| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projection from elevated point - angle below horizontal or horizontal |
| Difficulty | Standard +0.3 This is a standard projectile motion problem with straightforward application of kinematic equations. Given time, horizontal distance, and vertical displacement, students use s = ut + ½at² in both directions to find two unknowns (speed and angle). Part (ii) requires finding velocity components and calculating the angle, which is routine. The problem involves multiple steps but uses well-practiced techniques with no conceptual surprises, making it slightly easier than average. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks |
|---|---|
| 5 (i) | vcosθ = 8/2 |
| Answer | Marks |
|---|---|
| v = 5 ms−1 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | Accept with sign errors | |
| Page 5 | Mark Scheme | Syllabus |
| Cambridge International A Level – May/June 2016 | 9709 | 53 |
| Qu | Answer | Part |
| Marks | Marks | Notes |
| (ii) | θ = 36.9 |
| Answer | Marks |
|---|---|
| = 9.9 with the vertical | A1 |
| Answer | Marks |
|---|---|
| A1 | 6 |
| Answer | Marks |
|---|---|
| (ii) | 2cos45 + 2 × 3/5 = 0.4ω2× 0.3 |
| Answer | Marks |
|---|---|
| v = 1.61 ms−1 | M1 |
| Answer | Marks |
|---|---|
| A1 | 3 |
| Answer | Marks |
|---|---|
| 4 | Uses N2L with 2 components of T and |
| Answer | Marks |
|---|---|
| (iii) | 12(1.6–1.2)/1.2 = mgsin30 |
| Answer | Marks |
|---|---|
| v = 1.5 ms−1 | M1 |
| Answer | Marks |
|---|---|
| A1 | 2 |
| Answer | Marks |
|---|---|
| 5 | Uses T = λext/l |
Question 5:
--- 5 (i) ---
5 (i) | vcosθ = 8/2
–26 = –2vsinθ – g22 /2
vsinθ = 3
v2 = (+/–3)2 + 42 or tanθ = 3/4
v = 5 ms−1 | B1
M1
A1
M1
A1 | Accept with sign errors
Page 5 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – May/June 2016 | 9709 | 53
Qu | Answer | Part
Marks | Marks | Notes
(ii) | θ = 36.9
v = (+/–3) + 2g
v
tanα = v /(8/2)
v
Angle = 80.1 with the horizontal or angle
= 9.9 with the vertical | A1
M1
M1
A1 | 6
3
6 (i) (a)
(i) (b)
(ii) | 2cos45 + 2 × 3/5 = 0.4ω2× 0.3
rads−1
ω = 4.67
R + 2sin45 + 2 × 4/5 = 0.4 g
R = 0.986 N
Tsin45 + T(4/5) = 0.4 g
T = 2.65
Tcos45 + T(3/5) = 0.4v2 /0.3
v = 1.61 ms−1 | M1
A1
A1
M1
A1
M1
A1
M1
A1 | 3
2
4 | Uses N2L with 2 components of T and
0.3ω2
accn =
2.654
7 (i)
(ii)
(iii) | 12(1.6–1.2)/1.2 = mgsin30
m = 0.8 kg
PE change = 1.6
IKE + 12 × 0.42 /2.4 =
1.6 × 0.2gsin30 + 12 × 0.22 /2.4
IKE = 1 J AG
12e/1.2 = 1.6 g sin30
e = 0.8
1.6v2 /2 + 12 × 0.82 /2.4 =
1.6g × 0.6sin30 + 12 × 0.22 /2.4
v = 1.5 ms−1 | M1
A1
B1
B1
M1
A1
M1
A1
M1
A1
A1 | 2
4
5 | Uses T = λext/l
2 × ans(i)
Both EE terms correct
KE/PE/EE balance
Both EE terms correct
λe × t/l = new weight component
May be stated without explanation
Must use new equilibrium positionA particle is projected at an angle of $θ°$ below the horizontal from a point at the top of a vertical cliff $26$ m high. The particle strikes horizontal ground at a distance $8$ m from the foot of the cliff $2$ s after the instant of projection. Find
\begin{enumerate}[label=(\roman*)]
\item the speed of projection of the particle and the value of $θ$, [6]
\item the direction of motion of the particle immediately before it strikes the ground. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2016 Q5 [9]}}