Question 1 - A-Level Maths

CAIE FP2 2017 June — Question 1 3 marks

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2017
Session	June
Marks	3
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Momentum and Collisions
Type	Bullet penetration with resistance
Difficulty	Moderate -0.5 This is a straightforward application of the impulse-momentum theorem (F·t = m·Δv) with clearly stated values. It requires only one equation and basic algebraic manipulation, making it slightly easier than average despite being from Further Maths, as the mechanics content is standard A-level.
Spec	6.03f Impulse-momentum: relation 6.03g Impulse in 2D: vector form

A bullet of mass 0.08 kg is fired horizontally into a fixed vertical barrier. It enters the barrier horizontally with speed 300 m s\(^{-1}\) and emerges horizontally after 0.02 s. There is a constant horizontal resisting force of magnitude 1000 N. Find the speed with which the bullet emerges from the barrier. [3]

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Question 1:

Answer	Marks	Guidance
1	0⋅08 × (300 – v) = 1000 × 0⋅02 (AEF)	M1 A1

momentum = Ft

(if 300 + v or equivalent, can allow M1 only)

Answer	Marks
v = 300 – 250 = 50 [m s–1]	A1
Total:	3

Question 1:
1 | 0⋅08 × (300 – v) = 1000 × 0⋅02 (AEF) | M1 A1 | Find eqn for exit speed v from e.g. change in
momentum = Ft
(if 300 + v or equivalent, can allow M1 only)
v = 300 – 250 = 50 [m s–1] | A1
Total: | 3

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A bullet of mass 0.08 kg is fired horizontally into a fixed vertical barrier. It enters the barrier horizontally with speed 300 m s$^{-1}$ and emerges horizontally after 0.02 s. There is a constant horizontal resisting force of magnitude 1000 N. Find the speed with which the bullet emerges from the barrier.
[3]

\hfill \mbox{\textit{CAIE FP2 2017 Q1 [3]}}

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Q1 3 Q2 8 Q3 10 Q4 10 Q5 12 Q6 5 Q7 7 Q8 10 Q9 10 Q10 11 Q11 28