron is 1830
times the acceleration of the proton) | Could be either way round
Reciprocals not acceptable
in final answer
2 | (c) | Because N is a dimensionless quantity | E1
[1] | 3.5b | Must see ‘dimensionless’ or ‘no
dimensions’
2 A solenoid is a device formed by winding a wire tightly around a hollow cylinder so that the wire forms (approximately) circular loops along the cylinder (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{9bc86277-9e6b-41f6-a2c3-94c85e7b1360-2_161_691_1681_246}
When the wire carries an electrical current a magnetic field is created inside the solenoid which can cause a particle which is moving inside the solenoid to accelerate.
A student is carrying out experiments on particles moving inside solenoids. His professor suggests that, for a particle of mass $m$ moving with speed $v$ inside a solenoid of length $h$, the acceleration $a$ of the particle can be modelled by a relationship of the form $a = \mathrm { km } ^ { \alpha } \mathrm { v } ^ { \beta } \mathrm { h } ^ { \gamma }$, where $k$ is a constant. The professor tells the student that $[ k ] = \mathrm { MLT } ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Use dimensional analysis to find $\alpha , \beta$ and $\gamma$.
\item The mass of an electron is $9.11 \times 10 ^ { - 31 } \mathrm {~kg}$ and the mass of a proton is $1.67 \times 10 ^ { - 27 } \mathrm {~kg}$.
For an electron and a proton moving inside the same solenoid with the same speed, use the model to find the ratio of the acceleration of the electron to the acceleration of the proton. [3]
\item The professor tells the student that $a$ also depends on the number of turns or loops of wire, $N$, that the solenoid has.
Explain why dimensional analysis cannot be used to determine the dependence of $a$ on $N$. [1
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2019 Q2 [10]}}