| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Year | 2022 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Projectile energy - basic KE/PE calculation |
| Difficulty | Moderate -0.3 Part (a) is direct substitution into KE = ½mv² (2 marks). Part (b) requires understanding energy conservation (KE converts to PE) and recognizing that air resistance causes energy loss, making actual height less than theoretical—standard A-level mechanics reasoning with straightforward calculations, slightly easier than average due to minimal problem-solving demand. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2}(0.75)(12)^2 = 54 \text{ J}\) | B1 | Recalls formula for kinetic energy and calculates initial KE; condone missing units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(mgh = 54\), so \(h = \frac{54}{(0.75)(9.8)} = 7.34\ldots\) | M1 | Uses conservation of energy to form equation with PE and KE from part (a) |
| \(h = 7.3\) (AWRT 7.3) | A1 | Must have clearly rearranged to find \(h\); correct value (7.3469...) to at least 3 sf; AG |
| Jeff has assumed no air resistance to obtain \(h = 7.3\); Gurjas includes air resistance so knows ball will not reach 7.3 m | E1 | Makes inference about one or more assumptions for both Jeff and Gurjas; e.g. Jeff assumes all energy conserved, Gurjas does not |
## Question 3(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2}(0.75)(12)^2 = 54 \text{ J}$ | B1 | Recalls formula for kinetic energy and calculates initial KE; condone missing units |
**Subtotal: 1 mark**
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## Question 3(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $mgh = 54$, so $h = \frac{54}{(0.75)(9.8)} = 7.34\ldots$ | M1 | Uses conservation of energy to form equation with PE and KE from part (a) |
| $h = 7.3$ (AWRT 7.3) | A1 | Must have clearly rearranged to find $h$; correct value (7.3469...) to at least 3 sf; AG |
| Jeff has assumed no air resistance to obtain $h = 7.3$; Gurjas includes air resistance so knows ball will not reach 7.3 m | E1 | Makes inference about one or more assumptions for both Jeff and Gurjas; e.g. Jeff assumes all energy conserved, Gurjas does not |
**Subtotal: 3 marks**
**Question 3 Total: 4 marks**3 In this question use $g = 9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$
A ball of mass of 0.75 kg is thrown vertically upwards with an initial speed of $12 \mathrm {~ms} ^ { - 1 }$ The ball is thrown from ground level.
3
\begin{enumerate}[label=(\alph*)]
\item Calculate the initial kinetic energy of the ball.
3
\item The maximum height of the ball above the ground is $h$ metres.\\
Jeff and Gurjas use an energy method to find $h$\\
Jeff concludes that $h = 7.3$
Gurjas concludes that $h < 7.3$\\
Explain the reasoning that they have used, showing any calculations that you make.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2022 Q3 [4]}}