| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Moderate -0.3 This is a comprehensive but standard M1 projectiles question covering all basic techniques: deriving kinematic equations from initial conditions (given), finding maximum height, solving quadratic equations for time, calculating velocity components, and eliminating the parameter to find trajectory. All parts follow routine procedures with no novel problem-solving required, though the multi-part nature and final algebraic manipulation make it slightly more substantial than the most basic exercises. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vertical: \(y = 25\sin\theta \cdot t - \frac{1}{2}(9.8)t^2 = 25(0.28)t - 4.9t^2 = 7t - 4.9t^2\) | B1 | |
| Horizontal: \(x = 25\cos\theta \cdot t = 25(0.96)t = 24t\) | B1 | |
| Both shown correctly | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Maximum height when \(\frac{dy}{dt} = 7 - 9.8t = 0 \Rightarrow t = \frac{5}{7}\) | M1 | |
| \(y_{max} = 7\cdot\frac{5}{7} - 4.9\cdot\frac{25}{49} = 5 - 2.5 = 2.5\) m | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \frac{1}{2}(2.5) = 1.25\): \(7t - 4.9t^2 = 1.25\) | M1 | |
| \(4.9t^2 - 7t + 1.25 = 0\) | A1 | |
| \(t = \frac{7 \pm \sqrt{49 - 4(4.9)(1.25)}}{9.8} = \frac{7\pm\sqrt{24.5}}{9.8}\) | M1 | |
| \(t \approx 0.208\) s and \(t \approx 1.22\) s | A1 | |
| Horizontal distance \(= 24(1.22 - 0.208) = 24 \times \frac{\sqrt{24.5}}{4.9}\) | M1 | |
| \(\approx 24.5\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (A) \(v_y = 7 - 9.8(1.25) = 7 - 12.25 = -5.25\) m s\(^{-1}\) | B1 | |
| (B) Negative vertical velocity, so ball is falling | B1 | Must give reason |
| (C) \(v_x = 24\) m s\(^{-1}\); speed \(= \sqrt{24^2 + 5.25^2} = \sqrt{576+27.5625} = \sqrt{603.56} \approx 24.6\) m s\(^{-1}\) | M1 A1 | |
| Total 5 marks for (iv) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(t = \frac{x}{24}\), substitute into \(y = 7t-4.9t^2\) | M1 | |
| \(y = \frac{7x}{24} - 4.9\frac{x^2}{576} = \frac{7x}{24} - \frac{4.9x^2}{576}\) | A1 | |
| \(= \frac{0.7x}{576}(240 - 7x)\) shown | A1 | Completion |
| Range: \(y=0 \Rightarrow x=0\) or \(x = \frac{240}{7} \approx 34.3\) m | M1 A1 |
# Question 8:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertical: $y = 25\sin\theta \cdot t - \frac{1}{2}(9.8)t^2 = 25(0.28)t - 4.9t^2 = 7t - 4.9t^2$ | B1 | |
| Horizontal: $x = 25\cos\theta \cdot t = 25(0.96)t = 24t$ | B1 | |
| Both shown correctly | B1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Maximum height when $\frac{dy}{dt} = 7 - 9.8t = 0 \Rightarrow t = \frac{5}{7}$ | M1 | |
| $y_{max} = 7\cdot\frac{5}{7} - 4.9\cdot\frac{25}{49} = 5 - 2.5 = 2.5$ m | A1 | cao |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{1}{2}(2.5) = 1.25$: $7t - 4.9t^2 = 1.25$ | M1 | |
| $4.9t^2 - 7t + 1.25 = 0$ | A1 | |
| $t = \frac{7 \pm \sqrt{49 - 4(4.9)(1.25)}}{9.8} = \frac{7\pm\sqrt{24.5}}{9.8}$ | M1 | |
| $t \approx 0.208$ s and $t \approx 1.22$ s | A1 | |
| Horizontal distance $= 24(1.22 - 0.208) = 24 \times \frac{\sqrt{24.5}}{4.9}$ | M1 | |
| $\approx 24.5$ m | A1 | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| (A) $v_y = 7 - 9.8(1.25) = 7 - 12.25 = -5.25$ m s$^{-1}$ | B1 | |
| (B) Negative vertical velocity, so ball is falling | B1 | Must give reason |
| (C) $v_x = 24$ m s$^{-1}$; speed $= \sqrt{24^2 + 5.25^2} = \sqrt{576+27.5625} = \sqrt{603.56} \approx 24.6$ m s$^{-1}$ | M1 A1 | |
| | | Total 5 marks for (iv) |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $t = \frac{x}{24}$, substitute into $y = 7t-4.9t^2$ | M1 | |
| $y = \frac{7x}{24} - 4.9\frac{x^2}{576} = \frac{7x}{24} - \frac{4.9x^2}{576}$ | A1 | |
| $= \frac{0.7x}{576}(240 - 7x)$ shown | A1 | Completion |
| Range: $y=0 \Rightarrow x=0$ or $x = \frac{240}{7} \approx 34.3$ m | M1 A1 | |8 A ball is kicked from ground level over horizontal ground. It leaves the ground at a speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and at an angle $\theta$ to the horizontal such that $\cos \theta = 0.96$ and $\sin \theta = 0.28$.
\begin{enumerate}[label=(\roman*)]
\item Show that the height, $y \mathrm {~m}$, of the ball above the ground $t$ seconds after projection is given by $y = 7 t - 4.9 t ^ { 2 }$. Show also that the horizontal distance, $x \mathrm {~m}$, travelled by this time is given by $x = 24 t$.
\item Calculate the maximum height reached by the ball.
\item Calculate the times at which the ball is at half its maximum height.
Find the horizontal distance travelled by the ball between these times.
\item Determine the following when $t = 1.25$.\\
(A) The vertical component of the velocity of the ball.\\
(B) Whether the ball is rising or falling. (You should give a reason for your answer.)\\
(C) The speed of the ball.
\item Show that the equation of the trajectory of the ball is
$$y = \frac { 0.7 x } { 576 } ( 240 - 7 x )$$
Hence, or otherwise, find the range of the ball.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 2007 Q8 [19]}}