| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Projectile motion: trajectory equation |
| Difficulty | Standard +0.3 This is a standard projectile motion question requiring routine application of SUVAT equations in 2D. Parts (i)-(iii) involve direct substitution into standard formulas (s = ut + ½at², symmetry of parabolic motion), while part (iv) requires finding velocity components and calculating an angle. The given values and 'show that' structure reduce problem-solving demand, making this slightly easier than average but still requiring multiple coordinated steps. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Horiz \((40 \cos 50)t\) | B1 | |
| Vert \((40 \sin 50)t - 4.9t^2\) | 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 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Need \((40 \sin 50)t - 4.9t^2 = 0\) | M1 | Equating their \(y\) to zero. Allow quadratic \(y\) only |
| so \(t = \frac{40 \sin 50}{4.9} = 6.2534...\) so 6.253 s (3 d. p.) | M1 | Dep on 1st M1. Attempt to solve. |
| E1 | Clearly shown [or M1 (allow \(u = 40\) and \(s \leftrightarrow c\)) A1 time to greatest height: E1] | |
| Range is \((40 \cos 50) \times 6.2534... = 160.78...\) so 161 m (3 s. f.) | M1 | Use of their horiz expression |
| A1 | Any reasonable accuracy | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Time AB is given by \((40 \cos 50)T = 30\) so \(T = 1.16679...\) so 1.17 s | M1 | Equating their linear \(x\) to 30. |
| A1 | ||
| either By symmetry, time AC is time AD – time AB so time AC is \(6.2534... - \frac{30}{40 \cos 50} = 5.086...\) so 5.09 s (3 s. f.) | M1 | Symmetry need not be explicit. Method may be implied. Any valid method using symmetry. |
| A1 | cao | |
| or height is \((40 \sin 50)T - 4.9T^2\) and we need \((40 \sin 50)T - 4.9T^2 = (40 \sin 50)T - 4.9T^2\) solved for larger root i.e. solve \(4.9T^2 - (40 \sin 50)t + 29.0871...= 0\) for larger root giving 5.086... | M1 | Complete method to find time to second occasion at that height |
| A1 | cao | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(k = 40 \cos 50\) | B1 | Must be part of a method using velocities. |
| \(\phi = 40 \sin 50 - 9.8 \times 5.086...\) | M1 | Use of vert cpt of vel. Allow only sign error. FT use of their 5.086.. |
| A1 | ||
| Need \(\arctan \frac{\phi}{k}\) | M1 | May be implied. Accept \(\arctan \frac{\phi}{k}\) but not use of \(\phi_0\). |
| So \(-36.761...°\) so 36.8° below horizontal (3 s.f.) | A1 | Accept \(\pm 36.8\) or equivalent. Condone direction not clear. |
| Total: 5 |
**Part (i)**
| Horiz $(40 \cos 50)t$ | B1 | |
| Vert $(40 \sin 50)t - 4.9t^2$ | 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 |
| | | **Total: 3** |
**Part (ii)**
| Need $(40 \sin 50)t - 4.9t^2 = 0$ | M1 | Equating their $y$ to zero. Allow quadratic $y$ only |
| so $t = \frac{40 \sin 50}{4.9} = 6.2534...$ so 6.253 s (3 d. p.) | M1 | Dep on 1st M1. Attempt to solve. |
| | E1 | Clearly shown [or M1 (allow $u = 40$ and $s \leftrightarrow c$) A1 time to greatest height: E1] |
| Range is $(40 \cos 50) \times 6.2534... = 160.78...$ so 161 m (3 s. f.) | M1 | Use of their horiz expression |
| | A1 | Any reasonable accuracy |
| | | **Total: 5** |
**Part (iii)**
| Time AB is given by $(40 \cos 50)T = 30$ so $T = 1.16679...$ so 1.17 s | M1 | Equating their linear $x$ to 30. |
| | A1 | |
| **either** By symmetry, time AC is time AD – time AB so time AC is $6.2534... - \frac{30}{40 \cos 50} = 5.086...$ so 5.09 s (3 s. f.) | M1 | Symmetry need not be explicit. Method may be implied. Any valid method using symmetry. |
| | A1 | cao |
| **or** height is $(40 \sin 50)T - 4.9T^2$ and we need $(40 \sin 50)T - 4.9T^2 = (40 \sin 50)T - 4.9T^2$ solved for larger root i.e. solve $4.9T^2 - (40 \sin 50)t + 29.0871...= 0$ for larger root giving 5.086... | M1 | Complete method to find time to second occasion at that height |
| | A1 | cao |
| | | **Total: 4** |
**Part (iv)**
| $k = 40 \cos 50$ | B1 | Must be part of a method using velocities. |
| $\phi = 40 \sin 50 - 9.8 \times 5.086...$ | M1 | Use of vert cpt of vel. Allow only sign error. FT use of their 5.086.. |
| | A1 | |
| Need $\arctan \frac{\phi}{k}$ | M1 | May be implied. Accept $\arctan \frac{\phi}{k}$ but not use of $\phi_0$. |
| So $-36.761...°$ so 36.8° below horizontal (3 s.f.) | A1 | Accept $\pm 36.8$ or equivalent. Condone direction not clear. |
| | | **Total: 5** |
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**Grand Total: 100 marks**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]{c84a748a-a6f4-48c5-b864-fe543569bdf5-4_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 2005 Q7 [17]}}