| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Session | Specimen |
| Marks | 10 |
| 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 application of standard restitution formulas with routine kinematics. Part (a) is a 'show that' with the answer given, part (b) repeats the same calculation, and part (c) requires only basic observation. No problem-solving insight needed beyond direct formula application. |
| Spec | 6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Using the model and \(v^2 = u^2 + 2as\) to find \(v\) | M1 | 3.4 |
| \(v^2 = 2as = 2g \times 2.4 = 4.8g \Rightarrow v = \sqrt{4.8g}\) | A1 | 1.1b |
| Using the model and \(v^2 = u^2 + 2as\) to find \(u\) | M1 | 3.4 |
| \(0^2 = u^2 - 2g \times 0.6 \Rightarrow u = \sqrt{1.2g}\) | A1 | 1.1b |
| Using the correct strategy to solve the problem by finding the separation speed and approach speed and applying Newton's Law of Restitution | M1 | 3.1b |
| \(e = \sqrt{1.2g} / \sqrt{4.8g} = 0.5\) | A1* | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Using the model and \(e = \text{separation speed} / \text{approach speed}\), \(v = 0.5\sqrt{1.2g}\) | M1 | 3.4 |
| Using the model and \(v^2 = u^2 + 2as\) | M1 | 3.4 |
| \(0^2 = 0.25(1.2g) - 2gh \Rightarrow h = 0.15\) (m) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Ball continues to bounce with the height of each bounce being a quarter of the previous one | B1 | 2.2b |
# Question 1
## 1(a)
Using the model and $v^2 = u^2 + 2as$ to find $v$ | M1 | 3.4
$v^2 = 2as = 2g \times 2.4 = 4.8g \Rightarrow v = \sqrt{4.8g}$ | A1 | 1.1b
Using the model and $v^2 = u^2 + 2as$ to find $u$ | M1 | 3.4
$0^2 = u^2 - 2g \times 0.6 \Rightarrow u = \sqrt{1.2g}$ | A1 | 1.1b
Using the correct strategy to solve the problem by finding the separation speed and approach speed and applying Newton's Law of Restitution | M1 | 3.1b
$e = \sqrt{1.2g} / \sqrt{4.8g} = 0.5$ | A1* | 1.1b
**(6 marks)**
**Guidance for 1(a):**
- M1: For a complete method to find $v$
- A1: For a correct value (may be numerical)
- M1: For a complete method to find $u$
- A1: For a correct value (may be numerical)
- M1: For finding both $v$ and $u$ and use of Newton's Law of Restitution
- A1*: For the given answer
## 1(b)
Using the model and $e = \text{separation speed} / \text{approach speed}$, $v = 0.5\sqrt{1.2g}$ | M1 | 3.4
Using the model and $v^2 = u^2 + 2as$ | M1 | 3.4
$0^2 = 0.25(1.2g) - 2gh \Rightarrow h = 0.15$ (m) | A1 | 1.1b
**(3 marks)**
**Guidance for 1(b):**
- M1: For use of Newton's Law of Restitution to find rebound speed
- M1: For a complete method to find $h$
- A1: For 0.15 (m) or equivalent
## 1(c)
Ball continues to bounce with the height of each bounce being a quarter of the previous one | B1 | 2.2b
**(1 mark)**
**Guidance for 1(c):**
- B1: For a clear description including reference to a quarter
**(Total: 10 marks)**\begin{enumerate}
\item A small ball of mass 0.1 kg is dropped from a point which is 2.4 m above a horizontal floor. The ball falls freely under gravity, strikes the floor and bounces to a height of 0.6 m above the floor. The ball is modelled as a particle.\\
(a) Show that the coefficient of restitution between the ball and the floor is 0.5\\
(b) Find the height reached by the ball above the floor after it bounces on the floor for the second time.\\
(c) By considering your answer to (b), describe the subsequent motion of the ball.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM1 AS Q1 [10]}}