| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Successive collisions, three particles in line |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics problem involving multiple collisions with coefficient of restitution. Part (i) is routine application of conservation of momentum and Newton's experimental law. Parts (ii) and (iii) require systematic application of the same principles with careful bookkeeping of directions, but follow predictable patterns. The algebra is straightforward and the problem structure is typical of FP2 collision questions, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Conservation of momentum (1st collision): \(mu_A + 2mu_B = 2mu\) | B1 | |
| Newton's law of restitution: \(u_A - u_B = -e \cdot 2u\) | B1 | |
| Eliminate \(u_A\) to find \(u_B\): \(u_B = 2u(1+e)/3\) | M1 A1 | A.G. Total part: [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Conservation of momentum (2nd collision): \(2mv_B + mv_C = 2mu_B - mu\) | M1 | |
| Newton's law of restitution: \(v_B - v_C = -e(u_B + u)\) | M1 | |
| Substitute and solve for \(v_B\): \(v_B = u(1+e)(1-2e)/9\) | A1 | (A.E.F.) Total part: [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Find \(u_A\): \(u_A = \frac{2}{3}u(1-2e)\) | B1 | |
| State/imply directions in which \(A\), \(B\) move: \(e > \frac{1}{2}\) so \(A/B\) change direction in 1st/2nd collision | B1 | (A.E.F.), needs \(u_A\), \(v_B\) correct |
| Show \( | u_A | > |
| \(= 6/(1+e) > 1\) | M1 A1 | (A.E.F.) Total part: [4], Question total: [11] |
## Question 4(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Conservation of momentum (1st collision): $mu_A + 2mu_B = 2mu$ | B1 | |
| Newton's law of restitution: $u_A - u_B = -e \cdot 2u$ | B1 | |
| Eliminate $u_A$ to find $u_B$: $u_B = 2u(1+e)/3$ | M1 A1 | **A.G.** Total part: [4] |
## Question 4(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Conservation of momentum (2nd collision): $2mv_B + mv_C = 2mu_B - mu$ | M1 | |
| Newton's law of restitution: $v_B - v_C = -e(u_B + u)$ | M1 | |
| Substitute and solve for $v_B$: $v_B = u(1+e)(1-2e)/9$ | A1 | (A.E.F.) Total part: [3] |
## Question 4(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Find $u_A$: $u_A = \frac{2}{3}u(1-2e)$ | B1 | |
| State/imply directions in which $A$, $B$ move: $e > \frac{1}{2}$ so $A/B$ change direction in 1st/2nd collision | B1 | (A.E.F.), needs $u_A$, $v_B$ correct |
| Show $|u_A| > |v_B|$: $|u_A|/|v_B| = \frac{2}{3}/(1+e)/9$ | | needs $u_A$, $v_B$ correct |
| $= 6/(1+e) > 1$ | M1 A1 | (A.E.F.) Total part: [4], Question total: [11] |
---4 Three particles $A , B$ and $C$ have masses $m , 2 m$ and $m$ respectively. The particles are able to move on a smooth horizontal surface in a straight line, and $B$ is between $A$ and $C$. Initially $A$ is moving towards $B$ with speed $2 u$ and $C$ is moving towards $B$ with speed $u$. The particle $B$ is at rest. The coefficient of restitution between any pair of particles is $e$. The first collision is between $A$ and $B$.\\
(i) Show that the speed of $B$ immediately before its collision with $C$ is $\frac { 2 } { 3 } u ( 1 + e )$.\\
(ii) Find the velocity of $B$ immediately after its collision with $C$.\\
(iii) Given that $e > \frac { 1 } { 2 }$, show that there are no further collisions between the particles.
\hfill \mbox{\textit{CAIE FP2 2012 Q4 [11]}}