| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile clearing obstacle |
| Difficulty | Standard +0.3 This is a standard two-part projectile question requiring routine application of kinematic equations and trajectory derivation. Part (i) involves straightforward substitution to derive the trajectory equation (a common textbook exercise), while part (ii) requires substituting x=1.5 into the trajectory equation and solving a quadratic for the range. The calculations are mechanical with no novel problem-solving insight required, making it slightly easier than average. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = (15\cos41)t\) | B1 | |
| \(y = (15\sin41)t - gt^2/2\) | B1 | |
| \(y = (15\sin41)x/(15\cos41) - 5[x/(15\cos41)]^2\) | M1 | |
| \(y = 0.869x - 0.0390x^2\) | A1 [4] | \(y = 0.86928..x - 0.03901..x^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H = 0.869\times1.5 - 0.039\times1.5^2 + 1.6\) | M1 | Must add height of \(O\) |
| \(H = 2.82\text{ m}\) | A1 | |
| \(0.039x^2 - 0.869x - 1.6 = 0\) | M1 | Uses \(y = -1.6\) and tries to solve |
| \(D = 23.99 - 1.5\) | DM1 | Solve a 3 term quadratic equation and minus \(1.5\) |
| \(D = 22.5\text{ m}\) | A1 [5] | Accept \(22.4\) |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = (15\cos41)t$ | B1 | |
| $y = (15\sin41)t - gt^2/2$ | B1 | |
| $y = (15\sin41)x/(15\cos41) - 5[x/(15\cos41)]^2$ | M1 | |
| $y = 0.869x - 0.0390x^2$ | A1 [4] | $y = 0.86928..x - 0.03901..x^2$ |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H = 0.869\times1.5 - 0.039\times1.5^2 + 1.6$ | M1 | Must add height of $O$ |
| $H = 2.82\text{ m}$ | A1 | |
| $0.039x^2 - 0.869x - 1.6 = 0$ | M1 | Uses $y = -1.6$ and tries to solve |
| $D = 23.99 - 1.5$ | DM1 | Solve a 3 term quadratic equation and minus $1.5$ |
| $D = 22.5\text{ m}$ | A1 [5] | Accept $22.4$ |7 A small ball $B$ is projected with speed $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $41 ^ { \circ }$ above the horizontal from a point $O$ which is 1.6 m above horizontal ground. At time $t \mathrm {~s}$ after projection the horizontal and vertically upward displacements of $B$ 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 $B$ is
$$y = 0.869 x - 0.0390 x ^ { 2 }$$
where the coefficients are correct to 3 significant figures.
A vertical fence is 1.5 m from $O$ and perpendicular to the plane in which $B$ moves. $B$ just passes over the fence and subsequently strikes the ground at the point $A$.\\
(ii) Calculate the height of the fence, and the distance from the fence to $A$.
\hfill \mbox{\textit{CAIE M2 2012 Q7 [9]}}