OCR M2 2005 June — Question 2 6 marks

Exam BoardOCR
ModuleM2 (Mechanics 2)
Year2005
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeHorizontal projection from height
DifficultyModerate -0.8 This is a straightforward projectile motion problem requiring standard SUVAT equations in two dimensions. Students apply v² = u² + 2as vertically to find the vertical component, then use Pythagoras and trigonometry to find speed and direction—routine mechanics with no problem-solving insight needed.
Spec3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model
2 A particle is projected horizontally with a speed of \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point 10 m above horizontal ground. The particle moves freely under gravity. Calculate the speed and direction of motion of the particle at the instant it hits the ground.
AnswerMarks Guidance
\(v^2 = 2 \times 9.8 \times 10\)M1 energy: \(\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mgh\)
\(v = 14\)A1 \(\frac{1}{2}v^2 = \frac{1}{2}.36 + 9.8 \times 10\)
speed \(= \sqrt{(14^2 + 6^2)}\)M1 (must be \(6^2\)) \(v^2 = 36 + 196 = 232\)
speed \(= 15.2 \text{ ms}^{-1}\)A1
\(\tan\theta = 14/6\)M1 \(\cos^{-1}(6/15.2)\) etc
\(\theta = 66.8°\) (below) horiz.A1 6 
$v^2 = 2 \times 9.8 \times 10$ | M1 | energy: $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mgh$
$v = 14$ | A1 | $\frac{1}{2}v^2 = \frac{1}{2}.36 + 9.8 \times 10$
speed $= \sqrt{(14^2 + 6^2)}$ | M1 | (must be $6^2$) $v^2 = 36 + 196 = 232$
speed $= 15.2 \text{ ms}^{-1}$ | A1 | 
$\tan\theta = 14/6$ | M1 | $\cos^{-1}(6/15.2)$ etc
$\theta = 66.8°$ (below) horiz. | A1 | 6 | or $23.2°$ to the vertical
2 A particle is projected horizontally with a speed of $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point 10 m above horizontal ground. The particle moves freely under gravity. Calculate the speed and direction of motion of the particle at the instant it hits the ground.

\hfill \mbox{\textit{OCR M2 2005 Q2 [6]}}
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