| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.8 This is a multi-part projectiles question requiring algebraic manipulation to find k from two simultaneous conditions, conversion between trajectory parameters and angle of projection, and then kinematic calculations. While the individual techniques are standard M2 content, the question requires careful algebraic work with the trajectory equation, solving a system to find k, and multiple follow-up calculations involving parametric equations of motion. The 11 marks and three-part structure with non-trivial algebra place it moderately above average difficulty. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks |
|---|---|
| (ii) | a = 14k – 0.8(1 + k2) and |
| Answer | Marks |
|---|---|
| t (= 0.89442..) = 0.894s | M1 |
| Answer | Marks |
|---|---|
| A1 | 5 |
| 2 | Creates 2 simultaneous equations |
| Answer | Marks | Guidance |
|---|---|---|
| Page 6 | Mark Scheme | Syllabus |
| Cambridge International A Level – October/November 2016 | 9709 | 53 |
| (iii) | V = 35sin63.4 – g[42/(35cos63.4)] |
| Answer | Marks |
|---|---|
| α = 15.9° | M1 |
| Answer | Marks |
|---|---|
| A1 | 4 |
| 4 | V = 4.495 |
Question 7:
--- 7 (i)
(ii) ---
7 (i)
(ii) | a = 14k – 0.8(1 + k2) and
2a = 42k – 7.2(1 + k2)
42k – 7.2(1 + k2) = 2[14k – 0.8(1 + k2)]
k = 1/2 and 2
θ = tan–1k
θ = 63.435 AG
t = 14/(35cos63.435)
t (= 0.89442..) = 0.894s | M1
M1
B1
M1
A1
M1
A1 | 5
2 | Creates 2 simultaneous equations
Creates a single equation in k
Both values
With 1 of the candidates value of k
Page 6 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2016 | 9709 | 53
(iii) | V = 35sin63.4 – g[42/(35cos63.4)]
v
tanα = 4.495/(35cos63.4)
α = 15.9° above the horizontal
V2 = 4.4952 + (35cos63.4)2
V = 16.3 ms–1
OR
2a = 48
V2 = 352 – 2g x 48
V = 16.3 ms–1
cosα = 35cos63.435/16.3
α = 15.9° | M1
A1
M1
A1
M1
A1
M1
A1 | 4
4 | V = 4.495
v
Accept 16(.0)°
42 x 2 – 7.2(1 + 22)A particle $P$ is projected with speed 35 m s$^{-1}$ from a point $O$ on a horizontal plane. In the subsequent motion, the horizontal and vertically upwards displacements of $P$ from $O$ are $x$ m and $y$ m respectively. The equation of the trajectory of $P$ is
$$y = kx - \frac{(1 + k^2)x^2}{245},$$
where $k$ is a constant. $P$ passes through the points $A(14, a)$ and $B(42, 2a)$, where $a$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Calculate the two possible values of $k$ and hence show that the larger of the two possible angles of projection is 63.435°, correct to 3 decimal places. [5]
\end{enumerate}
For the larger angle of projection, calculate
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item the time after projection when $P$ passes through $A$, [2]
\item the speed and direction of motion of $P$ when it passes through $B$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2016 Q7 [11]}}