| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Horizontal projection from height |
| Difficulty | Moderate -0.8 This is a straightforward horizontal projectile motion question requiring only standard kinematic equations (s = ut + ½at²) with no problem-solving insight. Students apply memorized formulas directly: vertical motion uses h = -½gt² (or with initial vertical velocity if given), horizontal uses x = ut. Part (ii) involves setting vertical displacement equal to the height and solving for t, then substituting into horizontal equation. This is easier than average A-level mechanics as it's a textbook exercise with clear setup and routine application of SUVAT equations. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Vertically: \(y = 8t - 4.9t^2\) | M1 | Use of \(s = ut + 0.5at^2\) with \(g = \pm 9.8, \pm 10\). Accept \(u = 0\) or \(14.4...\) or \(14.4\sin\theta\) or \(u\sin\theta\) but not 12. Allow use of \(+3.6\). |
| A1 | Accept derivation of \(-4.9\) not clear. cao. | |
| Horizontally: \(x = 12t\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| either: Require \(y = -3.6\), so \(-3.6 = 8t - 4.9t^2\) | M1 | Equating their \(y\) to \(\pm 3.6\) or equiv. Any form. |
| Use of formula or \(4.9(t-2)(t + \frac{18}{49}) = 0\) | M1 | A method for solving a 3 term quadratic to give at least 1 root. Allow their \(y\) and re-arrangement errors. |
| Roots are \(2\) and \(-\frac{18}{49}\) \((= -0.367346...)\) | A1 | WWW. Accept no reference to 2nd root. [Award SC3 for \(t = 2\) seen WWW] |
| Horizontal distance is \(12 \times 2 = 24\) | M1 | FT their \(x\) and \(t\). |
| so \(24 \text{ m}\) | F1 | FT only their \(t\) (as long as it is +ve and is not obtained with sign error(s) e.g. \(-\)ve sign just dropped) |
| or: Require \(y = -3.6\), eliminate \(t\) between \(x = 12t\) and \(-3.6 = 8t - 4.9t^2\) | M1 | Equating their \(y\) to \(\pm 3.6\) or equiv. Any form. |
| M1 | Expressions in any form. Elimination must be complete. | |
| so \(0 = 3.6 + \frac{8x}{12} - \frac{4.9x^2}{144}\) | A1 | Accept in any form. May be implied. |
| Use of formula or factorise | M1 | A method for solving a 3 term quadratic to give at least 1 root. Allow their \(y\) and re-arrangement errors. |
| +ve root is \(24\) so \(24\text{ m}\) | F1 | FT from their quadratic after re-arrangement. Must be +ve. |
| or (sections method): (A) \(0.8163...\text{ s}\); \(9.7959..\text{ m}\); (B) \(0.816...\text{ s}\); \(9.7959..\text{ m}\); (C) \(0.3673...\text{ s}\); \(4.4081...\text{ m}\) | M1 | Attempt to find times or distances for sections that give the total horizontal distance travelled |
| M1 | Correct method for one section to find time or distance | |
| A1 | Any time or distance for a section correct | |
| A1 | 2nd time or distance correct (the two sections must not be A and B) | |
| A1 | cao |
# Question 6:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Vertically: $y = 8t - 4.9t^2$ | M1 | Use of $s = ut + 0.5at^2$ with $g = \pm 9.8, \pm 10$. Accept $u = 0$ or $14.4...$ or $14.4\sin\theta$ or $u\sin\theta$ but not 12. Allow use of $+3.6$. |
| | A1 | Accept derivation of $-4.9$ not clear. cao. |
| Horizontally: $x = 12t$ | B1 | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| **either**: Require $y = -3.6$, so $-3.6 = 8t - 4.9t^2$ | M1 | Equating **their** $y$ to $\pm 3.6$ or equiv. Any form. |
| Use of formula or $4.9(t-2)(t + \frac{18}{49}) = 0$ | M1 | A method for solving a 3 term quadratic to give at least 1 root. Allow **their** $y$ and re-arrangement errors. |
| Roots are $2$ and $-\frac{18}{49}$ $(= -0.367346...)$ | A1 | WWW. Accept no reference to 2nd root. [Award SC3 for $t = 2$ seen WWW] |
| Horizontal distance is $12 \times 2 = 24$ | M1 | FT **their** $x$ and $t$. |
| so $24 \text{ m}$ | F1 | FT only **their** $t$ (as long as it is +ve and is not obtained with sign error(s) e.g. $-$ve sign just dropped) |
| **or**: Require $y = -3.6$, eliminate $t$ between $x = 12t$ and $-3.6 = 8t - 4.9t^2$ | M1 | Equating **their** $y$ to $\pm 3.6$ or equiv. Any form. |
| | M1 | Expressions in any form. Elimination must be complete. |
| so $0 = 3.6 + \frac{8x}{12} - \frac{4.9x^2}{144}$ | A1 | Accept in any form. May be implied. |
| Use of formula or factorise | M1 | A method for solving a 3 term quadratic to give at least 1 root. Allow **their** $y$ and re-arrangement errors. |
| +ve root is $24$ so $24\text{ m}$ | F1 | FT from **their** quadratic after re-arrangement. Must be +ve. |
| **or** (sections method): (A) $0.8163...\text{ s}$; $9.7959..\text{ m}$; (B) $0.816...\text{ s}$; $9.7959..\text{ m}$; (C) $0.3673...\text{ s}$; $4.4081...\text{ m}$ | M1 | Attempt to find times or distances for sections that give the total horizontal distance travelled |
| | M1 | Correct method for one section to find time or distance |
| | A1 | Any time or distance for a section correct |
| | A1 | 2nd time or distance correct (the two sections must not be A and B) |
| | A1 | cao |
---6 A small ball is kicked off the edge of a jetty over a calm sea. Air resistance is negligible. Fig. 6 shows
\begin{itemize}
\item the point of projection, O,
\item the initial horizontal and vertical components of velocity,
\item the point A on the jetty vertically below O and at sea level,
\item the height, OA, of the jetty above the sea.
\end{itemize}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6cca1e5e-82b0-487d-8048-b9db7745dea6-3_458_1008_1786_571}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
The time elapsed after the ball is kicked is $t$ seconds.\\
(i) Find an expression in terms of $t$ for the height of the ball above O at time $t$. Find also an expression for the horizontal distance of the ball from O at this time.\\
(ii) Determine how far the ball lands from A .
\hfill \mbox{\textit{OCR MEI M1 2010 Q6 [8]}}