| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Relative velocity: find resultant velocity (magnitude and/or direction) |
| Difficulty | Challenging +1.8 This M4 relative velocity problem requires setting up and solving simultaneous equations from two vector magnitude conditions, involving algebraic manipulation of dot products and the cosine rule. It demands conceptual understanding beyond routine application, but follows a systematic approach once the vector relationships are established. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((\mathbf{v}_P - \mathbf{v}_Q)^2 = \mathbf{v}_P^2\) ① | M1 A1 | |
| \((\mathbf{v}_P + \mathbf{v}_Q)^2 = 4\mathbf{v}_P^2\) ② | M1 A1 | |
| \(4\mathbf{v}_P \cdot \mathbf{v}_Q = 3\mathbf{v}_P^2\) ② \(-\) ① | M1 A1 | |
| From ①: \(2\mathbf{v}_P \cdot \mathbf{v}_Q = \mathbf{v}_Q^2\) ③ | M1 A1 | |
| \(\therefore \sqrt{\dfrac{2}{3}} = \dfrac{ | \mathbf{v}_P | }{ |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| From ③: \(2 | \mathbf{v}_P | |
| \(\cos\theta = \dfrac{1}{2}\sqrt{\dfrac{3}{2}} = \dfrac{\sqrt{6}}{4}\) | A1 | (3), (12) |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(\mathbf{v}_P - \mathbf{v}_Q)^2 = \mathbf{v}_P^2$ ① | M1 A1 | |
| $(\mathbf{v}_P + \mathbf{v}_Q)^2 = 4\mathbf{v}_P^2$ ② | M1 A1 | |
| $4\mathbf{v}_P \cdot \mathbf{v}_Q = 3\mathbf{v}_P^2$ ② $-$ ① | M1 A1 | |
| From ①: $2\mathbf{v}_P \cdot \mathbf{v}_Q = \mathbf{v}_Q^2$ ③ | M1 A1 | |
| $\therefore \sqrt{\dfrac{2}{3}} = \dfrac{|\mathbf{v}_P|}{|\mathbf{v}_Q|}$ | A1 | **(9)** |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| From ③: $2|\mathbf{v}_P||\mathbf{v}_Q|\cos\theta = |\mathbf{v}_Q|^2$ | M1 A1 | |
| $\cos\theta = \dfrac{1}{2}\sqrt{\dfrac{3}{2}} = \dfrac{\sqrt{6}}{4}$ | A1 | **(3), (12)** |
---6. Two particles $P$ and $Q$ have constant velocity vectors $\mathbf { v } _ { P }$ and $\mathbf { v } _ { Q }$ respectively. The magnitude of the velocity of $P$ relative to $Q$ is equal to the speed of $P$. If the direction of motion of one of the particles is reversed, the magnitude of the velocity of $P$ relative to $Q$ is doubled.
Find
\begin{enumerate}[label=(\alph*)]
\item the ratio of the speeds of $P$ and $Q$,
\item the cosine of the angle between the directions of motion of $P$ and $Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 Q6 [12]}}