| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, find velocities/angles |
| Difficulty | Challenging +1.2 This is a standard M3 oblique collision problem requiring resolution of velocities, application of conservation of momentum along the line of centres, and Newton's law of restitution. While it involves multiple steps and careful component work, it follows a well-established procedure taught in all M3 courses with no novel insight required. The given cos θ = 0.6 simplifies calculations. Slightly above average difficulty due to the two-sphere oblique setup and multi-part nature, but remains a textbook exercise. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.4(7x0.6) - 0.3x2.8 = 0.4a + 0.3b\) | M1 | For using the principle of conservation of momentum |
| \(0.7(7x0.6 + 2.8) = b - a\) | M1 | For using \(e(\Delta u) = \Delta v\) |
| M1 | For eliminating \(a\) from equations | |
| Speed of \(B\) is \(4ms^{-1}\) | A1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = (-) 0.9\) | B1 | |
| Component perp. to l.o.c. is \(5.6\) | B1 | |
| \(\tan\alpha = 5.6/0.9\) | M1 | For attempting to find \(\alpha\) - the angle between the direction of motion of \(A\) after collision and the l.o.c. to the left, or \(90° - \alpha\) |
| \(\alpha = 80.9°\) | A1 | |
| Angle turned through is \(46.0°\) (0.803°) | A1ft | \(126.9° - \alpha\) |
| [5] |
**Part i**
$0.4(7x0.6) - 0.3x2.8 = 0.4a + 0.3b$ | M1 | For using the principle of conservation of momentum
$0.7(7x0.6 + 2.8) = b - a$ | M1 | For using $e(\Delta u) = \Delta v$
| M1 | For eliminating $a$ from equations
Speed of $B$ is $4ms^{-1}$ | A1 |
| [6] |
**Part ii**
$a = (-) 0.9$ | B1 |
Component perp. to l.o.c. is $5.6$ | B1 |
$\tan\alpha = 5.6/0.9$ | M1 | For attempting to find $\alpha$ - the angle between the direction of motion of $A$ after collision and the l.o.c. to the left, or $90° - \alpha$
$\alpha = 80.9°$ | A1 |
Angle turned through is $46.0°$ (0.803°) | A1ft | $126.9° - \alpha$
| [5] |4\\
\includegraphics[max width=\textwidth, alt={}, center]{67af8d98-85af-42b1-9e7f-c6380a1f8a3f-3_497_1157_255_493}
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 0.3 kg . Immediately before the collision $A$ is moving with speed $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an acute angle $\theta$ to the line of centres, where $\cos \theta = 0.6$, and $B$ is moving with speed $2.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ along the line of centres (see diagram). The coefficient of restitution between the spheres is 0.7. Find\\
(i) the speed of $B$ immediately after the collision,\\
(ii) the angle turned through by the direction of motion of $A$ as a result of the collision.
\hfill \mbox{\textit{OCR M3 2011 Q4 [11]}}