| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | September |
| Marks | 14 |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Standard +0.3 Part (i) is a standard 'show that' derivation of the trajectory equation using parametric equations and eliminating time—a core M1 skill. Parts (ii-iii) involve routine applications (completing the square, solving velocity component equations). Part (iv) requires stating modeling assumptions, which is straightforward recall. This is a typical textbook-style projectiles question with no novel insight required, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=13.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = (5\cos\theta)t\) | B1 | |
| \(y = (5\sin\theta)t - 5t^2\) | B1 | |
| \(y = (5\sin\theta)\left(\frac{x}{5\cos\theta}\right) - 5\left(\frac{x}{5\cos\theta}\right)^2\) | M1 | Eliminate \(t\) by substitution of \(x\) into \(y\) |
| \(y = x\tan\theta - \frac{5x^2}{25\cos^2\theta} \Rightarrow y = x\tan\theta - \frac{x^2}{5}\sec^2\theta\) | ||
| \(\Rightarrow y = x\tan\theta - \frac{x^2}{5}(1 + \tan^2\theta)\) | A1 | AG – sufficient working must be shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 3x - 2x^2\) | M1 | Substitute \(\tan\theta = 3\) |
| \(\Rightarrow \frac{dy}{dx} = 3 - 4x = 0 \Rightarrow x = \ldots\) | M1 | Set derivative equal to zero and attempt to solve for \(x\) |
| Max. height \(= 1.125\,\text{m}\) | A1 | oe, e.g. \(\frac{9}{8}\) metres |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 3x - 2x^2\) | M1 | Substitute \(\tan\theta = 3\) |
| Vertex of parabola is at \(x = -\frac{b}{2a} = -\frac{3}{2 \times (-2)} = \ldots\) | M1 | attempt to find \(x\) coordinate of vertex |
| Max. height \(= 1.125\,\text{m}\) | A1 | oe, e.g. \(\frac{9}{8}\) metres |
| Answer | Marks | Guidance |
|---|---|---|
| \(3 - 4x = \pm\frac{1}{3}\) | M1* | Set derivative equal to \(\pm\frac{1}{3}\) |
| \(x = \frac{1}{4}\left(3 \pm \frac{1}{3}\right)\) | A1ft | One correct value of \(x\) – follow through their derivative |
| \(x = \frac{2}{3}\) and \(x = \frac{5}{6}\) | A1 | Both values correct |
| \(t = \frac{x}{5\cos\theta}\) | dep*M1 | Used for at least one of their values of \(x\) and numerical attempt at \(\cos(\tan^{-1}3)\) |
| \(t = 0.422\) and \(t = 0.527\) | A1 | \(\frac{2}{15}\sqrt{10}\) and \(\frac{1}{6}\sqrt{10}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Vertical velocity \(= 5\sin\theta - 10t\) | B1 | \(\theta\) need not be evaluated here |
| \(\frac{5\sin\theta - 10t}{5\cos\theta} = \pm\frac{1}{3}\) | M1* | With numerical values for sin\(\theta\) and cos\(\theta\) (exact or decimal); allow + only on RHS for M mark |
| A1 | Correct equation with both + and − | |
| \(t = \frac{1}{10}\left(\frac{15}{\sqrt{10}} \pm \frac{5}{3\sqrt{10}}\right)\) | dep*M1 | Solve for \(t\) (exact or decimal values) |
| \(t = \frac{2}{15}\sqrt{10}\) and \(\frac{1}{6}\sqrt{10}\) | A1 | oe |
| 0.4216..., 0.5270... |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. air resistance not taken into account | B1 | Any valid reason |
| e.g. the value of \(g\) is not 10 | B1 | Two valid reasons |
### (i)
$x = (5\cos\theta)t$ | B1 |
$y = (5\sin\theta)t - 5t^2$ | B1 |
$y = (5\sin\theta)\left(\frac{x}{5\cos\theta}\right) - 5\left(\frac{x}{5\cos\theta}\right)^2$ | M1 | Eliminate $t$ by substitution of $x$ into $y$
$y = x\tan\theta - \frac{5x^2}{25\cos^2\theta} \Rightarrow y = x\tan\theta - \frac{x^2}{5}\sec^2\theta$ | |
$\Rightarrow y = x\tan\theta - \frac{x^2}{5}(1 + \tan^2\theta)$ | A1 | AG – sufficient working must be shown
**[4 marks total]**
### (ii)
$y = 3x - 2x^2$ | M1 | Substitute $\tan\theta = 3$
$\Rightarrow \frac{dy}{dx} = 3 - 4x = 0 \Rightarrow x = \ldots$ | M1 | Set derivative equal to zero and attempt to solve for $x$
Max. height $= 1.125\,\text{m}$ | A1 | oe, e.g. $\frac{9}{8}$ metres
