| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, find velocities/angles |
| Difficulty | Challenging +1.2 This is an oblique collision problem requiring conservation of momentum parallel and perpendicular to the line of centres, plus Newton's experimental law. While it involves multiple unknowns and several equations, the techniques are standard M3 material with clear structure. Part (iii) requires algebraic manipulation to derive an exact equation, which adds some challenge, but overall this is a typical exam question testing routine application of collision mechanics rather than requiring novel insight. |
| Spec | 6.03c Momentum in 2D: vector form6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \([u\sin30° = 3]\) | M1 | For momentum equation for B, normal to line of centres |
| \(u = 6\) | A1 | 2 |
| (ii) \([4\sin88.1° = v\sin45°]\) | M1 | For momentum equation for A, normal to line of centres |
| \(v = 5.65\) | A1 | |
| M1 | For momentum equation along line of centres | |
| \(0.4(4\cos88.1°) - mu\cos30° = -0.4v\cos45°\) | A1 | |
| \(m = 0.318\) | A1 | 5 |
| (iii) | M1 | For using NEL |
| \(0.75(4\cos\theta + u\cos30°) = v\cos45°\) | A1 | |
| \(4\sin\theta = v\sin45°\) | B1 | |
| \([3\cos\theta + 4.5\cos30° = 4\sin\theta]\) | M1 | For eliminating v |
| \(8\sin\theta - 6\cos\theta = 9\cos30°\) | A1 | 5 AG |
# Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $[u\sin30° = 3]$ | M1 | For momentum equation for B, normal to line of centres |
| $u = 6$ | A1 | **2** |
| **(ii)** $[4\sin88.1° = v\sin45°]$ | M1 | For momentum equation for A, normal to line of centres |
| $v = 5.65$ | A1 | |
| | M1 | For momentum equation along line of centres |
| $0.4(4\cos88.1°) - mu\cos30° = -0.4v\cos45°$ | A1 | |
| $m = 0.318$ | A1 | **5** |
| **(iii)** | M1 | For using NEL |
| $0.75(4\cos\theta + u\cos30°) = v\cos45°$ | A1 | |
| $4\sin\theta = v\sin45°$ | B1 | |
| $[3\cos\theta + 4.5\cos30° = 4\sin\theta]$ | M1 | For eliminating v |
| $8\sin\theta - 6\cos\theta = 9\cos30°$ | A1 | **5** AG |
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Two uniform smooth spheres $A$ and $B$ of equal radius are moving on a horizontal surface when they collide. $A$ has mass 0.4 kg , and $B$ has mass $m \mathrm {~kg}$. Immediately before the collision, $A$ is moving with speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an acute angle $\theta$ to the line of centres, and $B$ is moving with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $30 ^ { \circ }$ to the line of centres. Immediately after the collision $A$ is moving with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $45 ^ { \circ }$ to the line of centres, and $B$ is moving with speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ perpendicular to the line of centres (see diagram).\\
(i) Find $u$.\\
(ii) Given that $\theta = 88.1 ^ { \circ }$ correct to 1 decimal place, calculate the approximate values of $v$ and $m$.\\
(iii) The coefficient of restitution is 0.75 . Show that the exact value of $\theta$ is a root of the equation $8 \sin \theta - 6 \cos \theta = 9 \cos 30 ^ { \circ }$.
\hfill \mbox{\textit{OCR M3 2007 Q6 [12]}}