| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Vertical drop and bounce |
| Difficulty | Standard +0.3 This is a straightforward two-part mechanics question requiring standard application of SUVAT equations to find impact speed, then direct use of coefficient of restitution and impulse formulas. All steps are routine M2 techniques with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 6.03c Momentum in 2D: vector form6.03f Impulse-momentum: relation6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 = 21^2 + 2x40x9.8\) | M1 | |
| \(x = 35\) | A1 | |
| \(0 = y^2 - 2x40x9.8\) | M1 | |
| \(y = 28\) | A1 | |
| \(e = 28/35\) | M1 | |
| \(e = 0.8\) | A1 | aef |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.2x28 - 0.2x35\) | M1 | must be double negative |
| \(I = 12.6\) | A1 | 2 |
**(i)**
$x^2 = 21^2 + 2x40x9.8$ | M1 |
$x = 35$ | A1 |
$0 = y^2 - 2x40x9.8$ | M1 |
$y = 28$ | A1 |
$e = 28/35$ | M1 |
$e = 0.8$ | A1 | aef | 6 |
**(ii)**
$0.2x28 - 0.2x35$ | M1 | must be double negative
$I = 12.6$ | A1 | 2 | 8 |3 A small sphere of mass 0.2 kg is projected vertically downwards with speed $21 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point at a height of 40 m above horizontal ground. It hits the ground and rebounds vertically upwards, coming to instantaneous rest at its initial point of projection. Ignoring air resistance, calculate\\
(i) the coefficient of restitution between the sphere and the ground,\\
(ii) the magnitude of the impulse which the ground exerts on the sphere.
\hfill \mbox{\textit{OCR M2 2007 Q3 [8]}}