CAIE M2 2015 November — Question 1 4 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2015
SessionNovember
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 application of constant acceleration equations. Students must find when the speed equals 18 m/s using v_x² + v_y² = 18², where v_x is constant and v_y = u_y - gt. The 'rising' condition simply means taking the earlier time solution. It's slightly above average difficulty due to requiring algebraic manipulation of the speed equation, but remains a standard M2 exercise with no novel insight required.
Spec3.02i 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
WorkingMark Guidance
\(V_v^2 = 18^2 - (25\cos50)^2\)M1 Finds vertical component of velocity
\(V_v = 8.1095656...\)A1
\(8.1095656... = 25\sin50 - gt\)M1 \(v = u - gt\) vertically
\(t = 1.1(0)\) sA1 [4]
## Question 1:

| Working | Mark | Guidance |
|---------|------|----------|
| $V_v^2 = 18^2 - (25\cos50)^2$ | M1 | Finds vertical component of velocity |
| $V_v = 8.1095656...$ | A1 | |
| $8.1095656... = 25\sin50 - gt$ | M1 | $v = u - gt$ vertically |
| $t = 1.1(0)$ s | A1 | **[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 2015 Q1 [4]}}