| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Three-particle sequential collisions |
| Difficulty | Standard +0.8 This is a sequential collision problem requiring conservation of momentum and Newton's restitution law applied twice, with algebraic manipulation to prove both that directions are unchanged and that a second collision occurs. It requires systematic multi-step reasoning with parameters (e and m), going beyond routine single-collision problems, but uses standard FM1 techniques without requiring novel geometric or conceptual insight. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| CLM: \(6mu+4mu(=10mu)=3mv+4mw\) \((10u=3v+4w)\) | M1, A1 | All terms required. Condone sign errors; correct unsimplified equation |
| Impact Law: \(w-v=e(2u-u)(=eu)\) | M1, A1 | Law used correctly. Condone sign errors; correct unsimplified equation |
| Solve for \(v\) or \(w\) | M1 | Use their correctly formed equations to solve for \(v\) or \(w\) |
| \(w=\frac{u}{7}(10+3e)\) | A1 | Either velocity correct |
| \(v=\frac{u}{7}(10-4e)\) | A1 | Both velocities correct |
| \(0\le e\le1 \Rightarrow 10+3e>0\) and \(10-4e>0\), hence both particles still travelling in original direction | A1* | Use possible values of \(e\) to justify given result from correct working |
| Total: (8) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| CLM: \(4mw=4mx+2my\) \((2w=2x+y)\) | M1 | All terms required. Condone sign errors |
| Impact: \(y-x=ew\) | M1 | Correct use of impact law. Condone sign errors |
| \(\Rightarrow w(2-e)=3x\), \(x=\frac{u}{21}(10+3e)(2-e)\) | M1 | Use their correctly formed equations to find velocity of \(B\) \((x)\) |
| Consider \(v-x\) i.e. \(\frac{u}{7}(10-4e)-\frac{u}{21}(10+3e)(2-e)\) \(=(3e^2-8e+10)\) | M1 | Form relevant difference for a second collision |
| Show that \(v-x>0\ \forall e\) | M1 | Complete correct method (e.g. differentiation or completing the square or discriminant) to determine when inequality is true |
| Complete correct argument and conclusion | A1* | Reach correct conclusion from correct work |
| Total: (6) |
## Question 3:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| CLM: $6mu+4mu(=10mu)=3mv+4mw$ $(10u=3v+4w)$ | M1, A1 | All terms required. Condone sign errors; correct unsimplified equation |
| Impact Law: $w-v=e(2u-u)(=eu)$ | M1, A1 | Law used correctly. Condone sign errors; correct unsimplified equation |
| Solve for $v$ or $w$ | M1 | Use their correctly formed equations to solve for $v$ or $w$ |
| $w=\frac{u}{7}(10+3e)$ | A1 | Either velocity correct |
| $v=\frac{u}{7}(10-4e)$ | A1 | Both velocities correct |
| $0\le e\le1 \Rightarrow 10+3e>0$ and $10-4e>0$, hence both particles still travelling in original direction | A1* | Use possible values of $e$ to justify given result from correct working |
| **Total: (8)** | | |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| CLM: $4mw=4mx+2my$ $(2w=2x+y)$ | M1 | All terms required. Condone sign errors |
| Impact: $y-x=ew$ | M1 | Correct use of impact law. Condone sign errors |
| $\Rightarrow w(2-e)=3x$, $x=\frac{u}{21}(10+3e)(2-e)$ | M1 | Use their correctly formed equations to find velocity of $B$ $(x)$ |
| Consider $v-x$ i.e. $\frac{u}{7}(10-4e)-\frac{u}{21}(10+3e)(2-e)$ $=(3e^2-8e+10)$ | M1 | Form relevant difference for a second collision |
| Show that $v-x>0\ \forall e$ | M1 | Complete correct method (e.g. differentiation or completing the square or discriminant) to determine when inequality is true |
| Complete correct argument and conclusion | A1* | Reach correct conclusion from correct work |
| **Total: (6)** | | |
---\begin{enumerate}
\item Two particles, $A$ and $B$, have masses $3 m$ and $4 m$ respectively. The particles are moving in the same direction along the same straight line on a smooth horizontal surface when they collide directly. Immediately before the collision the speed of $A$ is $2 u$ and the speed of $B$ is $u$.
\end{enumerate}
The coefficient of restitution between $A$ and $B$ is $e$.\\
(a) Show that the direction of motion of each of the particles is unchanged by the collision.\\
(8)
After the collision with $A$, particle $B$ collides directly with a third particle, $C$, of mass $2 m$, which is at rest on the surface.
The coefficient of restitution between $B$ and $C$ is also $e$.\\
(b) Show that there will be a second collision between $A$ and $B$.
\hfill \mbox{\textit{Edexcel FM1 2020 Q3 [14]}}