| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with speed relationships |
| Difficulty | Standard +0.8 This is a multi-part M2 collision problem requiring conservation of momentum, coefficient of restitution equations, and calculus optimization. Part (i) involves algebraic manipulation with two equations and two unknowns in terms of parameter e. Part (ii) requires differentiation to minimize speed. Part (iii) involves impulse calculations with a quadratic equation. The problem requires systematic application of standard mechanics principles across multiple steps, but follows predictable M2 patterns without requiring novel geometric insight or proof techniques. |
| Spec | 3.02i Projectile motion: constant acceleration model6.03b Conservation of momentum: 1D two particles6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4 - 4(1 - e + e^2) = -e(u - 4)\) | M1, A1 | Use of restitution, may have sign errors, must be correct ratio \((v/u)\) |
| \(u = 4e\) | A1 | oe |
| \(mu + 0.2 \times 4 = 0.2 \times 4(1 - e + e^2) + 4m\) | M1, A1 | Use of conservation of momentum |
| \(m = 0.2e\) | A1 | oe |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Valid method to find \(e\) that gives the least speed | M1 | Differentiate \(v_A\) and equate to 0 or complete the square on \(v_A\) |
| Get \(e = \frac{1}{2}\) | A1 | www |
| \(\frac{1}{2} \times 0.2 \times 4^2 + \frac{1}{2} \times 0.1 \times 2^2 - (\frac{1}{2} \times 0.2 \times 3^2 + \frac{1}{2} \times 0.1 \times 4^2)\) | M1, A1 | Difference of KE with 4 terms; must have found \(e\) from legitimate method |
| \((\pm)\, 0.1\) J | A1 | SCM1A1 Loss of KE \(= 8e(1-e)^3/5\) or \(8e(1-3e+3e^2-e^3)/5\) or \(8e/5 - 24e^2/5 + 24e^3/5 - 8e^4/5\) |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.2e(4 - 4e) = 0.192\) or \(0.2(4 - (4 - 4e + 4e^2)) = 0.192\) | *M1, A1 | Attempt to use impulse = change in momentum on one particle |
| Solve three term QE in \(e\) | dep*M1 | Method should lead to 2 real values for \(e\) |
| \(e = 0.4\) or \(0.6\) | A1 | For both |
| [4] |
## Question 6:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4 - 4(1 - e + e^2) = -e(u - 4)$ | M1, A1 | Use of restitution, may have sign errors, must be correct ratio $(v/u)$ |
| $u = 4e$ | A1 | oe |
| $mu + 0.2 \times 4 = 0.2 \times 4(1 - e + e^2) + 4m$ | M1, A1 | Use of conservation of momentum |
| $m = 0.2e$ | A1 | oe |
| **[6]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Valid method to find $e$ that gives the least speed | M1 | Differentiate $v_A$ and equate to 0 or complete the square on $v_A$ |
| Get $e = \frac{1}{2}$ | A1 | www |
| $\frac{1}{2} \times 0.2 \times 4^2 + \frac{1}{2} \times 0.1 \times 2^2 - (\frac{1}{2} \times 0.2 \times 3^2 + \frac{1}{2} \times 0.1 \times 4^2)$ | M1, A1 | Difference of KE with 4 terms; must have found $e$ from legitimate method |
| $(\pm)\, 0.1$ J | A1 | SCM1A1 Loss of KE $= 8e(1-e)^3/5$ or $8e(1-3e+3e^2-e^3)/5$ or $8e/5 - 24e^2/5 + 24e^3/5 - 8e^4/5$ |
| **[5]** | | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.2e(4 - 4e) = 0.192$ or $0.2(4 - (4 - 4e + 4e^2)) = 0.192$ | *M1, A1 | Attempt to use impulse = change in momentum on one particle |
| Solve three term QE in $e$ | dep*M1 | Method should lead to 2 real values for $e$ |
| $e = 0.4$ or $0.6$ | A1 | For both |
| **[4]** | | |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{74eaa61a-1507-4cef-8f97-5c1860bdc36a-4_243_1179_1580_443}
The masses of two particles $A$ and $B$ are 0.2 kg and $m \mathrm {~kg}$ respectively. The particles are moving with constant speeds $4 \mathrm {~ms} ^ { - 1 }$ and $u \mathrm {~ms} ^ { - 1 }$ in the same horizontal line and in the same direction (see diagram). The two particles collide and the coefficient of restitution between the particles is $e$. After the collision, $A$ and $B$ continue in the same direction with speeds $4 \left( 1 - e + e ^ { 2 } \right) \mathrm { ms } ^ { - 1 }$ and $4 \mathrm {~ms} ^ { - 1 }$ respectively.\\
(i) Find $u$ and $m$ in terms of $e$.\\
(ii) Find the value of $e$ for which the speed of $A$ after the collision is least and find, in this case, the total loss in kinetic energy due to the collision.\\
(iii) Find the possible values of $e$ for which the magnitude of the impulse that $B$ exerts on $A$ is 0.192 Ns .\\
\includegraphics[max width=\textwidth, alt={}, center]{74eaa61a-1507-4cef-8f97-5c1860bdc36a-5_744_887_264_589}
The diagram shows a surface consisting of a horizontal part $O A$ and a plane $A B$ inclined at an angle of $70 ^ { \circ }$ to the horizontal. A particle is projected from the point $O$ with speed $u \mathrm {~ms} ^ { - 1 }$ at an angle of $\theta ^ { \circ }$ above the horizontal $O A$. The particle hits the plane $A B$ at the point $P$, with speed $14 \mathrm {~ms} ^ { - 1 }$ and at right angles to the plane, 1.4 s after projection.\\
(i) Show that the value of $u$ is 15.9 , correct to 3 significant figures, and find the value of $\theta$.\\
(ii) Find the height of $P$ above the level of $A$.
The particle rebounds with speed $v \mathrm {~ms} ^ { - 1 }$. The particle next lands at $A$.\\
(iii) Find the value of $v$.\\
(iv) Find the coefficient of restitution between the particle and the plane at $P$.
\hfill \mbox{\textit{OCR M2 2013 Q6 [15]}}