| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with direction reversal |
| Difficulty | Standard +0.3 This is a standard A-level mechanics collision problem requiring conservation of momentum and Newton's restitution law. While it involves algebraic manipulation with multiple masses and the restitution coefficient, the techniques are routine for Further Maths students. Part (i) is guided ('show that'), part (ii) is straightforward substitution, and part (iii) requires calculating kinetic energies—all standard procedures with no novel insight required. The 9 marks and multi-part structure place it slightly above average difficulty, but it remains a textbook-style exercise. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| 2(i) | 5mv + 2mv = 5mu – 4mu = mu (AEF) | |
| A B | M1 | Use momentum (allow m omitted) |
| Answer | Marks | Guidance |
|---|---|---|
| B A | M1 | Use Newton’s law (M0 if LHS signs inconsistent) |
| Answer | Marks | Guidance |
|---|---|---|
| A | A1 | Combine to find/verify speeds of A and B after colln. |
| Answer | Marks | Guidance |
|---|---|---|
| B | A1 | (ignore signs) |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 2(ii) | (u/7) (1 – 6e) = – ½ u, e = ¾ or 0.75 | M1A1 |
| Answer | Marks |
|---|---|
| 2(iii) | KE = ½ × 5m {u2 – (½ u)2} and [= (15/8)mu2] |
| Answer | Marks | Guidance |
|---|---|---|
| B | M1A1 | Find loss of KE for A and B |
| Answer | Marks | Guidance |
|---|---|---|
| A B | A1 | Combine to find ratio |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 2:
--- 2(i) ---
2(i) | 5mv + 2mv = 5mu – 4mu = mu (AEF)
A B | M1 | Use momentum (allow m omitted)
v – v = e(u + 2u) = 3eu
B A | M1 | Use Newton’s law (M0 if LHS signs inconsistent)
v = (u/7) (1 – 6e)
A | A1 | Combine to find/verify speeds of A and B after colln.
v = (u/7) (1 + 15e) AG
B | A1 | (ignore signs)
4
Question | Answer | Marks | Guidance
--- 2(ii) ---
2(ii) | (u/7) (1 – 6e) = – ½ u, e = ¾ or 0.75 | M1A1 | Combine to find e from v = – ½ u
A
(M0 if dirn. of motion not reversed)
2
--- 2(iii) ---
2(iii) | KE = ½ × 5m {u2 – (½ u)2} and [= (15/8)mu2]
A
KE = ½ × 2m {(2u)2 – (7u/4)2} [= (15/16)mu2]
B | M1A1 | Find loss of KE for A and B
KE / KE = (15/8) / (15/16) = 2:1 or 2/1 or 2
A B | A1 | Combine to find ratio
3
Question | Answer | Marks | GuidanceTwo uniform small smooth spheres $A$ and $B$ have equal radii and masses $5m$ and $2m$ respectively. Sphere $A$ is moving with speed $u$ on a smooth horizontal surface when it collides directly with sphere $B$ which is moving towards it with speed $2u$. The coefficient of restitution between the spheres is $e$.
\begin{enumerate}[label=(\roman*)]
\item Show that the speed of $B$ after the collision is $\frac{1}{7}u(1 + 15e)$ and find an expression for the speed of $A$. [4]
\end{enumerate}
In the collision, the speed of $A$ is halved and its direction of motion is reversed.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the value of $e$. [2]
\item For this collision, find the ratio of the loss of kinetic energy of $A$ to the loss of kinetic energy of $B$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2018 Q2 [9]}}