CAIE M2 2017 November — Question 2 4 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2017
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeFinding angle given constraints
DifficultyStandard +0.8 This projectiles question requires working backwards from an intermediate point to find initial conditions. Students must decompose velocity at the given point, apply kinematic equations in both horizontal and vertical directions, and solve simultaneous equations involving trigonometry. While the mechanics concepts are standard M2 material, the reverse-engineering approach and algebraic manipulation required elevate this above routine projectile problems.
Spec3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model
2 A small ball is projected from a point 1.5 m above horizontal ground. At a point 9 m above the ground the ball is travelling at \(45 ^ { \circ }\) above the horizontal and its velocity is \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find the angle of projection of the ball.
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(V\cos\theta = 4\cos45\)B1 Using horizontal motion with \(V\) = velocity of projection and \(\theta\) = angle of projection
\((4\sin45)^2 = (V\sin\theta)^2 - 2g(9-1.5)\) (leads to \(V\sin\theta = \sqrt{158}\))M1 Uses \(v^2 = u^2 + 2as\) vertically
\(\tan\theta = \sqrt{158}/(4\cos45)\)M1 Uses trigonometry
\(\theta = 77.3°\)A1 
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $V\cos\theta = 4\cos45$ | B1 | Using horizontal motion with $V$ = velocity of projection and $\theta$ = angle of projection |
| $(4\sin45)^2 = (V\sin\theta)^2 - 2g(9-1.5)$ (leads to $V\sin\theta = \sqrt{158}$) | M1 | Uses $v^2 = u^2 + 2as$ vertically |
| $\tan\theta = \sqrt{158}/(4\cos45)$ | M1 | Uses trigonometry |
| $\theta = 77.3°$ | A1 | |

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2 A small ball is projected from a point 1.5 m above horizontal ground. At a point 9 m above the ground the ball is travelling at $45 ^ { \circ }$ above the horizontal and its velocity is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the angle of projection of the ball.\\

\hfill \mbox{\textit{CAIE M2 2017 Q2 [4]}}
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