| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | November |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Pendulum particle collision at lowest point |
| Difficulty | Challenging +1.2 This is a multi-part mechanics problem combining circular motion, energy conservation, collision mechanics, and tension analysis. While it requires several connected steps (energy conservation to find speed at lowest point, momentum conservation for collision, then energy/circular motion for the slack condition), each individual technique is standard for Further Maths mechanics. The algebraic manipulation is straightforward, and the problem structure guides students through each stage clearly. It's moderately harder than average A-level due to being Further Maths content with multiple integrated concepts, but doesn't require exceptional insight. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}mv_1^2 = \frac{1}{2}mu^2 + 2mga\) | M1 A1 | Find \(v_1^2\) at lowest point from conservation of energy |
| \(Mv_2 = mv_1\) with \(M = (\lambda + 1)m\) | M1 A1 | Verify new \(v_2\) from conservation of momentum |
| \(v_2 = v_1/(\lambda + 1) = \{5/(\lambda+1)\}\sqrt{(\frac{1}{3}ag)}\) | A.G. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(Mv_3^2/a = Mg\cos\theta\) | B1 | Use \(F = ma\) radially at slack point with \(T = 0\); \(\theta\) is angle of string with upward vertical |
| \(\frac{1}{2}Mv_3^2 = \frac{1}{2}Mv_2^2 - Mg(4a/3)\) | M1 A1 | Find \(v_3^2\) at slack point from conservation of energy; SR: lose max 1 mark if mass is \(m\) in either equation |
| \(ag/3 = v_2^2 - 8ag/3\) | M1 | Eliminate \(v_3^2\ [= ag/3]\) using \(\cos\theta = \frac{1}{3}\) |
| \(3ag = \{25/(\lambda+1)^2\}\ ag/3\) | M1 A1 | Substitute for \(v_2\) to find \(\lambda\) |
| \((\lambda+1)^2 = 25/9,\ \lambda = \frac{2}{3}\) [rejecting \(-8/3\)] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T_1 = mv_1^2/a + mg [= (25/3 + 1)mg = 28mg/3]\) | B1 | Use \(F = ma\) radially just before collision |
| \(T_2 = Mv_2^2/a + Mg [= (3+1)(5m/3)g = 20mg/3]\) | B1 | Use \(F = ma\) radially just after collision |
| \((25mg/3)\{\lambda/(\lambda+1)\} - \lambda mg = 8mg/3\) or \(2.67mg\) | A1 A1 | Find change in tension (either sign, AEF) |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}mv_1^2 = \frac{1}{2}mu^2 + 2mga$ | M1 A1 | Find $v_1^2$ at lowest point from conservation of energy |
| $Mv_2 = mv_1$ with $M = (\lambda + 1)m$ | M1 A1 | Verify new $v_2$ from conservation of momentum |
| $v_2 = v_1/(\lambda + 1) = \{5/(\lambda+1)\}\sqrt{(\frac{1}{3}ag)}$ | | A.G. |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Mv_3^2/a = Mg\cos\theta$ | B1 | Use $F = ma$ radially at slack point with $T = 0$; $\theta$ is angle of string with upward vertical |
| $\frac{1}{2}Mv_3^2 = \frac{1}{2}Mv_2^2 - Mg(4a/3)$ | M1 A1 | Find $v_3^2$ at slack point from conservation of energy; SR: lose max 1 mark if mass is $m$ in either equation |
| $ag/3 = v_2^2 - 8ag/3$ | M1 | Eliminate $v_3^2\ [= ag/3]$ using $\cos\theta = \frac{1}{3}$ |
| $3ag = \{25/(\lambda+1)^2\}\ ag/3$ | M1 A1 | Substitute for $v_2$ to find $\lambda$ |
| $(\lambda+1)^2 = 25/9,\ \lambda = \frac{2}{3}$ [rejecting $-8/3$] | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T_1 = mv_1^2/a + mg [= (25/3 + 1)mg = 28mg/3]$ | B1 | Use $F = ma$ radially just before collision |
| $T_2 = Mv_2^2/a + Mg [= (3+1)(5m/3)g = 20mg/3]$ | B1 | Use $F = ma$ radially just after collision |
| $(25mg/3)\{\lambda/(\lambda+1)\} - \lambda mg = 8mg/3$ or $2.67mg$ | A1 A1 | Find change in tension (either sign, AEF) |
---4 A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point $O$. The particle is held vertically above $O$ with the string taut and then projected horizontally with speed $\sqrt { } \left( \frac { 13 } { 3 } a g \right)$. It begins to move in a vertical circle with centre $O$. When $P$ is at its lowest point, it collides with a stationary particle of mass $\lambda m$. The two particles coalesce.\\
(i) Show that the speed of the combined particle immediately after the impact is $\frac { 5 } { \lambda + 1 } \sqrt { } \left( \frac { 1 } { 3 } a g \right)$.
In the subsequent motion, the string becomes slack when the combined particle is at a height of $\frac { 1 } { 3 } a$ above the level of $O$.\\
(ii) Find the value of $\lambda$.\\
(iii) Find, in terms of $m$ and $g$, the instantaneous change in the tension in the string as a result of the collision.
\hfill \mbox{\textit{CAIE FP2 2016 Q4 [14]}}