| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Moderate -0.3 This is a standard M1 projectile question requiring routine application of SUVAT equations and elimination of parameter t. All steps are textbook procedures with given values, though part (iv) requires substitution and comparison. Slightly easier than average due to straightforward setup and clear numerical values. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
# Question 1
(i) $x=10t$
$y=10\sqrt{3}t-4.9t^2$
**B1**
**B1**
Allow $x = 20\cos 60° t$
Allow $y=20\sin 60° t - \frac{g}{2}t^2$ or $y=17.3 t - \frac{g}{2}t^2$
where $g = 9.8$
(ii) Substitute $t = \frac{x}{10}$ in equation for $y$
$\Rightarrow y= \sqrt{3}x-0.049x^2$
**M1** Substitution of a correct expression for $t$.
**A1** Notice that this is a given result
(iii) When $y=0$, $x= \frac{1.732}{0.049}$ (or $0$)
The range is 35.3 m
**M1** Use of $y = 0$, or $2 \times$ Time to maximum height
**A1**
(iv) When $x=20$, $y=1.732 \times 20 - 0.049 \times 20^2$
Height is 15.04 m so passes below the bird whose height is 16 m
**M1** Use of equation of trajectory
**A1**
**Special Case:** Allow SC2 for substituting $y = 16$ in the trajectory, showing the equation for $x$ has no real roots and concluding the height of the ball is always less than 16 m. This can also be done with the equation for vertical motion.
(iv) **Alternative: Using time**
When $x = 20$, $t = 2$
$y=10\sqrt{3} \times 2 - 4.9 \times 2^2$
Height is 15.04 m so passes below the bird whose height is 16 m
**M1** Use of equation for the height
**A1**
(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**1 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 $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. 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 Q1 [8]}}