Question 5 - A-Level Maths

OCR MEI M1 2015 June — Question 5 8 marks

Exam Board	OCR MEI
Module	M1 (Mechanics 1)
Year	2015
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Projectiles
Type	Deriving trajectory equation
Difficulty	Moderate -0.3 This is a standard projectile motion question requiring routine application of SUVAT equations and elimination of the parameter t. All steps are textbook procedures: writing parametric equations, deriving the trajectory by substitution, finding range by setting y=0, and checking a point. The calculations are straightforward with given values (60° angle, 20 m/s speed), making it slightly easier than average despite being multi-part.
Spec	3.02h Motion under gravity: vector form 3.02i Projectile motion: constant acceleration model

5 A golf ball is hit at an angle of $60 ^ { \circ }$ to the horizontal from a point, O , on level horizontal ground. Its initial speed is $20 \mathrm {~ms} ^ { - 1 }$. The standard projectile model, in which air resistance is neglected, is used to describe the subsequent motion of the golf ball. At time $t \mathrm {~s}$ the horizontal and vertical components of its displacement from O are denoted by $x \mathrm {~m}$ and $y \mathrm {~m}$.

Write down equations for $x$ and $y$ in terms of $t$.
Hence show that the equation of the trajectory is $$y = \sqrt { 3 } x - 0.049 x ^ { 2 } .$$
Find the range of the golf ball.
A bird is hovering at position $( 20,16 )$. Find whether the golf ball passes above it, passes below it or hits it.

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Question 5:

Part (i):

Answer	Marks	Guidance
$x = 10t$	B1	Allow $x = 20\cos 60°\, t$
$y = 10\sqrt{3}\,t - 4.9t^2$	B1	Allow $y = 20\sin 60°\, t - \frac{g}{2}t^2$ or $y = 17.3\,t - \frac{9.8}{2}t^2$

Part (ii):

Answer	Marks	Guidance
Substitute $t = \frac{x}{10}$ in equation for $y$	M1	Substitution of a correct expression for $t$
$\Rightarrow y = \sqrt{3}\,x - 0.049x^2$	A1	Note that this is a given result

Part (iii):

Answer	Marks	Guidance
When $y = 0$, $x = \frac{1.732}{0.049}$ (or 0)	M1	Use of $y = 0$, or $2 \times$ time to maximum height
The range is 35.3 m	A1

Part (iv):

Answer	Marks	Guidance
When $x = 20$, $y = 1.732 \times 20 - 0.049 \times 20^2$	M1	Use of equation of trajectory
Height is 15.04 m so passes below the bird whose height is 16 m	A1	Special Case: Allow SC2 for substituting $y = 16$ in the trajectory, showing no real roots and concluding height always less than 16 m. Can also be done with equation for vertical motion.

Part (iv) Alternative: Using time:

Answer	Marks	Guidance
When $x = 20$, $t = 2$
$y = 10\sqrt{3} \times 2 - 4.9 \times 2^2$	M1	Use of equation for the height
Height is 15.04 m so passes below the bird whose height is 16 m	A1

Part (iv) Alternative: Maximum height:

Answer	Marks	Guidance
The maximum height of the ball is 15.3 m	M1	A valid method for finding the maximum height
Since $15.3 < 16$, it is always below the bird	A1

# Question 5:

## Part (i):
| $x = 10t$ | B1 | Allow $x = 20\cos 60°\, t$ |
| $y = 10\sqrt{3}\,t - 4.9t^2$ | B1 | Allow $y = 20\sin 60°\, t - \frac{g}{2}t^2$ or $y = 17.3\,t - \frac{9.8}{2}t^2$ |

## Part (ii):
| Substitute $t = \frac{x}{10}$ in equation for $y$ | M1 | Substitution of a correct expression for $t$ |
| $\Rightarrow y = \sqrt{3}\,x - 0.049x^2$ | A1 | Note that this is a given result |

