| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Collision with unchanged direction |
| Difficulty | Standard +0.3 This is a standard A-level mechanics collision problem with three routine parts: (a) straightforward momentum calculation, (b) applying conservation of momentum and restitution equation (show-that format guides the answer), (c) finding constraint on e from direction reversal. While it requires multiple steps and careful algebra, it follows a completely standard template with no novel insight required. Slightly easier than average due to the structured guidance. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03d Conservation in 2D: vector momentum6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a) | Obtains 10mu | 1.1b |
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 8(b) | Forms a conservation of |
| Answer | Marks | Guidance |
|---|---|---|
| expression from part (a). | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| equation. | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Newton’s law of restitution. | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | 2.1 | R1 |
| Subtotal | 4 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 8(c) | Uses a method to find the | |
| velocity of A | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ACF | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone use of v A = 0 or v A ≤ 0 | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | 2.2a | R1 |
| Subtotal | 5 | |
| Question total | 10 | |
| Q | Marking instructions | AO |
Question 8:
--- 8(a) ---
8(a) | Obtains 10mu | 1.1b | B1 | 4mu+6mu=10mu
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 8(b) ---
8(b) | Forms a conservation of
momentum equation using their
expression from part (a). | 3.1b | M1 | Speed of A after collision = v A
Speed of B after collision = v B
C of M
mv + 6mv = 10mu
A B
v + 6v = 10u
A B
NLR
v – v = 3ue
B A
7v = 3ue + 10u
B
u(3e+10)
v =
B
7
Obtains a correct momentum
equation. | 1.1b | A1
Forms a correct equation using
Newton’s law of restitution. | 1.1b | B1
Completes a reasoned
argument using both
conservation of momentum and
Newton’s law of restitution to
obtain the correct speed of B
AG | 2.1 | R1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 8(c) ---
8(c) | Uses a method to find the
velocity of A | 1.1a | M1 | v + 6v = 10u
A B
6u(3e+10)
v = 10u –
A
7
10u–18ue
v =
A
7
10u–18ue
< 0
7
5
< e
9
5
< e ≤ 1
9
10u–18ue
Obtains
7
ACF | 1.1b | A1
Uses their expression for v A and
forms the inequality v A < 0
Condone use of v A = 0 or v A ≤ 0 | 3.1b | M1
5
Obtains oe or AWRT 0.56
9 | 1.1b | B1
5
Deduces < e ≤ 1
9
5
Condone < e < 1
9 | 2.2a | R1
Subtotal | 5
Question total | 10
Q | Marking instructions | AO | Marks | Typical solutionTwo spheres, $A$ and $B$, of equal size are moving in the same direction along a straight line on a smooth horizontal surface.
Sphere $A$ has mass $m$ and is moving with speed $4u$
Sphere $B$ has mass $6m$ and is moving with speed $u$
The diagram shows the spheres and their velocities.
\includegraphics{figure_8}
Subsequently $A$ collides directly with $B$
The coefficient of restitution between $A$ and $B$ is $e$
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $m$ and $u$, the total momentum of the spheres before the collision. [1 mark]
\item Show that the speed of $B$ immediately after the collision is $\frac{u(3e + 10)}{7}$ [4 marks]
\item After the collision sphere $A$ moves in the opposite direction.
Find the range of possible values for $e$ [5 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2024 Q8 [10]}}