| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: released from rest at natural length or above (string initially slack) |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics energy conservation problem with elastic strings. While it requires setting up energy equations with elastic potential energy, gravitational potential energy, and kinetic energy, the method is routine for FM students. The single-step energy conservation approach (initial KE = final GPE + final EPE at maximum height) is straightforward once the extension is calculated, though the algebra requires care with the modulus and natural length values. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | e.g. Air resistance is ignored |
| Answer | Marks | Guidance |
|---|---|---|
| no need to model spin | B1 | |
| [1] | 3.3 | Allow correct sensible |
| Answer | Marks |
|---|---|
| particle (so COM is at it’s centre) | Accept: |
| Answer | Marks |
|---|---|
| (b) | Final GPE is 2.5g8.1 (J) (=198.45) |
Total final energy = “198.45” + “320.62...”
| Answer | Marks |
|---|---|
| u = awrt 20.4 (ms–1 ) | B1 |
| Answer | Marks |
|---|---|
| [4] | 3.4 |
| Answer | Marks |
|---|---|
| 1.1 | DR |
| Answer | Marks |
|---|---|
| Positive answer only | ALT Method |
Question 2:
2 | (a) | e.g. Air resistance is ignored
or B is modelled as a particle or e.g. comment
about dimensions – no size or shape
gravity is a constant value
no need to model spin | B1
[1] | 3.3 | Allow correct sensible
alternatives e.g. elastic limit of the
string is not exceeded. Ignore
extra comments as long as they
are not incorrect e.g. it is a
particle (so COM is at it’s centre) | Accept:
Condone point(-like) mass?
Do not accept:
B has no mass; assume B doesn’t
collide with O; B is uniform; String
is inextensible; Centre of mass is at
it’s centre
(b) | Final GPE is 2.5g8.1 (J) (=198.45)
20(8.1−1.4)2
Final EPE is (J) (=320.642…)
21.4
1
Initial KE is 2 .5 u 2
2
Total final energy = “198.45” + “320.62...”
1
= 2 .5 u 2
2
u = awrt 20.4 (ms–1 ) | B1
B1
M1
A1
[4] | 3.4
3.4
3.4
1.1 | DR
For any one of these terms correct
For any two of these terms correct
519.09...
Using cons of energy with their
EPE, KE and GPE terms of
correct form (three terms –
implicitly final KE=0) Signs must
be correct. Condone m seen in
energy equation. Do not accept
EPE with 1.4 or 8.1 as extension.
Positive answer only | ALT Method
If candidate finds velocity when
string is taut and uses this as the
initial velocity, will see the energy
terms/equation:
1
2.5(𝑢2 −27.44)
2
20×6.72
= 6.7𝑔×2.5+
2×1.42 One end of a light elastic string of natural length 1.4 m and modulus of elasticity 20 N is attached to a small object $B$ of mass 2.5 kg . The other end of the string is attached to a fixed point $O$.
Object $B$ is projected vertically upwards from $O$ with a speed of $u \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item State one assumption required to model the motion of $B$.
The greatest height above $O$ achieved by $B$ is 8.1 m .
\item Determine the value of $u$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2024 Q2 [5]}}