CAIE FP2 2013 June — Question 1 4 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJune
Marks4
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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, the single-step calculation and direct substitution make it easier than average A-level questions.
Spec6.03f Impulse-momentum: relation
1 A bullet of mass \(m \mathrm {~kg}\) is fired into a fixed vertical barrier. It enters the barrier horizontally with speed \(280 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and emerges horizontally after 0.01 s with speed \(30 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). There is a constant horizontal resisting force of magnitude 1500 N . Find \(m\).
Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(I = 1500 \times 0.01 = 15\)M1 A1 Use \(I = Ft\) to find impulse \(I\)
\(m = I/250 = 0.06\)M1 A1 Use \(I = m(v_1 - v_2)\) to find mass \(m\)
\(a = 250/0.01 = 250000\)(M1 A1) OR: Use \(v_2 = v_1 - at\) to find deceleration \(a\)
\(m = 1500/a = 0.06\)(M1 A1) Use \(F = ma\) to find mass \(m\)
\(s = \frac{1}{2} \times 310 \times 0.01 = 1.55\)(M1 A1) OR: Use \(s = \frac{1}{2}(v_1 + v_2)t\) to find distance \(s\)
\(m = 2 \times 1500s/77500 = 0.06\)(M1 A1) Use \(Fs = \frac{1}{2}m(v_1^2 - v_2^2)\) to find mass \(m\)
\(m = I/310 = 0.048[3]\) (max 2/4)(M1 A1) S.R Taking \(v_1 - v_2 = 280 + 30\) in any method
Total: 4 marks
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $I = 1500 \times 0.01 = 15$ | M1 A1 | Use $I = Ft$ to find impulse $I$ |
| $m = I/250 = 0.06$ | M1 A1 | Use $I = m(v_1 - v_2)$ to find mass $m$ |
| $a = 250/0.01 = 250000$ | (M1 A1) | OR: Use $v_2 = v_1 - at$ to find deceleration $a$ |
| $m = 1500/a = 0.06$ | (M1 A1) | Use $F = ma$ to find mass $m$ |
| $s = \frac{1}{2} \times 310 \times 0.01 = 1.55$ | (M1 A1) | OR: Use $s = \frac{1}{2}(v_1 + v_2)t$ to find distance $s$ |
| $m = 2 \times 1500s/77500 = 0.06$ | (M1 A1) | Use $Fs = \frac{1}{2}m(v_1^2 - v_2^2)$ to find mass $m$ |
| $m = I/310 = 0.048[3]$ (max 2/4) | (M1 A1) | **S.R** Taking $v_1 - v_2 = 280 + 30$ in any method |

**Total: 4 marks**

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1 A bullet of mass $m \mathrm {~kg}$ is fired into a fixed vertical barrier. It enters the barrier horizontally with speed $280 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and emerges horizontally after 0.01 s with speed $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. There is a constant horizontal resisting force of magnitude 1500 N . Find $m$.

\hfill \mbox{\textit{CAIE FP2 2013 Q1 [4]}}
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