| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Velocity direction at specific time/point |
| Difficulty | Standard +0.3 This is a standard projectiles question requiring routine application of SUVAT equations and basic trigonometry. Parts (i)-(iii) involve straightforward substitution into kinematic formulas, while part (iv) requires finding velocity components and using arctan—all standard M1 techniques with no novel problem-solving required. Slightly easier than average due to the structured guidance and given information. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Triangle with three forces: 250 N downward, 25 N horizontal, tension \(F\) along rope | B1 B1 B1 | Correct triangle, all forces labelled with values and directions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Horizontal: \(F\sin\theta = 25\) | M1 | |
| Vertical: \(F\cos\theta = 250\) | M1 | |
| \(\tan\theta = \frac{25}{250}=0.1\), so \(\theta = 5.71°\) | A1 | |
| \(F = \frac{250}{\cos\theta}=251.2\) N | A1 | |
| Horizontal distance \(= 30\tan\theta = 30\times0.1=3\) m ✓ | A1 | Shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Vertical: \(S\cos\alpha = 250 + T\sin\beta\) ... wait — from diagram: \(S\cos\alpha - T\sin\beta = 250\) | B1 | |
| Horizontal: \(S\sin\alpha = T\cos\beta\) | B1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\alpha=8.5°\), \(\beta=35°\): \(S\sin8.5° = T\cos35°\) | M1 | |
| \(S\cos8.5° - T\sin35° = 250\) | ||
| Solving simultaneously | M1 | |
| \(T \approx 30.4\) N, \(S \approx 168\) N | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At \(\alpha=60°\): \(S\cos60°=0.5S\), so \(S=500\) N | M1 | |
| Abi's weight \(= 40\times10=400\) N | M1 | |
| \(S=500 > 400\) N, so Abi cannot exert sufficient force | A1 |
# Question 7:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Triangle with three forces: 250 N downward, 25 N horizontal, tension $F$ along rope | B1 B1 B1 | Correct triangle, all forces labelled with values and directions |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Horizontal: $F\sin\theta = 25$ | M1 | |
| Vertical: $F\cos\theta = 250$ | M1 | |
| $\tan\theta = \frac{25}{250}=0.1$, so $\theta = 5.71°$ | A1 | |
| $F = \frac{250}{\cos\theta}=251.2$ N | A1 | |
| Horizontal distance $= 30\tan\theta = 30\times0.1=3$ m ✓ | A1 | Shown |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Vertical: $S\cos\alpha = 250 + T\sin\beta$ ... wait — from diagram: $S\cos\alpha - T\sin\beta = 250$ | B1 | |
| Horizontal: $S\sin\alpha = T\cos\beta$ | B1 B1 | |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\alpha=8.5°$, $\beta=35°$: $S\sin8.5° = T\cos35°$ | M1 | |
| $S\cos8.5° - T\sin35° = 250$ | | |
| Solving simultaneously | M1 | |
| $T \approx 30.4$ N, $S \approx 168$ N | A1 A1 | |
## Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| At $\alpha=60°$: $S\cos60°=0.5S$, so $S=500$ N | M1 | |
| Abi's weight $= 40\times10=400$ N | M1 | |
| $S=500 > 400$ N, so Abi cannot exert sufficient force | A1 | |
---7 The trajectory ABCD of a small stone moving with negligible air resistance is shown in Fig. 7. AD is horizontal and BC is parallel to AD .
The stone is projected from A with speed $40 \mathrm {~ms} ^ { - 1 }$ at $50 ^ { \circ }$ to the horizontal.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9a79f274-1a3f-4d11-9775-313d82075035-004_341_1107_484_498}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
(i) Write down an expression for the horizontal displacement from A of the stone $t$ seconds after projection. Write down also an expression for the vertical displacement at time $t$.\\
(ii) Show that the stone takes 6.253 seconds (to three decimal places) to travel from A to D . Calculate the range of the stone.
You are given that $X = 30$.\\
(iii) Calculate the time it takes the stone to reach B . Hence determine the time for it to travel from A to C.\\
(iv) Calculate the direction of the motion of the stone at $\mathbf { C }$.
Section B (36 marks)
\hfill \mbox{\textit{OCR MEI M1 Q7}}