| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | December |
| Marks | 16 |
| Topic | Projectiles |
| Type | Maximum range or optimal angle |
| Difficulty | Standard +0.3 This is a standard projectile motion question requiring derivation of familiar formulae (H and R), algebraic manipulation to reach a given result, and solving a quadratic equation. While part (b) requires some algebraic skill to reach the given form, all steps follow routine M1 techniques with no novel insight needed. The question is slightly easier than average due to its structured guidance and use of standard projectile results. |
| Spec | 1.05p Proof involving trig: functions and identities3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(0 = (V\sin\alpha)^2 + 2(-g)H\) | M1 | Use of \(v^2 = u^2 + 2as\) vertically |
| \(H = \frac{V^2}{2g}\sin^2\alpha\) | A1 | AG – sufficient working must be shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = (V\cos\alpha)t\) and \(0 = (V\sin\alpha)t + \frac{1}{2}(-g)t^2\) | M1* | Use of \(s = ut + \frac{1}{2}at^2\) horizontally and vertically |
| \((V\sin\alpha) - \frac{1}{2}g(\frac{R}{V\cos\alpha}) = 0\) | M1dep* | Re-arranging and eliminating \(t\) |
| \(gR = 2V^2\sin\alpha\cos\alpha \Rightarrow R = \frac{V^2}{g}\sin 2\alpha\) | A1 | AG – sufficient working must be shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(R_0 = \frac{V^2}{g}\) | B1 | Correct expression for maximum range |
| \(\sin^2\alpha = \frac{2gH}{V^2}, \quad \cos^2\alpha = \frac{V^2 - 2gH}{V^2}\) | M1* | Obtain expressions for \(\sin^2\alpha\) and \(\cos^2\alpha\) using \(\sin^2\theta + \cos^2\theta = 1\) |
| \(R = \frac{2V^2}{g} \times \frac{2gH}{V} \times \frac{\sqrt{V^2 - 2gH}}{V}\) | M1dep* | Eliminate \(\alpha\) |
| \(R = \frac{2}{g}\sqrt{2gH}\sqrt{V^2 - 2gH}\) | M1 | Eliminate \(V\) using maximum range and remove square roots – dependent on both previous M marks |
| \(\Rightarrow 16H^2 - 8R_0 H + R^2 = 0\) | A1 | AG – sufficient working must be shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(16H^2 - 1600H + 36864 = 0\) | M1 | Substitute given values to obtain a quadratic in \(H\) |
| \(H = 64\text{ m or } 36\text{ m}\) | A1 | BC |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin 2\alpha = \frac{192}{200}\) | M1 | Use given values to obtain a trigonometric equation in \(\alpha\) |
| \(\alpha = 36.9°\) | A1 | \(0.644\) rad |
| \(\alpha = 53.1°\) | A1 | \(0.927\) rad |
| Answer | Marks | Guidance |
|---|---|---|
| • The possible spin of the ball | B1 | Any one correct limitation identified |
## Part (a)(i)
$0 = (V\sin\alpha)^2 + 2(-g)H$ | M1 | Use of $v^2 = u^2 + 2as$ vertically
$H = \frac{V^2}{2g}\sin^2\alpha$ | A1 | AG – sufficient working must be shown
## Part (a)(ii)
$R = (V\cos\alpha)t$ and $0 = (V\sin\alpha)t + \frac{1}{2}(-g)t^2$ | M1* | Use of $s = ut + \frac{1}{2}at^2$ horizontally and vertically | Alternatively: Finding $t$ from $0 = V\sin\alpha - gt$ and using double the value, oe
$(V\sin\alpha) - \frac{1}{2}g(\frac{R}{V\cos\alpha}) = 0$ | M1dep* | Re-arranging and eliminating $t$
$gR = 2V^2\sin\alpha\cos\alpha \Rightarrow R = \frac{V^2}{g}\sin 2\alpha$ | A1 | AG – sufficient working must be shown
## Part (b)
$R_0 = \frac{V^2}{g}$ | B1 | Correct expression for maximum range
$\sin^2\alpha = \frac{2gH}{V^2}, \quad \cos^2\alpha = \frac{V^2 - 2gH}{V^2}$ | M1* | Obtain expressions for $\sin^2\alpha$ and $\cos^2\alpha$ using $\sin^2\theta + \cos^2\theta = 1$
$R = \frac{2V^2}{g} \times \frac{2gH}{V} \times \frac{\sqrt{V^2 - 2gH}}{V}$ | M1dep* | Eliminate $\alpha$
$R = \frac{2}{g}\sqrt{2gH}\sqrt{V^2 - 2gH}$ | M1 | Eliminate $V$ using maximum range and remove square roots – dependent on both previous M marks
$\Rightarrow 16H^2 - 8R_0 H + R^2 = 0$ | A1 | AG – sufficient working must be shown
## Part (c)(i)
$16H^2 - 1600H + 36864 = 0$ | M1 | Substitute given values to obtain a quadratic in $H$
$H = 64\text{ m or } 36\text{ m}$ | A1 | BC
## Part (c)(ii)
$\sin 2\alpha = \frac{192}{200}$ | M1 | Use given values to obtain a trigonometric equation in $\alpha$
$\alpha = 36.9°$ | A1 | $0.644$ rad
$\alpha = 53.1°$ | A1 | $0.927$ rad
## Part (d)
The model has not considered the possibility of:
• Air resistance
• The ball would have dimensions
• Wind
• The possible spin of the ball | B1 | Any one correct limitation identifiedA ball $B$ is projected with speed $V$ at an angle $\alpha$ above the horizontal from a point $O$ on horizontal ground. The greatest height of $B$ above $O$ is $H$ and the horizontal range of $B$ is $R$. The ball is modelled as a particle moving freely under gravity.
\begin{enumerate}[label=(\alph*)]
\item Show that
\begin{enumerate}[label=(\roman*)]
\item $H = \frac{V^2}{2g}\sin^2 \alpha$, [2]
\item $R = \frac{V^2}{g}\sin 2\alpha$. [3]
\end{enumerate}
\item Hence show that $16H^2 - 8R_0 H + R^2 = 0$, where $R_0$ is the maximum range for the given speed of projection. [5]
\item Given that $R_0 = 200\text{m}$ and $R = 192\text{m}$, find
\begin{enumerate}[label=(\roman*)]
\item the two possible values of the greatest height of $B$, [2]
\item the corresponding values of the angle of projection. [3]
\end{enumerate}
\item State one limitation of the model that could affect your answers to part (iii). [1]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q11 [16]}}