| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Finding angle given constraints |
| Difficulty | Standard +0.3 This is a straightforward projectile motion question requiring standard SUVAT equations. Part (i) is direct recall (horizontal distance = ut cos α), part (ii) involves simple substitution and arithmetic, and part (iii) uses a standard formula for maximum height. The question is slightly easier than average as it guides students through each step with no problem-solving insight required. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks |
|---|---|
| \(32\cos\alpha\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(32\cos\alpha \times 5 = 44.8\) | M1 | FT their \(x\) |
| so \(160\cos\alpha = 44.8\) and \(\cos\alpha = 0.28\) | E1 | Shown. Must see some working e.g. \(\cos\alpha = 44.8/160\) or \(160\cos\alpha = 44.8\). If \(32 \times 0.28 \times 5 = 44.8\) seen then this needs a statement that 'hence \(\cos\alpha = 0.28\)' |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin\alpha = 0.96\) | B1 | Need not be explicit e.g. accept \(\sin(73.73...)\) seen |
| either: \(0 = (32 \times 0.96)^2 - 2 \times 9.8 \times s\) | M1 | Allow use of \(u = 32\), \(g = \pm(10, 9.8, 9.81)\) |
| A1 | Correct substitution | |
| \(s = 48.1488...\) so 48.1 m (3 s.f.) | A1 | cao |
| or: Time to max height: \(32 \times 0.96 - 9.8T = 0\) so \(T = 3.1349...\) | B1 | Could use \(\frac{1}{2}\) total time of flight to the horizontal |
| \(y = 32 \times 0.96\, t - 4.9\, t^2\) | M1 | Allow use of \(u = 32\), \(g = \pm(10, 9.8, 9.81)\). May use \(s = \frac{(u+v)}{2}t\) |
| putting \(t = T\), \(y = 48.1488\) so 48.1 m (3 s.f.) | A1 | cao |
# Question 4:
## Part (i)
| $32\cos\alpha$ | B1 | |
## Part (ii)
| $32\cos\alpha \times 5 = 44.8$ | M1 | FT their $x$ |
| so $160\cos\alpha = 44.8$ and $\cos\alpha = 0.28$ | E1 | Shown. Must see some working e.g. $\cos\alpha = 44.8/160$ or $160\cos\alpha = 44.8$. If $32 \times 0.28 \times 5 = 44.8$ seen then this needs a statement that 'hence $\cos\alpha = 0.28$' |
## Part (iii)
| $\sin\alpha = 0.96$ | B1 | Need not be explicit e.g. accept $\sin(73.73...)$ seen |
| **either:** $0 = (32 \times 0.96)^2 - 2 \times 9.8 \times s$ | M1 | Allow use of $u = 32$, $g = \pm(10, 9.8, 9.81)$ |
| | A1 | Correct substitution |
| $s = 48.1488...$ so 48.1 m (3 s.f.) | A1 | cao |
| **or:** Time to max height: $32 \times 0.96 - 9.8T = 0$ so $T = 3.1349...$ | B1 | Could use $\frac{1}{2}$ total time of flight to the horizontal |
| $y = 32 \times 0.96\, t - 4.9\, t^2$ | M1 | Allow use of $u = 32$, $g = \pm(10, 9.8, 9.81)$. May use $s = \frac{(u+v)}{2}t$ |
| putting $t = T$, $y = 48.1488$ so 48.1 m (3 s.f.) | A1 | cao |
---4 Fig. 4 shows a particle projected over horizontal ground from a point O at ground level. The particle initially has a speed of $32 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\alpha$ to the horizontal. The particle is a horizontal distance of 44.8 m from O after 5 seconds. Air resistance should be neglected.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d6e78f93-ac2c-4053-87e4-5e5537d6dc3d-4_570_757_447_694}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
(i) Write down an expression, in terms of $\alpha$ and $t$, for the horizontal distance of the particle from O at time $t$ seconds after it is projected.\\
(ii) Show that $\cos \alpha = 0.28$.\\
(iii) Calculate the greatest height reached by the particle.
\hfill \mbox{\textit{OCR MEI M1 2009 Q4 [7]}}