CAIE M2 2015 November — Question 3 5 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2015
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeProjectile passing through given point
DifficultyStandard +0.3 This is a standard two-equation projectile problem requiring students to resolve the position at a given time into horizontal and vertical components, then solve simultaneous equations. While it involves trigonometry and algebraic manipulation, it follows a well-practiced method with no novel insight required, making it slightly easier than average.
Spec3.02i Projectile motion: constant acceleration model

3 A particle \(P\) is projected with speed \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(\theta ^ { \circ }\) above the horizontal from a point \(O\) on horizontal ground. At the instant 4 s after projection the particle passes through the point \(A\), where \(O A = 40 \mathrm {~m}\) and the line \(O A\) makes an angle of \(30 ^ { \circ }\) with the horizontal. Calculate \(V\) and \(\theta\).

Question 3:
AnswerMarks Guidance
WorkingMark Guidance
\(4V\cos\theta = 40\cos30\)B1
\(4V\sin\theta - \frac{4^2 g}{2} = 40\sin30\)B1
\(V^2 = (10\cos30)^2 + 25^2\) or \(\tan\theta = \frac{25}{10\cos30}\)M1
\(V = 26.5\)A1
\(\theta = 70.9\)A1 [5]
## Question 3:

| Working | Mark | Guidance |
|---------|------|----------|
| $4V\cos\theta = 40\cos30$ | B1 | |
| $4V\sin\theta - \frac{4^2 g}{2} = 40\sin30$ | B1 | |
| $V^2 = (10\cos30)^2 + 25^2$ or $\tan\theta = \frac{25}{10\cos30}$ | M1 | |
| $V = 26.5$ | A1 | |
| $\theta = 70.9$ | A1 | **[5]** |

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3 A particle $P$ is projected with speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $\theta ^ { \circ }$ above the horizontal from a point $O$ on horizontal ground. At the instant 4 s after projection the particle passes through the point $A$, where $O A = 40 \mathrm {~m}$ and the line $O A$ makes an angle of $30 ^ { \circ }$ with the horizontal. Calculate $V$ and $\theta$.

\hfill \mbox{\textit{CAIE M2 2015 Q3 [5]}}