| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, vector velocity form |
| Difficulty | Standard +0.8 This is an M4 oblique collision problem requiring vector momentum conservation, impulse calculation, and Newton's experimental law in 2D. While methodical, it demands careful vector decomposition along the line of centres and understanding that impulses act along this line—more sophisticated than standard 1D collisions but follows established M4 techniques without requiring novel insight. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| \(C\) is due south of \(S\); \(S\) moves on bearing \(140°\) | B1 | Correct understanding of geometry |
| Minimum \(V\) when course of \(C\) is perpendicular to velocity of \(S\) relative to \(C\) | M1 | Correct method — \(V_{\min} = 12\sin 40°\) |
| \(V_{\min} = 12\sin 40° = 7.71\) km h\(^{-1}\) | A1 | Correct answer to 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Using sine rule in velocity triangle: \(\frac{\sin\theta}{12} = \frac{\sin 40°}{14}\) | M1 | Correct velocity triangle |
| \(\sin\theta = \frac{12\sin 40°}{14}\) | A1 | Correct equation |
| \(\theta = 33.1°\) or \(\theta = 146.9°\) | A1 | One or both values |
| Bearing \(= 360° - (50° - 33.1°) \) or equivalent consideration | M1 | Correct bearing calculation method |
| Bearing \(= 343°\) (or \(\approx 343°\)) | A1 | Correct final bearing |
## Question 5:
**(a)** Minimum value of $V$
| $C$ is due south of $S$; $S$ moves on bearing $140°$ | B1 | Correct understanding of geometry |
| Minimum $V$ when course of $C$ is perpendicular to velocity of $S$ relative to $C$ | M1 | Correct method — $V_{\min} = 12\sin 40°$ |
| $V_{\min} = 12\sin 40° = 7.71$ km h$^{-1}$ | A1 | Correct answer to 3 s.f. |
**(b)** Bearing when $V = 14$
| Using sine rule in velocity triangle: $\frac{\sin\theta}{12} = \frac{\sin 40°}{14}$ | M1 | Correct velocity triangle |
| $\sin\theta = \frac{12\sin 40°}{14}$ | A1 | Correct equation |
| $\theta = 33.1°$ or $\theta = 146.9°$ | A1 | One or both values |
| Bearing $= 360° - (50° - 33.1°) $ or equivalent consideration | M1 | Correct bearing calculation method |
| Bearing $= 343°$ (or $\approx 343°$) | A1 | Correct final bearing |\begin{enumerate}
\item Two small smooth spheres $A$ and $B$, of mass 2 kg and 1 kg respectively, are moving on a smooth horizontal plane when they collide. Immediately before the collision the velocity of $A$ is $( \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ and the velocity of $B$ is $- 2 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Immediately after the collision the velocity of $A$ is $\mathbf { j } \mathrm { m } \mathrm { s } ^ { - 1 }$.\\
(a) Show that the velocity of $B$ immediately after the collision is $2 \mathbf { j } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(b) Find the impulse of $B$ on $A$ in the collision, giving your answer as a vector, and hence show that the line of centres is parallel to $\mathbf { i } + \mathbf { j }$.\\
(c) Find the coefficient of restitution between $A$ and $B$.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 Q5 [8]}}