| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.3 This is a standard M1 projectiles question requiring routine application of range formula and inequality solving. Part (a) tests standard modelling assumptions, part (b) is a show-that using the standard range formula (minimal algebraic manipulation), and part (c) requires simple inequality solving. Slightly easier than average due to the guided 'show that' structure and straightforward final calculation. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| It is a particle /No air resistance / lift forces act on the ball. | B1, B1 (Total: 2) | Particle. Other acceptable assumption. Deduct one mark for each additional incorrect assumption. |
| Answer | Marks | Guidance |
|---|---|---|
| \(V \sin 40°t - \frac{1}{2} \times 9.8t^2 = 0\) → \(t = \frac{V \sin 40°}{4.9}\) → \(s = V \cos 40° \times \frac{V \sin 40°}{4.9} = \frac{V^2 \cos 40° \sin 40°}{4.9}\) | M1, A1, dM1, A1 (Total: 6) | Vertical equation to find \(t\). Correct equation (Equals zero may be implied). Solving for \(t\). Correct \(t\). Finding range with their \(t\). Correct range from correct working. SC Quoting the formula for the range 2 marks. |
| Answer | Marks | Guidance |
|---|---|---|
| \(76 < \frac{V^2 \cos 40° \sin 40°}{4.9} \leq 82\) → \(\sqrt{\frac{76 \times 4.9}{\cos 40° \sin 40°}} < V < \sqrt{\frac{82 \times 4.9}{\cos 40° \sin 40°}}\) → \(27.5 < V < 28.6\) | M1, A1, A1, A1 (Total: 4) | An equation to find one value of \(V\). Correct value for \(V\). Other value of \(V\) correct. Correct range of values. Accept 27.5 – 28.6 but not 28.6-27.5. For using values close to 76 and 82 deduct one mark. |
### Part (a)
It is a particle /No air resistance / lift forces act on the ball. | B1, B1 (Total: 2) | Particle. Other acceptable assumption. Deduct one mark for each additional incorrect assumption.
### Part (b)
$V \sin 40°t - \frac{1}{2} \times 9.8t^2 = 0$ → $t = \frac{V \sin 40°}{4.9}$ → $s = V \cos 40° \times \frac{V \sin 40°}{4.9} = \frac{V^2 \cos 40° \sin 40°}{4.9}$ | M1, A1, dM1, A1 (Total: 6) | Vertical equation to find $t$. Correct equation (Equals zero may be implied). Solving for $t$. Correct $t$. Finding range with their $t$. Correct range from correct working. SC Quoting the formula for the range 2 marks.
### Part (c)
$76 < \frac{V^2 \cos 40° \sin 40°}{4.9} \leq 82$ → $\sqrt{\frac{76 \times 4.9}{\cos 40° \sin 40°}} < V < \sqrt{\frac{82 \times 4.9}{\cos 40° \sin 40°}}$ → $27.5 < V < 28.6$ | M1, A1, A1, A1 (Total: 4) | An equation to find one value of $V$. Correct value for $V$. Other value of $V$ correct. Correct range of values. Accept 27.5 – 28.6 but not 28.6-27.5. For using values close to 76 and 82 deduct one mark.
---7 A golfer hits a ball which is on horizontal ground. The ball initially moves with speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $40 ^ { \circ }$ above the horizontal. There is a pond further along the horizontal ground. The diagram below shows the initial position of the ball and the position of the pond.\\
\includegraphics[max width=\textwidth, alt={}, center]{217f0e3e-9d1b-41f1-8299-f56d073fbbeb-5_387_1230_502_395}
\begin{enumerate}[label=(\alph*)]
\item State two assumptions that you should make in order to model the motion of the ball.\\
(2 marks)
\item Show that the horizontal distance, in metres, travelled by the ball when it returns to ground level is
$$\frac { V ^ { 2 } \sin 40 ^ { \circ } \cos 40 ^ { \circ } } { 4.9 }$$
\item Find the range of values of $V$ for which the ball lands in the pond.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2008 Q7 [12]}}