| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2021 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with energy loss |
| Difficulty | Standard +0.3 This is a standard Further Mechanics collision problem requiring conservation of momentum, coefficient of restitution formula, and energy considerations. While it involves multiple steps and algebraic manipulation across three parts, the techniques are routine for FM1 students with no novel insight required—slightly easier than average A-level maths overall due to its structured, methodical nature. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of CLM | M1 | |
| \(5mv + 6mv(= 11mv) = 5mx - 3my \quad (11v = 5x - 3y)\) | A1 | |
| Use of impact law | M1 | |
| \(v = e(x + y)\) | A1 | |
| \(\begin{cases} 11ev = 5ex - 3ey \\ 3v = 3ex - 3ey \end{cases} \Rightarrow x = \frac{v}{8e}(11e+3)\) | M1 | |
| \(y = \frac{v}{8e}(5 - 11e)\) | A1 | |
| \(e > 0 \Rightarrow x > 0 \Rightarrow 5 - 11e > 0\) | M1 | |
| \(\Rightarrow 0 < e < \frac{5}{11}\) | A1 | |
| (8 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Form equation for KE | M1 | |
| \(\frac{1}{2}\times5m\times v^2 = \frac{16}{100}\times\frac{1}{2}\times5m\times\frac{v^2}{64e^2}(11e+3)^2\) | A1ft | |
| \((4(11e+3) = (\pm)80e) \quad e = \frac{1}{3}\) | A1 | |
| Impulse \(= -5m(v - x)\) | M1 | |
| \(= -5m\left(v - \frac{11v}{8} - \frac{3v}{8e}\right)\) or \(3m\left(2v + \frac{5v}{8e} - \frac{11v}{8}\right)\) | A1ft | |
| Magnitude \(= \frac{15}{2}mv\) | A1 | |
| (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Form equation for KE | M1 | |
| \(\frac{1}{2}\times5m\times v^2 = \frac{16}{100}\times\frac{1}{2}\times5m\times x^2\) | A1 | |
| \(\Rightarrow x = \frac{5v}{2},\ y = \frac{v}{2} \Rightarrow e = \frac{1}{3}\) | A1 |
## Question 2:
### Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of CLM | M1 | |
| $5mv + 6mv(= 11mv) = 5mx - 3my \quad (11v = 5x - 3y)$ | A1 | |
| Use of impact law | M1 | |
| $v = e(x + y)$ | A1 | |
| $\begin{cases} 11ev = 5ex - 3ey \\ 3v = 3ex - 3ey \end{cases} \Rightarrow x = \frac{v}{8e}(11e+3)$ | M1 | |
| $y = \frac{v}{8e}(5 - 11e)$ | A1 | |
| $e > 0 \Rightarrow x > 0 \Rightarrow 5 - 11e > 0$ | M1 | |
| $\Rightarrow 0 < e < \frac{5}{11}$ | A1 | |
| **(8 marks)** | | |
### Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Form equation for KE | M1 | |
| $\frac{1}{2}\times5m\times v^2 = \frac{16}{100}\times\frac{1}{2}\times5m\times\frac{v^2}{64e^2}(11e+3)^2$ | A1ft | |
| $(4(11e+3) = (\pm)80e) \quad e = \frac{1}{3}$ | A1 | |
| Impulse $= -5m(v - x)$ | M1 | |
| $= -5m\left(v - \frac{11v}{8} - \frac{3v}{8e}\right)$ or $3m\left(2v + \frac{5v}{8e} - \frac{11v}{8}\right)$ | A1ft | |
| Magnitude $= \frac{15}{2}mv$ | A1 | |
| **(6 marks)** | | |
### Alt(b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Form equation for KE | M1 | |
| $\frac{1}{2}\times5m\times v^2 = \frac{16}{100}\times\frac{1}{2}\times5m\times x^2$ | A1 | |
| $\Rightarrow x = \frac{5v}{2},\ y = \frac{v}{2} \Rightarrow e = \frac{1}{3}$ | A1 | |\begin{enumerate}
\item Two particles, $A$ and $B$, are moving in opposite directions along the same straight line on a smooth horizontal surface when they collide directly.
\end{enumerate}
Particle $A$ has mass $5 m$ and particle $B$ has mass $3 m$.\\
The coefficient of restitution between $A$ and $B$ is $e$, where $e > 0$\\
Immediately after the collision the speed of $A$ is $v$ and the speed of $B$ is $2 v$.\\
Given that $A$ and $B$ are moving in the same direction after the collision,\\
(a) find the set of possible values of $e$.
Given also that the kinetic energy of $A$ immediately after the collision is $16 \%$ of the kinetic energy of $A$ immediately before the collision,\\
(b) find\\
(i) the value of $e$,\\
(ii) the magnitude of the impulse received by $A$ in the collision, giving your answer in terms of $m$ and $v$.
\hfill \mbox{\textit{Edexcel FM1 2021 Q2 [14]}}