CAIE M2 2018 June — Question 1 4 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2018
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeVelocity direction at specific time/point
DifficultyModerate -0.8 This is a straightforward projectile motion question requiring only standard kinematic equations to find velocity components at a given time, then combining them for speed and direction. It's below average difficulty as it involves direct application of v = u + at with no problem-solving or geometric insight required.
Spec3.02i Projectile motion: constant acceleration model

A small ball \(B\) is projected from a point \(O\) on horizontal ground. The initial velocity of \(B\) has horizontal and vertically upwards components of \(18 \text{ ms}^{-1}\) and \(25 \text{ ms}^{-1}\) respectively. For the instant \(4 \text{ s}\) after projection, find the speed and direction of motion of \(B\). [4]

Question 1:
AnswerMarks Guidance
1Vertical component of velocity = 25 – 4g M1
( )
25−4g
v2 =182 +(25−4g)2 or tanθ =
AnswerMarks
18M1
v=23.4 ms−1A1
θ=39.8° below the horizontalA1
4
AnswerMarks Guidance
QuestionAnswer Marks
Question 1:
1 | Vertical component of velocity = 25 – 4g | M1 | Use v = u + at
( )
25−4g
v2 =182 +(25−4g)2 or tanθ =
18 | M1
v=23.4 ms−1 | A1
θ=39.8° below the horizontal | A1
4
Question | Answer | Marks | Guidance
A small ball $B$ is projected from a point $O$ on horizontal ground. The initial velocity of $B$ has horizontal and vertically upwards components of $18 \text{ ms}^{-1}$ and $25 \text{ ms}^{-1}$ respectively. For the instant $4 \text{ s}$ after projection, find the speed and direction of motion of $B$. [4]

\hfill \mbox{\textit{CAIE M2 2018 Q1 [4]}}