| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Energy loss in collision |
| Difficulty | Standard +0.3 This is a standard two-part collision problem requiring conservation of momentum and the restitution formula. Part (i) involves straightforward algebraic manipulation of two equations to show a given result. Part (ii) requires an additional energy equation but follows a routine procedure. The problem is slightly easier than average as it's a direct application of standard mechanics formulas with no geometric complications or novel insights required. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(mv_A = mu - mku\) (AEF) | M1 | Use conservation of momentum (\(m\) may be omitted) |
| \(v_A = -e(u + ku)\) | M1 | Use Newton's restitution law |
| \(-e(1+k) = 1-k\), \(e = \frac{(k-1)}{(k+1)}\) AG | A1 | Combine to find/verify \(e\) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(KE_I - KE_F = 0.6\, KE_I\) *or* \(0.4\, KE_I = KE_F\) | M1 | Attempt to relate initial and final KEs |
| \(0.4 \times \frac{1}{2}mu^2(1+k^2) = \frac{1}{2}mv_A^2 = \frac{1}{2}mu^2(1-k)^2\) | M1 | Substitute for KEs (\(\frac{1}{2}m\) may be omitted) |
| \(0.6k^2 - 2k + 0.6 = 0\) *or* \(3k^2 - 10k + 3 = 0\) (AEF) | A1 | Simplify to quadratic eqn in \(k\) |
| \((3k-1)(k-3) = 0\) so solutions are \(3\) and \(\frac{1}{3}\) | A1 | Find both solutions of quadratic eqn |
| [Reject \(k = \frac{1}{3}\)] so \(k = 3\) | A1 | (Implicitly) reject one solution to find possible value of \(k\) since \(0 < e < 1\) or \(v_A < 0\) |
| \(e = \frac{2}{4} = \frac{1}{2}\) (FT on \(k\)) | A1\(\sqrt{}\) | Find single corresponding value of \(e\) |
| 6 |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $mv_A = mu - mku$ (AEF) | M1 | Use conservation of momentum ($m$ may be omitted) |
| $v_A = -e(u + ku)$ | M1 | Use Newton's restitution law |
| $-e(1+k) = 1-k$, $e = \frac{(k-1)}{(k+1)}$ **AG** | A1 | Combine to find/verify $e$ |
| | **3** | |
---
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $KE_I - KE_F = 0.6\, KE_I$ *or* $0.4\, KE_I = KE_F$ | M1 | Attempt to relate initial and final KEs |
| $0.4 \times \frac{1}{2}mu^2(1+k^2) = \frac{1}{2}mv_A^2 = \frac{1}{2}mu^2(1-k)^2$ | M1 | Substitute for KEs ($\frac{1}{2}m$ may be omitted) |
| $0.6k^2 - 2k + 0.6 = 0$ *or* $3k^2 - 10k + 3 = 0$ (AEF) | A1 | Simplify to quadratic eqn in $k$ |
| $(3k-1)(k-3) = 0$ so solutions are $3$ and $\frac{1}{3}$ | A1 | Find both solutions of quadratic eqn |
| [Reject $k = \frac{1}{3}$] so $k = 3$ | A1 | (Implicitly) reject one solution to find possible value of $k$ since $0 < e < 1$ or $v_A < 0$ |
| $e = \frac{2}{4} = \frac{1}{2}$ (FT on $k$) | A1$\sqrt{}$ | Find single corresponding value of $e$ |
| | **6** | |
---3 Two identical uniform small spheres $A$ and $B$, each of mass $m$, are moving towards each other in a straight line on a smooth horizontal surface. Their speeds are $u$ and $k u$ respectively, and they collide directly. The coefficient of restitution between the spheres is $e$. Sphere $B$ is brought to rest by the collision.\\
(i) Show that $e = \frac { k - 1 } { k + 1 }$.\\
(ii) Given that $60 \%$ of the total initial kinetic energy is lost in the collision, find the values of $k$ and $e$.\\
\hfill \mbox{\textit{CAIE FP2 2018 Q3 [9]}}