| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Vertical drop and bounce |
| Difficulty | Moderate -0.8 This is a straightforward two-part mechanics question requiring standard application of impulse-momentum theorem and coefficient of restitution formula. Students need to find velocities using v² = u² + 2as (SUVAT), then apply I = m(v - u) and e = speed after/speed before. All steps are routine with no problem-solving insight required, making it easier than average. |
| Spec | 6.03f Impulse-momentum: relation6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
2 A small sphere of mass 0.2 kg is dropped from rest at a height of 3 m above horizontal ground. It falls vertically, hits the ground and rebounds vertically upwards, coming to instantaneous rest at a height of 1.8 m above the ground.\\
(i) Calculate the magnitude of the impulse which the ground exerts on the sphere.\\
(ii) Calculate the coefficient of restitution between the sphere and the ground.
\hfill \mbox{\textit{OCR M2 2010 Q2 [7]}}