| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Velocity direction at specific time/point |
| Difficulty | Standard +0.3 This is a straightforward projectiles question requiring standard application of SUVAT equations and basic trigonometry. Parts (i)-(iii) involve routine calculations with given values, 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 | 1.05g Exact trigonometric values: for standard angles1.10c Magnitude and direction: of vectors3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Horizontal: \((40\cos 50)t\) | B1 | |
| Vertical: \((40\sin 50)t - 4.9t^2\) | M1 | Use of \(s = ut + 0.5at^2\) with \(a = \pm 9.8\) or \(\pm 10\); allow \(u = 40\); condone \(s \leftrightarrow c\) |
| A1 | Any form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Need \((40\sin 50)t - 4.9t^2 = 0\) | M1 | Equating their \(y\) to zero; allow quadratic \(y\) only |
| So \(t = \dfrac{40\sin 50}{4.9}\) | M1 | Dep on 1st M1; attempt to solve |
| \(= 6.2534\ldots\) so 6.253 s (3 d.p.) | E1 | Clearly shown; [or M1 (allow \(u=40\) and \(s\leftrightarrow c\)) A1 time to greatest height; E1] |
| Range is \((40\cos 50) \times 6.2534\ldots\) | M1 | Use of their horizontal expression |
| \(= 160.78\ldots\) so 161 m (3 s.f.) | A1 | Any reasonable accuracy |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Time AB: \((40\cos 50)T = 30\), so \(T = 1.16679\ldots\) so 1.17 s | M1, A1 | Equating their linear \(x\) to 30 |
| Either: By symmetry, time AC = time AD \(-\) time AB | M1 | Symmetry need not be explicit; method may be implied; any valid method using symmetry |
| So time AC is \(6.2534\ldots - \dfrac{30}{40\cos 50} = 5.086\ldots\) so 5.09 s (3 s.f.) | A1 | cao |
| Or: Height is \((40\sin 50)T - 4.9T^2\); solve \((40\sin 50)t - 4.9t^2 = (40\sin 50)T - 4.9T^2\) for larger root | M1 | Complete method to find time to second occasion at that height |
| i.e. solve \(4.9t^2 - (40\sin 50)t + 29.08712\ldots = 0\) for larger root giving 5.086… | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dot{x} = 40\cos 50\) | B1 | Must be part of a method using velocities |
| \(\dot{y} = 40\sin 50 - 9.8 \times 5.086\ldots\) | M1, A1 | Use of vert cpt of vel; allow only sign error; FT use of their 5.086… |
| Need \(\arctan\dfrac{\dot{y}}{\dot{x}}\) | M1 | May be implied; accept \(\arctan\dfrac{\dot{x}}{\dot{y}}\) but not use of \(\dot{s}_0\) |
| So \(-36.761\ldots°\), so 36.8° below horizontal (3 s.f.) | A1 | Accept \(\pm 36.8\) or equivalent; condone direction not clear |
# Question 3:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal: $(40\cos 50)t$ | B1 | |
| Vertical: $(40\sin 50)t - 4.9t^2$ | M1 | Use of $s = ut + 0.5at^2$ with $a = \pm 9.8$ or $\pm 10$; allow $u = 40$; condone $s \leftrightarrow c$ |
| | A1 | Any form |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Need $(40\sin 50)t - 4.9t^2 = 0$ | M1 | Equating their $y$ to zero; allow quadratic $y$ only |
| So $t = \dfrac{40\sin 50}{4.9}$ | M1 | Dep on 1st M1; attempt to solve |
| $= 6.2534\ldots$ so 6.253 s (3 d.p.) | E1 | Clearly shown; [or M1 (allow $u=40$ and $s\leftrightarrow c$) A1 time to greatest height; E1] |
| Range is $(40\cos 50) \times 6.2534\ldots$ | M1 | Use of their horizontal expression |
| $= 160.78\ldots$ so 161 m (3 s.f.) | A1 | Any reasonable accuracy |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Time AB: $(40\cos 50)T = 30$, so $T = 1.16679\ldots$ so 1.17 s | M1, A1 | Equating their linear $x$ to 30 |
| **Either:** By symmetry, time AC = time AD $-$ time AB | M1 | Symmetry need not be explicit; method may be implied; any valid method using symmetry |
| So time AC is $6.2534\ldots - \dfrac{30}{40\cos 50} = 5.086\ldots$ so 5.09 s (3 s.f.) | A1 | cao |
| **Or:** Height is $(40\sin 50)T - 4.9T^2$; solve $(40\sin 50)t - 4.9t^2 = (40\sin 50)T - 4.9T^2$ for larger root | M1 | Complete method to find time to second occasion at that height |
| i.e. solve $4.9t^2 - (40\sin 50)t + 29.08712\ldots = 0$ for larger root giving 5.086… | A1 | cao |
## Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dot{x} = 40\cos 50$ | B1 | Must be part of a method using velocities |
| $\dot{y} = 40\sin 50 - 9.8 \times 5.086\ldots$ | M1, A1 | Use of vert cpt of vel; allow only sign error; FT use of their 5.086… |
| Need $\arctan\dfrac{\dot{y}}{\dot{x}}$ | M1 | May be implied; accept $\arctan\dfrac{\dot{x}}{\dot{y}}$ but not use of $\dot{s}_0$ |
| So $-36.761\ldots°$, so 36.8° below horizontal (3 s.f.) | A1 | Accept $\pm 36.8$ or equivalent; condone direction not clear |3 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]{4e0ddc86-c340-4057-bf3a-1c98587c3110-3_316_1032_583_504}
\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 C .
\hfill \mbox{\textit{OCR MEI M1 Q3 [17]}}