CAIE FP2 2012 June — Question 2 6 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2012
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImpulse and momentum (advanced)
TypeLoss of kinetic energy
DifficultyStandard +0.8 This is a moderately challenging mechanics problem requiring systematic application of impulse-momentum principles, careful sign convention handling, and algebraic manipulation through multiple steps. While the techniques are standard for Further Maths, the need to work with impulse magnitude (requiring consideration of two cases), find both final velocities, calculate initial and final KE, and simplify to the given answer makes this more demanding than a routine collision question.
Spec6.02d Mechanical energy: KE and PE concepts6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form

Two particles, of masses \(3m\) and \(m\), are moving in the same straight line towards each other with speeds \(2u\) and \(u\) respectively. When they collide, the impulse acting on each particle has magnitude \(4mu\). Show that the total loss in kinetic energy is \(\frac{4}{5}mu^2\). [6]

Question 2:
AnswerMarks
2Find two momentum eqns, e.g.: 3mv 3m = 3m × 2u – 4mu or
mvm = – mu + 4mu or
3mv3m + mvm = 3m × 2u – mu M1 M1
Solve to find both speeds after colln.: v3m = 2u/3 and vm = 3u A1
Find total loss in KE: ½m {3(2u) 2 + u 2 – 3v3m 2 – vm 2 } M1 A1
2
= ½m (12 + 1 – 4/3 – 9) u
2
= (13/2 – 31/6 or 16/3 – 4) mu
AnswerMarks Guidance
= (4/3) mu 2 A.G. A16 [6]
Question 2:
2 | Find two momentum eqns, e.g.: 3mv 3m = 3m × 2u – 4mu or
mvm = – mu + 4mu or
3mv3m + mvm = 3m × 2u – mu M1 M1
Solve to find both speeds after colln.: v3m = 2u/3 and vm = 3u A1
Find total loss in KE: ½m {3(2u) 2 + u 2 – 3v3m 2 – vm 2 } M1 A1
2
= ½m (12 + 1 – 4/3 – 9) u
2
= (13/2 – 31/6 or 16/3 – 4) mu
= (4/3) mu 2 A.G. A1 | 6 | [6]
Two particles, of masses $3m$ and $m$, are moving in the same straight line towards each other with speeds $2u$ and $u$ respectively. When they collide, the impulse acting on each particle has magnitude $4mu$. Show that the total loss in kinetic energy is $\frac{4}{5}mu^2$. [6]

\hfill \mbox{\textit{CAIE FP2 2012 Q2 [6]}}