| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| 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 with routine application of SUVAT equations. Part (a) involves symmetric projectile motion (standard textbook exercise), while part (b) requires solving a quadratic equation for asymmetric motion. The 'show that' in (a)(i) guides students through the calculation, and all parts follow predictable methods with no novel problem-solving required. Slightly easier than average due to its structured, methodical nature. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(0 = 40 \sin 35° t - 4.9t^2\) | M1, A1 | Equation to find time of flight with 40, \(\sin/\cos 35°\) and \(-4.9\) or \(-\frac{g}{2}\). Correct equation. |
| \(t = \frac{40 \sin 35°}{4.9} = 4.68\) s | m1, A1 | Solving for \(t\). Correct time from correct working. Note: candidates must have a method for the complete time of flight before any marks can be awarded. Condone the use of a formula for the time of flight. |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB = 40 \cos 35° \times 4.682 = 153\) m | M1, A1 | Calculating the range using 40, cos/sin 35° and 4.68 and acceleration zero. Correct range. Accept AWRT 153. |
| Answer | Marks | Guidance |
|---|---|---|
| \(-1 = 40 \sin 35° t - 4.9t^2\) | M1 | Equation to find time of flight with a ±1, 40, sin/cos 35° and \(-4.9\) or \(-\frac{g}{2}\). |
| \(4.9t^2 - 40 \sin 35° t - 1 = 0\) | A1, A1 | Correct terms. Correct signs. |
| \(t = \frac{40 \sin 35° \pm \sqrt{(40 \sin 35°)^2 - 4 \times 4.9 \times (-1)}}{2 \times 4.9}\) | m1 | Solving quadratic equation. |
| \(t = 4.73\) or \(t = -0.0432\) | A1 | Accept AWRT 4.73 or 4.72. |
| \(t = 4.73\) | A1 | Rejection of negative solution indicated (Only 4.73 or 4.72 given award 5/6 marks). |
| Answer | Marks | Guidance |
|---|---|---|
| For example, \(t = 4.682 + 0.044 = 4.73\) or \(t = 2.341 + 2.384 = 4.73\) | M1, A1, m1, A1, A1, A1 | Addition of two times. Use of AWRT 4.68 or AWRT 2.34. Calculation of time for 'second' part. Correct expression for time for 'second' part. Correct time (Allow AWRT 0.04 or AWRT 2.38). Correct total time. Accept 4.72. |
**7(a)(i)**
| $0 = 40 \sin 35° t - 4.9t^2$ | M1, A1 | Equation to find time of flight with 40, $\sin/\cos 35°$ and $-4.9$ or $-\frac{g}{2}$. Correct equation. |
| $t = \frac{40 \sin 35°}{4.9} = 4.68$ s | m1, A1 | Solving for $t$. Correct time from correct working. Note: candidates must have a method for the complete time of flight before any marks can be awarded. Condone the use of a formula for the time of flight. |
**7(a)(ii)**
| $AB = 40 \cos 35° \times 4.682 = 153$ m | M1, A1 | Calculating the range using 40, cos/sin 35° and 4.68 and acceleration zero. Correct range. Accept AWRT 153. |
**7(b)**
| $-1 = 40 \sin 35° t - 4.9t^2$ | M1 | Equation to find time of flight with a ±1, 40, sin/cos 35° and $-4.9$ or $-\frac{g}{2}$. |
| $4.9t^2 - 40 \sin 35° t - 1 = 0$ | A1, A1 | Correct terms. Correct signs. |
| $t = \frac{40 \sin 35° \pm \sqrt{(40 \sin 35°)^2 - 4 \times 4.9 \times (-1)}}{2 \times 4.9}$ | m1 | Solving quadratic equation. |
| $t = 4.73$ or $t = -0.0432$ | A1 | Accept AWRT 4.73 or 4.72. |
| $t = 4.73$ | A1 | Rejection of negative solution indicated (Only 4.73 or 4.72 given award 5/6 marks). |
**Alternative methods based on finding two times:**
| For example, $t = 4.682 + 0.044 = 4.73$ or $t = 2.341 + 2.384 = 4.73$ | M1, A1, m1, A1, A1, A1 | Addition of two times. Use of AWRT 4.68 or AWRT 2.34. Calculation of time for 'second' part. Correct expression for time for 'second' part. Correct time (Allow AWRT 0.04 or AWRT 2.38). Correct total time. Accept 4.72. |7 A ball is hit by a bat so that, when it leaves the bat, its velocity is $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $35 ^ { \circ }$ above the horizontal. Assume that the ball is a particle and that its weight is the only force that acts on the ball after it has left the bat.
\begin{enumerate}[label=(\alph*)]
\item A simple model assumes that the ball is hit from the point $A$ and lands for the first time at the point $B$, which is at the same level as $A$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{a381686b-0b1e-41ba-b88f-be1601e42098-4_321_1063_1370_484}
\begin{enumerate}[label=(\roman*)]
\item Show that the time that it takes for the ball to travel from $A$ to $B$ is 4.68 seconds, correct to three significant figures.
\item Find the horizontal distance from $A$ to $B$.
\end{enumerate}\item A revised model assumes that the ball is hit from the point $C$, which is 1 metre above $A$. The ball lands at the point $D$, which is at the same level as $A$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{a381686b-0b1e-41ba-b88f-be1601e42098-4_431_1177_2181_420}
Find the time that it takes for the ball to travel from $C$ to $D$.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2008 Q7 [12]}}