| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.8 This is a straightforward M1 projectiles question requiring standard SUVAT equations and Pythagoras. Part (i) is basic vector magnitude, part (ii) uses standard formulas for time of flight (2u sin θ/g) and range, and part (iii) involves simple verification and a conceptual observation about complementary angles. All techniques are routine textbook exercises with no problem-solving insight required, making it easier than average A-level maths. |
| Spec | 1.10c Magnitude and direction: of vectors3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Initial speed is \(25 \text{ ms}^{-1}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vertical motion: \(y = 20t - 4.9t^2\) | M1 | Forming an equation or expression for vertical motion |
| When \(y = 0\), \(T = 0\) or \(\frac{20}{4.9} = 4.08\) s | A1 | |
| \(R = 15 \times 4.08... = 61.22\) | F1 | Allow \(15 \times\) their \(T\). Note: if horizontal and vertical components interchanged, treat as misread; if no other errors gives 3 marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vertical motion: \(v = 20 - 9.8t\) | M1 | Forming an equation or expression for vertical motion |
| Flight time \(= 2 \times\) Time to top | M1 | Using flight time is twice time to maximum height or equivalent for range |
| \(T = 2 \times \frac{20}{9.8} = 4.08\) s | A1 | |
| \(R = 15 \times 4.08... = 61.22\) | F1 | Allow \(15 \times\) their \(T\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha = \arctan\left(\frac{20}{15}\right) = 53.1°\) | M1 | Only award if clear intention to use this method |
| \(R = \frac{2u^2 \sin\alpha \cos\alpha}{g} = \frac{2 \times 25^2 \times \sin 53.1° \times \cos 53.1°}{9.8}\) | M1 | Allow alternative form \(R = \frac{u^2 \sin 2\alpha}{g}\) with substitution |
| \(R = 61.2\) | A1 | |
| \(T = \frac{2u \sin\alpha}{g} = 4.08\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Flight time \(= \frac{15}{4.9}\), Range \(= 20 \times \frac{15}{4.9} = 61.22\) | B1 | Allow FT from part (ii) for a correct argument that they should be the same |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| No | M1 | Attempt at disproof or counter-example. Must reference the angle. |
| e.g. angle of projection \(45°\) | A1 | Complete argument |
## Question 2:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Initial speed is $25 \text{ ms}^{-1}$ | B1 | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertical motion: $y = 20t - 4.9t^2$ | M1 | Forming an equation or expression for vertical motion |
| When $y = 0$, $T = 0$ or $\frac{20}{4.9} = 4.08$ s | A1 | |
| $R = 15 \times 4.08... = 61.22$ | F1 | Allow $15 \times$ their $T$. Note: if horizontal and vertical components interchanged, treat as misread; if no other errors gives 3 marks. |
**Alternative: Using time to maximum height**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertical motion: $v = 20 - 9.8t$ | M1 | Forming an equation or expression for vertical motion |
| Flight time $= 2 \times$ Time to top | M1 | Using flight time is twice time to maximum height or equivalent for range |
| $T = 2 \times \frac{20}{9.8} = 4.08$ s | A1 | |
| $R = 15 \times 4.08... = 61.22$ | F1 | Allow $15 \times$ their $T$ |
**Alternative: Using formulae**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha = \arctan\left(\frac{20}{15}\right) = 53.1°$ | M1 | Only award if clear intention to use this method |
| $R = \frac{2u^2 \sin\alpha \cos\alpha}{g} = \frac{2 \times 25^2 \times \sin 53.1° \times \cos 53.1°}{9.8}$ | M1 | Allow alternative form $R = \frac{u^2 \sin 2\alpha}{g}$ with substitution |
| $R = 61.2$ | A1 | |
| $T = \frac{2u \sin\alpha}{g} = 4.08$ | A1 | |
### Part (iii)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Flight time $= \frac{15}{4.9}$, Range $= 20 \times \frac{15}{4.9} = 61.22$ | B1 | Allow FT from part (ii) for a correct argument that they should be the same |
### Part (iii)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| No | M1 | Attempt at disproof or counter-example. Must reference the angle. |
| e.g. angle of projection $45°$ | A1 | Complete argument |
---2 In this question, air resistance should be neglected.\\
Fig. 2 illustrates the flight of a golf ball. The golf ball is initially on the ground, which is horizontal.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b9e41fac-9f4b-4165-af03-67ebdcb326de-1_285_1117_1450_497}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
It is hit and given an initial velocity with components of $15 \mathrm {~ms} ^ { - 1 }$ in the horizontal direction and $20 \mathrm {~ms} ^ { - 1 }$ in the vertical direction.
\begin{enumerate}[label=(\roman*)]
\item Find its initial speed.
\item Find the ball's flight time and range, $R \mathrm {~m}$.
\item (A) Show that the range is the same if the components of the initial velocity of the ball are $20 \mathrm {~ms} ^ { - 1 }$ in the horizontal direction and $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the vertical direction.\\
(B) State, justifying your answer, whether the range is the same whenever the ball is hit with the same initial speed.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 Q2 [8]}}