| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Moderate -0.8 This is a straightforward projectiles question requiring standard application of kinematic equations (x = ut, y = ½gt²) and elimination of t to derive the trajectory. Part (ii) involves routine Pythagoras and speed calculation. All steps are textbook-standard with no problem-solving insight required, making it easier than average. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(y = gt^2/2\) | B1 | Must be positive |
| \(x = 5t\) | M1 | |
| \(y [= 10(x/5)^2/2] = 0.2x^2\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(x^2 + y^2 = 18^2\) | M1* | Sets up and tries to solve a 3 term quadratic equation |
| \(x^2 + (0.2x^2)^2 = 18^2\) | D*M1 | |
| \(0.04(x^2)^2 + x^2 - 324 = 0\) | A1 | |
| \(x = 8.85\) | B1 | \(8.8523\ldots\) from \(x^2 = 78.3639\ldots\) |
| \(y = 0.2 \times 8.8523\ldots^2\) | M1 | \(y = 15.67\ldots\) |
| \(v^2 = 5^2 + 2g \times 15.67\ldots\) | ||
| \(v = 18.4 \text{ ms}^{-1}\) | A1 | [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(V^2 = 2g \times 0.2 \times 8.8523\ldots^2\) | B1 | \(V = 17.7046\) |
| \(v^2 = 5^2 + 17.7\ldots^2\) | M1 | |
| \(v = 18.4 \text{ ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(t = 8.85/5 = 1.77,\ V = 10 \times 1.77 = 17.7\) | B1 | |
| \(v^2 = 17.7^2 + 5^2\) | M1 | |
| \(V = 18.4 \text{ ms}^{-1}\) | A1 |
## Question 5:
### Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $y = gt^2/2$ | B1 | Must be positive |
| $x = 5t$ | M1 | |
| $y [= 10(x/5)^2/2] = 0.2x^2$ | A1 | [3] |
### Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $x^2 + y^2 = 18^2$ | M1* | Sets up and tries to solve a 3 term quadratic equation |
| $x^2 + (0.2x^2)^2 = 18^2$ | D*M1 | |
| $0.04(x^2)^2 + x^2 - 324 = 0$ | A1 | |
| $x = 8.85$ | B1 | $8.8523\ldots$ from $x^2 = 78.3639\ldots$ |
| $y = 0.2 \times 8.8523\ldots^2$ | M1 | $y = 15.67\ldots$ |
| $v^2 = 5^2 + 2g \times 15.67\ldots$ | | |
| $v = 18.4 \text{ ms}^{-1}$ | A1 | [6] |
**OR**
| Working | Mark | Guidance |
|---------|------|----------|
| $V^2 = 2g \times 0.2 \times 8.8523\ldots^2$ | B1 | $V = 17.7046$ |
| $v^2 = 5^2 + 17.7\ldots^2$ | M1 | |
| $v = 18.4 \text{ ms}^{-1}$ | A1 | |
**OR**
| Working | Mark | Guidance |
|---------|------|----------|
| $t = 8.85/5 = 1.77,\ V = 10 \times 1.77 = 17.7$ | B1 | |
| $v^2 = 17.7^2 + 5^2$ | M1 | |
| $V = 18.4 \text{ ms}^{-1}$ | A1 | |
---5 A small ball is thrown horizontally with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $O$ on the roof of a building. At time $t \mathrm {~s}$ after projection, the horizontal and vertically downwards displacements of the ball 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 trajectory of the ball is $y = 0.2 x ^ { 2 }$.
The ball strikes the horizontal ground which surrounds the building at a point $A$.\\
(ii) Given that $O A = 18 \mathrm {~m}$, calculate the value of $x$ at $A$, and the speed of the ball immediately before it strikes the ground at $A$.
\hfill \mbox{\textit{CAIE M2 2014 Q5 [9]}}