CAIE M2 2014 November — Question 1 3 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2014
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeFinding angle given constraints
DifficultyModerate -0.5 This is a straightforward projectiles question requiring application of standard SUVAT equations in two dimensions. Students need to use horizontal motion (48 = u cos θ × 2.4) and vertical motion (0 = u sin θ × 2.4 - ½g × 2.4²) to find two equations in two unknowns, then solve for θ. While it requires careful algebraic manipulation, it's a standard textbook exercise with no novel insight needed, making it slightly easier than average.
Spec3.02i Projectile motion: constant acceleration model
1 A golf ball \(B\) is projected from a point \(O\) on horizontal ground. \(B\) hits the ground for the first time at a point 48 m away from \(O\) at time 2.4 s after projection. Calculate the angle of projection.
Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(V\sin\theta = \frac{2.4g}{2}\) or \(48 = V\cos\theta \times 2.4\)B1 \(-V\sin\theta = V\sin\theta - 2.4g\)
\(\tan\theta = \frac{12}{\frac{48}{2.4}}\)M1
\(\theta = 31(.0)°\)A1 [3] 
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $V\sin\theta = \frac{2.4g}{2}$ or $48 = V\cos\theta \times 2.4$ | B1 | $-V\sin\theta = V\sin\theta - 2.4g$ |
| $\tan\theta = \frac{12}{\frac{48}{2.4}}$ | M1 | |
| $\theta = 31(.0)°$ | A1 | [3] |

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1 A golf ball $B$ is projected from a point $O$ on horizontal ground. $B$ hits the ground for the first time at a point 48 m away from $O$ at time 2.4 s after projection. Calculate the angle of projection.

\hfill \mbox{\textit{CAIE M2 2014 Q1 [3]}}
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