| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with direction reversal |
| Difficulty | Standard +0.3 This is a standard Further Maths mechanics collision problem requiring conservation of momentum and Newton's restitution law. The multi-part structure guides students through the solution systematically, with part (b)(i) being a 'show that' which provides the answer. While it requires careful sign conventions for opposite directions, it's a routine application of well-practiced techniques with no novel insight needed—slightly easier than average for Further Maths. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Momentum for \(A = 4(0.4) = 1.6\); Momentum for \(B = -2(0.2) = -0.4\) | M1 | Forms a term using \(mv\) for either particle |
| Total momentum \(= 1.2\) Ns | A1 | Obtains \(1.2\) kg ms\(^{-1}\) or \(1.2\) Ns. Condone missing units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1.2 = 0.4u + 0.2v\), i.e. \(12 = 4u + 2v\) | M1 | Forms a conservation of momentum equation using their answer from (a) |
| NLR: \(v - u = 6e\) | B1 | Forms a correct equation using Newton's law of restitution |
| \(12 = 4(v-6e) + 2v\), \(12 = 6v - 24e\), \(2 = v - 4e\), \(v = 4e + 2\) | R1 | Completes a rigorous argument using both conservation of momentum and Newton's law of restitution to verify the correct speed of \(B\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4e + 2 - u = 6e\) | M1 | Substitutes the speed of \(B\) back into either of their equations |
| \(u = 2 - 2e\) | A1 | Rearranges their equation to obtain the correct speed of \(A\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Perfectly elastic collision so \(e = 1\) | M1 | States that \(e = 1\) |
| \(u = 2 - 2(1) = 0\), hence particle \(A\) comes to rest | R1 | Deduces that \(A\) comes to rest |
## Question 7(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Momentum for $A = 4(0.4) = 1.6$; Momentum for $B = -2(0.2) = -0.4$ | M1 | Forms a term using $mv$ for either particle |
| Total momentum $= 1.2$ Ns | A1 | Obtains $1.2$ kg ms$^{-1}$ or $1.2$ Ns. Condone missing units |
## Question 7(b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1.2 = 0.4u + 0.2v$, i.e. $12 = 4u + 2v$ | M1 | Forms a conservation of momentum equation using their answer from (a) |
| NLR: $v - u = 6e$ | B1 | Forms a correct equation using Newton's law of restitution |
| $12 = 4(v-6e) + 2v$, $12 = 6v - 24e$, $2 = v - 4e$, $v = 4e + 2$ | R1 | Completes a rigorous argument using both conservation of momentum and Newton's law of restitution to verify the correct speed of $B$ |
## Question 7(b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4e + 2 - u = 6e$ | M1 | Substitutes the speed of $B$ back into either of their equations |
| $u = 2 - 2e$ | A1 | Rearranges their equation to obtain the correct speed of $A$ |
## Question 7(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Perfectly elastic collision so $e = 1$ | M1 | States that $e = 1$ |
| $u = 2 - 2(1) = 0$, hence particle $A$ comes to rest | R1 | Deduces that $A$ comes to rest |7 The particles $A$ and $B$ are moving on a smooth horizontal surface directly towards each other.
Particle $A$ has mass 0.4 kg and particle $B$ has mass 0.2 kg\\
Particle $A$ has speed $4 \mathrm {~ms} ^ { - 1 }$ and particle $B$ has speed $2 \mathrm {~ms} ^ { - 1 }$ when they collide, as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{ec39a757-5867-4798-b26c-73cd5746581c-08_392_1064_625_488}
The coefficient of restitution between the particles is $e$\\
7
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the total momentum of the particles before the collision.\\[0pt]
[2 marks]
7
\item (i) Show that the speed of $B$ immediately after the collision is $( 4 e + 2 ) \mathrm { ms } ^ { - 1 }$\\[0pt]
[3 marks]\\
7 (b) (ii) Find an expression, in terms of $e$, for the speed of $A$ immediately after the collision.\\
7
\item Explain what happens to particle $A$ when the collision is perfectly elastic.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2022 Q7 [9]}}