| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with direction reversal |
| Difficulty | Moderate -0.3 This is a straightforward momentum and restitution problem requiring standard application of conservation of momentum and Newton's experimental law. Part (a) involves routine algebraic manipulation with given values, and part (b) is a standard modeling critique question. While it's Further Maths content, the mechanics is entirely procedural with no novel insight required. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| CoM: \(0.8 \times 6 + 0.4 \times (-6) = 0.8 \times (-1.2) + 0.4v_B\) → \(2.4 = -0.96 + 0.4v_B\) | M1 | Uses conservation of momentum to form equation, with correct terms (condone sign errors) |
| \(v_B = 8.4\) | A1 | Obtains correct velocity for \(B\) after collision |
| Newton's law of restitution: \(-1.2 - 8.4 = -e(6-(-6))\) → \(e = \frac{9.6}{12} = 0.8\) | M1 | Uses law of restitution to obtain equation for velocity of \(A\), with correct terms and 'their' velocity for \(B\) |
| Correct equation for 'their' values | A1F | Follow through their other velocity, provided all M1 marks awarded |
| Completes rigorous argument to reach correct coefficient of restitution AG; must use correct signs with velocities | R1 | Reasoning mark for complete rigorous argument |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Include friction, which will reduce the speed of the discs as they move towards each other before the collision | E1 | States a valid refinement |
# Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| CoM: $0.8 \times 6 + 0.4 \times (-6) = 0.8 \times (-1.2) + 0.4v_B$ → $2.4 = -0.96 + 0.4v_B$ | M1 | Uses conservation of momentum to form equation, with correct terms (condone sign errors) |
| $v_B = 8.4$ | A1 | Obtains correct velocity for $B$ after collision |
| Newton's law of restitution: $-1.2 - 8.4 = -e(6-(-6))$ → $e = \frac{9.6}{12} = 0.8$ | M1 | Uses law of restitution to obtain equation for velocity of $A$, with correct terms and 'their' velocity for $B$ |
| Correct equation for 'their' values | A1F | Follow through their other velocity, provided all M1 marks awarded |
| Completes rigorous argument to reach correct coefficient of restitution **AG**; must use correct signs with velocities | R1 | Reasoning mark for complete rigorous argument |
**Total for 4(a): 5 marks**
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# Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Include friction, which will reduce the speed of the discs as they move towards each other before the collision | E1 | States a valid refinement |
**Total for 4(b): 1 mark**
**Question 4 Total: 6 marks**4 Two discs, $A$ and $B$, have equal radii and masses 0.8 kg and 0.4 kg respectively. The discs are placed on a horizontal surface.
The discs are set in motion when they are 3 metres apart, so that they move directly towards each other, each travelling at a speed of $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The discs collide directly with each other.
After the collision $A$ moves in the opposite direction with a speed of $1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
The coefficient of restitution between the two discs is $e$.
4
\begin{enumerate}[label=(\alph*)]
\item Assuming that the surface is smooth, show that $e = 0.8$\\
4
\item Describe one way in which the model you have used could be refined.
Turn over for the next question
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics Q4 [6]}}