| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Maximum range or optimal angle |
| Difficulty | Standard +0.3 Part (a) is a standard derivation of the range formula that appears in most M2 textbooks. Parts (b-d) involve straightforward application of this formula with routine trigonometry and optimization. While multi-part, each step follows directly from standard projectile motion theory with no novel insight required, making it slightly easier than average. |
| Spec | 1.05o Trigonometric equations: solve in given intervals3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(s_y = (ut\sin\alpha - \frac{1}{2}gt^2) = t(u\sin\alpha - \frac{1}{2}gt)\) | M1 A1 | |
| \(s_y = 0\) when \(t = 0\) (at A) and when \(t = \frac{2u}{g}\sin\alpha\) (at B) | A1 | |
| \(s_x = ut\cos\alpha = u(\frac{2u}{g}\sin\alpha)\cos\alpha\) (at B) | M1 | |
| \(= \frac{u^2}{g}(2\sin\alpha\cos\alpha) = \frac{u^2}{g}\sin 2\alpha\) | M1 A1 | |
| (b) \(\frac{u^2}{g}\sin 2\alpha = 80 \therefore \frac{45^2}{9.8}\sin 2\alpha = 80\) | M1 A1 | |
| \(\sin 2\alpha = 0.387\) giving \(\alpha = 11.4°, 78.6°\) (1dp) | M2 A1 | |
| (c) \(11.4°\) as larger horiz. component of vel. | B2 | |
| (d) \(t = \frac{2s45}{g}\sin(11.4°) = 1.8\) seconds (1dp) | M1 A1 | (15) |
**(a)** $s_y = (ut\sin\alpha - \frac{1}{2}gt^2) = t(u\sin\alpha - \frac{1}{2}gt)$ | M1 A1 |
$s_y = 0$ when $t = 0$ (at A) and when $t = \frac{2u}{g}\sin\alpha$ (at B) | A1 |
$s_x = ut\cos\alpha = u(\frac{2u}{g}\sin\alpha)\cos\alpha$ (at B) | M1 |
$= \frac{u^2}{g}(2\sin\alpha\cos\alpha) = \frac{u^2}{g}\sin 2\alpha$ | M1 A1 |
**(b)** $\frac{u^2}{g}\sin 2\alpha = 80 \therefore \frac{45^2}{9.8}\sin 2\alpha = 80$ | M1 A1 |
$\sin 2\alpha = 0.387$ giving $\alpha = 11.4°, 78.6°$ (1dp) | M2 A1 |
**(c)** $11.4°$ as larger horiz. component of vel. | B2 |
**(d)** $t = \frac{2s45}{g}\sin(11.4°) = 1.8$ seconds (1dp) | M1 A1 | (15)6. A particle $P$ is projected from a point $A$ on horizontal ground with speed $u$ at an angle of elevation $\alpha$ and moves freely under gravity. $P$ hits the ground at the point $B$.
\begin{enumerate}[label=(\alph*)]
\item Show that $A B = \frac { u ^ { 2 } } { g } \sin 2 \alpha$.
An archer fires an arrow with an initial speed of $45 \mathrm {~ms} ^ { - 1 }$ at a target which is level with the point of projection and at a distance of 80 m .
Given that the arrow hits the target,
\item find in degrees, correct to 1 decimal place, the two possible angles of projection.
\item Write down, with a reason, which of the two possible angles of projection would give the shortest time of flight.\\
(2 marks)
\item Show that the minimum time of flight is 1.8 seconds, correct to 1 decimal place.\\
(2 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q6 [15]}}