CAIE M2 2019 March — Question 1 5 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2019
SessionMarch
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeSpeed at specific time or position
DifficultyModerate -0.5 This is a straightforward projectile motion problem requiring resolution of initial velocity into components, application of constant acceleration equations (v = u + at for vertical component, horizontal unchanged), then recombination using Pythagoras and trigonometry. It's slightly easier than average as it involves direct application of standard formulas with no problem-solving insight required, though the multi-step calculation and vector recombination elevate it above pure recall.
Spec3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model
1 A particle is projected with speed \(24 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(30 ^ { \circ }\) above the horizontal. Find the speed and direction of motion of the particle at the instant 4 s after projection.
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(x' = 24\cos30 = 12\sqrt{3}\)B1 Use horizontal motion
\(y' = 24\sin30 - 4g = -28\)B1 Use vertical motion
\(V^2 = (24\cos30)^2 + (24\sin30 - 4g)^2 = (12\sqrt{3})^2 + (-28)^2\) OR \(\tan\alpha = (24\sin30 - 4g)/(24\cos30) = -28/(12\sqrt{3})^2\)M1 Where \(V\) is the required speed and \(\alpha\) is the angle below the horizontal
\(V = 34.9 \text{ ms}^{-1}\)A1
\(\alpha = 53.4°\) below the horizontalA1
Total: 5 
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x' = 24\cos30 = 12\sqrt{3}$ | B1 | Use horizontal motion |
| $y' = 24\sin30 - 4g = -28$ | B1 | Use vertical motion |
| $V^2 = (24\cos30)^2 + (24\sin30 - 4g)^2 = (12\sqrt{3})^2 + (-28)^2$ OR $\tan\alpha = (24\sin30 - 4g)/(24\cos30) = -28/(12\sqrt{3})^2$ | M1 | Where $V$ is the required speed and $\alpha$ is the angle below the horizontal |
| $V = 34.9 \text{ ms}^{-1}$ | A1 | |
| $\alpha = 53.4°$ below the horizontal | A1 | |
| **Total: 5** | | |

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1 A particle is projected with speed $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ above the horizontal. Find the speed and direction of motion of the particle at the instant 4 s after projection.\\

\hfill \mbox{\textit{CAIE M2 2019 Q1 [5]}}
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