| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| 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 linear force function and solving a quadratic equation. The integration is elementary (polynomial), and the impulse-momentum theorem application is standard. While it's Further Maths content, the mathematical techniques are routine A-level calculus and algebra, making it slightly easier than average overall. |
| Spec | 6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Impulse \(= \int_0^T (2t+3)\,dt\) | B1 | Forms a correct definite integral for impulse |
| \(= [t^2 + 3t]_0^T\) | M1 | Integrates with at least one term correct |
| \(= T^2 + 3T\), so \(a=1\) and \(b=3\) | A1 | Obtains \(T^2 + 3T\) or \(a=1\) and \(b=3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I = mv - mu\) | M1 | Uses \(mv - mu\) |
| \(I = 0.2(4) - 0.2(1) = 0.6\), so \(T^2 + 3T = 0.6\) | A1 | Obtains 0.6 |
| \(T = 0.188\) or \(-3.19\); as \(0 \leq t \leq T\), \(T = 0.188\) | M1 | Equates their answer to part (a) to their change in momentum and solves their quadratic equation |
| \(T = 0.188\) | A1 | Obtains \(T = 0.188\) and clearly rejects the negative value |
## Question 6(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Impulse $= \int_0^T (2t+3)\,dt$ | B1 | Forms a correct definite integral for impulse |
| $= [t^2 + 3t]_0^T$ | M1 | Integrates with at least one term correct |
| $= T^2 + 3T$, so $a=1$ and $b=3$ | A1 | Obtains $T^2 + 3T$ or $a=1$ and $b=3$ |
## Question 6(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I = mv - mu$ | M1 | Uses $mv - mu$ |
| $I = 0.2(4) - 0.2(1) = 0.6$, so $T^2 + 3T = 0.6$ | A1 | Obtains 0.6 |
| $T = 0.188$ or $-3.19$; as $0 \leq t \leq T$, $T = 0.188$ | M1 | Equates their answer to part (a) to their change in momentum and solves their quadratic equation |
| $T = 0.188$ | A1 | Obtains $T = 0.188$ and clearly rejects the negative value |6 An ice hockey puck, of mass 0.2 kg , is moving in a straight line on a horizontal ice rink under the action of a single force which acts in the direction of motion.
At time $t$ seconds, the force has magnitude ( $2 t + 3$ ) newtons.\\
The force acts on the puck from $t = 0$ to $t = T$\\
6
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the impulse of the force is $a T ^ { 2 } + b T$, where $a$ and $b$ are integers to be found.\\[0pt]
[3 marks]\\
6
\item While the force acts on the puck, its speed increases from $1 \mathrm {~ms} ^ { - 1 }$ to $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
Use your answer from part (a) to find $T$, giving your answer to three significant figures.\\
Fully justify your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2022 Q6 [7]}}