| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projection from elevated point - angle above horizontal |
| Difficulty | Moderate -0.3 This is a straightforward projectile motion problem requiring standard SUVAT equations applied in two dimensions. Part (i) involves solving a quadratic for time of flight and calculating horizontal range—routine mechanics calculations with given values. Part (ii) tests conceptual understanding of how g affects range, requiring only qualitative reasoning rather than computation. Slightly easier than average due to the verification format and explicit guidance. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vertical component of initial velocity \(=20\sin30°\ (=10)\) | B1 | |
| Vertical motion \(s=s_0+ut+\frac{1}{2}at^2\) | M1 | Substitution required. Sign of \(g\) must be correct. Condone no \(s_0\) |
| When it hits the sea: \(0=75+10t-5t^2\) | A1 | |
| \(75+10\times5-5\times5^2=0\), satisfied when \(t=5\) | E1 | Or equivalent, e.g. solving the quadratic equation |
| Alternative: Vertical motion \(v=u+at\); at top \(0=10-10t\Rightarrow t=1\) | B1, M1 | Complete method for finding \(t=5\) required |
| Takes another 1 second to reach cliff top level; speed \(10\ \text{ms}^{-1}\) downwards | ||
| \(-75=-10t-5t^2\); \(t^2+2t-15=0\Rightarrow t=3\); Total \(=1+1+3=5\) seconds | A1, E1 | Or equivalent finding time (4 seconds) from top (height 80 m) to hitting the sea |
| Horizontal motion \(x=20\times\cos30°\times t\); \(t=5\Rightarrow 86.6\) | M1 | |
| It is \(3.4\) m from the ship so within \(5\) m | E1 | Condone \(3.5\) m |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| It is longer in the air so it goes further | B1 | Justification for travelling further is required for this mark |
| [1] |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertical component of initial velocity $=20\sin30°\ (=10)$ | B1 | |
| Vertical motion $s=s_0+ut+\frac{1}{2}at^2$ | M1 | Substitution required. Sign of $g$ must be correct. Condone no $s_0$ |
| When it hits the sea: $0=75+10t-5t^2$ | A1 | |
| $75+10\times5-5\times5^2=0$, satisfied when $t=5$ | E1 | Or equivalent, e.g. solving the quadratic equation |
| **Alternative:** Vertical motion $v=u+at$; at top $0=10-10t\Rightarrow t=1$ | B1, M1 | Complete method for finding $t=5$ required |
| Takes another 1 second to reach cliff top level; speed $10\ \text{ms}^{-1}$ downwards | | |
| $-75=-10t-5t^2$; $t^2+2t-15=0\Rightarrow t=3$; Total $=1+1+3=5$ seconds | A1, E1 | Or equivalent finding time (4 seconds) from top (height 80 m) to hitting the sea |
| Horizontal motion $x=20\times\cos30°\times t$; $t=5\Rightarrow 86.6$ | M1 | |
| It is $3.4$ m from the ship so within $5$ m | E1 | Condone $3.5$ m |
| **[6]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| It is longer in the air so it goes further | B1 | Justification for travelling further is required for this mark |
| **[1]** | | |
---4 Fig. 4 illustrates a situation in which a film is being made. A cannon is fired from the top of a vertical cliff towards a ship out at sea. The director wants the cannon ball to fall just short of the ship so that it appears to be a near-miss. There are actors on the ship so it is important that it is not hit by mistake.
The cannon ball is fired from a height 75 m above the sea with an initial velocity of $20 \mathrm {~ms} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ above the horizontal. The ship is 90 m from the bottom of the cliff.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{63a2dc41-5e8b-4275-8653-ece5067c4306-3_337_1242_1717_406}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
(i) The director calculates where the cannon ball will hit the sea, using the standard projectile model and taking the value of $g$ to be $10 \mathrm {~ms} ^ { - 2 }$.
Verify that according to this model the cannon ball is in the air for 5 seconds. Show that it hits the water less than 5 m from the ship.\\
(ii) Without doing any further calculations state, with a brief reason, whether the cannon ball would be predicted to travel further from the cliff if the value of $g$ were taken to be $9.8 \mathrm {~ms} ^ { - 2 }$.
\hfill \mbox{\textit{OCR MEI M1 2014 Q4 [7]}}