CAIE M2 2016 March — Question 1 5 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2016
SessionMarch
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeFinding angle given constraints
DifficultyStandard +0.3 This is a straightforward projectiles question requiring standard kinematic equations. Given that the particle is at maximum height at t=2s (where vertical velocity = 0) and horizontal distance = 30m, students apply v_y = u sin θ - gt = 0 and x = u cos θ × t to find two equations in two unknowns. The setup is clear with no geometric insight required, making it slightly easier than average.
Spec3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model
1 A particle is projected from a point on horizontal ground. At the instant 2 s after projection, the particle has travelled a horizontal distance of 30 m and is at its greatest height above the ground. Find the initial speed and the angle of projection of the particle.
Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(V\sin\theta = 2g\ (= 20)\)B1 Using vertical motion to greatest height
\(V\cos\theta = 30/2\ (= 15)\)B1 Using horizontal motion
\(V^2 = 15^2 + 20^2\) or \(\tan\theta = 20/15\)M1 Using Pythagoras or trigonometry
\(V = 25\ \text{ms}^{-1}\)A1
\(\theta = 53.1°\)A1 Total: 5 
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $V\sin\theta = 2g\ (= 20)$ | B1 | Using vertical motion to greatest height |
| $V\cos\theta = 30/2\ (= 15)$ | B1 | Using horizontal motion |
| $V^2 = 15^2 + 20^2$ or $\tan\theta = 20/15$ | M1 | Using Pythagoras or trigonometry |
| $V = 25\ \text{ms}^{-1}$ | A1 | |
| $\theta = 53.1°$ | A1 | **Total: 5** |

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1 A particle is projected from a point on horizontal ground. At the instant 2 s after projection, the particle has travelled a horizontal distance of 30 m and is at its greatest height above the ground. Find the initial speed and the angle of projection of the particle.

\hfill \mbox{\textit{CAIE M2 2016 Q1 [5]}}
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