| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.8 This is a straightforward projectiles question requiring standard SUVAT equations and basic trigonometry. Part (i) is a 'show that' using v² = u² + 2as with given maximum height, part (ii) uses Pythagoras to find the angle, and part (iii) applies range formula. All steps are routine textbook exercises with no problem-solving insight required, making it easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(0^2 = V^2 - 2 \times 9.8 \times 22.5\) | M1 | Use of appropriate *uvast*. Give for correct expression |
| \(V = 21\) so \(21\) m s\(^{-1}\) | E1 | Clearly shown. Do not allow \(v^2 = 0 + 2gs\) without explanation. Accept using \(V = 21\) to show \(s = 22.5\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(28\sin\theta = 21\) so \(\theta = 48.59037...\) | M1, A1 | Attempt to find angle of projection. Allow \(\sin \leftrightarrow \cos\). |
| Answer | Marks | Guidance |
|---|---|---|
| Time to highest point is \(\frac{21}{9.8} = \frac{15}{7}\) | B1 | Or equivalent (time of whole flight) |
| Distance is \(2 \times \frac{15}{7} \times 28 \times \cos(\textbf{their}\ \theta)\) | M1 | Valid method for horizontal distance. Accept \(\frac{1}{2}\) time. Do not accept 28 used for horizontal speed or vertical speed when calculating time. |
| B1 | Horizontal speed correct | |
| \(79.3725...\) so \(79.4\) m (3 s.f.) | A1 | cao. Accept answers rounding to 79 or 80. [If angle with vertical found in (ii) allow up to full marks in (iii). If \(\sin \leftrightarrow \cos\) allow up to B1 B1 M0 A1] [If \(u^2\sin 2\theta / g\) used then M1* correct formula, M1 dep correct subst. FT their angle. A2 cao] |
# Question 5:
## Part (i)
| $0^2 = V^2 - 2 \times 9.8 \times 22.5$ | M1 | Use of appropriate *uvast*. Give for correct expression |
| $V = 21$ so $21$ m s$^{-1}$ | E1 | Clearly shown. Do not allow $v^2 = 0 + 2gs$ without explanation. Accept using $V = 21$ to show $s = 22.5$. |
## Part (ii)
| $28\sin\theta = 21$ so $\theta = 48.59037...$ | M1, A1 | Attempt to find angle of projection. Allow $\sin \leftrightarrow \cos$. |
## Part (iii)
| Time to highest point is $\frac{21}{9.8} = \frac{15}{7}$ | B1 | Or equivalent (time of whole flight) |
| Distance is $2 \times \frac{15}{7} \times 28 \times \cos(\textbf{their}\ \theta)$ | M1 | Valid method for horizontal distance. Accept $\frac{1}{2}$ time. Do not accept 28 used for horizontal speed or vertical speed when calculating time. |
| | B1 | Horizontal speed correct |
| $79.3725...$ so $79.4$ m (3 s.f.) | A1 | cao. Accept answers rounding to 79 or 80. [If angle with vertical found in (ii) allow up to full marks in (iii). If $\sin \leftrightarrow \cos$ allow up to B1 B1 M0 A1] [If $u^2\sin 2\theta / g$ used then M1* correct formula, M1 dep correct subst. FT their angle. A2 cao] |
---5 You should neglect air resistance in this question.\\
A small stone is projected from ground level. The maximum height of the stone above horizontal ground is 22.5 m .\\
(i) Show that the vertical component of the initial velocity of the stone is $21 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
The speed of projection is $28 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) Find the angle of projection of the stone.\\
(iii) Find the horizontal range of the stone.
Section B (36 marks)\\
\hfill \mbox{\textit{OCR MEI M1 2006 Q5 [8]}}