| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.3 This is a comprehensive but standard M1 projectiles question covering all basic trajectory calculations (equations of motion, maximum height, time of flight, speed at a point, and trajectory equation). While multi-part with several marks, each component uses routine application of SUVAT equations and standard projectile formulas with no novel problem-solving required. The 'show that' parts provide the answers, making it slightly easier than average. |
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 25\sin\theta t + 0.5 \times (-9.8)t^2\) | M1 | Use of \(s = ut + \frac{1}{2}at^2\). Accept sin, cos, 0.96, 0.28, \(\pm9.8\), \(\pm10\), \(u=25\) and derivation of \(-4.9\) not clear |
| \(= 7t - 4.9t^2\) | E1 | Shown including deriv of \(-4.9\). Accept \(25\sin\theta t = 7t\) WW |
| \(x = 25\cos\theta t = 25 \times 0.96t = 24t\) | B1 | Accept \(25 \times 0.96t\) or \(25\cos\theta t\) seen WW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0 = 7^2 - 19.6s\) | M1 | Accept sequence of *uvast*. Accept \(u=24\) but not 25. Allow \(u \leftrightarrow v\) and \(\pm9.8\) and \(\pm10\) |
| \(s = 2.5\) so \(2.5\) m | A1 | +ve answer obtained by correct manipulation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Need \(7t - 4.9t^2 = 1.25\), so \(4.9t^2 - 7t + 1.25 = 0\) | M1 | Equate \(y\) to their (ii)/2 or equivalent |
| M1 | Correct sub into quad formula of their 3 term quadratic being solved (i.e. allow manipulation errors before using the formula) | |
| \(t = 0.209209\ldots\) and \(1.219361\ldots\) | A1 | Both. cao. [Award M1 A1 for two correct roots WW] |
| need \(24 \times (1.219\ldots - 0.209209\ldots) = 24 \times 1.01\ldots\) so \(24.2\) m (3 s.f.) | B1 | FT their roots (only if both positive) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dot{y} = 7 - 9.8t\) | M1 | Attempt at \(\dot{y}\). Accept sign errors and \(u=24\) but not 25 |
| \(\dot{y}(1.25) = 7 - 9.8 \times 1.25 = -5.25\) m s\(^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Falling as velocity is negative | E1 | Reason must be clear. FT their \(\dot{y}\) even if not a velocity. Could use an argument involving time |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Speed is \(\sqrt{24^2 + (-5.25)^2}\) | M1 | Use of Pythagoras and 24 or 7 with their \(\dot{y}\) |
| \(= 24.5675\ldots\) so \(24.6\) m s\(^{-1}\) (3 s.f.) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 7t - 4.9t^2\), \(x = 24t\) | M1 | Elimination of \(t\) |
| so \(y = \frac{7x}{24} - 4.9\left(\frac{x}{24}\right)^2\) | A1 | Elimination correct. Condone wrong notation with interpretation correct for the problem |
| \(y = \frac{7x}{24} - 4.9 \times \frac{x^2}{576} = \frac{0.7x}{576}(240 - 7x)\) | E1 | If not wrong accept as long as \(24^2 = 576\) seen. Condone wrong notation with interpretation correct |
| Need \(y = 0\) | M1 | |
| so \(x = 0\) or \(\frac{240}{7}\), so \(\frac{240}{7}\) m | A1 | Accept \(x=0\) not mentioned. Condone \(0 \leq X \leq \frac{240}{7}\) |
| or Time of flight \(\frac{10}{7}\) s | B1 | |
| Range \(\frac{240}{7}\) m. Condone \(0 \leq X \leq \frac{240}{7}\) | B1 |
# Question 2:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 25\sin\theta t + 0.5 \times (-9.8)t^2$ | M1 | Use of $s = ut + \frac{1}{2}at^2$. Accept sin, cos, 0.96, 0.28, $\pm9.8$, $\pm10$, $u=25$ and derivation of $-4.9$ not clear |
| $= 7t - 4.9t^2$ | E1 | Shown including deriv of $-4.9$. Accept $25\sin\theta t = 7t$ WW |
| $x = 25\cos\theta t = 25 \times 0.96t = 24t$ | B1 | Accept $25 \times 0.96t$ or $25\cos\theta t$ seen WW |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = 7^2 - 19.6s$ | M1 | Accept sequence of *uvast*. Accept $u=24$ but not 25. Allow $u \leftrightarrow v$ and $\pm9.8$ and $\pm10$ |
| $s = 2.5$ so $2.5$ m | A1 | +ve answer obtained by correct manipulation |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Need $7t - 4.9t^2 = 1.25$, so $4.9t^2 - 7t + 1.25 = 0$ | M1 | Equate $y$ to **their** (ii)/2 or equivalent |
| | M1 | Correct sub into quad formula of their 3 term quadratic being solved (i.e. allow manipulation errors before using the formula) |
| $t = 0.209209\ldots$ and $1.219361\ldots$ | A1 | Both. cao. [Award M1 A1 for two correct roots WW] |
| need $24 \times (1.219\ldots - 0.209209\ldots) = 24 \times 1.01\ldots$ so $24.2$ m (3 s.f.) | B1 | FT **their** roots (only if both positive) |
## Part (iv)(A)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dot{y} = 7 - 9.8t$ | M1 | Attempt at $\dot{y}$. Accept sign errors and $u=24$ but not 25 |
| $\dot{y}(1.25) = 7 - 9.8 \times 1.25 = -5.25$ m s$^{-1}$ | A1 | |
## Part (iv)(B)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Falling as velocity is negative | E1 | Reason must be clear. FT **their** $\dot{y}$ even if not a velocity. Could use an argument involving time |
## Part (iv)(C)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Speed is $\sqrt{24^2 + (-5.25)^2}$ | M1 | Use of Pythagoras and 24 or 7 with **their** $\dot{y}$ |
| $= 24.5675\ldots$ so $24.6$ m s$^{-1}$ (3 s.f.) | A1 | cao |
## Part (v)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 7t - 4.9t^2$, $x = 24t$ | M1 | Elimination of $t$ |
| so $y = \frac{7x}{24} - 4.9\left(\frac{x}{24}\right)^2$ | A1 | Elimination correct. Condone wrong notation with interpretation correct for the problem |
| $y = \frac{7x}{24} - 4.9 \times \frac{x^2}{576} = \frac{0.7x}{576}(240 - 7x)$ | E1 | If not wrong accept as long as $24^2 = 576$ seen. Condone wrong notation with interpretation correct |
| Need $y = 0$ | M1 | |
| so $x = 0$ or $\frac{240}{7}$, so $\frac{240}{7}$ m | A1 | Accept $x=0$ not mentioned. Condone $0 \leq X \leq \frac{240}{7}$ |
| **or** Time of flight $\frac{10}{7}$ s | B1 | |
| Range $\frac{240}{7}$ m. Condone $0 \leq X \leq \frac{240}{7}$ | B1 | |
---2 A ball is kicked from ground level over horizontal ground. It leaves the ground at a speed of 25 ms 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 Q2 [19]}}