## Part (iii):
| When $y = 0$, $x = \frac{1.732}{0.049}$ (or 0) | M1 | Use of $y = 0$, or $2 \times$ time to maximum height |
| The range is 35.3 m | A1 | |

## Part (iv):
| When $x = 20$, $y = 1.732 \times 20 - 0.049 \times 20^2$ | M1 | Use of equation of trajectory |
| Height is 15.04 m so passes below the bird whose height is 16 m | A1 | Special Case: Allow SC2 for substituting $y = 16$ in the trajectory, showing no real roots and concluding height always less than 16 m. Can also be done with equation for vertical motion. |

### Part (iv) Alternative: Using time:
| When $x = 20$, $t = 2$ | | |
| $y = 10\sqrt{3} \times 2 - 4.9 \times 2^2$ | M1 | Use of equation for the height |
| Height is 15.04 m so passes below the bird whose height is 16 m | A1 | |

### Part (iv) Alternative: Maximum height:
| The maximum height of the ball is 15.3 m | M1 | A valid method for finding the maximum height |
| Since $15.3 < 16$, it is always below the bird | A1 | |

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Show LaTeX source

5 A golf ball is hit at an angle of $60 ^ { \circ }$ to the horizontal from a point, O , on level horizontal ground. Its initial speed is $20 \mathrm {~ms} ^ { - 1 }$. The standard projectile model, in which air resistance is neglected, is used to describe the subsequent motion of the golf ball. At time $t \mathrm {~s}$ the horizontal and vertical components of its displacement from O are denoted by $x \mathrm {~m}$ and $y \mathrm {~m}$.\\
(i) Write down equations for $x$ and $y$ in terms of $t$.\\
(ii) Hence show that the equation of the trajectory is

$$y = \sqrt { 3 } x - 0.049 x ^ { 2 } .$$

(iii) Find the range of the golf ball.\\
(iv) A bird is hovering at position $( 20,16 )$.

Find whether the golf ball passes above it, passes below it or hits it.

\hfill \mbox{\textit{OCR MEI M1 2015 Q5 [8]}}

This paper (6 questions)

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Q1 5 Q2 8 Q3 7 Q4 8 Q5 8 Q6 18

Answer	Marks	Guidance
\(x = 10t\)	B1	Allow \(x = 20\cos 60°\, t\)
\(y = 10\sqrt{3}\,t - 4.9t^2\)	B1	Allow \(y = 20\sin 60°\, t - \frac{g}{2}t^2\) or \(y = 17.3\,t - \frac{9.8}{2}t^2\)

Answer	Marks	Guidance
Substitute \(t = \frac{x}{10}\) in equation for \(y\)	M1	Substitution of a correct expression for \(t\)
\(\Rightarrow y = \sqrt{3}\,x - 0.049x^2\)	A1	Note that this is a given result

Answer	Marks	Guidance
When \(y = 0\), \(x = \frac{1.732}{0.049}\) (or 0)	M1	Use of \(y = 0\), or \(2 \times\) time to maximum height
The range is 35.3 m	A1

Answer	Marks	Guidance
When \(x = 20\), \(y = 1.732 \times 20 - 0.049 \times 20^2\)	M1	Use of equation of trajectory
Height is 15.04 m so passes below the bird whose height is 16 m	A1	Special Case: Allow SC2 for substituting \(y = 16\) in the trajectory, showing no real roots and concluding height always less than 16 m. Can also be done with equation for vertical motion.

Answer	Marks	Guidance
When \(x = 20\), \(t = 2\)
\(y = 10\sqrt{3} \times 2 - 4.9 \times 2^2\)	M1	Use of equation for the height
Height is 15.04 m so passes below the bird whose height is 16 m	A1

Answer	Marks	Guidance
The maximum height of the ball is 15.3 m	M1	A valid method for finding the maximum height
Since \(15.3 < 16\), it is always below the bird	A1