| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | March |
| Marks | 7 |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string with compression (spring) |
| Difficulty | Challenging +1.2 This is a multi-part energy conservation problem with elastic springs requiring calculation of whether the ball reaches the ceiling, followed by qualitative reasoning about model modifications. While it involves several steps (finding equilibrium position, applying energy conservation with elastic PE and gravitational PE), the mathematical techniques are standard for Further Mechanics. The conceptual understanding needed for part (ii) about strings vs springs and air resistance is moderately challenging but well within typical FM scope. |
| 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 principle |
| Answer | Marks | Guidance |
|---|---|---|
| Initial energy = \(\frac{1}{2} \times 1.7 \times 0.5^2 + \frac{50 \times (3.2 - 1.2)^2}{2 \times 1.2}\) | M1 | May include a gravitational PE term |
| 83.5 | A1 | |
| Minimum energy required to reach O is | B1 | |
| \(1.7 \times 9.8 \times 3.2 + \frac{50 \times 1.2^2}{2 \times 1.2} = 83.3(12)\) | ||
| So according to the model B reaches O | E1 [4] |
| Answer | Marks |
|---|---|
| Yes because work would have to be done against air resistance so more energy would be required to reach O and so the conclusion could change | E1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| No because a string has no elastic energy after it goes slack so less initial energy would be required to enable B to reach O and so the conclusion does not change | E1 [1] | Or work has to be done against the thrust as the spring goes into compression |
| Answer | Marks | Guidance |
|---|---|---|
| Either: The direction of projection has no effect on energy calculations, so (if nothing else interferes with the motion) the conclusion does not change | E1 [1] | Or: If the downward motion means that the ball hits the floor some energy will be lost, so the conclusion could change |
## (i)
Initial energy = $\frac{1}{2} \times 1.7 \times 0.5^2 + \frac{50 \times (3.2 - 1.2)^2}{2 \times 1.2}$ | M1 | May include a gravitational PE term
83.5 | A1 |
Minimum energy required to reach O is | B1 |
$1.7 \times 9.8 \times 3.2 + \frac{50 \times 1.2^2}{2 \times 1.2} = 83.3(12)$ | |
So according to the model B reaches O | E1 [4] |
## (ii)(a)
Yes because work would have to be done against air resistance so more energy would be required to reach O and so the conclusion could change | E1 [1] |
## (ii)(b)
No because a string has no elastic energy after it goes slack so less initial energy would be required to enable B to reach O and so the conclusion does not change | E1 [1] | Or work has to be done against the thrust as the spring goes into compression
## (ii)(c)
Either: The direction of projection has no effect on energy calculations, so (if nothing else interferes with the motion) the conclusion does not change | E1 [1] | Or: If the downward motion means that the ball hits the floor some energy will be lost, so the conclusion could change | Or: any other sensible discussion about how the model might break down, e.g. the downward motion causing the spring to break or Hooke's law to fail, with an appropriate conclusion stated
---4 A ball $B$ of mass 1.7 kg is connected to one end of a light elastic spring of natural length 1.2 m . The other end of the spring is attached to a point $O$ on the ceiling of a large room. The modulus of elasticity of the spring is 50 N . The ball is held 3.2 m vertically below $O$ and projected upwards with an initial speed of $0.5 \mathrm {~ms} ^ { - 1 }$. In order to model the motion of $B$ (before any collision with the ceiling) the following assumptions are made.
\begin{itemize}
\item Air resistance is ignored.
\item $B$ is small.
\item The fully compressed length of the spring is negligible.
\begin{enumerate}[label=(\roman*)]
\item Determine whether, according to the model, $B$ reaches $O$.
\item Without doing any further calculations, explain whether the answer to part (i) could change in each of the following different cases.
\begin{enumerate}[label=(\alph*)]
\end{enumerate}\item A new model is used in which air resistance is taken into account.
\item The spring is replaced by an elastic string with the same natural length and modulus of elasticity.
\item $\quad B$ is initially projected downwards rather than upwards.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q4 [7]}}