| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with direction reversal |
| Difficulty | Challenging +1.2 This is a standard SHM problem requiring knowledge of the acceleration formula a = ω²x and integration to find the period. While it involves multiple steps and algebraic manipulation (finding OM from the ratio, then using calculus for the period), the techniques are routine for Further Maths students and follow predictable patterns without requiring novel insight or complex problem-solving strategies. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| 2(i) | OL/OM = ¾, OM = 2 [m] | M1 A1 |
| Total: | 2 |
| Answer | Marks |
|---|---|
| 2(ii) | EITHER: |
| Answer | Marks | Guidance |
|---|---|---|
| so 2ω = sin–1 0⋅6 + sin–1 0⋅8 [= 0⋅6435 + 0⋅9273] | (*M1 A1) | Find eqn. (AEF) for ω, for example using x = a sin ωt |
| Answer | Marks | Guidance |
|---|---|---|
| or π – cos–1 0⋅8 – cos–1 0⋅6 | (*M1 A1) | or x = a cos ωt |
| Answer | Marks | Guidance |
|---|---|---|
| ω = π / 4 or 0⋅785 (M1 dep *M1) | DM1 A1 | Simplify to find ω (may be implied by T) |
| T = 2π / ω = 8 [s] | B1 FT | Find period T (FT on ω) |
| Total: | 5 | |
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 2(iii) | v = ω√(2⋅52 – 1⋅52) = 2ω = π / 2 or 1⋅57 [m s–1] | |
| L | M1 A1 | Find speed v at L |
| Answer | Marks |
|---|---|
| Total: | 2 |
Question 2:
--- 2(i) ---
2(i) | OL/OM = ¾, OM = 2 [m] | M1 A1 | Find OM by equating ratios of distances and acceln.
Total: | 2
--- 2(ii) ---
2(ii) | EITHER:
ω t = sin–1 (OL/2⋅5) + sin–1(OM/2⋅5)
LM
or ω t = sin–1 (OL/2⋅5), ω (2 – t ) = sin–1 OM/2⋅5)
OL OL
so 2ω = sin–1 0⋅6 + sin–1 0⋅8 [= 0⋅6435 + 0⋅9273] | (*M1 A1) | Find eqn. (AEF) for ω, for example using x = a sin ωt
OR:
ω t = cos–1 (–OM/2⋅5) – cos–1(OL/2⋅5)
LM
or ω t = cos–1 (OL/2⋅5), ω (2 + t ) = cos–1 (–OM/2⋅5)
AL AL
so 2ω = cos–1 (–0⋅8) – cos–1 0⋅6 [= 2⋅498 – 0⋅927]
or π – cos–1 0⋅8 – cos–1 0⋅6 | (*M1 A1) | or x = a cos ωt
(A is at 2⋅5 from O, near L)
ω = π / 4 or 0⋅785 (M1 dep *M1) | DM1 A1 | Simplify to find ω (may be implied by T)
T = 2π / ω = 8 [s] | B1 FT | Find period T (FT on ω)
Total: | 5
Question | Answer | Marks | Guidance
--- 2(iii) ---
2(iii) | v = ω√(2⋅52 – 1⋅52) = 2ω = π / 2 or 1⋅57 [m s–1]
L | M1 A1 | Find speed v at L
L
Total: | 2A particle $P$ moves on a straight line in simple harmonic motion. The centre of the motion is $O$, and the amplitude of the motion is $2.5$ m. The points $L$ and $M$ are on the line, on opposite sides of $O$, with $OL = 1.5$ m. The magnitudes of the accelerations of $P$ at $L$ and at $M$ are in the ratio $3 : 4$.
\begin{enumerate}[label=(\roman*)]
\item Find the distance $OM$. [2]
\end{enumerate}
The time taken by $P$ to travel directly from $L$ to $M$ is $2$ s.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the period of the motion. [5]
\item Find the speed of $P$ when it passes through $L$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2017 Q2 [9]}}