CAIE M2 2011 November — Question 1 3 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2011
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeSpeed at specific time or position
DifficultyModerate -0.8 This is a straightforward projectile motion question requiring resolution of initial velocity into components, application of constant acceleration equations (v = u + at for vertical component), and Pythagoras to find resultant speed. It's a standard single-concept application with clear given values and a routine method, making it easier than average but not trivial since it requires multiple steps and understanding of vector components.
Spec3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model

1 A particle is projected with speed \(17 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(50 ^ { \circ }\) above the horizontal from a point on horizontal ground. Calculate the speed of the particle 2 s after the instant of projection.

AnswerMarks Guidance
\(17\sin50 - 2g\)B1 Vertical component of velocity
\(v^2 = (17\sin50 - 2g)^2 + (17\cos50)^2\)M1 Pythagoras with 2 perpendicular components
\(v = 13(0) \text{ ms}^{-1}\)A1 [3]
$17\sin50 - 2g$ | B1 | Vertical component of velocity
$v^2 = (17\sin50 - 2g)^2 + (17\cos50)^2$ | M1 | Pythagoras with 2 perpendicular components
$v = 13(0) \text{ ms}^{-1}$ | A1 | [3]
1 A particle is projected with speed $17 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $50 ^ { \circ }$ above the horizontal from a point on horizontal ground. Calculate the speed of the particle 2 s after the instant of projection.

\hfill \mbox{\textit{CAIE M2 2011 Q1 [3]}}