| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Horizontal projection from height |
| Difficulty | Moderate -0.3 This is a standard projectiles question requiring routine application of kinematic equations (x = Vt, y = -½gt²), elimination of parameter t to get trajectory equation, then substitution of given points to find constants. All steps are mechanical with no novel insight required, making it slightly easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = Vt\) or \(y = -\frac{1}{2}gt^2\) | B1 | Use horizontal motion or vertical motion |
| M1 | Attempt to eliminate \(t\) | |
| \(y = -5x^2/V^2\) AG | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-a = -5a^2/V^2\) or \(-16a = -5(a^2)^2/V^2\) | B1 | |
| M1 | Attempt to eliminate \(V\) | |
| \(a = 4\) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = -5(5\times4)^2/(5\times4)\ (= -100)\) | M1 | Use the value of \(a\) found in part (ii) |
| Height \(= 100\) m | A1 | |
| 2 |
## Question 4:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = Vt$ **or** $y = -\frac{1}{2}gt^2$ | B1 | Use horizontal motion or vertical motion |
| | M1 | Attempt to eliminate $t$ |
| $y = -5x^2/V^2$ **AG** | A1 | |
| | **3** | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-a = -5a^2/V^2$ **or** $-16a = -5(a^2)^2/V^2$ | B1 | |
| | M1 | Attempt to eliminate $V$ |
| $a = 4$ | A1 | |
| | **3** | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = -5(5\times4)^2/(5\times4)\ (= -100)$ | M1 | Use the value of $a$ found in part (ii) |
| Height $= 100$ m | A1 | |
| | **2** | |
---4 A small object is projected horizontally with speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $O$ above horizontal ground. At time $t \mathrm {~s}$ after projection, the horizontal and vertically upwards displacements of the object from $O$ are $x \mathrm {~m}$ and $y \mathrm {~m}$ respectively.\\
(i) Express $x$ and $y$ in terms of $t$ and hence show that the equation of the path of the object is $y = - \frac { 5 x ^ { 2 } } { V ^ { 2 } }$.\\
The object passes through points with coordinates $( a , - a )$ and $\left( a ^ { 2 } , - 16 a \right)$, where $a$ is a positive constant.\\
(ii) Find the value of $a$.\\
(iii) Given that the object strikes the ground at the point where $x = 5 a$, find the height of $O$ above the ground .\\
\hfill \mbox{\textit{CAIE M2 2018 Q4 [8]}}