| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.3 This is a standard projectile motion question requiring routine application of kinematic equations, elimination of parameter t to find trajectory, then substitution of given coordinates to find V and velocity direction. All steps are textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02e Calculate KE and PE: using formulae6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = Vt\cos60^\circ\) | B1 | |
| \(y = Vt\sin60^\circ - \frac{1}{2}gt^2\) | B1 | |
| M1 | For eliminating \(t\) | |
| \(y = x\sin60^\circ/\cos60^\circ - \frac{1}{2}gx^2/(V^2\cos^260^\circ)\) | A1 | For any correct form |
| \(y = \sqrt{3}\,x - 20x^2/V^2\) | A1 | AG [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(10 = \sqrt{3}\times70 - 20\times70^2/V^2\) | M1 | For substituting \(x=70\), \(y=10\) and attempting to solve for \(V\) |
| \(V = 29.7\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(dy/dx = \sqrt{3} - 40x/V^2\) | M1 | For differentiating \(y(x)\) |
| Gradient is \(-1.44\ldots\) at \(A\) | A1ft | |
| Direction is \(55.3^\circ\) downwards from the horizontal | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For attempting to find either \(\dot{x}\) or \(\dot{y}\) at \(A\) | |
| \([x = 14.8...,\, \dot{y} = -21.42...]\) | A1ft | |
| Direction is \(55.3^\circ\) downwards from the horizontal | A1 | [3] |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = Vt\cos60^\circ$ | B1 | |
| $y = Vt\sin60^\circ - \frac{1}{2}gt^2$ | B1 | |
| | M1 | For eliminating $t$ |
| $y = x\sin60^\circ/\cos60^\circ - \frac{1}{2}gx^2/(V^2\cos^260^\circ)$ | A1 | For any correct form |
| $y = \sqrt{3}\,x - 20x^2/V^2$ | A1 | AG **[5]** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $10 = \sqrt{3}\times70 - 20\times70^2/V^2$ | M1 | For substituting $x=70$, $y=10$ and attempting to solve for $V$ |
| $V = 29.7$ | A1 | **[2]** |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $dy/dx = \sqrt{3} - 40x/V^2$ | M1 | For differentiating $y(x)$ |
| Gradient is $-1.44\ldots$ at $A$ | A1ft | |
| Direction is $55.3^\circ$ downwards from the horizontal | A1 | **[3]** |
### Alternative for Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For attempting to find either $\dot{x}$ or $\dot{y}$ at $A$ |
| $[x = 14.8...,\, \dot{y} = -21.42...]$ | A1ft | |
| Direction is $55.3^\circ$ downwards from the horizontal | A1 | **[3]** |7 A particle $P$ is projected from a point $O$ on horizontal ground with speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and direction $60 ^ { \circ }$ upwards from the horizontal. At time $t \mathrm {~s}$ later the horizontal and vertical displacements of $P$ from $O$ are $x \mathrm {~m}$ and $y \mathrm {~m}$ respectively.\\
(i) Write down expressions for $x$ and $y$ in terms of $V$ and $t$ and hence show that the equation of the trajectory of $P$ is
$$y = ( \sqrt { } 3 ) x - \frac { 20 x ^ { 2 } } { V ^ { 2 } }$$
$P$ passes through the point $A$ at which $x = 70$ and $y = 10$. Find\\
(ii) the value of $V$,\\
(iii) the direction of motion of $P$ at the instant it passes through $A$.
\hfill \mbox{\textit{CAIE M2 2008 Q7 [10]}}