CAIE M2 Specimen — Question 1 4 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
SessionSpecimen
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeTime when specific condition met
DifficultyStandard +0.3 This is a straightforward projectile motion problem requiring resolution of velocity components and solving a quadratic equation. Students must find when |v| = 18 using v_x = 25cos(50°) (constant) and v_y = 25sin(50°) - gt, then verify the particle is rising. Standard M2 technique with no novel insight required, slightly above average due to the 'rising' condition check.
Spec3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model

1 A particle is projected with speed \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(50 ^ { \circ }\) above the horizontal. Calculate the time after projection when the particle has speed \(18 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and is rising.

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(V_v^2 = 18^2 - (25\cos50)^2\)M1 Finds vertical comp of velocity
\(V_v = 8.1095656...\)A1
\(8.1095656... = 25\sin50 - gt\)M1 \(v = u - gt\) vertically
\(t = 1.1(0)\) sA1
Total: 4
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $V_v^2 = 18^2 - (25\cos50)^2$ | M1 | Finds vertical comp of velocity |
| $V_v = 8.1095656...$ | A1 | |
| $8.1095656... = 25\sin50 - gt$ | M1 | $v = u - gt$ vertically |
| $t = 1.1(0)$ s | A1 | |
| **Total: 4** | | |

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1 A particle is projected with speed $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $50 ^ { \circ }$ above the horizontal. Calculate the time after projection when the particle has speed $18 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and is rising.\\

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