CAIE FP2 2011 June — Question 1 8 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2011
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeCondition for further/no further collision
DifficultyStandard +0.8 This is a multi-stage collision problem requiring systematic application of conservation of momentum and Newton's restitution law across two collisions, with the added complexity of finding constraints on parameters. While the techniques are standard for Further Maths mechanics, the problem requires careful algebraic manipulation across multiple stages and checking inequality conditions for no further collisions, placing it moderately above average difficulty.
Spec6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts

1 Three small spheres, \(A , B\) and \(C\), of masses \(m , k m\) and \(6 m\) respectively, have the same radius. They are at rest on a smooth horizontal surface, in a straight line with \(B\) between \(A\) and \(C\). The coefficient of restitution between \(A\) and \(B\) is \(\frac { 1 } { 2 }\) and the coefficient of restitution between \(B\) and \(C\) is \(e\). Sphere \(A\) is projected towards \(B\) with speed \(u\) and is brought to rest by the subsequent collision. Show that \(k = 2\). Given that there are no further collisions after \(B\) has collided with \(C\), show that \(e \leqslant \frac { 1 } { 3 }\).

Question 1:
AnswerMarks Guidance
Working/AnswerMark Guidance
Use conservation of momentum for 1st collision: \(kmu_B = mu\)B1
\(u_B = \frac{1}{2}u\)B1
\(k = 2\)B1 A.G.
Use conservation of momentum for 2nd collision: \(kmv_B + 6mv_C = kmu_B\)M1
Use Newton's law of restitution: \(v_B - v_C = -eu_B\)M1
Substitute and solve for \(v_B\): \(2v_B + 6v_C = u,\ v_B - v_C = -\frac{1}{2}eu\)
\(v_B = \frac{(1-3e)u}{8}\ \left[v_C = \frac{(1+e)u}{8}\right]\)M1 A1
Use \(v_B \geq 0\) for no further collisions: \(1 - 3e \geq 0,\ e \leq \frac{1}{3}\)B1 A.G.
S.R. Taking \(v_B = 0\) throughout: \(e = \frac{1}{3}\)(M1 A1) (2)
Total: 8
## Question 1:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Use conservation of momentum for 1st collision: $kmu_B = mu$ | B1 | |
| $u_B = \frac{1}{2}u$ | B1 | |
| $k = 2$ | B1 | **A.G.** | [Subtotal: 3] |
| Use conservation of momentum for 2nd collision: $kmv_B + 6mv_C = kmu_B$ | M1 | |
| Use Newton's law of restitution: $v_B - v_C = -eu_B$ | M1 | |
| Substitute and solve for $v_B$: $2v_B + 6v_C = u,\ v_B - v_C = -\frac{1}{2}eu$ | | |
| $v_B = \frac{(1-3e)u}{8}\ \left[v_C = \frac{(1+e)u}{8}\right]$ | M1 A1 | |
| Use $v_B \geq 0$ for no further collisions: $1 - 3e \geq 0,\ e \leq \frac{1}{3}$ | B1 | **A.G.** | [Subtotal: 5] |
| **S.R.** Taking $v_B = 0$ throughout: $e = \frac{1}{3}$ | (M1 A1) | (2) |

**Total: 8**

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1 Three small spheres, $A , B$ and $C$, of masses $m , k m$ and $6 m$ respectively, have the same radius. They are at rest on a smooth horizontal surface, in a straight line with $B$ between $A$ and $C$. The coefficient of restitution between $A$ and $B$ is $\frac { 1 } { 2 }$ and the coefficient of restitution between $B$ and $C$ is $e$. Sphere $A$ is projected towards $B$ with speed $u$ and is brought to rest by the subsequent collision. Show that $k = 2$.

Given that there are no further collisions after $B$ has collided with $C$, show that $e \leqslant \frac { 1 } { 3 }$.

\hfill \mbox{\textit{CAIE FP2 2011 Q1 [8]}}