| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Session | Specimen |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Impulse from variable force (then find velocity) |
| Difficulty | Standard +0.3 This is a straightforward impulse-momentum question requiring integration of a polynomial force function and application of the impulse-momentum theorem. Part (a) is immediate from boundary conditions, part (b) involves routine polynomial integration, and part (c) applies a standard formula. While it's Further Maths content, the mathematical techniques are elementary calculus with no conceptual challenges or novel problem-solving required. |
| Spec | 6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form6.03h Variable force impulse: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b = 0.02\) | B1 | Deduces correct value for \(b\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I = \int_0^{0.02} kt^2(t-0.02)^2 \, dt\) | M1 | Forms an integral to find the impulse |
| \(= k\int_0^{0.02}(t^4 - 0.04t^3 + 0.0004t^2) \, dt\) | M1 | Integrates terms and uses limits or uses a calculator for definite integral |
| \(= k\left[\frac{t^5}{5} - \frac{t^4}{100} + \frac{t^3}{7500}\right]_0^{0.02}\) | ||
| \(= k \times 1.07 \times 10^{-10} \text{ Ns}\) | A1 | Obtains correct value for impulse (AWRT \(1.1 \times 10^{-10}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k \times 1.07 \times 10^{-10} = \ | 0.15 \times 4 - 0.15 \times (-2)\ | \) |
| \(k = \frac{0.9}{1.07 \times 10^{-10}}\) | A1F | Obtains a correct equation for 'their' impulse |
| \(= 8.4 \times 10^9\) | A1 | Obtains the correct value for \(k\). CAO |
## Question 7:
### Part 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b = 0.02$ | B1 | Deduces correct value for $b$ |
### Part 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I = \int_0^{0.02} kt^2(t-0.02)^2 \, dt$ | M1 | Forms an integral to find the impulse |
| $= k\int_0^{0.02}(t^4 - 0.04t^3 + 0.0004t^2) \, dt$ | M1 | Integrates terms and uses limits or uses a calculator for definite integral |
| $= k\left[\frac{t^5}{5} - \frac{t^4}{100} + \frac{t^3}{7500}\right]_0^{0.02}$ | | |
| $= k \times 1.07 \times 10^{-10} \text{ Ns}$ | A1 | Obtains correct value for impulse (AWRT $1.1 \times 10^{-10}$) |
### Part 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k \times 1.07 \times 10^{-10} = \|0.15 \times 4 - 0.15 \times (-2)\|$ | M1 | Uses 'impulse equals change in momentum' to form an equation with 'their' impulse from (a) |
| $k = \frac{0.9}{1.07 \times 10^{-10}}$ | A1F | Obtains a correct equation for 'their' impulse |
| $= 8.4 \times 10^9$ | A1 | Obtains the correct value for $k$. CAO |
---7 A disc, of mass 0.15 kg , slides across a smooth horizontal table and collides with a vertical wall which is perpendicular to the path of the disc.
The disc is in contact with the wall for 0.02 seconds and then rebounds.\\
A possible model for the force, $F$ newtons, exerted on the disc by the wall, whilst in contact, is given by
$$F = k t ^ { 2 } ( t - b ) ^ { 2 } \quad \text { for } \quad 0 \leq t \leq 0.020$$
where $k$ and $b$ are constants.\\
The force is initially zero and becomes zero again as the disc loses contact with the wall.
7
\begin{enumerate}[label=(\alph*)]
\item State the value of $b$.\\
7
\item Find the magnitude of the impulse on the disc, giving your answer in terms of $k$.\\
7
\item The disc is travelling at $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it hits the wall.\\
The disc rebounds with a speed of $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
Find $k$.\\[0pt]
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics Q7 [3]}}