| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Direct collision, find impulse magnitude |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics question testing basic impulse-momentum principles. Students apply impulse = change in momentum to particle P (given impulse magnitude directly), then use conservation of momentum to find Q's final speed. All steps are routine applications of standard formulas with no problem-solving insight required, making it easier than average. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(5mu = 2m(v_P - -2u)\) | M1 A1 | Complete method for \(v_P\); \(5mu\) = change in momentum of \(P\); must have \(2m\) in both terms |
| \(v_P = \frac{1}{2}u\) | A1 (3) | A0 if negative |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Reversed | B1 (1) | Only allow if \(\frac{1}{2}u\) or \(-\frac{1}{2}u\) correctly obtained in (a); allow "(Yes) it has" but NOT just "Yes" or "opposite" |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(5mu = 3m(v_Q - -u)\) | M1 A1 | Complete method for \(v_Q\); \(5mu\) = change in momentum of \(Q\); must have \(3m\) in both terms |
| \(v_Q = \frac{2}{3}u\) | A1 (3) | A0 if negative |
| OR \(2m(2u) - 3mu = -2m\frac{1}{2}u + 3mv_Q\) | M1 A1 | CLM with correct no. of terms; condone sign errors |
| \(v_Q = \frac{2}{3}u\) | A1 (3) |
## Question 1:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|---------|
| $5mu = 2m(v_P - -2u)$ | M1 A1 | Complete method for $v_P$; $5mu$ = change in momentum of $P$; must have $2m$ in both terms |
| $v_P = \frac{1}{2}u$ | A1 (3) | A0 if negative |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|---------|
| Reversed | B1 (1) | Only allow if $\frac{1}{2}u$ or $-\frac{1}{2}u$ correctly obtained in (a); allow "(Yes) it has" but NOT just "Yes" or "opposite" |
### Part (c)
| Working | Marks | Guidance |
|---------|-------|---------|
| $5mu = 3m(v_Q - -u)$ | M1 A1 | Complete method for $v_Q$; $5mu$ = change in momentum of $Q$; must have $3m$ in both terms |
| $v_Q = \frac{2}{3}u$ | A1 (3) | A0 if negative |
| **OR** $2m(2u) - 3mu = -2m\frac{1}{2}u + 3mv_Q$ | M1 A1 | CLM with correct no. of terms; condone sign errors |
| $v_Q = \frac{2}{3}u$ | A1 (3) | |
---\begin{enumerate}
\item Two particles, $P$ and $Q$, have masses $2 m$ and $3 m$ respectively. They are moving towards each other, in opposite directions, along the same straight line, on a smooth horizontal plane. The particles collide. Immediately before they collide the speed of $P$ is $2 u$ and the speed of $Q$ is $u$. In the collision the magnitude of the impulse exerted on $P$ by $Q$ is $5 m u$.\\
(a) Find the speed of $P$ immediately after the collision.\\
(b) State whether the direction of motion of $P$ has been reversed by the collision.\\
(c) Find the speed of $Q$ immediately after the collision.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2016 Q1 [7]}}