| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with energy loss |
| Difficulty | Standard +0.3 This is a standard M2 collision problem requiring systematic application of conservation of momentum, Newton's law of restitution, and impulse-momentum theorem. While it has three parts and involves algebraic manipulation with the parameter m, each step follows a well-established procedure taught in mechanics courses. The calculations are straightforward with no conceptual surprises, making it slightly easier than the average A-level question. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation6.03j Perfectly elastic/inelastic: collisions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((2m)(4) - (3m)(2) = 2mv_A + 3mv_B\) | *M1 | Attempt at use of conservation of momentum |
| A1 | ||
| \((v_B - v_A)/(4 - -2) = 0.4\) | *M1 | Attempt at use of coefficient of restitution |
| A1 | ||
| Speed \(A = 1.04\) m s\(^{-1}\), Speed \(B = 1.36\) m s\(^{-1}\) | Dep**M1 | Solving for \(v_A\) and \(v_B\) |
| A1 | Final answers must be positive | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Energy before \(= \frac{1}{2}(2m)(4^2) + \frac{1}{2}(3m)(2^2)\) | B1ft | Energy before or Loss in \(A\)'s KE |
| Energy after \(= \frac{1}{2}(2m)(1.04^2) + \frac{1}{2}(3m)(1.36^2)\) | B1ft | Energy after or Loss in \(B\)'s KE |
| \(22m - 3.856m\) | M1 | Difference of total OR sum of differences (total kinetic energy must decrease) |
| \(18.1m\) | A1 | \(18.144m\) (Exact) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{1}{2}\dfrac{m_1 m_2}{m_1+m_2}(1-e^2)A^2\) | *B1 | Loss of kinetic energy formula, where \(A\) = approach speed |
| \(\dfrac{1}{2}\dfrac{(2m)(3m)}{2m+3m}(1-0.4^2)(4+2)^2\) | Dep*M1 | Substitution of values into quoted formula |
| \(18.1m\) | A1 | \(18.144m\) (Exact) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2m(4) - 2m(-1.04) = 2.52\) | M1 | Attempt at change in momentum and equate to impulse. Must use \(2m\) or \(3m\) |
| A1ft | Or \(3m(2) - 3m(-1.36) = 2.52\) | |
| \(m = 0.25\) | A1 | Exact |
| [3] |
## Question 6:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2m)(4) - (3m)(2) = 2mv_A + 3mv_B$ | *M1 | Attempt at use of conservation of momentum |
| | A1 | |
| $(v_B - v_A)/(4 - -2) = 0.4$ | *M1 | Attempt at use of coefficient of restitution |
| | A1 | |
| Speed $A = 1.04$ m s$^{-1}$, Speed $B = 1.36$ m s$^{-1}$ | Dep**M1 | Solving for $v_A$ and $v_B$ |
| | A1 | Final answers must be positive |
| **[6]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Energy before $= \frac{1}{2}(2m)(4^2) + \frac{1}{2}(3m)(2^2)$ | B1ft | Energy before or Loss in $A$'s KE |
| Energy after $= \frac{1}{2}(2m)(1.04^2) + \frac{1}{2}(3m)(1.36^2)$ | B1ft | Energy after or Loss in $B$'s KE |
| $22m - 3.856m$ | M1 | Difference of total OR sum of differences (total kinetic energy must decrease) |
| $18.1m$ | A1 | $18.144m$ (Exact) |
| **[4]** | | |
**OR**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{1}{2}\dfrac{m_1 m_2}{m_1+m_2}(1-e^2)A^2$ | *B1 | Loss of kinetic energy formula, where $A$ = approach speed |
| $\dfrac{1}{2}\dfrac{(2m)(3m)}{2m+3m}(1-0.4^2)(4+2)^2$ | Dep*M1 | Substitution of values into quoted formula |
| $18.1m$ | A1 | $18.144m$ (Exact) |
| **[4]** | | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2m(4) - 2m(-1.04) = 2.52$ | M1 | Attempt at change in momentum and equate to impulse. Must use $2m$ or $3m$ |
| | A1ft | Or $3m(2) - 3m(-1.36) = 2.52$ |
| $m = 0.25$ | A1 | Exact |
| **[3]** | | |
---6 Two small spheres $A$ and $B$, of masses $2 m \mathrm {~kg}$ and $3 m \mathrm {~kg}$ respectively, are moving in opposite directions along the same straight line towards each other on a smooth horizontal surface. $A$ has speed $4 \mathrm {~ms} ^ { - 1 }$ and $B$ has speed $2 \mathrm {~ms} ^ { - 1 }$ before they collide. The coefficient of restitution between $A$ and $B$ is 0.4 .\\
(i) Find the speed of each sphere after the collision.\\
(ii) Find, in terms of $m$, the loss of kinetic energy during the collision.\\
(iii) Given that the magnitude of the impulse exerted on $A$ by $B$ during the collision is 2.52 Ns , find $m$.
\hfill \mbox{\textit{OCR M2 2014 Q6 [13]}}