| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Moderate -0.3 This is a standard M3 projectiles question requiring derivation of the trajectory equation from parametric equations (a routine 'show that' proof), followed by straightforward substitution to solve for range. The derivation uses well-practiced techniques (eliminating time parameter), and part (b) is simple equation solving. Slightly easier than average due to the structured guidance and standard method, though the multi-part nature and modeling assumptions add minor complexity. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Component along line of centres: A: \(4\cos\alpha = 4 \times \frac{3}{5} = \frac{12}{5}\); B: \(-2.6\cos\beta = -2.6 \times \frac{5}{13} = -1\) | M1 A1 | Resolving along line of centres (note \(\cos\alpha = \frac{3}{5}\), \(\cos\beta = \frac{5}{13}\)) |
| CLM along line of centres: \(2 \times \frac{12}{5} + 1\times(-1) = 2v_A' + 1 \times v_B'\) | M1 | |
| \(\frac{19}{5} = 2v_A' + v_B'\) | A1 | |
| NEL: \(v_B' - v_A' = \frac{4}{7}(\frac{12}{5}+1) = \frac{4}{7} \times \frac{17}{5} = \frac{68}{35}\) | M1 A1 | |
| Solving: \(v_B' = \frac{19}{5} \times \frac{1}{3} + \frac{2}{3} \times \frac{68}{35} = \frac{19}{15} + \frac{136}{105} = \frac{133+136}{105} = \frac{269}{105}\) ... leading to speeds | DM1 A1 | |
| Speed of A after: combine along-line component with perpendicular component \(4\sin\alpha = \frac{12}{5}\) | M1 | |
| Speed of B after: combine along-line component with perpendicular \(2.6\sin\beta = 2.6 \times \frac{12}{13} = 2.4\) | M1 A1 | Final speeds evaluated |
# Question 5:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Component along line of centres: A: $4\cos\alpha = 4 \times \frac{3}{5} = \frac{12}{5}$; B: $-2.6\cos\beta = -2.6 \times \frac{5}{13} = -1$ | M1 A1 | Resolving along line of centres (note $\cos\alpha = \frac{3}{5}$, $\cos\beta = \frac{5}{13}$) |
| CLM along line of centres: $2 \times \frac{12}{5} + 1\times(-1) = 2v_A' + 1 \times v_B'$ | M1 | |
| $\frac{19}{5} = 2v_A' + v_B'$ | A1 | |
| NEL: $v_B' - v_A' = \frac{4}{7}(\frac{12}{5}+1) = \frac{4}{7} \times \frac{17}{5} = \frac{68}{35}$ | M1 A1 | |
| Solving: $v_B' = \frac{19}{5} \times \frac{1}{3} + \frac{2}{3} \times \frac{68}{35} = \frac{19}{15} + \frac{136}{105} = \frac{133+136}{105} = \frac{269}{105}$ ... leading to speeds | DM1 A1 | |
| Speed of A after: combine along-line component with perpendicular component $4\sin\alpha = \frac{12}{5}$ | M1 | |
| Speed of B after: combine along-line component with perpendicular $2.6\sin\beta = 2.6 \times \frac{12}{13} = 2.4$ | M1 A1 | Final speeds evaluated |
---5 A football is kicked from a point $O$ on a horizontal football ground with a velocity of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of elevation of $30 ^ { \circ }$. During the motion, the horizontal and upward vertical displacements of the football from $O$ are $x$ metres and $y$ metres respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that $x$ and $y$ satisfy the equation
$$y = x \tan 30 ^ { \circ } - \frac { g x ^ { 2 } } { 800 \cos ^ { 2 } 30 ^ { \circ } }$$
\item On its downward flight the ball hits the horizontal crossbar of the goal at a point which is 2.5 m above the ground. Using the equation given in part (a), find the horizontal distance from $O$ to the goal.\\
(4 marks)\\
\includegraphics[max width=\textwidth, alt={}, center]{fc5bfc4b-68bb-4a23-874b-87e9558dc990-04_330_1411_1902_303}
\item State two modelling assumptions that you have made.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 Q5}}