CAIE M2 2011 June — Question 1 2 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2011
SessionJune
Marks2
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeBasic trajectory calculations
DifficultyEasy -1.2 This is a straightforward single-step projectile motion problem requiring only the vertical motion equation with standard SUVAT. Students need to recall that vertical displacement is zero when the particle returns to ground level, then solve 0 = ut sin(40°) - ½gt² for t. It's more routine than average A-level questions as it requires no problem-solving insight, just direct application of a memorized formula.
Spec3.02i Projectile motion: constant acceleration model

1 A particle is projected with speed \(15 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(40 ^ { \circ }\) above the horizontal from a point on horizontal ground. Calculate the time taken for the particle to hit the ground.

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(0 = (15\sin 40°)t - gt^2/2\)M1 Accept quoting the formula \(T = 2V\sin\theta/g\) for the time of flight
\(t = 1.93\)A1 [2]
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $0 = (15\sin 40°)t - gt^2/2$ | M1 | Accept quoting the formula $T = 2V\sin\theta/g$ for the time of flight |
| $t = 1.93$ | A1 [2] | |

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1 A particle is projected with speed $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $40 ^ { \circ }$ above the horizontal from a point on horizontal ground. Calculate the time taken for the particle to hit the ground.

\hfill \mbox{\textit{CAIE M2 2011 Q1 [2]}}