| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Horizontal projection from height |
| Difficulty | Moderate -0.8 This is a straightforward two-part projectile motion question requiring only standard SUVAT equations applied separately to horizontal and vertical components. Part (i) uses s=ut+½at² with convenient numbers (1.225m gives t=0.5s cleanly), and part (ii) requires Pythagoras and basic trigonometry to find resultant speed and angle. No problem-solving insight needed—pure routine application of memorized techniques. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vertical motion: \(s = ut + \frac{1}{2}at^2\) | ||
| At water: \(-1.225 = 0\times t + \frac{1}{2}\times(-9.8)\times t^2\) | M1 | Condone sign errors |
| \(\Rightarrow t = 0.5\text{ s}\) | A1 | Signs must be consistent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Horizontal component of velocity \(= 20\text{ m s}^{-1}\) | B1 | |
| Vertical component \(= 0.5\times9.8 = 4.9\text{ m s}^{-1}\) | B1 | Follow through for "their \(t\times9.8\)" |
| Speed \(= \sqrt{20^2 + 4.9^2} = 20.6\) | M1 | Use of Pythagoras on previous two answers |
| \(\tan\alpha = \frac{4.9}{20}\) | M1 | Use of appropriate trig ratio with their figures for \(\mathbf{v}\). Must be explicit if final answer is incorrect |
| \(\alpha = 13.8°\) | A1 | Cao |
## Question 5:
**Part (i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertical motion: $s = ut + \frac{1}{2}at^2$ | | |
| At water: $-1.225 = 0\times t + \frac{1}{2}\times(-9.8)\times t^2$ | M1 | Condone sign errors |
| $\Rightarrow t = 0.5\text{ s}$ | A1 | Signs must be consistent |
**Part (ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Horizontal component of velocity $= 20\text{ m s}^{-1}$ | B1 | |
| Vertical component $= 0.5\times9.8 = 4.9\text{ m s}^{-1}$ | B1 | Follow through for "their $t\times9.8$" |
| Speed $= \sqrt{20^2 + 4.9^2} = 20.6$ | M1 | Use of Pythagoras on previous two answers |
| $\tan\alpha = \frac{4.9}{20}$ | M1 | Use of appropriate trig ratio with their figures for $\mathbf{v}$. Must be explicit if final answer is incorrect |
| $\alpha = 13.8°$ | A1 | Cao |
---5 Ali is throwing flat stones onto water, hoping that they will bounce, as illustrated in Fig. 5.\\
Ali throws one stone from a height of 1.225 m above the water with initial speed $20 \mathrm {~ms} ^ { - 1 }$ in a horizontal direction. Air resistance should be neglected.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{13f555cc-d506-48e5-a0e4-225cae4251dc-5_229_953_434_557}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}
(i) Find the time it takes for the stone to reach the water.\\
(ii) Find the speed of the stone when it reaches the water and the angle its trajectory makes with the horizontal at this time.
\hfill \mbox{\textit{OCR MEI M1 2013 Q5 [7]}}