| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2020 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Particle brought to rest by collision |
| Difficulty | Standard +0.3 This is a straightforward Further Mechanics collision problem requiring standard application of conservation of momentum and Newton's restitution law. Part (a) uses impulse-momentum theorem, part (b) applies the restitution formula with given conditions, and part (c) calculates energy loss using KE formula. All steps are routine for FM1 with no novel insight required, making it slightly easier than average A-level difficulty. |
| Spec | 6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Scheme | Marks | Guidance |
| Use of CLM: \(m \times \frac{l}{m} = 4mw\) | M1 | Correct no. of terms, condone extra \(g\) \(s\), sign errors (must be equation in \(l\), \(m\) and \(w\) only) |
| \(l = 4mw\) | A1 | Correct equation |
| \(e = \frac{w'}{4w} = \frac{1}{4}\) * | B1* | Use of NLR to obtain given answer |
| \(\text{KE Loss} = \frac{1}{2}m(4w)^2 - \frac{1}{2}4mw^2\) | M1 | Allow negative loss |
| \(= 6mw^2\) | A1 | cao |
| Scheme | Marks | Guidance |
|--------|-------|----------|
| Use of CLM: $m \times \frac{l}{m} = 4mw$ | M1 | Correct no. of terms, condone extra $g$ $s$, sign errors (must be equation in $l$, $m$ and $w$ only) |
| $l = 4mw$ | A1 | Correct equation |
| $e = \frac{w'}{4w} = \frac{1}{4}$ * | B1* | Use of NLR to obtain given answer |
| $\text{KE Loss} = \frac{1}{2}m(4w)^2 - \frac{1}{2}4mw^2$ | M1 | Allow negative loss |
| $= 6mw^2$ | A1 | cao |
**Notes:**
- Answer not given, so a correct answer with no clear error seen will score M1A1
- An answer that relies on an impulse-momentum equation using $4m$ will score M0\begin{enumerate}
\item Two particles $P$ and $Q$ have masses $m$ and $4 m$ respectively. The particles are at rest on a smooth horizontal plane. Particle $P$ is given a horizontal impulse, of magnitude $I$, in the direction $P Q$. Particle $P$ then collides directly with $Q$. Immediately after this collision, $P$ is at rest and $Q$ has speed $w$. The coefficient of restitution between the particles is $e$.\\
(a) Find $I$ in terms of $m$ and $w$.\\
(b) Show that $e = \frac { 1 } { 4 }$\\
(c) Find, in terms of $m$ and $w$, the total kinetic energy lost in the collision between $P$ and $Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM1 AS 2020 Q1 [5]}}