| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, find velocities/angles |
| Difficulty | Standard +0.8 This M3 oblique collision problem requires resolving velocities into components along and perpendicular to the line of centres, applying both momentum conservation and the restitution equation, then finding the resultant angle. While systematic, it demands careful vector decomposition, multiple equations, and trigonometric manipulation beyond standard direct-impact questions, placing it moderately above average difficulty. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using conservation of momentum along loc: \(0.1\times 2.8 + 0.4\times 1\times 0.8 = 0.4\times b\) | M1, A1 | 3 (or 4) terms, correct dimensions; allow sign errors, (sin/cos) |
| Using NEL: \(b - 0 = -e(1\times 0.8 - 2.8)\) | M1, A1 | Vel diff after \(= e\times\) vel diff before; may see \(b = 1.5\); allow \(\pm e\) |
| \(e = 0.75\) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(b(\text{perp}) = 0.6\); \(\tan\beta = \frac{b(\text{perp})}{\text{their } 1.5}\) | B1, M1* | \(\beta = 21.8°\); ft \(1.5\) from (i); may be on diagram; \(21.8014...(0.381\ \text{rad})\) |
| Angle turned through is \(36.9° - \beta = 15.1°\ (0.262\ \text{rad})\) | *M1, A1 [4] | Must be \(36.9°-\) their \(\beta\) (soi); \(36.86989\); \(15.068\) scB1 for \(165°\) after B1M1 |
## Question 3:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using conservation of momentum along loc: $0.1\times 2.8 + 0.4\times 1\times 0.8 = 0.4\times b$ | M1, A1 | 3 (or 4) terms, correct dimensions; allow sign errors, (sin/cos) |
| Using NEL: $b - 0 = -e(1\times 0.8 - 2.8)$ | M1, A1 | Vel diff after $= e\times$ vel diff before; may see $b = 1.5$; allow $\pm e$ |
| $e = 0.75$ | A1 [5] | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $b(\text{perp}) = 0.6$; $\tan\beta = \frac{b(\text{perp})}{\text{their } 1.5}$ | B1, M1* | $\beta = 21.8°$; ft $1.5$ from (i); may be on diagram; $21.8014...(0.381\ \text{rad})$ |
| Angle turned through is $36.9° - \beta = 15.1°\ (0.262\ \text{rad})$ | *M1, A1 [4] | Must be $36.9°-$ their $\beta$ (soi); $36.86989$; $15.068$ scB1 for $165°$ after B1M1 |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{3243c326-a51c-462f-a57c-a150d0044ea9-2_403_951_1247_559}
Two uniform smooth spheres $A$ and $B$ of equal radius are moving on a horizontal surface when they collide. $A$ has mass 0.1 kg and $B$ has mass 0.4 kg . Immediately before the collision $A$ is moving with speed $2.8 \mathrm {~ms} ^ { - 1 }$ along the line of centres, and $B$ is moving with speed $1 \mathrm {~ms} ^ { - 1 }$ at an angle $\theta$ to the line of centres, where $\cos \theta = 0.8$ (see diagram). Immediately after the collision $A$ is stationary. Find\\
(i) the coefficient of restitution between $A$ and $B$,\\
(ii) the angle turned through by the direction of motion of $B$ as a result of the collision.
\section*{$\mathrm { OCR } ^ { \text {勾 } }$}
\hfill \mbox{\textit{OCR M3 2014 Q3 [9]}}