| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Calculate impulse from force-time data |
| Difficulty | Standard +0.3 This is a straightforward impulse-momentum question requiring integration of an exponential function and application of the impulse-momentum theorem. Part (a) involves routine integration of e^t and e^{2t}, while part (b) applies a standard formula. The exponential functions and ln 8 limit add minor complexity but the techniques are standard Further Maths mechanics content with no novel problem-solving required. |
| Spec | 6.03f Impulse-momentum: relation6.03h Variable force impulse: using integration |
| Answer | Marks |
|---|---|
| 7(a) | Forms a correct definite integral |
| Answer | Marks | Guidance |
|---|---|---|
| Condone one error | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone missing units. | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 7(b) | States that I = mv – mu |
| Answer | Marks | Guidance |
|---|---|---|
| States a =3et + e2t | 1.2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone missing units. | 1.1b | A1F |
| Subtotal | 3 | |
| Question total | 5 | |
| Q | Marking instructions | AO |
Question 7:
--- 7(a) ---
7(a) | Forms a correct definite integral
for impulse.
Condone one error | 3.4 | M1 | ln 8
Impulse = ∫ (6et + 2e2t)dt
0
= 105 Ns
Obtains 105 Ns
Condone missing units. | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 7(b) ---
7(b) | States that I = mv – mu
Or
States a =3et + e2t | 1.2 | B1 | I = mv – mu
2(v) – 2(5) = 105
v = 57.5 m s–1
Forms the equation
2(v) – 2(5) = their impulse from
part (a).
Or
1 3
Obtains v=3et + e2t +
2 2 | 1.1a | M1
Obtains 57.5
Follow through their answer to
part (a).
Condone missing units. | 1.1b | A1F
Subtotal | 3
Question total | 5
Q | Marking instructions | AO | Marks | Typical solutionA single force, $F$ newtons, acts on a particle moving on a straight, smooth, horizontal line.
The force $F$ acts in the direction of motion of the particle.
At time $t$ seconds, $F = 6e^t + 2e^{2t}$ where $0 \leq t \leq \ln 8$
\begin{enumerate}[label=(\alph*)]
\item Find the impulse of $F$ over the interval $0 \leq t \leq \ln 8$ [2 marks]
\item The particle has a mass of 2 kg and at time $t = 0$ has velocity $5 \text{ m s}^{-1}$
Find the velocity of the particle when $t = \ln 8$ [3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2024 Q7 [5]}}