| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| 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) requires finding maximum height using v² = u² + 2as. All steps are routine applications of memorized formulas with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(32\cos\alpha \cdot t\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(32\cos\alpha \times 5 = 44.8\) | M1 | FT their \(x\). Shown. Must see some working. |
| so \(160\cos\alpha = 44.8\) and \(\cos\alpha = 0.28\) | E1 | 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sin\alpha = 0.96\) | B1 | Need not be explicit e.g. accept \(\sin(73.73...)\) seen. |
| \(0 = (32 \times 0.96)^2 - 2 \times 9.8 \times s\) | 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. |
| \(y = 32 \times 0.96\, t - 4.9t^2\) | 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 | ca |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $32\cos\alpha \cdot t$ | B1 | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $32\cos\alpha \times 5 = 44.8$ | M1 | FT **their** $x$. Shown. Must see some working. |
| so $160\cos\alpha = 44.8$ and $\cos\alpha = 0.28$ | E1 | 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):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin\alpha = 0.96$ | B1 | Need not be explicit e.g. accept $\sin(73.73...)$ seen. |
| $0 = (32 \times 0.96)^2 - 2 \times 9.8 \times s$ | 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. |
| $y = 32 \times 0.96\, t - 4.9t^2$ | 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 | ca |7 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]{bb65e726-a5e0-4060-81a6-6837dea82e64-5_562_757_389_729}
\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 Q7 [7]}}