CAIE M2 2017 March — Question 1 5 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2017
SessionMarch
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeBasic trajectory calculations
DifficultyModerate -0.3 This is a standard projectile motion problem requiring students to recognize that 'travelling horizontally' means vertical velocity is zero (at maximum height), then apply standard SUVAT equations. It's slightly easier than average because it's a direct application of well-practiced techniques with no problem-solving insight required, though the 5 marks indicate multiple calculation steps.
Spec3.02i Projectile motion: constant acceleration model

A small ball is projected with speed \(15 \text{ m s}^{-1}\) at an angle of \(60°\) above the horizontal. Find the distance from the point of projection of the ball at the instant when it is travelling horizontally. [5]

Question 1:
AnswerMarks Guidance
10 = 15sin60 – gt (t = 3 3 /4 = 1.30) M1
x = 15cos60 x 1.3 (= 9.7428..)A1
y = (15sin60)2/(2g) (= 8.4375..)B1 Uses v2 = u2 + 2as vertically
D = (9.742 + 8.442)M1 Applies Pythagoras's theorem
D = 12.9mA1
Total:5
QuestionAnswer Marks
Question 1:
1 | 0 = 15sin60 – gt (t = 3 3 /4 = 1.30) | M1 | Uses v = u + at vertically
x = 15cos60 x 1.3 (= 9.7428..) | A1
y = (15sin60)2/(2g) (= 8.4375..) | B1 | Uses v2 = u2 + 2as vertically
D = (9.742 + 8.442) | M1 | Applies Pythagoras's theorem
D = 12.9m | A1
Total: | 5
Question | Answer | Marks | Guidance
A small ball is projected with speed $15 \text{ m s}^{-1}$ at an angle of $60°$ above the horizontal. Find the distance from the point of projection of the ball at the instant when it is travelling horizontally. [5]

\hfill \mbox{\textit{CAIE M2 2017 Q1 [5]}}