| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Direct collision, find velocities |
| Difficulty | Standard +0.3 This is a standard two-part mechanics question combining momentum/collisions with projectiles. Part (i) requires routine application of conservation of momentum and Newton's restitution law with straightforward algebra. Part (ii) involves standard projectile motion using SUVAT equations. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = 3\) m s\(^{-1}\) | B1 | |
| \(6 = 2x + 3y\) | M1, A1 | |
| \(e = (y-x)/3\) | M1, A1 | (\(e = \frac{2}{3}\)) (equs must be consistent) |
| \(y = 2\) | A1 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v_h = 2\) | B1 | or (B1) \(\frac{1}{2}mx2^2\) |
| \(v_v^2 = 2 \times 9.8 \times 4\) | M1 | (B1) \(\frac{1}{2}mxv^2\) |
| \(v_v = 8.85 \quad (14\sqrt{10/5})\) | A1 | (B1) \(mx9.8x4\) |
| speed \(= \sqrt{(8.85^2 + 2^2)}\) | M1 | \(v = \sqrt{(2^2 + 2\times9.8\times4)}\) |
| \(9.08\) m s\(^{-1}\) | A1 | |
| \(\tan^{-1}(8.85/2)\) | M1 | or \(\cos^{-1}(2/9.08)\) |
| \(77.3°\) to horizontal | A1 | 7 |
# Question 7:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = 3$ m s$^{-1}$ | B1 | |
| $6 = 2x + 3y$ | M1, A1 | |
| $e = (y-x)/3$ | M1, A1 | ($e = \frac{2}{3}$) (equs must be consistent) |
| $y = 2$ | A1 | **6** | **AG** |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v_h = 2$ | B1 | or (B1) $\frac{1}{2}mx2^2$ |
| $v_v^2 = 2 \times 9.8 \times 4$ | M1 | (B1) $\frac{1}{2}mxv^2$ |
| $v_v = 8.85 \quad (14\sqrt{10/5})$ | A1 | (B1) $mx9.8x4$ |
| speed $= \sqrt{(8.85^2 + 2^2)}$ | M1 | $v = \sqrt{(2^2 + 2\times9.8\times4)}$ |
| $9.08$ m s$^{-1}$ | A1 | |
| $\tan^{-1}(8.85/2)$ | M1 | or $\cos^{-1}(2/9.08)$ |
| $77.3°$ to horizontal | A1 | **7** | $12.7°$ to vertical |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{6ae57fe9-3b6f-46c2-95b8-d48903ed796b-4_305_1301_1708_424}
Two small spheres $A$ and $B$ of masses 2 kg and 3 kg respectively lie at rest on a smooth horizontal platform which is fixed at a height of 4 m above horizontal ground (see diagram). Sphere $A$ is given an impulse of 6 N s towards $B$, and $A$ then strikes $B$ directly. The coefficient of restitution between $A$ and $B$ is $\frac { 2 } { 3 }$.\\
(i) Show that the speed of $B$ after it has been hit by $A$ is $2 \mathrm {~ms} ^ { - 1 }$.
Sphere $B$ leaves the platform and follows the path of a projectile.\\
(ii) Calculate the speed and direction of motion of $B$ at the instant when it hits the ground.
\hfill \mbox{\textit{OCR M2 2008 Q7 [13]}}