| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2018 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Collision followed by wall impact |
| Difficulty | Standard +0.8 This is a multi-stage collision problem requiring conservation of momentum, Newton's restitution law, and tracking multiple collisions with algebraic manipulation. Part (a) is a standard show-that requiring two equations, part (b) requires inequality reasoning about direction reversal, and part (c) involves tracking two particles through multiple collisions to find a meeting point. The problem requires careful bookkeeping and multiple applications of collision principles, making it moderately challenging but still within standard FM1 scope. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete strategy: use CLM, impact law, solve simultaneous equations | M1 | Complete strategy |
| Use of CLM: \(6mu - 5mu\,(= mu) = 3mv + 5mw\) | M1 | All terms required; dimensionally correct; condone sign errors |
| \(6mu - 5mu = 3mv + 5mw\) | A1 | Correct unsimplified equation |
| Use of impact law | M1 | Must be used the right way round; condone sign error |
| \(w - v = 3ue\) | A1 | Correct unsimplified equation; signs consistent with CLM equation |
| \(3v + 5w = u \\ 3w - 3v = 9ue\) \(\Rightarrow 8w = u + 9ue,\quad w = \frac{u}{8}(9e+1)\) | A1* | Obtain given answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = w - 3ue = \frac{u}{8}(1 - 15e)\) and \(v > 0\) | M1 | Find speed of \(P\) and form correct inequality consistent with their directions |
| \(\Rightarrow (0 \leq)\, e < \frac{1}{15}\) | A1 | Correct solution; need not mention lower limit |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete strategy to find time for \(Q\) to reach second collision | M1 | Complete strategy |
| Speed of \(Q\) after impact with wall \(= \frac{u}{16}\) | B1 | Correct use of impact law |
| Time for \(Q\): \(\frac{16d}{3u} + \frac{16x}{u}\) following their \(\frac{u}{16}\) and \(\frac{16d}{3u}\) | A1ft | Correct unsimplified expression using time \(=\frac{\text{distance}}{\text{speed}}\) |
| Time for \(P = \frac{48(d-x)}{u}\) following their \(\frac{u}{48}\) | B1ft | Correct use of time \(= \frac{\text{distance}}{\text{speed}}\) |
| Set both times equal (same place, same time) | M1 | Must be valid expressions for the times |
| \(x = \frac{128d}{192} = \frac{2d}{3}\) | A1 | Correct answer or exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete strategy to find position of second collision | M1 | e.g. considering distances and relative velocities |
| Speed of \(Q\) after impact with wall \(= \frac{u}{16}\) | B1 | Correct use of impact law |
| Distance apart when \(Q\) strikes wall \(= \frac{8d}{9}\) | B1ft | Follow their values |
| Gap closing at \(\frac{u}{16} + \frac{u}{48}\) | A1ft | Follow their \(\frac{u}{16}\) and \(\frac{u}{48}\) |
| \(t = \dfrac{\frac{8d}{9}}{\frac{u}{16}+\frac{u}{48}} = \frac{32d}{3u}\) | M1 | Correct use of time \(= \frac{\text{distance}}{\text{speed}}\) |
| \(x = \frac{u}{16} \times \frac{32d}{3u} = \frac{2d}{3}\) | A1 | Correct answer |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete strategy: use CLM, impact law, solve simultaneous equations | M1 | Complete strategy |
| Use of CLM: $6mu - 5mu\,(= mu) = 3mv + 5mw$ | M1 | All terms required; dimensionally correct; condone sign errors |
| $6mu - 5mu = 3mv + 5mw$ | A1 | Correct unsimplified equation |
| Use of impact law | M1 | Must be used the right way round; condone sign error |
| $w - v = 3ue$ | A1 | Correct unsimplified equation; signs consistent with CLM equation |
| $3v + 5w = u \\ 3w - 3v = 9ue$ $\Rightarrow 8w = u + 9ue,\quad w = \frac{u}{8}(9e+1)$ | A1* | Obtain given answer from correct working |
**Total: 6 marks**
---
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = w - 3ue = \frac{u}{8}(1 - 15e)$ and $v > 0$ | M1 | Find speed of $P$ and form correct inequality consistent with their directions |
| $\Rightarrow (0 \leq)\, e < \frac{1}{15}$ | A1 | Correct solution; need not mention lower limit |
**Total: 2 marks**
---
## Question 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete strategy to find time for $Q$ to reach second collision | M1 | Complete strategy |
| Speed of $Q$ after impact with wall $= \frac{u}{16}$ | B1 | Correct use of impact law |
| Time for $Q$: $\frac{16d}{3u} + \frac{16x}{u}$ following their $\frac{u}{16}$ and $\frac{16d}{3u}$ | A1ft | Correct unsimplified expression using time $=\frac{\text{distance}}{\text{speed}}$ |
| Time for $P = \frac{48(d-x)}{u}$ following their $\frac{u}{48}$ | B1ft | Correct use of time $= \frac{\text{distance}}{\text{speed}}$ |
| Set both times equal (same place, same time) | M1 | Must be valid expressions for the times |
| $x = \frac{128d}{192} = \frac{2d}{3}$ | A1 | Correct answer or exact equivalent |
**Total: 6 marks**
---
### Alternative method for 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete strategy to find position of second collision | M1 | e.g. considering distances and relative velocities |
| Speed of $Q$ after impact with wall $= \frac{u}{16}$ | B1 | Correct use of impact law |
| Distance apart when $Q$ strikes wall $= \frac{8d}{9}$ | B1ft | Follow their values |
| Gap closing at $\frac{u}{16} + \frac{u}{48}$ | A1ft | Follow their $\frac{u}{16}$ and $\frac{u}{48}$ |
| $t = \dfrac{\frac{8d}{9}}{\frac{u}{16}+\frac{u}{48}} = \frac{32d}{3u}$ | M1 | Correct use of time $= \frac{\text{distance}}{\text{speed}}$ |
| $x = \frac{u}{16} \times \frac{32d}{3u} = \frac{2d}{3}$ | A1 | Correct answer |
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\item A particle $P$ of mass $3 m$ is moving in a straight line on a smooth horizontal floor. A particle $Q$ of mass $5 m$ is moving in the opposite direction to $P$ along the same straight line.
\end{enumerate}
The particles collide directly.\\
Immediately before the collision, the speed of $P$ is $2 u$ and the speed of $Q$ is $u$. The coefficient of restitution between $P$ and $Q$ is $e$.\\
(a) Show that the speed of $Q$ immediately after the collision is $\frac { u } { 8 } ( 9 e + 1 )$\\
(b) Find the range of values of $e$ for which the direction of motion of $P$ is not changed as a result of the collision.
When $P$ and $Q$ collide they are at a distance $d$ from a smooth fixed vertical wall, which is perpendicular to their direction of motion. After the collision with $P$, particle $Q$ collides directly with the wall and rebounds so that there is a second collision between $P$ and $Q$. This second collision takes place at a distance $x$ from the wall.
Given that $e = \frac { 1 } { 18 }$ and the coefficient of restitution between $Q$ and the wall is $\frac { 1 } { 3 }$\\
(c) find $x$ in terms of $d$.
\hfill \mbox{\textit{Edexcel FM1 AS 2018 Q4 [14]}}