| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Angle between two vectors |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring differentiation of position vectors to find velocity and acceleration, then applying dot product formulas. The steps are routine: differentiate twice, substitute t=2, use cos θ = (a·b)/(|a||b|), and set dot product to zero. While it involves multiple techniques, each is standard M2 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10h Vectors in kinematics: uniform acceleration in vector form3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks |
|---|---|
| 5 (i) | xtanα = 0 so α = 0 |
| Answer | Marks |
|---|---|
| V = 10 m s | B1 |
| Answer | Marks |
|---|---|
| [3] | Justification needed |
| Answer | Marks |
|---|---|
| (ii) | dy |
| Answer | Marks |
|---|---|
| y = 15 (below) or –15 | M1 |
| Answer | Marks |
|---|---|
| A1 | Uses Pythagoras |
Question 5:
--- 5 (i) ---
5 (i) | xtanα = 0 so α = 0
gx2
=0.05x2
2V2cos20
–1
V = 10 m s | B1
M1
A1
[3] | Justification needed
Comparison with standard eqn
(ii) | dy
= –0.1x
dx
–0.1x = –tan60
2
y (= –0.05(10tan60) ) = –15
2
v =
2
10 + 2g15
–1
v = 20 m s
OR
y' = 10tan 60
2
(10√3) =2gh
y = –15
2
v =
2 2
10 + (10√3)
–1
v = 20 m s
OR
vcos60 = 10
–1
v = 20 m s
10 3 = 10t
t = 3
3
y = 10 3×
2
y = 15 (below) or –15 | M1
M1
A1
M1
A1
A1
[6]
M1
M1
A1
M1
A1
A1
M1
A1
M1
A1
M1
A1 | Uses Pythagoras
ft candidate’s value (V(i), y)
y' = B’s downward velocity =10√3
Negative, y = –h
Uses Pythagoras
ft candidate’s value (V(i))The position vector of a particle at time $t$ is given by $\mathbf{r} = t^2\mathbf{i} + (3t - 1)\mathbf{j}$, where $\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors.
Find the velocity and acceleration of the particle when $t = 2$.
\begin{enumerate}[label=(\roman*)]
\item Hence find the angle between the velocity and acceleration vectors when $t = 2$. [3]
\item Find the value of $t$ for which the velocity and acceleration vectors are perpendicular. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2014 Q5 [7]}}