| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Three-particle sequential collisions |
| Difficulty | Standard +0.3 This is a standard M2 sequential collision problem requiring conservation of momentum and Newton's restitution law applied twice. Part (i) is routine calculation (showing a given result). Part (ii) requires setting up an inequality condition but follows a well-established method. Slightly above average due to the two-collision setup and inequality reasoning, but this is a textbook M2 question type. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.2 \times 1.8 = 0.2v_A + 0.4v_B\) | *M1 | Attempt at conservation of momentum |
| A1 | ||
| \(v_B - v_A = \frac{1}{3} \times 1.8\) | *M1 | Attempt at restitution |
| A1 | aef | |
| Solve for \(v_A\) or \(v_B\) | Dep*M1 | |
| \(v_B = 0.8 \text{ ms}^{-1}\) and \(v_A = 0.2 \text{ ms}^{-1}\) AG | A1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.4 \times 0.8 + 0.6 \times 0.2 = 0.4v_{B'} + 0.6v_C\) | B1 | |
| \(v_C - v_{B'} = e(0.8 - 0.2)\) | B1 | aef |
| Use two relevant equations to eliminate \(v_C\) | *M1 | |
| State \(v_{B'} \geq 0.2\) | B1 | soi, Allow \(v_{B'} > 0.2\) |
| Set up (in)equality in \(e\) and their \(v_A\) | dep*M1 | Condone incorrect inequality sign for M1 only |
| \(0.44 - 0.36e \geq 0.2\) or \(0.44 - 0.36e = 0.2\) | A1 | Allow \(0.44 - 0.36e > 0.2\) |
| \(e \leq \frac{2}{3}\) or \(0.667\) | A1 | |
| [7] | ||
| OR: \(0.4 \times 0.8 + 0.6 \times 0.2 = 0.4v_{B'} + 0.6v_C\) | B1 | |
| \(v_C - v_{B'} = e(0.8 - 0.2)\) | B1 | aef |
| State \(v_{B'} \geq 0.2\) | B1 | soi, Allow \(v_{B'} > 0.2\) |
| Sub \(v_{B'}\) in momentum equation & solve for \(v_C\) | *M1 | |
| \((v_C =) 0.6\) | A1 | |
| Set up (in)equality in \(e\) and their \(v_A\) | dep*M1 | e.g. \(0.6 - e(0.8-0.2) \geq 0.2\), Condone incorrect inequality sign for M1 only |
| \(e \leq \frac{2}{3}\) or \(0.667\) | A1 | |
| [7] |
## Question 6:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.2 \times 1.8 = 0.2v_A + 0.4v_B$ | *M1 | Attempt at conservation of momentum |
| | A1 | |
| $v_B - v_A = \frac{1}{3} \times 1.8$ | *M1 | Attempt at restitution |
| | A1 | aef |
| Solve for $v_A$ or $v_B$ | Dep*M1 | |
| $v_B = 0.8 \text{ ms}^{-1}$ and $v_A = 0.2 \text{ ms}^{-1}$ **AG** | A1 | |
| **[6]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.4 \times 0.8 + 0.6 \times 0.2 = 0.4v_{B'} + 0.6v_C$ | B1 | |
| $v_C - v_{B'} = e(0.8 - 0.2)$ | B1 | aef |
| Use two relevant equations to eliminate $v_C$ | *M1 | |
| State $v_{B'} \geq 0.2$ | B1 | soi, Allow $v_{B'} > 0.2$ |
| Set up (in)equality in $e$ and their $v_A$ | dep*M1 | Condone incorrect inequality sign for M1 only |
| $0.44 - 0.36e \geq 0.2$ or $0.44 - 0.36e = 0.2$ | A1 | Allow $0.44 - 0.36e > 0.2$ |
| $e \leq \frac{2}{3}$ or $0.667$ | A1 | |
| **[7]** | | |
| OR: $0.4 \times 0.8 + 0.6 \times 0.2 = 0.4v_{B'} + 0.6v_C$ | B1 | |
| $v_C - v_{B'} = e(0.8 - 0.2)$ | B1 | aef |
| State $v_{B'} \geq 0.2$ | B1 | soi, Allow $v_{B'} > 0.2$ |
| Sub $v_{B'}$ in momentum equation & solve for $v_C$ | *M1 | |
| $(v_C =) 0.6$ | A1 | |
| Set up (in)equality in $e$ and their $v_A$ | dep*M1 | e.g. $0.6 - e(0.8-0.2) \geq 0.2$, Condone incorrect inequality sign for M1 only |
| $e \leq \frac{2}{3}$ or $0.667$ | A1 | |
| **[7]** | | |
---6 Three particles $A , B$ and $C$ are in a straight line on a smooth horizontal surface. The particles have masses $0.2 \mathrm {~kg} , 0.4 \mathrm {~kg}$ and 0.6 kg respectively. $B$ is at rest. $A$ is projected towards $B$ with a speed of $1.8 \mathrm {~ms} ^ { - 1 }$ and collides with $B$. The coefficient of restitution between $A$ and $B$ is $\frac { 1 } { 3 }$.\\
(i) Show that the speed of $B$ after the collision is $0.8 \mathrm {~ms} ^ { - 1 }$ and find the speed of $A$ after the collision.\\
$C$ is moving with speed $0.2 \mathrm {~ms} ^ { - 1 }$ in the same direction as $B$. Particle $B$ subsequently collides with $C$. The coefficient of restitution between $B$ and $C$ is $e$.\\
(ii) Find the set of values for $e$ such that $B$ does not collide again with $A$.
\hfill \mbox{\textit{OCR M2 2012 Q6 [13]}}