**Alternative method:**
$y = 3x - 2x^2$ | M1 | Substitute $\tan\theta = 3$
Vertex of parabola is at $x = -\frac{b}{2a} = -\frac{3}{2 \times (-2)} = \ldots$ | M1 | attempt to find $x$ coordinate of vertex
Max. height $= 1.125\,\text{m}$ | A1 | oe, e.g. $\frac{9}{8}$ metres
**[3 marks total]**
### (iii)
$3 - 4x = \pm\frac{1}{3}$ | M1* | Set derivative equal to $\pm\frac{1}{3}$
$x = \frac{1}{4}\left(3 \pm \frac{1}{3}\right)$ | A1ft | One correct value of $x$ – follow through their derivative
$x = \frac{2}{3}$ and $x = \frac{5}{6}$ | A1 | Both values correct
$t = \frac{x}{5\cos\theta}$ | dep*M1 | Used for at least one of their values of $x$ and numerical attempt at $\cos(\tan^{-1}3)$
$t = 0.422$ and $t = 0.527$ | A1 | $\frac{2}{15}\sqrt{10}$ and $\frac{1}{6}\sqrt{10}$
**Alternative solution:**
Vertical velocity $= 5\sin\theta - 10t$ | B1 | $\theta$ need not be evaluated here
$\frac{5\sin\theta - 10t}{5\cos\theta} = \pm\frac{1}{3}$ | M1* | With numerical values for sin$\theta$ and cos$\theta$ (exact or decimal); allow + only on RHS for M mark
| A1 | Correct equation with both + and −
$t = \frac{1}{10}\left(\frac{15}{\sqrt{10}} \pm \frac{5}{3\sqrt{10}}\right)$ | dep*M1 | Solve for $t$ (exact or decimal values)
$t = \frac{2}{15}\sqrt{10}$ and $\frac{1}{6}\sqrt{10}$ | A1 | oe
0.4216..., 0.5270... | |
**[5 marks total]**
### (iv)
e.g. air resistance not taken into account | B1 | Any valid reason
e.g. the value of $g$ is not 10 | B1 | Two valid reasons
**[2 marks total]**
---10 A small ball $P$ is projected with speed $5 \mathrm {~ms} ^ { - 1 }$ at an angle $\theta$ above the horizontal from a point $O$ and moves freely under gravity. The horizontal and vertically upwards displacements of the ball from $O$ at any subsequent time $t$ seconds are $x \mathrm {~m}$ and $y \mathrm {~m}$ respectively.
The ball is modelled as a particle and the acceleration due to gravity is taken to be $10 \mathrm {~ms} ^ { - 2 }$.\\
(i) Show that the equation of the trajectory of $P$ is
$$y = x \tan \theta - \frac { x ^ { 2 } } { 5 } \left( 1 + \tan ^ { 2 } \theta \right)$$
It is given that $\tan \theta = 3$.\\
(ii) Using part (i), find the maximum height above the level of $O$ of $P$ in the subsequent motion.\\
(iii) Find the values of $t$ when $P$ is moving at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac { 1 } { 3 }$.\\
(iv) Give two possible reasons why the values of $t$ found in part (iii) may not be accurate.\\
\includegraphics[max width=\textwidth, alt={}, center]{28beb431-45d5-4300-88fe-00d05d78790b-09_435_714_267_678}
Two particles $P$ and $Q$, with masses 2 kg and 8 kg respectively, are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed at a point on the intersection of two fixed inclined planes. The string lies in a vertical plane that contains a line of greatest slope of each of the two inclined planes. Plane $\Pi _ { 1 }$ is inclined at an angle of $30 ^ { \circ }$ to the horizontal and plane $\Pi _ { 2 }$ is inclined at an angle of $\theta$ to the horizontal. Particle $P$ is on $\Pi _ { 1 }$ and $Q$ is on $\Pi _ { 2 }$ with the string taut (see diagram).\\
$\Pi _ { 1 }$ is rough and the coefficient of friction between $P$ and $\Pi _ { 1 }$ is $\frac { \sqrt { 3 } } { 3 }$.\\
$\Pi _ { 2 }$ is smooth.\\
The particles are released from rest and $P$ begins to move towards the pulley with an acceleration of $g \cos \theta \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Show that $\theta$ satisfies the equation
$$4 \sin \theta - 5 \cos \theta = 1 .$$
(ii) By expressing $4 \sin \theta - 5 \cos \theta$ in the form $R \sin ( \theta - \alpha )$, where $R > 0$ and $0 < \alpha < 90 ^ { \circ }$, find, correct to 3 significant figures, the tension in the string.
\hfill \mbox{\textit{OCR H240/03 2018 Q10 [14]}}