| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.3 Part (i) is a standard bookwork derivation of the trajectory equation that appears in every M2 textbook. Parts (ii) and (iii) apply this to a contextual problem requiring substitution of given values and solving a quadratic, which is routine for M2 students. The algebra is straightforward and the problem-solving is minimal—this is slightly easier than a typical exam question. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
A particle is projected with speed $u$ ms$^{-1}$ at an angle of $\theta$ above the horizontal from a point $O$. At time $t$ s after projection, the horizontal and vertically upwards displacements of the particle from $O$ are $x$ m and $y$ m respectively.
\begin{enumerate}[label=(\roman*)]
\item Express $x$ and $y$ in terms of $t$ and $\theta$ and hence obtain the equation of trajectory
$$y = x \tan \theta - \frac{gx^2 \sec^2 \theta}{2u^2}.$$ [4]
\end{enumerate}
In a shot put competition, a shot is thrown from a height of 2.1 m above horizontal ground. It has initial velocity of 14 ms$^{-1}$ at an angle of $\theta$ above the horizontal. The shot travels a horizontal distance of 22 m before hitting the ground.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that $12.1 \tan^2 \theta - 22 \tan \theta + 10 = 0$, and find the value of $\theta$. [5]
\item Find the time of flight of the shot. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR M2 2013 Q7 [11]}}