| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.3 This is a standard M3 projectile question requiring trajectory equations and quadratic solving. Part (a) is routine algebraic manipulation of projectile equations, part (b) is straightforward quadratic solving, and part (c) involves evaluating horizontal velocity components. While multi-step, it follows a predictable template with no novel insight required, making it slightly easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(5 = 10\cos\alpha\cdot t\) | M1 | |
| \(t = \frac{5}{10\cos\alpha}\) | A1 | |
| \(1 = -\frac{1}{2}(9.8)t^2+10\sin\alpha\cdot t\) | M1A1 | |
| \(1 = -\frac{1}{2}(9.8)\frac{25}{100\cos^2\alpha}+10\sin\alpha\frac{5}{10\cos\alpha}\) | m1 | Dependent on both M1s |
| \(1 = -\frac{1}{2}(9.8)\frac{25}{100}(1+\tan^2\alpha)+10\sin\alpha\frac{5}{10\cos\alpha}\) | A1 | |
| \(49\tan^2\alpha - 200\tan\alpha + 89 = 0\) | A1 | Answer given. Total: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan\alpha = \frac{200\pm\sqrt{40000-4(49)(89)}}{2\times49}\) | M1 | |
| \(= 3.57,\ 0.508\) | A1 | AWRT |
| \(\alpha = 74.4°,\ 26.9°\) | A1F | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(10\cos26.9° = 8.92\ (\text{or } 8.91) > 8 \Rightarrow\) The can will be knocked off the wall | M1 | Both values checked |
| \(10\cos74.4° = 2.69 < 8 \Rightarrow\) The can will not be knocked off the wall | A1F | Acc. of both results |
| E1 | Correct conclusions. Total: 3 | |
| Alternative: Can knocked off if \(10\cos\alpha > 8\), i.e. \(\cos\alpha > 0.8\), \(\alpha < 36.9°\) | M1A1 | |
| So for \(\alpha=26.9°\) can will be knocked off; for \(\alpha=74.4°\) can will not be knocked off | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = ut\), \(t = \frac{5}{10\cos26.9°}\) | ||
| \(v = 10\sin26.9° - 9.8\left(\frac{5}{10\cos26.9°}\right)\) | M1 | Any correct use of equations |
| \(v = -0.970\) | A1F | |
| \(\tan\theta = \frac{-0.970}{8.92}\) | M1 | |
| \(\theta = -6.2°\) | ||
| At an angle of depression of \(6.2°\) | A1F | AWRT \(6°\). Total: 4 |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $5 = 10\cos\alpha\cdot t$ | M1 | |
| $t = \frac{5}{10\cos\alpha}$ | A1 | |
| $1 = -\frac{1}{2}(9.8)t^2+10\sin\alpha\cdot t$ | M1A1 | |
| $1 = -\frac{1}{2}(9.8)\frac{25}{100\cos^2\alpha}+10\sin\alpha\frac{5}{10\cos\alpha}$ | m1 | Dependent on both M1s |
| $1 = -\frac{1}{2}(9.8)\frac{25}{100}(1+\tan^2\alpha)+10\sin\alpha\frac{5}{10\cos\alpha}$ | A1 | |
| $49\tan^2\alpha - 200\tan\alpha + 89 = 0$ | A1 | Answer given. **Total: 7** |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\alpha = \frac{200\pm\sqrt{40000-4(49)(89)}}{2\times49}$ | M1 | |
| $= 3.57,\ 0.508$ | A1 | AWRT |
| $\alpha = 74.4°,\ 26.9°$ | A1F | **Total: 3** |
### Part (c)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $10\cos26.9° = 8.92\ (\text{or } 8.91) > 8 \Rightarrow$ The can will be knocked off the wall | M1 | Both values checked |
| $10\cos74.4° = 2.69 < 8 \Rightarrow$ The can will not be knocked off the wall | A1F | Acc. of both results |
| | E1 | Correct conclusions. **Total: 3** |
| **Alternative:** Can knocked off if $10\cos\alpha > 8$, i.e. $\cos\alpha > 0.8$, $\alpha < 36.9°$ | M1A1 | |
| So for $\alpha=26.9°$ can will be knocked off; for $\alpha=74.4°$ can will not be knocked off | E1 | |
### Part (c)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = ut$, $t = \frac{5}{10\cos26.9°}$ | | |
| $v = 10\sin26.9° - 9.8\left(\frac{5}{10\cos26.9°}\right)$ | M1 | Any correct use of equations |
| $v = -0.970$ | A1F | |
| $\tan\theta = \frac{-0.970}{8.92}$ | M1 | |
| $\theta = -6.2°$ | | |
| At an angle of depression of $6.2°$ | A1F | AWRT $6°$. **Total: 4** |5 A boy throws a small ball from a height of 1.5 m above horizontal ground with initial velocity $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\alpha$ above the horizontal. The ball hits a small can placed on a vertical wall of height 2.5 m , which is at a horizontal distance of 5 m from the initial position of the ball, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{eed9842d-cd89-4eb7-b5ba-9380971be196-3_499_1180_1283_424}
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha$ satisfies the equation
$$49 \tan ^ { 2 } \alpha - 200 \tan \alpha + 89 = 0$$
\item Find the two possible values of $\alpha$, giving your answers to the nearest $0.1 ^ { \circ }$.
\item \begin{enumerate}[label=(\roman*)]
\item To knock the can off the wall, the horizontal component of the velocity of the ball must be greater than $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Show that, for one of the possible values of $\alpha$ found in part (b), the can will be knocked off the wall, and for the other, it will not be knocked off the wall.\\
(3 marks)
\item Given that the can is knocked off the wall, find the direction in which the ball is moving as it hits the can.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M3 2008 Q5 [17]}}