CAIE M2 2019 June — Question 2 5 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2019
SessionJune
Marks5
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Mark schemeDownload PDF ↗
TopicProjectiles
TypeSpeed at specific time or position
DifficultyStandard +0.3 This is a standard two-equation projectile problem requiring resolution of velocity components at a specific time, using v_x = V cos θ (constant) and v_y = V sin θ - gt. With two unknowns and two conditions (speed and angle at t=4s), it's straightforward algebra after setting up the equations. Slightly above average due to the simultaneous equation solving, but still a routine M2 exercise.
Spec3.02i Projectile motion: constant acceleration model
2 A particle is projected with speed \(\mathrm { V } \mathrm { m } \mathrm { s } ^ { - 1 }\) at an angle of \(\theta ^ { \circ }\) above the horizontal. At the instant 4 s after projection the speed of the particle is \(16 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and its direction of motion is \(30 ^ { \circ }\) above the horizontal. Find \(V\) and \(\theta\).
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(V\cos\theta=16\cos30\left(=8\sqrt{3}=13.856...\right)\)B1 Use horizontal motion
\(V\sin\theta=16\sin30+4g\ (=48)\)B1 Use vertical motion
\(V^2=(16\cos30)^2+(16\sin30+4g)^2\) OR \(\tan\theta=\frac{(16\sin30+4g)}{16\cos30}\)M1 Use Pythagoras's theorem or trigonometry of a right angled triangle
\(V=50(.0)\)A1
\(\theta=73.9°\)A1
Total: 5 
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V\cos\theta=16\cos30\left(=8\sqrt{3}=13.856...\right)$ | B1 | Use horizontal motion |
| $V\sin\theta=16\sin30+4g\ (=48)$ | B1 | Use vertical motion |
| $V^2=(16\cos30)^2+(16\sin30+4g)^2$ **OR** $\tan\theta=\frac{(16\sin30+4g)}{16\cos30}$ | M1 | Use Pythagoras's theorem or trigonometry of a right angled triangle |
| $V=50(.0)$ | A1 | |
| $\theta=73.9°$ | A1 | |
| **Total: 5** | | |
2 A particle is projected with speed $\mathrm { V } \mathrm { m } \mathrm { s } ^ { - 1 }$ at an angle of $\theta ^ { \circ }$ above the horizontal. At the instant 4 s after projection the speed of the particle is $16 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and its direction of motion is $30 ^ { \circ }$ above the horizontal. Find $V$ and $\theta$.\\

\hfill \mbox{\textit{CAIE M2 2019 Q2 [5]}}
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Q1 5 Q2 5 Q3 5 Q4 8 Q5 8 Q6 7 Q7 12