| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string with compression (spring) |
| Difficulty | Standard +0.3 This is a standard elastic potential energy problem requiring application of energy conservation in two stages (spring release, then projectile motion). The calculations are straightforward with given values, and part (c) requires basic critical thinking about modeling assumptions. Slightly easier than average due to clear structure and routine energy methods, though the modeling discussion in (c) adds minor complexity. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{EPE} = \frac{1}{2}kx^2 = \frac{1}{2} \times 60 \times 0.03^2 = 0.027\) J | B1 | Uses EPE formula and calculates initial stored energy, substituting appropriate values |
| EPE lost \(=\) PE gained \(+\) KE gained: \(0.027 = \frac{1}{2}(0.018)v^2 + 0.018g(0.03)\) | M1 | Forms conservation of energy equation containing EPE, KE and PE, substituting appropriate values |
| \(v^2 = 2.412\) (fully correct three term equation) | A1 | |
| \(v = 1.6\) ms\(^{-1}\) | A1F | Must be rounded correctly to 2 significant figures; condone missing units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| EPE lost \(=\) PE gained, \(0.027 = 0.018(9.8)h\), \(h = 0.153\) m | M1 | Translates problem into finding maximum height using energy equation with \(KE = 0\) |
| \(h = 0.15\) m (2sf) | A1 | AWRT 0.15; condone missing units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The rocket has been modelled as a particle but has size / no air resistance assumed | E1 | Identifies at least two assumptions which limit the model |
| If air resistance taken into account this would reduce the height reached | E1F | Evaluates impact of at least one assumption in context; condone one assumption; follow through their answer to (b) |
| As predicted height is only just enough to win the prize, it is unlikely the prize will be won because of effects of air resistance, even when size is taken into account | R1F | Uses at least one identified limitation to infer whether or not Jo-Jo wins a prize |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{EPE} = \frac{1}{2}kx^2 = \frac{1}{2} \times 60 \times 0.03^2 = 0.027$ J | B1 | Uses EPE formula and calculates initial stored energy, substituting appropriate values |
| EPE lost $=$ PE gained $+$ KE gained: $0.027 = \frac{1}{2}(0.018)v^2 + 0.018g(0.03)$ | M1 | Forms conservation of energy equation containing EPE, KE and PE, substituting appropriate values |
| $v^2 = 2.412$ (fully correct three term equation) | A1 | |
| $v = 1.6$ ms$^{-1}$ | A1F | Must be rounded correctly to 2 significant figures; condone missing units |
---
## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| EPE lost $=$ PE gained, $0.027 = 0.018(9.8)h$, $h = 0.153$ m | M1 | Translates problem into finding maximum height using energy equation with $KE = 0$ |
| $h = 0.15$ m (2sf) | A1 | AWRT 0.15; condone missing units |
---
## Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The rocket has been modelled as a particle but has size / no air resistance assumed | E1 | Identifies at least two assumptions which limit the model |
| If air resistance taken into account this would reduce the height reached | E1F | Evaluates impact of at least one assumption in context; condone one assumption; follow through their answer to (b) |
| As predicted height is only just enough to win the prize, it is unlikely the prize will be won because of effects of air resistance, even when size is taken into account | R1F | Uses at least one identified limitation to infer whether or not Jo-Jo wins a prize |
---7 In this question use $g = 9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$
As part of a competition, Jo-Jo makes a small pop-up rocket.\\
It is operated by pressing the rocket vertically downwards to compress a light spring, which is positioned underneath the rocket.
The rocket is released from rest and moves vertically upwards.\\
The mass of the rocket is 18 grams and the stiffness constant of the spring is $60 \mathrm { Nm } ^ { - 1 }$
Initially the spring is compressed by 3 cm\\
7
\begin{enumerate}[label=(\alph*)]
\item Find the speed of the rocket when the spring first reaches its natural length.\\
7
\item By considering energy find the distance that the rocket rises.
7
\item In order to win a prize in the competition, the rocket must reach a point which is 15 cm vertically above its starting position.
With reference to the assumptions you have made, determine if Jo-Jo wins a prize or not.
Fully justify your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2020 Q7 [9]}}