| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Three-particle sequential collisions |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring sequential collision analysis with general coefficients. Students must derive conditions for a second collision using both momentum and restitution equations, then track velocities through two collisions. The algebraic manipulation with parameters k and e, plus the inequality reasoning for part (a), elevates this significantly above standard single-collision questions. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of conservation of momentum | M1 | Correct no. of terms and dimensionally correct but condone sign errors |
| \(mu = -mv_Q + kmv_R\) | A1 | Correct equation |
| Use of NLR | M1 | Use of NLR with \(e\) on the correct side |
| \(eu = v_Q + v_R\) | A1 | Correct equation (any equivalent form); signs consistent with CLM equation |
| Using correct strategy to solve problem by finding \(v_Q\) | M1 | Solving for \(v_Q\) - complete correct strategy (i.e. correct use of CLM and NLR) |
| \(v_Q = \dfrac{u(ke-1)}{k+1}\) or \(v_Q = \dfrac{v_R(ke-1)}{1+e}\) | A1 | Correct expression for \(v_Q\); can be implied by a correct multiple of \(v_Q\) |
| For second collision, \(v_Q > 0\) | M1 | Use of appropriate condition for their \(v_Q\) |
| \(\dfrac{u(ke-1)}{k+1} > 0\) | M1 | Complete correct strategy to find values for \(k\) (i.e. set up and solve inequality) |
| \(k > \dfrac{1}{e}\) | A1 | cso |
| (9) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{u(ke-1)^2}{(k+1)^2}\) | B1 | Or equivalent; cao |
| (1) |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of conservation of momentum | M1 | Correct no. of terms and dimensionally correct but condone sign errors |
| $mu = -mv_Q + kmv_R$ | A1 | Correct equation |
| Use of NLR | M1 | Use of NLR with $e$ on the correct side |
| $eu = v_Q + v_R$ | A1 | Correct equation (any equivalent form); signs consistent with CLM equation |
| Using correct strategy to solve problem by finding $v_Q$ | M1 | Solving for $v_Q$ - complete correct strategy (i.e. correct use of CLM and NLR) |
| $v_Q = \dfrac{u(ke-1)}{k+1}$ or $v_Q = \dfrac{v_R(ke-1)}{1+e}$ | A1 | Correct expression for $v_Q$; can be implied by a correct multiple of $v_Q$ |
| For second collision, $v_Q > 0$ | M1 | Use of appropriate condition for their $v_Q$ |
| $\dfrac{u(ke-1)}{k+1} > 0$ | M1 | Complete correct strategy to find values for $k$ (i.e. set up and solve inequality) |
| $k > \dfrac{1}{e}$ | A1 | cso |
| | **(9)** | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{u(ke-1)^2}{(k+1)^2}$ | B1 | Or equivalent; cao |
| | **(1)** | |\begin{enumerate}
\item Three particles, $P , Q$ and $R$, are at rest on a smooth horizontal plane. The particles lie along a straight line with $Q$ between $P$ and $R$. The particles $Q$ and $R$ have masses $m$ and $k m$ respectively, where $k$ is a constant.
\end{enumerate}
Particle $Q$ is projected towards $R$ with speed $u$ and the particles collide directly.\\
The coefficient of restitution between each pair of particles is $e$.\\
(a) Find, in terms of $e$, the range of values of $k$ for which there is a second collision.
Given that the mass of $P$ is $k m$ and that there is a second collision,\\
(b) write down, in terms of $u , k$ and $e$, the speed of $Q$ after this second collision.
\hfill \mbox{\textit{Edexcel FM1 AS 2019 Q4 [10]}}