| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| 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 Further Maths mechanics problem involving successive collisions with coefficient of restitution. It requires systematic application of conservation of momentum and Newton's restitution law across two collisions, but follows a predictable template with no novel geometric or algebraic insight needed. The multi-part structure and algebraic manipulation place it slightly above average difficulty. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2mv_A + 4mv_B = 4mu - 4mu\) \([v_A + 2v_B = 0]\) | M1 | Use conservation of momentum for \(A\) and \(B\) (\(m\) may be omitted) |
| \(v_B - v_A = e(2u + u)\) \([v_B - v_A = 3eu]\) | M1 | Use Newton's restitution law with consistent LHS signs |
| \(v_A = -2eu\) and \(v_B = eu\) | A1 | Combine to find \(v_A\) and \(v_B\) (A0 if directions unclear) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([4mv_B'] + mv_C = 4mv_B - \frac{4}{3}mu\) | M1 | Use conservation of momentum for \(B\) & \(C\) (\(m\) may be omitted) |
| \(v_C[-v_B'] = e(v_B + \frac{4u}{3})\) | M1 | Use Newton's restitution law |
| \(4v_B - \frac{4}{3}u = ev_B + \frac{4eu}{3}\), \(4e - \frac{4}{3} = e^2 + \frac{4e}{3}\) | M1 | Combine to find quadratic equation for \(e\) using \(v_B' = 0\) |
| \(3e^2 - 8e + 4 = 0\), \(e = \frac{2}{3}\) \([v_C = \frac{4u}{3}]\) | A1 | Find value of \(e\), (implicitly) rejecting \(e = 2\) |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For A: \(\text{Loss} = \frac{1}{2}\cdot 2m(2u)^2 - \frac{1}{2}\cdot 2m(\frac{4}{3}u)^2 = \frac{20}{9}mu^2\) | M1 | \(v_A = -\frac{4}{3}u\), \([v_B = \frac{2}{3}u]\), \(v_C = \frac{4}{3}u\); one correct |
| For B: \(\text{Loss} = \frac{1}{2}\cdot 4m\cdot u^2 = \frac{1}{2}mu^2\); For C: \(\text{Loss} = 0\) | M1 | Other two correct |
| \(E_{initial} - E_{final}\) or \(L_1 + L_2 = \frac{38}{9}mu^2\) | A1 | Hence find loss in KE |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E_{initial} = \frac{1}{2}\cdot 2m(2u)^2 + \frac{1}{2}\cdot 4mu^2 + \frac{1}{2}m(\frac{4u}{3})^2 = 4mu^2 + 2mu^2 + \frac{8}{9}mu^2 = \frac{62}{9}mu^2\) | M1 | Find initial KE of 3 particles in terms of \(m\) and \(u\) |
| \(E_{final} = \frac{1}{2}\cdot 2mv_A^2 [+\frac{1}{2}\cdot 4mv_{B}'^2] + \frac{1}{2}mv_C^2 = \frac{16}{9}mu^2 + \frac{8}{9}mu^2 = \frac{24}{9}mu^2\) | M1 | Find final KE of 3 particles in terms of \(m\) and \(u\) |
| \(E_{initial} - E_{final}\) or \(L_1 + L_2 = \frac{38}{9}mu^2\) | A1 | Hence find loss in KE |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2mv_A + 4mv_B = 4mu - 4mu$ $[v_A + 2v_B = 0]$ | M1 | Use conservation of momentum for $A$ and $B$ ($m$ may be omitted) |
| $v_B - v_A = e(2u + u)$ $[v_B - v_A = 3eu]$ | M1 | Use Newton's restitution law with consistent LHS signs |
| $v_A = -2eu$ and $v_B = eu$ | A1 | Combine to find $v_A$ and $v_B$ (A0 if directions unclear) |
| | **3** | |
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[4mv_B'] + mv_C = 4mv_B - \frac{4}{3}mu$ | M1 | Use conservation of momentum for $B$ & $C$ ($m$ may be omitted) |
| $v_C[-v_B'] = e(v_B + \frac{4u}{3})$ | M1 | Use Newton's restitution law |
| $4v_B - \frac{4}{3}u = ev_B + \frac{4eu}{3}$, $4e - \frac{4}{3} = e^2 + \frac{4e}{3}$ | M1 | Combine to find quadratic equation for $e$ using $v_B' = 0$ |
| $3e^2 - 8e + 4 = 0$, $e = \frac{2}{3}$ $[v_C = \frac{4u}{3}]$ | A1 | Find value of $e$, (implicitly) rejecting $e = 2$ |
| | **4** | |
## Question 3(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For A: $\text{Loss} = \frac{1}{2}\cdot 2m(2u)^2 - \frac{1}{2}\cdot 2m(\frac{4}{3}u)^2 = \frac{20}{9}mu^2$ | M1 | $v_A = -\frac{4}{3}u$, $[v_B = \frac{2}{3}u]$, $v_C = \frac{4}{3}u$; one correct |
| For B: $\text{Loss} = \frac{1}{2}\cdot 4m\cdot u^2 = \frac{1}{2}mu^2$; For C: $\text{Loss} = 0$ | M1 | Other two correct |
| $E_{initial} - E_{final}$ or $L_1 + L_2 = \frac{38}{9}mu^2$ | A1 | Hence find loss in KE |
| | **3** | |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E_{initial} = \frac{1}{2}\cdot 2m(2u)^2 + \frac{1}{2}\cdot 4mu^2 + \frac{1}{2}m(\frac{4u}{3})^2 = 4mu^2 + 2mu^2 + \frac{8}{9}mu^2 = \frac{62}{9}mu^2$ | M1 | Find initial KE of 3 particles in terms of $m$ and $u$ |
| $E_{final} = \frac{1}{2}\cdot 2mv_A^2 [+\frac{1}{2}\cdot 4mv_{B}'^2] + \frac{1}{2}mv_C^2 = \frac{16}{9}mu^2 + \frac{8}{9}mu^2 = \frac{24}{9}mu^2$ | M1 | Find final KE of 3 particles in terms of $m$ and $u$ |
| $E_{initial} - E_{final}$ or $L_1 + L_2 = \frac{38}{9}mu^2$ | A1 | Hence find loss in KE |
---3 Three uniform small spheres $A , B$ and $C$ have equal radii and masses $2 m , 4 m$ and $m$ respectively. The spheres are moving in a straight line on a smooth horizontal surface, with $B$ between $A$ and $C$. The coefficient of restitution between each pair of spheres is $e$. Spheres $A$ and $B$ are moving towards each other with speeds $2 u$ and $u$ respectively. The first collision is between $A$ and $B$.\\
(i) Find the velocities of $A$ and $B$ after this collision.\\
Sphere $C$ is moving towards $B$ with speed $\frac { 4 } { 3 } u$ and now collides with it. As a result of this collision, $B$ is brought to rest.\\
(ii) Find the value of $e$.\\
(iii) Find the total kinetic energy lost by the three spheres as a result of the two collisions.\\
\hfill \mbox{\textit{CAIE FP2 2019 Q3 [10]}}