| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, direction deflected given angle |
| Difficulty | Challenging +1.2 This is a standard M3 oblique collision problem requiring systematic application of momentum conservation (parallel and perpendicular to line of centres) and Newton's experimental law. While it involves multiple steps and careful component resolution, the method is well-practiced and follows a predictable template. The given constraint (B's final direction perpendicular to initial) provides helpful structure. Slightly above average difficulty due to the algebraic manipulation required across three parts, but remains a textbook-style question. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Velocity unchanged perpendicular to line of centres | M1 | Stated or used |
| \(0.6\sin 30° = v\cos 30°\) | M1 | Allow 1 sign or trig error |
| \(0.2\sqrt{3}\) ms\(^{-1}\) | A1 | (0.34641) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use momentum equation | M1 | Allow their \(v\); allow sign errors / omission of \(m\) |
| \(0.3m - 0.6m\cos 30° = am + 0.2\sqrt{3}m\cos 60°\) | A1ft | Follow through on \(v\); \(m\)'s not necessary |
| \((a =)\ 0.393\) to left | A1 [3] | Direction must be clearly stated or implied from working. WWW; (0.39282); Away from B/opposite direction to before |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of NLR | M1 | Allow sign error and/or trig error |
| \((0.2\sqrt{3})\cos 60° - (-0.393) = e(0.6\cos 30° + 0.3)\) | A1ft | Follow through on \(a\) and \(v\) |
| \(0.691\) | A1 [3] | CAO (0.69082 or 0.6905679) |
# Question 2:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Velocity unchanged perpendicular to line of centres | M1 | Stated or used |
| $0.6\sin 30° = v\cos 30°$ | M1 | Allow 1 sign or trig error |
| $0.2\sqrt{3}$ ms$^{-1}$ | A1 | (0.34641) |
| **[3]** | | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use momentum equation | M1 | Allow their $v$; allow sign errors / omission of $m$ |
| $0.3m - 0.6m\cos 30° = am + 0.2\sqrt{3}m\cos 60°$ | A1ft | Follow through on $v$; $m$'s not necessary |
| $(a =)\ 0.393$ to left | A1 **[3]** | Direction must be clearly stated or implied from working. WWW; (0.39282); Away from B/opposite direction to before |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of NLR | M1 | Allow sign error and/or trig error |
| $(0.2\sqrt{3})\cos 60° - (-0.393) = e(0.6\cos 30° + 0.3)$ | A1ft | Follow through on $a$ and $v$ |
| $0.691$ | A1 **[3]** | CAO (0.69082 or 0.6905679) |
---2 Two uniform smooth spheres $A$ and $B$, of equal radius and equal mass, are moving towards each other on a horizontal surface. Immediately before they collide, $A$ has speed $0.3 \mathrm {~ms} ^ { - 1 }$ along the line of centres and $B$ has speed $0.6 \mathrm {~ms} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ to the line of centres (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{dfe477d4-eae6-40e1-b704-1a97485f4c7e-2_302_1013_1247_502}
After the collision, the direction of motion of $B$ is at right angles to its original direction of motion. Find\\
(i) the speed of $B$ after the collision,\\
(ii) the speed and direction of motion of $A$ after the collision,\\
(iii) the coefficient of restitution between $A$ and $B$.
\hfill \mbox{\textit{OCR M3 2013 Q2 [9]}}