| 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 downward projection requiring application of kinematic equations in two dimensions. While it involves solving simultaneous equations and working with components, the setup is straightforward with all given values clearly stated, and the methods are routine for M2 level. The multi-step nature and need to find both speed and angle elevates it slightly above average, but it remains a textbook-style question without requiring novel insight. |
| 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 | 51 |
| 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 | 51
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 $26 \text{ m}$ high. The particle strikes horizontal ground at a distance $8 \text{ m}$ from the foot of the cliff $2 \text{ s}$ 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]}}