| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| 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 collision problem requiring direct application of the impulse-momentum theorem. Given the impulse magnitude, students simply apply I = mv - mu to find final velocities. The problem involves careful sign handling and basic algebra, but requires no problem-solving insight beyond standard textbook methods. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| For \(P\): \(-\frac{21mu}{4} = 3m(v_P - 2u)\) | M1A1 | M1 for Impulse = Change in Momentum of \(P\) (must have \(3m\) in both terms); condone sign errors. A1 for correct equation (could have \(-v_P\) in place of \(v_P\)) |
| \(v_P = \frac{u}{4}\) | A1 (3) | Must be positive |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| For \(Q\): \(\frac{21mu}{4} = m(v_Q - {-4u})\) | M1A1 | M1 for Impulse = Change in Momentum of \(Q\) (must have \(m\) in both terms); condone sign errors. A1 for correct equation (could have \(-v_Q\) in place of \(v_Q\)) |
| \(v_Q = \frac{5u}{4}\) | A1 (3) | Must be positive |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(3m \times 2u - m \times 4u = 3m \times \frac{u}{4} + mv_Q\) | M1 A1 | M1 for CLM with correct no. of terms; condone missing \(m\)'s or extra \(g\)'s and sign errors. A1 for correct equation |
| \(v_Q = \frac{5u}{4}\) | A1 | Must be positive |
# Question 1:
## Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| For $P$: $-\frac{21mu}{4} = 3m(v_P - 2u)$ | M1A1 | M1 for Impulse = Change in Momentum of $P$ (must have $3m$ in both terms); condone sign errors. A1 for correct equation (could have $-v_P$ in place of $v_P$) |
| $v_P = \frac{u}{4}$ | A1 (3) | Must be positive |
## Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| For $Q$: $\frac{21mu}{4} = m(v_Q - {-4u})$ | M1A1 | M1 for Impulse = Change in Momentum of $Q$ (must have $m$ in both terms); condone sign errors. A1 for correct equation (could have $-v_Q$ in place of $v_Q$) |
| $v_Q = \frac{5u}{4}$ | A1 (3) | Must be positive |
**OR (CLM method):**
| Working | Marks | Guidance |
|---------|-------|----------|
| $3m \times 2u - m \times 4u = 3m \times \frac{u}{4} + mv_Q$ | M1 A1 | M1 for CLM with correct no. of terms; condone missing $m$'s or extra $g$'s and sign errors. A1 for correct equation |
| $v_Q = \frac{5u}{4}$ | A1 | Must be positive |
**Total: (6)**
---\begin{enumerate}
\item Two particles, $P$ and $Q$, have masses $3 m$ and $m$ respectively. They are moving in opposite directions towards each other along the same straight line on a smooth horizontal plane and collide directly. The speeds of $P$ and $Q$ immediately before the collision are $2 u$ and $4 u$ respectively. The magnitude of the impulse received by each particle in the collision is $\frac { 21 m u } { 4 }$.\\
(a) Find the speed of $P$ after the collision.\\
(b) Find the speed of $Q$ after the collision.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2018 Q1 [6]}}