| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, find velocities/angles |
| Difficulty | Standard +0.3 This is a standard M3 oblique collision problem requiring resolution of velocities parallel and perpendicular to the line of centres, application of conservation of momentum and Newton's restitution law. While it involves multiple steps and careful component work, it follows a well-established procedure taught in M3 with no novel insight required, making it slightly easier than average. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (implied method) | M1 | \(\Sigma mv\) conserved in \(\mathbf{i}\) direction |
| \(2 \times 12\cos 60° - 3 \times 8 = 2a + 3b\) | A1 | |
| (implied method) | M1 | For using NEL |
| For LHS of equation below | A1 | |
| \(0.5(12\cos 60° + 8) = b - a\) | A1 | Complete equation with signs of \(a\) and \(b\) consistent with previous equation |
| (implied method) | M1 | For eliminating \(a\) or \(b\) |
| Speed of B is \(0.4\text{ms}^{-1}\) in \(\mathbf{i}\) direction | A1 | |
| \(a = -6.6\) | A1 | |
| Component of A's velocity in \(\mathbf{j}\) direction is \(12\sin 60°\) | B1 | May be shown on diagram or implied in subsequent work |
| Speed of A is \(12.3\text{ms}^{-1}\) | B1ft | |
| (implied method) | M1 | For using \(\theta = \tan^{-1}(\mathbf{j}\text{comp}/\pm\,\mathbf{i}\text{ comp})\) |
| Direction is at \(122.4°\) to the \(\mathbf{i}\) direction | A1ft | 1, 2 |
# Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| (implied method) | M1 | $\Sigma mv$ conserved in $\mathbf{i}$ direction |
| $2 \times 12\cos 60° - 3 \times 8 = 2a + 3b$ | A1 | |
| (implied method) | M1 | For using NEL |
| For LHS of equation below | A1 | |
| $0.5(12\cos 60° + 8) = b - a$ | A1 | Complete equation with signs of $a$ and $b$ consistent with previous equation |
| (implied method) | M1 | For eliminating $a$ or $b$ |
| Speed of B is $0.4\text{ms}^{-1}$ in $\mathbf{i}$ direction | A1 | |
| $a = -6.6$ | A1 | |
| Component of A's velocity in $\mathbf{j}$ direction is $12\sin 60°$ | B1 | May be shown on diagram or implied in subsequent work |
| Speed of A is $12.3\text{ms}^{-1}$ | B1ft | |
| (implied method) | M1 | For using $\theta = \tan^{-1}(\mathbf{j}\text{comp}/\pm\,\mathbf{i}\text{ comp})$ |
| Direction is at $122.4°$ to the $\mathbf{i}$ direction | A1ft | 1, 2 | Accept $\theta = 57.6°$ with $\theta$ correctly identified |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{09d3e8ca-0062-4f62-8453-7acaff591db5-3_362_841_264_651}
Two uniform smooth spheres $A$ and $B$, of equal radius, have masses 2 kg and 3 kg respectively. They are moving on a horizontal surface when they collide. Immediately before the collision $A$ is moving with speed $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $60 ^ { \circ }$ to the line of centres, and $B$ is moving with speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ along the line of centres (see diagram). The coefficient of restitution between the spheres is 0.5 . Find the speed and direction of motion of each sphere after the collision.
\hfill \mbox{\textit{OCR M3 2006 Q5 [12]}}