CAIE FP2 2018 June — Question 5 11 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2018
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeCoalescing particles collision
DifficultyChallenging +1.8 This is a multi-stage mechanics problem requiring: (1) energy conservation to find speed at lowest point, (2) momentum conservation for the collision, (3) circular motion dynamics before and after collision, and (4) solving a quadratic equation. While each individual step uses standard A-level mechanics techniques, the combination of multiple concepts (energy, momentum, circular motion) and the need to relate tensions through the collision makes this significantly harder than routine exercises. It's a typical Further Maths mechanics problem requiring careful systematic work through connected stages.
Spec6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles

5 \includegraphics[max width=\textwidth, alt={}, center]{1b542910-a57e-4f58-a19f-92e67ee9bdf7-08_323_515_260_813} A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point \(O\). The particle is held with the string taut and horizontal. It is projected downwards with speed \(\sqrt { } ( 12 a g )\). At the lowest point of its motion, \(P\) collides directly with a particle \(Q\) of mass \(k m\) which is at rest (see diagram). In the collision, \(P\) and \(Q\) coalesce. The tension in the string immediately after the collision is half of its value immediately before the collision. Find the possible values of \(k\).

Question 5:
AnswerMarks Guidance
\(\frac{1}{2}mv_1^2 = \frac{1}{2}mu^2 + mga\)M1 Find \(v_1^2\) at lowest point from conservation of energy (M0 if no \(m\))
\(v_1^2 = (12 + 2)ag = 14ag\)A1
\(Mv_2 = mv_1\) with \(M = (1+k)m\)M1 Find new \(v_2\) from conservation of momentum
\(v_2 = v_1/(1+k)\) [= \(\sqrt{(14ag)}/(1+k)\)]A1
\(T_1 = mv_1^2/a + mg = (14+1)mg = 15mg\)M1 A1 Find tension \(T_1\) just before collision by using \(F = ma\) radially
\(T_2 = Mv_2^2/a + Mg = (1+k)\{14/(1+k)^2 + 1\}mg\)M1 Find tension \(T_2\) just after collision by using \(F = ma\) radially (M1 needs \(M\), not \(m\), throughout)
or \(\{14/(1+k) + (1+k)\}mg\) (AEF)A1
\(14 + (1+k)^2 = 15(1+k)/2\)M1 Equate \(T_2\) and \(\frac{1}{2}T_1\) to give any eqn in \(k\)
\(2k^2 - 11k + 15 = 0\), \(k = 2.5\) or \(3\) (M1 dep)M1 A1 Solve resulting quadratic for \(k\) (M1 dep on all previous M1s, and requires quadratic eqn)
## Question 5:
$\frac{1}{2}mv_1^2 = \frac{1}{2}mu^2 + mga$ | M1 | Find $v_1^2$ at lowest point from conservation of energy (M0 if no $m$)

$v_1^2 = (12 + 2)ag = 14ag$ | A1 |

$Mv_2 = mv_1$ with $M = (1+k)m$ | M1 | Find new $v_2$ from conservation of momentum

$v_2 = v_1/(1+k)$ [= $\sqrt{(14ag)}/(1+k)$] | A1 |

$T_1 = mv_1^2/a + mg = (14+1)mg = 15mg$ | M1 A1 | Find tension $T_1$ just before collision by using $F = ma$ radially

$T_2 = Mv_2^2/a + Mg = (1+k)\{14/(1+k)^2 + 1\}mg$ | M1 | Find tension $T_2$ just after collision by using $F = ma$ radially (M1 needs $M$, not $m$, throughout)

or $\{14/(1+k) + (1+k)\}mg$ (AEF) | A1 |

$14 + (1+k)^2 = 15(1+k)/2$ | M1 | Equate $T_2$ and $\frac{1}{2}T_1$ to give any eqn in $k$

$2k^2 - 11k + 15 = 0$, $k = 2.5$ or $3$ (M1 dep) | M1 A1 | Solve resulting quadratic for $k$ (M1 dep on all previous M1s, and requires quadratic eqn)

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5\\
\includegraphics[max width=\textwidth, alt={}, center]{1b542910-a57e-4f58-a19f-92e67ee9bdf7-08_323_515_260_813}

A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point $O$. The particle is held with the string taut and horizontal. It is projected downwards with speed $\sqrt { } ( 12 a g )$. At the lowest point of its motion, $P$ collides directly with a particle $Q$ of mass $k m$ which is at rest (see diagram). In the collision, $P$ and $Q$ coalesce. The tension in the string immediately after the collision is half of its value immediately before the collision. Find the possible values of $k$.\\

\hfill \mbox{\textit{CAIE FP2 2018 Q5 [11]}}