| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Sphere rebounds off fixed wall obliquely |
| Difficulty | Standard +0.3 This is a straightforward two-part mechanics question requiring standard application of impulse-momentum theorem (using Pythagoras since perpendicular components) and coefficient of restitution formula. The perpendicular impulse adds a nice geometric element but the calculations are routine for M3 level with no conceptual surprises or extended reasoning required. |
| Spec | 6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(example = \frac{1}{2}mv^2\) | M1 | Method for... |
| \(= 12.5\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Triangle of velocities/momentum | M1 | For right angled triangle with at least one side correctly shown \((2.5, 2, 20I\) or \(0.125, 0.1, I)\) or vector equation \((v_1, v_2) = (0, 20I) + (2, 0)\) with at least 3 of the 4 components on RHS correct |
| All correct | A1 | |
| Use of Pythagoras' theorem to find \(I\) | M1 | \(400I^2 + 2^2 = 2.5^2\) or \(I^2 = 0.125^2 - 0.1^2\) |
| \(I = 0.075\) | A1 | May be implied by \(v_1^2 + v_2^2 = 2.5^2\) or \(\sin\alpha = 0.6\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Components of velocity parallel to wall before and after are 2 and 2 | B1 | May be implied |
| Components of velocity perpendicular to wall before and after are \((-)\,1.5\) and \(1.5e\) | B1 | |
| \([2^2 + (1.5e)^2 = 5]\) | M1 | For using \(v_1^2 + v_2^2 = 5\). Must be perp to wall |
| Coefficient is \(\frac{2}{3}\) or \(0.667\) | A1 | |
| [4] |
**Question 1:**
| Working/Answer | Marks | Guidance |
|---|---|---|
| $example = \frac{1}{2}mv^2$ | M1 | Method for... |
| $= 12.5$ | A1 | cao |
Please share the remaining pages and I'll extract and format them accordingly.
# OCR 4730 January 2012 Mark Scheme
---
## Question 1:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Triangle of velocities/momentum | M1 | For right angled triangle with at least one side correctly shown $(2.5, 2, 20I$ or $0.125, 0.1, I)$ or vector equation $(v_1, v_2) = (0, 20I) + (2, 0)$ with at least 3 of the 4 components on RHS correct |
| All correct | A1 | |
| Use of Pythagoras' theorem to find $I$ | M1 | $400I^2 + 2^2 = 2.5^2$ or $I^2 = 0.125^2 - 0.1^2$ |
| $I = 0.075$ | A1 | May be implied by $v_1^2 + v_2^2 = 2.5^2$ or $\sin\alpha = 0.6$ |
| **[4]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Components of velocity parallel to wall before and after are 2 and 2 | B1 | May be implied |
| Components of velocity perpendicular to wall before and after are $(-)\,1.5$ and $1.5e$ | B1 | |
| $[2^2 + (1.5e)^2 = 5]$ | M1 | For using $v_1^2 + v_2^2 = 5$. Must be perp to wall |
| Coefficient is $\frac{2}{3}$ or $0.667$ | A1 | |
| **[4]** | | |
---1 A particle $P$ of mass 0.05 kg is moving on a smooth horizontal surface with speed $2 \mathrm {~ms} ^ { - 1 }$, when it is struck by a horizontal blow in a direction perpendicular to its direction of motion. The magnitude of the impulse of the blow is $I$ Ns. The speed of $P$ after the blow is $2.5 \mathrm {~ms} ^ { - 1 }$.\\
(i) Find the value of $I$.
Immediately before the blow $P$ is moving parallel to a smooth vertical wall. After the blow $P$ hits the wall and rebounds from the wall with speed $\sqrt { 5 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) Find the coefficient of restitution between $P$ and the wall.
\hfill \mbox{\textit{OCR M3 2012 Q1 [8]}}