| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Year | 2021 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Power from work done over time (P = W/t) |
| Difficulty | Standard +0.3 This is a straightforward work-energy problem requiring standard formulas (PE = mgh, KE = œmv², Power = Work/time). Part (a) is direct substitution, and part (b) involves setting up an energy equation with given power and time. All values are provided clearly, requiring only systematic application of mechanics principles with no conceptual challenges or novel problem-solving. |
| Spec | 6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02k Power: rate of doing work |
| Answer | Marks |
|---|---|
| 3(a) | Recalls the formula for |
| Answer | Marks | Guidance |
|---|---|---|
| Accept 14700 | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Total | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 3(b) | Forms an equation recalling that |
| Answer | Marks | Guidance |
|---|---|---|
| PE and power | 3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| correct gain in KE | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| work done over 50 seconds | 3.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| units | 1.1b | A1 |
| Total | 4 | |
| Question total | 5 | |
| Q | Marking instructions | AO |
Question 3:
--- 3(a) ---
3(a) | Recalls the formula for
gravitational potential energy
and calculates the energy
gained
Condone missing or incorrect
units
Accept 14700 | 1.1b | B1 | ððgâ = 300Ã 9J. 8Ã5
== 1154070000 J
Total | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 3(b) ---
3(b) | Forms an equation recalling that
power is the rate of doing work
containing expressions for KE,
PE and power | 3.3 | M1 | Work done = gain in KE + gain in
PE
Increase in KE =
1 2
2(300)ð£ð£
Work done over 50 sec = 400 Ã 50
= 20000 J
1 2v = 5.9 m s-1
(300)ð£ð£ +14700 = 400Ã50
2
Forms an expression for the
correct gain in KE | 1.1b | B1
Obtains correct value for total
work done over 50 seconds | 3.1b | B1
Solves the equation correctly to
obtain the value of v
Accept AWRT 5.94
Condone missing or incorrect
units | 1.1b | A1
Total | 4
Question total | 5
Q | Marking instructions | AO | Marks | Typical solutionUse $g$ as 9.8 m s$^{-2}$ in this question.
A pump is used to pump water out of a pool.
The pump raises the water through a vertical distance of 5 metres and then ejects it through a pipe.
The pump works at a constant rate of 400 W
Over a period of 50 seconds, 300 litres of water are pumped out of the pool and the water is ejected with speed $v$ m s$^{-1}$
The mass of 1 litre of water is 1 kg
\begin{enumerate}[label=(\alph*)]
\item Find the gain in the potential energy of the 300 litres of water. [1 mark]
\item Calculate $v$ [4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2021 Q3 [5]}}