| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| 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 question requiring conservation of momentum and Newton's restitution law. The algebra is straightforward, and the techniques are routine for this level. Part (b) requires understanding the physical constraint 0≤e≤1, and part (c) applies standard impulse-momentum theorem. While it's a multi-part question worth 12 marks, each step follows predictable patterns without requiring novel insight, making it slightly easier than average for Further Maths. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a)(i) | Forms an equation using | |
| conservation of momentum | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| equation – can be unsimplified | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Newton’s law of restitution | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| speed of P | 2.1 | R1 |
| Answer | Marks |
|---|---|
| 7(a)(ii) | Substitutes the speed of P back |
| Answer | Marks | Guidance |
|---|---|---|
| equations | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| must be fully simplified | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7(b) | Substitutes either e = 0 or e = 1 into | |
| the expression for v | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| end | 2.1 | R1 |
| Answer | Marks |
|---|---|
| 7(c)(i) | Uses the formula for impulse and |
| Answer | Marks | Guidance |
|---|---|---|
| into the impulse formula | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| impulse | 2.2a | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | 3.2a | A1 |
| Answer | Marks |
|---|---|
| 7(c)(ii) | Deduces that the impulse Q exerts |
| Answer | Marks | Guidance |
|---|---|---|
| value obtained in part (c)(i) | 2.2a | E1F |
| Answer | Marks |
|---|---|
| Total | 12 |
Question 7:
--- 7(a)(i) ---
7(a)(i) | Forms an equation using
conservation of momentum | 1.1a | M1 | CoM
3mu =3mv+mw
3u =3v+w
NLR
w−v=ue
Subtracting equations gives
4v=3u−ue
u(3−e)
v=
4
Obtains a correct momentum
equation – can be unsimplified | 1.1b | A1
Forms a correct equation using
Newton’s law of restitution | 1.1b | B1
Completes a rigorous argument
using both conservation of
momentum and the coefficient of
restitution to verify the correct
speed of P | 2.1 | R1
--- 7(a)(ii) ---
7(a)(ii) | Substitutes the speed of P back
into either of their equations
Or
Eliminates v from their pair of
equations | 1.1a | M1 | u(3−e)
w= +ue
4
3u(1+e)
w=
4
Obtains the correct speed for Q -
must be fully simplified | 1.1b | A1
--- 7(b) ---
7(b) | Substitutes either e = 0 or e = 1 into
the expression for v | 1.1a | M1 | When e = 0
u(3−0) 3u
v= =
4 4
When e = 1
u(3−1) u
v= =
4 2
Hence
u 3u
≤v≤
2 4
Completes a rigorous argument by
correctly substituting both e = 0 and
e = 1 into the expression for v to
confirm the corresponding stated
maximum and minimum values of v
If maximum and minimum values
are clearly identified then there is
no need to state inequality at the
end | 2.1 | R1
--- 7(c)(i) ---
7(c)(i) | Uses the formula for impulse and
substitutes a pair of corresponding
velocities and consistent masses
into the impulse formula | 1.1a | M1 | u
I =3m −3mu
2
3mu
I =−
2
3mu
Magnitude =
2
Deduces that e = 1 will result in the
maximum magnitude of the
impulse and uses this to find the
impulse | 2.2a | R1
3mu
Obtains
2 | 3.2a | A1
--- 7(c)(ii) ---
7(c)(ii) | Deduces that the impulse Q exerts
on P is equal in magnitude but
opposite in direction
Accept
Impulse will be the negative of the
impulse P exerts on Q (opposite)
or
Equal and opposite
FT any reference to their specific
value obtained in part (c)(i) | 2.2a | E1F | The impulse Q exerts on P is equal
in magnitude but opposite in
direction
Total | 12Two smooth spheres, $P$ and $Q$, of equal radius are free to move on a smooth horizontal surface.
The masses of $P$ and $Q$ are $3m$ and $m$ respectively.
$P$ is set in motion with speed $u$ directly towards $Q$, which is initially at rest.
$P$ subsequently collides with $Q$.
\includegraphics{figure_7}
Immediately after the collision, $P$ moves with speed $v$ and $Q$ moves with speed $w$.
The coefficient of restitution between the spheres is $e$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that
$$v = \frac{u(3-e)}{4}$$
[4 marks]
\item Find $w$, in terms of $e$ and $u$, simplifying your answer.
[2 marks]
\end{enumerate}
\item Deduce that
$$\frac{u}{2} \leq v \leq \frac{3u}{4}$$
[2 marks]
\item \begin{enumerate}[label=(\roman*)]
\item Find, in terms of $m$ and $u$, the maximum magnitude of the impulse that $P$ exerts on $Q$.
[3 marks]
\item Describe the impulse that $Q$ exerts on $P$.
[1 mark]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2019 Q7 [12]}}