| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| 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 standard SUVAT equations. Part (i) uses vertical motion with s=1.225m, u=0, a=9.8 to find t=0.5s. Part (ii) combines horizontal and vertical components to find speed and angle using basic trigonometry. It's routine bookwork with no problem-solving insight required, making it easier than average but not trivial since it requires correct application of multiple equations. |
| Spec | 1.10b Vectors in 3D: i,j,k notation3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | 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 |
| \(\Rightarrow t = 0.5\) s | A1 | Signs must be consistent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Horizontal component of velocity \(= 20\) m s\(^{-1}\) | B1 | |
| Vertical component \(= 0.5 \times 9.8 = 4.9\) m s\(^{-1}\) | B1 | Follow through for "their \(t \times 9.8\)" |
| 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 an appropriate trig ratio with their figures for \(v\). Must be explicit if final answer is incorrect |
| \(\alpha = 13.8°\) | A1 | cao |
# Question 6:
## Part (i)
| Answer/Working | Mark | 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 |
| $\Rightarrow t = 0.5$ s | A1 | Signs must be consistent |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal component of velocity $= 20$ m s$^{-1}$ | B1 | |
| Vertical component $= 0.5 \times 9.8 = 4.9$ m s$^{-1}$ | B1 | Follow through for "their $t \times 9.8$" |
| 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 an appropriate trig ratio with their figures for $v$. Must be explicit if final answer is incorrect |
| $\alpha = 13.8°$ | A1 | cao |6 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]{9eab8ba4-d97b-4e3a-b36d-53f4bc7a80c2-4_233_959_482_575}
\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 Q6 [7]}}