CAIE M2 2010 June — Question 1 3 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2010
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeHorizontal projection from height
DifficultyModerate -0.8 This is a straightforward projectile motion question requiring only basic kinematic equations. Students need to find horizontal and vertical velocity components after 2s (v_x = 12, v_y = gt = 19.6), then calculate the angle using tan θ = v_y/v_x. It's a single-step application of standard formulas with no problem-solving insight required, making it easier than average.
Spec3.02i Projectile motion: constant acceleration model

1 A particle is projected horizontally with speed \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from the top of a high cliff. Find the direction of motion of the particle after 2 s .

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(v_{down} = 2g\)B1
\(\tan\theta = \frac{2g}{12}\)M1 \(\tan\alpha = \frac{12}{2g}\)
\(\theta = 59.0°\)A1
[3]
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $v_{down} = 2g$ | B1 | |
| $\tan\theta = \frac{2g}{12}$ | M1 | $\tan\alpha = \frac{12}{2g}$ |
| $\theta = 59.0°$ | A1 | |
| | **[3]** | |

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1 A particle is projected horizontally with speed $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from the top of a high cliff. Find the direction of motion of the particle after 2 s .

\hfill \mbox{\textit{CAIE M2 2010 Q1 [3]}}