| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 20 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile clearing obstacle |
| Difficulty | Standard +0.3 This is a standard M1 projectiles question with straightforward application of SUVAT equations and trajectory formulas. All parts follow routine procedures: deriving position equations, finding maximum height time, calculating range, and solving a quadratic. The multi-part structure and final calculation add some length but require no novel insight beyond textbook methods, making it slightly easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.05g Exact trigonometric values: for standard angles3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Eliminate \(t\) between \(y = 7\sqrt{3}\) and \(= 7t\) | M1 | Must see their \(t = x/7\) fully substituted in quadratic \(y\); accept bracket errors |
| So \(y = 7\sqrt{3}\dfrac{x}{7} - 4.9\left(\dfrac{x}{7}\right)^2 + 1\) | F1 | FT their \(x\) and 3 term quadratic \(y\) (neither using \(u = 14\)) |
| So \(y = \sqrt{3}x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Need \(6 = 7\sqrt{3}t - 4.9t^2 + 1\) | M1 | Their quadratic \(y\) from (i) \(= 6\), or equivalent |
| So \(4.9t^2 - 7\sqrt{3}t + 5 = 0\) | M1 | Dep. Attempt to solve this 3 term quadratic (allow \(u = 14\)) |
| \(t = \dfrac{5(\sqrt{3} \pm 1)}{7}\) (0.52289… or 1.95146…) | A1 | For either root |
| Moves by \(\left(\dfrac{5(\sqrt{3}+1)}{7} - \dfrac{5\sqrt{3}}{7}\right) \times 7\) | M1 | Moves by \ |
| \([(1.95146\ldots - 1.23717\ldots) \times 7]\) | ||
| \(= 5\) m | A1 | cao; [If new distance to wall found must have larger of 2 +ve roots for 3rd M and award max 4/5 for 13.66] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6 = \sqrt{3}x - 0.1x^2 + 1\) | M1 | Equating their quadratic trajectory eqn to 6 |
| Solving \(x^2 - 10\sqrt{3}x + 50 = 0\) | M1 | Dep. Attempt to solve this 3 term quadratic (allow \(u = 14\)) |
| \(x = 5(\sqrt{3} \pm 1)\) (13.660… or 3.6602…) | A1 | For either root |
| Distance is \(5(\sqrt{3}+1) - 5\sqrt{3}\) | M1 | Distance is \ |
| \(= 5\) m | A1 | cao; [If new distance to wall found must have larger of 2 +ve roots for 3rd M and award max 4/5 for 13.66] |
# Question 1 (Projectile - parts iii and iv):
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Eliminate $t$ between $y = 7\sqrt{3}$ and $= 7t$ | M1 | Must see their $t = x/7$ fully substituted in quadratic $y$; accept bracket errors |
| So $y = 7\sqrt{3}\dfrac{x}{7} - 4.9\left(\dfrac{x}{7}\right)^2 + 1$ | F1 | FT their $x$ and 3 term quadratic $y$ (neither using $u = 14$) |
| So $y = \sqrt{3}x$ | | |
## Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Need $6 = 7\sqrt{3}t - 4.9t^2 + 1$ | M1 | Their quadratic $y$ from (i) $= 6$, or equivalent |
| So $4.9t^2 - 7\sqrt{3}t + 5 = 0$ | M1 | Dep. Attempt to solve this 3 term quadratic (allow $u = 14$) |
| $t = \dfrac{5(\sqrt{3} \pm 1)}{7}$ (0.52289… or 1.95146…) | A1 | For either root |
| Moves by $\left(\dfrac{5(\sqrt{3}+1)}{7} - \dfrac{5\sqrt{3}}{7}\right) \times 7$ | M1 | Moves by \|their root - their (ii)(A)\| $\times 7$ or equivalent; award for recognition of correct dist (no calc) |
| $[(1.95146\ldots - 1.23717\ldots) \times 7]$ | | |
| $= 5$ m | A1 | cao; [If new distance to wall found must have larger of 2 +ve roots for 3rd M and award max 4/5 for 13.66] |
**Or** using equation of trajectory with $y = 6$:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6 = \sqrt{3}x - 0.1x^2 + 1$ | M1 | Equating their quadratic trajectory eqn to 6 |
| Solving $x^2 - 10\sqrt{3}x + 50 = 0$ | M1 | Dep. Attempt to solve this 3 term quadratic (allow $u = 14$) |
| $x = 5(\sqrt{3} \pm 1)$ (13.660… or 3.6602…) | A1 | For either root |
| Distance is $5(\sqrt{3}+1) - 5\sqrt{3}$ | M1 | Distance is \|their root - their(ii)(B)\| |
| $= 5$ m | A1 | cao; [If new distance to wall found must have larger of 2 +ve roots for 3rd M and award max 4/5 for 13.66] |
---1 A girl throws a small stone with initial speed $14 \mathrm {~ms} { } ^ { 1 }$ at an angle of $60 ^ { \circ }$ to the horizontal from a point 1 m above the ground. She throws the stone directly towards a vertical wall of height 6 m standing on horizontal ground. The point O is on the ground directly below the point of projection, as shown in Fig. 8. Air resistance is negligible.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4e0ddc86-c340-4057-bf3a-1c98587c3110-1_666_757_416_679}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Write down an expression in terms of $t$ for the horizontal displacement of the stone from O , $t$ seconds after projection. Find also an expression for the height of the stone above O at this time.
The stone is at the top of its trajectory when it passes over the wall.
\item (A) Find the time it takes for the stone to reach its highest point.\\
(B) Calculate the distance of O from the base of the wall.\\
(C) Show that the stone passes over the wall with 2.5 m clearance.
\item Find the cartesian equation of the trajectory of the stone referred to the horizontal and vertical axes, $\mathrm { O } x$ and $\mathrm { O } y$. There is no need to simplify your answer.
The girl now moves away a further distance $d \mathrm {~m}$ from the wall. She throws a stone as before and it just passes over the wall.
\item Calculate $d$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 Q1 [20]}}