| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 19 |
| 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 multi-part projectile motion question covering routine SUVAT applications. Parts (i)-(iv) involve straightforward calculations using given information, while part (v) requires deriving a trajectory equation—a common A-level mechanics exercise. The question is slightly easier than average due to its structured guidance and use of standard techniques throughout. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Time to fall: \(40 - 10 = 0.5 \times 9.8 \times t^2\) | M1 | Considering time from explosion with \(u = 0\). Condone sign errors. |
| LHS allows \(\pm 30\) | A1 | |
| All correct | A1 | |
| \(t = 2.47435\ldots\) | A1 | cao |
| Need \(15.75 \times 2.47435\ldots = 38.971\ldots\) so \(39.0\) (3sf) | F1 | FT their (iii) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Need time so \(10 = 28t - 4.9t^2\) | M1 | Equating \(28t - 4.9t^2 = \pm 10\) |
| \(4.9t^2 - 28t + 10 = 0\) | M1* | Dep. Attempt to solve quadratic by method giving two roots |
| \(t = \frac{28 \pm \sqrt{28^2 - 4 \times 4.9 \times 10}}{9.8}\) | ||
| so \(0.382784\ldots\) or \(5.33150\ldots\) | A1 | Larger root correct to at least 2 s.f. [SC1 for either root seen WW] |
| Time required is \(5.33150\ldots - \frac{20}{7} = 2.47435\ldots\) | M1 | |
| Need \(15.75 \times 2.47435\ldots = 38.971\ldots\) so \(39.0\) (3sf) | F1 | FT their (iii) only |
| 5 | [SC1 if either and or methods mixed to give \(\pm 30 = 28t - 4.9t^2\) or \(\pm 10 = 4.9t^2\)] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Horiz: \(x = 21t\) | B1 | |
| Elim \(t\) between \(x = 21t\) and \(y = 28t - 4.9t^2\) | M1 | Intention must be clear, with some attempt made |
| \(y = 28\left(\frac{x}{21}\right) - 4.9\left(\frac{x}{21}\right)^2\) | A1 | \(t\) completely and correctly eliminated. Accept wrong notation if subsequently explicitly given correct value e.g. \(\frac{x^2}{21}\) seen as \(\frac{x^2}{441}\) |
| \(y = \frac{4x}{3} - \frac{0.1x^2}{9} = \frac{1}{90}(120x - x^2)\) | E1 | Some simplification must be shown. [SC2 for 3 points shown to be on curve, if made clear (a) trajectory is parabola (b) 3 points define a parabola] |
| 4 |
# Question 1 (continued):
## Part (B)
**Either method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Time to fall: $40 - 10 = 0.5 \times 9.8 \times t^2$ | M1 | Considering time from explosion with $u = 0$. Condone sign errors. |
| LHS allows $\pm 30$ | A1 | |
| All correct | A1 | |
| $t = 2.47435\ldots$ | A1 | cao |
| Need $15.75 \times 2.47435\ldots = 38.971\ldots$ so $39.0$ (3sf) | F1 | FT **their** (iii) only |
**Or method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Need time so $10 = 28t - 4.9t^2$ | M1 | Equating $28t - 4.9t^2 = \pm 10$ |
| $4.9t^2 - 28t + 10 = 0$ | M1* | Dep. Attempt to solve quadratic by method giving two roots |
| $t = \frac{28 \pm \sqrt{28^2 - 4 \times 4.9 \times 10}}{9.8}$ | | |
| so $0.382784\ldots$ or $5.33150\ldots$ | A1 | Larger root correct to at least 2 s.f. [SC1 for either root seen WW] |
| Time required is $5.33150\ldots - \frac{20}{7} = 2.47435\ldots$ | M1 | |
| Need $15.75 \times 2.47435\ldots = 38.971\ldots$ so $39.0$ (3sf) | F1 | FT **their** (iii) only |
| | **5** | [SC1 if either and or methods mixed to give $\pm 30 = 28t - 4.9t^2$ or $\pm 10 = 4.9t^2$] |
## Part (v)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Horiz: $x = 21t$ | B1 | |
| Elim $t$ between $x = 21t$ and $y = 28t - 4.9t^2$ | M1 | Intention must be clear, with some attempt made |
| $y = 28\left(\frac{x}{21}\right) - 4.9\left(\frac{x}{21}\right)^2$ | A1 | $t$ completely and correctly eliminated. Accept wrong notation if subsequently explicitly given correct value e.g. $\frac{x^2}{21}$ seen as $\frac{x^2}{441}$ |
| $y = \frac{4x}{3} - \frac{0.1x^2}{9} = \frac{1}{90}(120x - x^2)$ | E1 | Some simplification must be shown. [SC2 for 3 points shown to be on curve, if made clear (a) trajectory is parabola (b) 3 points define a parabola] |
| | **4** | |
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1 A small firework is fired from a point O at ground level over horizontal ground. The highest point reached by the firework is a horizontal distance of 60 m from O and a vertical distance of 40 m from O , as shown in Fig. 7. Air resistance is negligible.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{362d5995-bd39-4b07-b6a4-63eb1dd3e69d-1_611_1047_486_538}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
The initial horizontal component of the velocity of the firework is $21 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\roman*)]
\item Calculate the time for the firework to reach its highest point and show that the initial vertical component of its velocity is $28 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Show that the firework is $\left( 28 t - 4.9 t ^ { 2 } \right) \mathrm { m }$ above the ground $t$ seconds after its projection.
When the firework is at its highest point it explodes into several parts. Two of the parts initially continue to travel horizontally in the original direction, one with the original horizontal speed of $21 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the other with a quarter of this speed.
\item State why the two parts are always at the same height as one another above the ground and hence find an expression in terms of $t$ for the distance between the parts $t$ seconds after the explosion.
\item Find the distance between these parts of the firework\\
(A) when they reach the ground,\\
(B) when they are 10 m above the ground.
\item Show that the cartesian equation of the trajectory of the firework before it explodes is $y = \frac { 1 } { 90 } \left( 120 x - x ^ { 2 } \right)$, referred to the coordinate axes shown in Fig. 7.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 Q1 [19]}}