| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Collision followed by wall impact |
| Difficulty | Standard +0.8 This is a multi-part mechanics problem requiring conservation of momentum, Newton's restitution law, and careful tracking of multiple collision events. Part (i) is standard A-level mechanics, but parts (ii) and (iii) require setting up and solving equations involving two separate collisions and finding when spheres meet again—this demands more sophisticated problem-solving than typical single-collision questions. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Conservation of momentum: \(2mv_A + mv_B = 2mu\) | M1 | allow \(2v_A + v_B = 2u\) |
| Newton's law of restitution: \(v_B - v_A = eu\) | M1 | |
| \(v_A = (2-e)\,u/3,\quad v_B = 2(1+e)\,u/3\) | A1 A1 | |
| Find \(e\) from \(v_A = | v_{B'} | \) with \(v_{B'} = [-]0.4\,v_B\): \((2-e) = 0.8(1+e),\quad e = 2/3\) |
| Equate times in terms of required distance \(x\): \((d-x)/v_A = d/v_B + x/v_{B'}\) (AEF) \([\,v_A = v_B' = 4u/9,\; v_B = 10u/9\,]\) | M1 A1 | speeds need not be found |
| Use \(v_A = v_B' = 0.4\,v_B\) to solve for \(x\): \(d - x = 0.4d + x,\quad x = 0.3d\) | M1 A1 | |
| OR: Find dist. moved by \(A\) when \(B\) reaches wall: \(d_A = (d/v_B)\,v_A = 0.4d\) | (M1 A1) | |
| Find required distance \(x\): \(x = \frac{1}{2}(d - d_A) = 0.3d\) | (M1 A1) |
# Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Conservation of momentum: $2mv_A + mv_B = 2mu$ | M1 | allow $2v_A + v_B = 2u$ |
| Newton's law of restitution: $v_B - v_A = eu$ | M1 | |
| $v_A = (2-e)\,u/3,\quad v_B = 2(1+e)\,u/3$ | A1 A1 | |
| Find $e$ from $v_A = |v_{B'}|$ with $v_{B'} = [-]0.4\,v_B$: $(2-e) = 0.8(1+e),\quad e = 2/3$ | M1 A1 | |
| Equate times in terms of required distance $x$: $(d-x)/v_A = d/v_B + x/v_{B'}$ (AEF) $[\,v_A = v_B' = 4u/9,\; v_B = 10u/9\,]$ | M1 A1 | speeds need not be found |
| Use $v_A = v_B' = 0.4\,v_B$ to solve for $x$: $d - x = 0.4d + x,\quad x = 0.3d$ | M1 A1 | |
| OR: Find dist. moved by $A$ when $B$ reaches wall: $d_A = (d/v_B)\,v_A = 0.4d$ | (M1 A1) | |
| Find required distance $x$: $x = \frac{1}{2}(d - d_A) = 0.3d$ | (M1 A1) | |
---2 A small uniform sphere $A$, of mass $2 m$, is moving with speed $u$ on a smooth horizontal surface when it collides directly with a small uniform sphere $B$, of mass $m$, which is at rest. The spheres have equal radii and the coefficient of restitution between them is $e$.\\
(i) Find expressions for the speeds of $A$ and $B$ immediately after the collision.\\
Subsequently $B$ collides with a vertical wall which is perpendicular to the direction of motion of $B$. The coefficient of restitution between $B$ and the wall is 0.4 . After $B$ has collided with the wall, the speeds of $A$ and $B$ are equal.\\
(ii) Find $e$.\\
(iii) Initially $B$ is at a distance $d$ from the wall. Find the distance of $B$ from the wall when it next collides with $A$.\\
$3 A$ and $B$ are two fixed points on a smooth horizontal surface, with $A B = 3 a \mathrm {~m}$. One end of a light elastic string, of natural length $a$ m and modulus of elasticity $m g \mathrm {~N}$, is attached to the point $A$. The other end of this string is attached to a particle $P$ of mass $m \mathrm {~kg}$. One end of a second light elastic string, of natural length $k a \mathrm {~m}$ and modulus of elasticity $2 m g \mathrm {~N}$, is attached to $B$. The other end of this string is attached to $P$. It is given that the system is in equilibrium when $P$ is at $M$, the mid-point of $A B$.\\
\hfill \mbox{\textit{CAIE FP2 2017 Q2 [10]}}