CAIE FP2 2018 June — Question 1 3 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2018
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeBullet penetration with resistance
DifficultyModerate -0.5 This is a straightforward application of the impulse-momentum theorem with constant force. Students need to apply F×t = m(v-u) with all values given directly, requiring only algebraic rearrangement to find m. While it involves mechanics rather than pure maths, it's a standard single-step problem with no conceptual challenges beyond recalling the basic formula.
Spec6.03e Impulse: by a force6.03f Impulse-momentum: relation

1 A bullet of mass \(m \mathrm {~kg}\) is fired horizontally into a fixed vertical block of material. It enters the block horizontally with speed \(250 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and emerges horizontally with speed \(70 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) after 0.04 s . The block offers a constant horizontal resisting force of magnitude 450 N . Find the value of \(m\).

Question 1:
AnswerMarks Guidance
\(m(250 - 70) = 450 \times 0.04\)M1 A1 Find \(m\) from change in momentum \(= Ft\)
\(m = \frac{18}{180} = 0.1\)A1 (can allow M1 if \(250 + 70\) used; similarly below)
*OR:* \(a = (250 - 70)/0.04\ [= 4500]\)(M1) Find deceleration \(a\) (ignore sign)
\(m = \frac{450}{a} = 0.1\)(M1 A1) Find \(m\) from \(F = ma\)
Total: 3 marks
**Question 1:**

$m(250 - 70) = 450 \times 0.04$ | M1 A1 | Find $m$ from change in momentum $= Ft$

$m = \frac{18}{180} = 0.1$ | A1 | (can allow M1 if $250 + 70$ used; similarly below)

*OR:* $a = (250 - 70)/0.04\ [= 4500]$ | (M1) | Find deceleration $a$ (ignore sign)

$m = \frac{450}{a} = 0.1$ | (M1 A1) | Find $m$ from $F = ma$

**Total: 3 marks**

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1 A bullet of mass $m \mathrm {~kg}$ is fired horizontally into a fixed vertical block of material. It enters the block horizontally with speed $250 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and emerges horizontally with speed $70 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ after 0.04 s . The block offers a constant horizontal resisting force of magnitude 450 N . Find the value of $m$.\\

\hfill \mbox{\textit{CAIE FP2 2018 Q1 [3]}}