| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Easy -1.2 This is a straightforward projectiles question testing basic SUVAT equations and interpretation of a velocity-time graph. Part (a) requires standard application of v²=u²+2as to find maximum height (a routine 3-mark calculation). Parts (b) and (c) involve reading values from a given graph and using symmetry of projectile motion. All techniques are standard textbook exercises with no problem-solving insight required, making this easier than average for A-level. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02h Motion under gravity: vector form |
| Answer | Marks |
|---|---|
| (ii) | –2 |
| Answer | Marks |
|---|---|
| T = 0.9 | M1 |
| Answer | Marks |
|---|---|
| A1 | [4] |
| [3] | For using Newton’s second law for both |
| Answer | Marks | Guidance |
|---|---|---|
| Page 6 | Mark Scheme | Syllabus |
| GCE A LEVEL – May/June 2014 | 9709 | 41 |
| (iii) | 1 |
| Answer | Marks |
|---|---|
| h = 1.1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| B1 | [3] | For using area property in graph or |
| Answer | Marks |
|---|---|
| (iii) | 1 |
| Answer | Marks |
|---|---|
| a = 0.0262 | M1 |
| Answer | Marks |
|---|---|
| A1 | [3] |
| Answer | Marks |
|---|---|
| [4] | 1 |
Question 6:
--- 6 (i)
(ii) ---
6 (i)
(ii) | –2
Acceleration is 5ms
Distance is 0.9m
1
0.6 × V = 0.9 → V = 3
2
T = 0.9 | M1
A1
M1
A1
B1
M1
A1 | [4]
[3] | For using Newton’s second law for both
particles and eliminating T, or using
(M + m)a = (M – m)g
1
2
For using s = 0 + at
2
ft distance in (i)
For using
0 = V – g(T – 0.6)
Page 6 | Mark Scheme | Syllabus | Paper
GCE A LEVEL – May/June 2014 | 9709 | 41
(iii) | 1
[sup = 0.9 × 3 and
2
1
2
sdown = 0 + g(1.6 – 0.9) ]
2
Distance upwards is 1.35m and
distance downwards is 2.45m
h = 1.1 | M1
A1
B1 | [3] | For using area property in graph or
equivalent
ft sdown – sup
7 (i)
(ii)
(iii) | 1
2
AB = 3 × 400 + 0.005 × 400 = 1600m
2
(AG)
or
–1
vB = 3 + 0.005 × 400 = 5 ms
–1
vB = 3 + 0.005 × 400 = 5 ms
or
1
2
AB = 3 × 400 + 0.005 × 400 = 1600m
2
(AG)
2 3 400
[0.02t – 0.0001t /3 + kt]0 = 1600
400k = 1600 – 0.02 × 400 2 +
0.0001 × 400 3 ÷ 3 →
k = 4 – 8 + 16/3 = 4/3
[dv/dt = 0.04 – 0.0002t
(= 0 when t = 200)
2
vmax = 0.04 × 200 – 0.0001 × 200 + 4/3
–1
Maximum speed is 5.33ms
Time taken is 280s
1
2
[1400 = 4/3 × 280 + 280 a]
2
a = 0.0262 | M1
A1
B1
M1
A1
A1
M1
A1
A1
M1
A1
M1
A1 | [3]
[6]
[4] | 1
2
For using s = ut + at to find the
2
distance AB, or for using v = u + at to
find P’s speed at B
For using ʃ 0 400 v dt = 1600
For differentiating and solving dv/dt = 0
ft incorrect k or incorrect value of t from
dv/dt = 0
–1
For using constant speed 5ms = 1400/T
1
2
For using s = ut + at to find a
2\includegraphics{figure_6}
A particle is projected vertically upward from ground level with speed $u$ m s$^{-1}$. The particle moves under gravity alone.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the maximum height reached by the particle. [3]
\end{enumerate}
\includegraphics{figure_6b}
The diagram shows a velocity-time graph for the motion of the particle.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Use the graph to find the value of $u$. [2]
\item Find the time taken for the particle to return to ground level. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2014 Q6 [8]}}