AQA Further AS Paper 2 Mechanics 2018 June — Question 4 11 marks

Exam BoardAQA
ModuleFurther AS Paper 2 Mechanics (Further AS Paper 2 Mechanics)
Year2018
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeDirect collision, find velocities
DifficultyStandard +0.8 This is a standard Further Maths collision problem requiring conservation of momentum and Newton's restitution law, but with multiple parts including a proof, finding expressions in terms of parameters, directional analysis, and an inequality involving impulse bounds over the range of e. The multi-step nature, parameter manipulation, and the final part requiring insight about e ∈ [0,1] bounds elevate this above typical A-level questions.
Spec6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts
4 Two smooth spheres \(A\) and \(B\) of equal radius are free to move on a smooth horizontal surface. The masses of \(A\) and \(B\) are \(m\) and \(4 m\) respectively.
The coefficient of restitution between the spheres is \(e\).
The spheres are projected directly towards each other, each with speed \(u\), and subsequently collide. 4
  1. Show that the speed of \(B\) immediately after the impact with \(A\) is $$\frac { u ( 3 - 2 e ) } { 5 }$$ 4
  2. Find the speed of \(A\) in terms of \(u\) and \(e\).
    4
  3. Comment on the direction of motion of the spheres after the collision, justifying your answer.
    4
  4. The magnitude of the impulse on \(B\) due to the collision is \(I\).
    Deduce that $$\frac { 8 m u } { 5 } \leq I \leq \frac { 16 m u } { 5 }$$
Question 4(a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(4mu - mu = mv_A + 4mv_B\) leading to \(3u = v_A + 4v_B\)M1 Forms equation using conservation of momentum; condone sign errors with correct terms
Correct momentum equation (can be unsimplified)A1
\(v_A - v_B = 2ue\) (Newton's Law of Restitution)B1 Forms equation using Newton's law of restitution
Subtracting equations: \(5v_B = 3u - 2ue\), so \(v_B = \frac{u(3-2e)}{5}\)R1 Completes rigorous argument using both conservation of momentum and coefficient of restitution to verify correct speed of \(B\)
Question 4(b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(v_A = \frac{u(3-2e)}{5} + 2ue\)M1 Substitutes speed/velocity of \(B\) back into either equation
\(v_A = \frac{u(3+8e)}{5}\)A1 Must be fully simplified
Question 4(c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Since \(0 \le e \le 1\) then expressions for the speeds above are both positiveE1 Uses maximum and minimum values of \(e\) to consider effect on direction of motion
Hence the spheres both travel in the same directionR1 Deduces that spheres both travel in same direction after collision and justifies conclusion
Question 4(d):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(I = 4mv_B - 4mu_B = \frac{4mu(3-2e)}{5} - 4mu\)M1 Recalls formula for impulse and substitutes a pair of corresponding velocities
\(I = \frac{8mu(1+e)}{5}\); substitutes \(e = 0\) or \(e = 1\)M1 Substitutes 0 or 1 for \(e\) into their expression
\(\frac{8mu}{5} \le I \le \frac{16mu}{5}\)R1 Completes rigorous argument using algebraic expressions for velocities, impulse formula and range of \(e\) to verify stated inequality 
## Question 4(a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $4mu - mu = mv_A + 4mv_B$ leading to $3u = v_A + 4v_B$ | M1 | Forms equation using conservation of momentum; condone sign errors with correct terms |
| Correct momentum equation (can be unsimplified) | A1 | |
| $v_A - v_B = 2ue$ (Newton's Law of Restitution) | B1 | Forms equation using Newton's law of restitution |
| Subtracting equations: $5v_B = 3u - 2ue$, so $v_B = \frac{u(3-2e)}{5}$ | R1 | Completes rigorous argument using both conservation of momentum and coefficient of restitution to verify correct speed of $B$ |

## Question 4(b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $v_A = \frac{u(3-2e)}{5} + 2ue$ | M1 | Substitutes speed/velocity of $B$ back into either equation |
| $v_A = \frac{u(3+8e)}{5}$ | A1 | Must be fully simplified |

## Question 4(c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Since $0 \le e \le 1$ then expressions for the speeds above are both positive | E1 | Uses maximum and minimum values of $e$ to consider effect on direction of motion |
| Hence the spheres both travel in the same direction | R1 | Deduces that spheres both travel in same direction after collision and justifies conclusion |

## Question 4(d):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $I = 4mv_B - 4mu_B = \frac{4mu(3-2e)}{5} - 4mu$ | M1 | Recalls formula for impulse and substitutes a pair of corresponding velocities |
| $I = \frac{8mu(1+e)}{5}$; substitutes $e = 0$ or $e = 1$ | M1 | Substitutes 0 or 1 for $e$ into their expression |
| $\frac{8mu}{5} \le I \le \frac{16mu}{5}$ | R1 | Completes rigorous argument using algebraic expressions for velocities, impulse formula and range of $e$ to verify stated inequality |

---
4 Two smooth spheres $A$ and $B$ of equal radius are free to move on a smooth horizontal surface.

The masses of $A$ and $B$ are $m$ and $4 m$ respectively.\\
The coefficient of restitution between the spheres is $e$.\\
The spheres are projected directly towards each other, each with speed $u$, and subsequently collide.

4
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $B$ immediately after the impact with $A$ is

$$\frac { u ( 3 - 2 e ) } { 5 }$$

4
\item Find the speed of $A$ in terms of $u$ and $e$.\\

4
\item Comment on the direction of motion of the spheres after the collision, justifying your answer.\\

4
\item The magnitude of the impulse on $B$ due to the collision is $I$.\\
Deduce that

$$\frac { 8 m u } { 5 } \leq I \leq \frac { 16 m u } { 5 }$$
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2018 Q4 [11]}}
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