| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics B AS (Further Mechanics B AS) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string equilibrium and statics |
| Difficulty | Standard +0.3 This is a straightforward elastic string problem requiring application of Hooke's law with two equilibrium conditions to find natural length and stiffness, then using these to predict a third case. The calculations are routine and the 'suggest a reason' part has an obvious answer (string exceeds elastic limit). Slightly above average difficulty only due to the multi-step nature and need to work with two simultaneous equations. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | 𝜆𝑒 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑙 | M1 | 1.1a |
| or cm for m. | Or allow T = ke |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑙 | A1 | 3.4 |
| T=ke | g = k(0.7 – l) |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑙 | A1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Solve simultaneous equations | M1 | 1.1 |
| l = 0.6 (m) or 60 (cm); λ = 6g or 58.8 (N) | A1 | 1.1 |
| Answer | Marks |
|---|---|
| λ or k | Check carefully: k=0.98 |
| Answer | Marks | Guidance |
|---|---|---|
| Length of string = 1.1 (m) | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| EPE = 12.25 (J) | A1 | 1.2 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | String stretched beyond its elastic limit | B1 |
Question 1:
1 | (a) | 𝜆𝑒
Use of 𝑇 =
𝑙 | M1 | 1.1a | Accept attempt with g missed and /
or cm for m. | Or allow T = ke
𝜆(0.7−𝑙)
𝑔 =
𝑙 | A1 | 3.4 | Allow FT if cm used, even in if used
T=ke | g = k(0.7 – l)
2g = k(0.8 – l)
𝜆(0.8−𝑙)
2𝑔 =
𝑙 | A1 | 3.4 | Allow FT if cm used, even in if used
T=ke
Solve simultaneous equations | M1 | 1.1 | Eliminate one variable
l = 0.6 (m) or 60 (cm); λ = 6g or 58.8 (N) | A1 | 1.1 | Or l = 0.6, k = 10g or 98
Must have correct l and either
correct
λ or k | Check carefully: k=0.98
might be used if working in
cm. Do not allow k=0.98
Length of string = 1.1 (m) | B1 | 1.1 | Allow 110 cm | This answer may be found
by inspection.
EPE = 12.25 (J) | A1 | 1.2 | Could be found via k or λ | 1
𝑇 = 𝑘𝑒2
2
[7]
(b) | String stretched beyond its elastic limit | B1 | 2.2b | Or other valid reason
[1]1 The end O of a light elastic string OA is attached to a fixed point.\\
Fiona attaches a mass of 1 kg to the string at A . The system hangs vertically in equilibrium and the length of the stretched string is 70 cm .
Fiona removes the 1 kg mass and attaches a mass of 2 kg to the string at A . The system hangs vertically in equilibrium and the length of the stretched string is now 80 cm .
Fiona then removes the 2 kg mass and attaches a mass of 5 kg to the string at A . The system hangs vertically in equilibrium.
\begin{enumerate}[label=(\alph*)]
\item Use the information given in the question to determine expected values for
\begin{itemize}
\item the length of the stretched string when the 5 kg mass is attached,
\item the elastic potential energy stored in the string in this case.
\end{itemize}
Fiona discovers that, when the mass of 5 kg is attached to the string at A , the length of the stretched string is greater than the expected length.
\item Suggest a reason why this has happened.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics B AS 2021 Q1 [8]}}