CAIE FP2 2019 June — Question 1 4 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2019
SessionJune
Marks4
PaperDownload PDF ↗
TopicVariable Force
TypeEnergy method with work done
DifficultyStandard +0.3 This is a straightforward mechanics problem requiring application of Newton's second law or impulse-momentum theorem with constant force. Students need to find constant deceleration from F=ma (a=-6000 m/s²), then use v=u+at to find time. It's slightly above average difficulty due to being Further Maths content and requiring careful handling of signs, but the method is standard with no conceptual challenges.
Spec6.02b Calculate work: constant force, resolved component6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form

1 A bullet of mass 0.2 kg is fired into a fixed vertical barrier. It enters the barrier horizontally with speed \(250 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and emerges horizontally after a time \(T\) seconds with speed \(40 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). There is a constant horizontal resisting force of magnitude 1200 N . Find \(T\).

1 A bullet of mass 0.2 kg is fired into a fixed vertical barrier. It enters the barrier horizontally with speed $250 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and emerges horizontally after a time $T$ seconds with speed $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. There is a constant horizontal resisting force of magnitude 1200 N . Find $T$.\\

\hfill \mbox{\textit{CAIE FP2 2019 Q1 [4]}}