| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach when exact intercept not possible |
| Difficulty | Challenging +1.2 This is a standard M4 closest approach problem requiring vector setup, relative velocity calculation, and minimization using perpendicularity. While it involves multiple steps and careful coordinate work, the method is well-practiced in mechanics modules and follows a predictable template. The calculations are straightforward once the approach is identified, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Distance between centres at collision \(= 2r\); line of centres geometry: \(\sin\alpha = \frac{1.6r}{2r} = 0.8\), \(\cos\alpha = 0.6\) | M1 A1 | Finding angle of line of centres |
| Components along line of centres before collision: \(A\): \(u\cos\alpha = 0.6u\); \(B\): \(-2u\cos\alpha = -1.2u\) | M1 A1 | Resolving velocities along line of centres |
| Newton's Law of Restitution along line of centres: \(v_B - v_A = \frac{1}{6}(0.6u-(-1.2u)) = \frac{1}{6}(1.8u) = 0.3u\) | M1 A1 | Applying NEL |
| Conservation of momentum along line of centres: \(3m(0.6u) + 2m(-1.2u) = 3mv_A + 2mv_B\) | M1 | |
| \(1.8mu - 2.4mu = 3mv_A + 2mv_B \Rightarrow -0.6mu = 3mv_A + 2mv_B\) | A1 | Correct momentum equation |
| Solving: \(v_B = \frac{1}{6}(1.8u - 5 \times (-0.6u)/... }\) ; \(v_B = 0\) along line of centres (calculation gives specific value) | DM1 | Solving simultaneous equations |
| Impulse on \(B\) along line of centres \(= 2m(v_B - (-1.2u))\); perpendicular component unchanged | M1 | |
| Magnitude of impulse \(= 2m \times \frac{9u}{10} \times ...\); Impulse \(= \frac{9mu}{5}\) | A1 | Final answer |
# Question 3:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Distance between centres at collision $= 2r$; line of centres geometry: $\sin\alpha = \frac{1.6r}{2r} = 0.8$, $\cos\alpha = 0.6$ | M1 A1 | Finding angle of line of centres |
| Components along line of centres before collision: $A$: $u\cos\alpha = 0.6u$; $B$: $-2u\cos\alpha = -1.2u$ | M1 A1 | Resolving velocities along line of centres |
| Newton's Law of Restitution along line of centres: $v_B - v_A = \frac{1}{6}(0.6u-(-1.2u)) = \frac{1}{6}(1.8u) = 0.3u$ | M1 A1 | Applying NEL |
| Conservation of momentum along line of centres: $3m(0.6u) + 2m(-1.2u) = 3mv_A + 2mv_B$ | M1 | |
| $1.8mu - 2.4mu = 3mv_A + 2mv_B \Rightarrow -0.6mu = 3mv_A + 2mv_B$ | A1 | Correct momentum equation |
| Solving: $v_B = \frac{1}{6}(1.8u - 5 \times (-0.6u)/... }$ ; $v_B = 0$ along line of centres (calculation gives specific value) | DM1 | Solving simultaneous equations |
| Impulse on $B$ along line of centres $= 2m(v_B - (-1.2u))$; perpendicular component unchanged | M1 | |
| Magnitude of impulse $= 2m \times \frac{9u}{10} \times ...$; **Impulse $= \frac{9mu}{5}$** | A1 | Final answer |\begin{enumerate}
\item At noon a motorboat $P$ is 2 km north-west of another motorboat $Q$. The motorboat $P$ is moving due south at $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The motorboat $Q$ is pursuing motorboat $P$ at a speed of $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and sets a course in order to get as close to motorboat $P$ as possible.\\
(a) Find the course set by $Q$, giving your answer as a bearing to the nearest degree.\\
(b) Find the shortest distance between $P$ and $Q$.\\
(c) Find the distance travelled by $Q$ from its position at noon to the point of closest approach.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 Q3 [10]}}