| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| 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-stage collision problem requiring conservation of momentum and Newton's restitution law applied twice, followed by solving simultaneous equations to find e. While the individual principles are standard A-level mechanics, the extended chain of reasoning (first collision → barrier bounce → second collision with constraint) and algebraic manipulation across multiple stages elevates this above typical textbook exercises. The 7 marks for part (ii) reflect substantial working required. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| 3(i) | 3mv + mv = 3mu, v – v = eu (AEF) | |
| A B B A | M1 | Use momentum and Newton’s law |
| Answer | Marks | Guidance |
|---|---|---|
| A B | A1, A1 | Combine to find velocities of A and B after colln. |
| Answer | Marks |
|---|---|
| Total: | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| 3(ii) | v ′ = – ¾ v [= – (9/16) (1 + e)u] (AEF) | |
| B B | B1 | Relate velocity v ′ of B after colln. with wall to v |
| Answer | Marks | Guidance |
|---|---|---|
| A B A B B | M1 | Use momentum (allow m omitted and V = 0) |
| Answer | Marks | Guidance |
|---|---|---|
| B A B A B | M1 | Use Newton’s law |
| Answer | Marks | Guidance |
|---|---|---|
| ¼(3 – e) 2 – (9/16)(1 + e) 2 = 0 (AEF) | (M1 A1) | Eliminate V with V = 0 and substitute for v and v ′ |
| Answer | Marks | Guidance |
|---|---|---|
| 3 (9 – 7e) = e (21 + 5e) | (M1 A1) | |
| 5e2 + 42e – 27 = 0, e = 3/5 or 0⋅6 | M1 A1 | Form and solve quadratic for e, rejecting root –9 |
| Total: | 7 | |
| Question | Answer | Marks |
Question 3:
--- 3(i) ---
3(i) | 3mv + mv = 3mu, v – v = eu (AEF)
A B B A | M1 | Use momentum and Newton’s law
(M0 if inconsistent LHS signs; allow 3v + v = 3u)
A B
v = ¼ (3 – e) u, v = ¾ (1 + e) u
A B | A1, A1 | Combine to find velocities of A and B after colln.
(signs must be consistent with chosen direction)
Total: | 3
--- 3(ii) ---
3(ii) | v ′ = – ¾ v [= – (9/16) (1 + e)u] (AEF)
B B | B1 | Relate velocity v ′ of B after colln. with wall to v
B B
[3mV +] mV = 3mv + mv ′ [V = 3 (9 – 7e) u/16]
A B A B B | M1 | Use momentum (allow m omitted and V = 0)
A
V [– V ] = – e(v ′ – v ) [V = e (21 + 5e) u/16]
B A B A B | M1 | Use Newton’s law
EITHER:
[4V =] (3 – e) v + (1 + e) v ′ = 0
A A B
¼(3 – e) 2 – (9/16)(1 + e) 2 = 0 (AEF) | (M1 A1) | Eliminate V with V = 0 and substitute for v and v ′
B A A B
OR:
3 (9 – 7e) = e (21 + 5e) | (M1 A1)
5e2 + 42e – 27 = 0, e = 3/5 or 0⋅6 | M1 A1 | Form and solve quadratic for e, rejecting root –9
Total: | 7
Question | Answer | Marks | GuidanceTwo uniform small smooth spheres $A$ and $B$ have equal radii and masses $3m$ and $m$ respectively. Sphere $A$ is moving with speed $u$ on a smooth horizontal surface when it collides directly with sphere $B$ which is at rest. The coefficient of restitution between the spheres is $e$.
\begin{enumerate}[label=(\roman*)]
\item Find, in terms of $u$ and $e$, expressions for the velocities of $A$ and $B$ after the collision. [3]
\end{enumerate}
Sphere $B$ continues to move until it strikes a fixed smooth vertical barrier which is perpendicular to the direction of motion of $B$. The coefficient of restitution between $B$ and the barrier is $\frac{3}{4}$. When the spheres subsequently collide, $A$ is brought to rest.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the value of $e$. [7]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2017 Q3 [10]}}