| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | October |
| Paper | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Direct collision, find impulse magnitude |
| Difficulty | Moderate -0.3 This is a straightforward M1 momentum conservation problem with clearly defined masses and velocities. Part (a) requires setting up and solving a single conservation equation to show a given result, while part (b) applies the impulse-momentum theorem. The question involves standard bookwork techniques with no conceptual surprises, making it slightly easier than average for A-level. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of CLM or equating impulses | M1 | Dimensionally correct. Condone sign errors and allow missing \(m\)'s and extra \(g\)'s |
| \((4m \times 2x) - 3mx = (3m \times 5y) - 4my\) Or: \(3m(5y - -x) = 4m(y - -2x)\) | A1 | Correct unsimplified equation, must see all 4 terms but allow missing \(m\)'s |
| \(y = \frac{5}{11}x\) | A1* | Obtain given answer from correct working with \(m\)'s seen. Allow \(y = \frac{5x}{11}\) but A0 if \(x\) is clearly in denominator |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(I = mv - mu\) | M1 | Dimensionally correct. Condone sign errors and \(y\) not substituted but mass and velocities must match. M0 if \(m\) is used for mass or \(g\) is included or \(m\) is missing |
| \(\pm 3m\left(\frac{25}{11}x + x\right)\) or \(\left(\pm 4m\left(\frac{5}{11}x + 2x\right)\right)\) | A1 | Correct unsimplified expression |
| \(\frac{108}{11}mx\) | A1 | \(9.8mx\) or better \((9.8181...mx)\), must be positive |
| [3] |
## Question 1:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of CLM or equating impulses | M1 | Dimensionally correct. Condone sign errors and allow missing $m$'s and extra $g$'s |
| $(4m \times 2x) - 3mx = (3m \times 5y) - 4my$ Or: $3m(5y - -x) = 4m(y - -2x)$ | A1 | Correct unsimplified equation, must see all 4 terms but allow missing $m$'s |
| $y = \frac{5}{11}x$ | A1* | Obtain given answer from correct working with $m$'s seen. Allow $y = \frac{5x}{11}$ but A0 if $x$ is clearly in denominator |
| **[3]** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $I = mv - mu$ | M1 | Dimensionally correct. Condone sign errors and $y$ not substituted but mass and velocities must match. M0 if $m$ is used for mass or $g$ is included or $m$ is missing |
| $\pm 3m\left(\frac{25}{11}x + x\right)$ or $\left(\pm 4m\left(\frac{5}{11}x + 2x\right)\right)$ | A1 | Correct unsimplified expression |
| $\frac{108}{11}mx$ | A1 | $9.8mx$ or better $(9.8181...mx)$, must be positive |
| **[3]** | | |
---\begin{enumerate}
\item Particle $A$ has mass $4 m$ and particle $B$ has mass $3 m$.
\end{enumerate}
The particles are moving in opposite directions along the same straight line on a smooth horizontal surface when they collide directly.
Immediately before the collision, the speed of $A$ is $2 x$ and the speed of $B$ is $x$.\\
Immediately after the collision, the speed of $A$ is $y$ and the speed of $B$ is $5 y$.\\
The direction of motion of each particle is reversed as a result of the collision.\\
(a) Show that $y = \frac { 5 } { 11 } x$.\\
(b) Find, in terms of $m$ and $x$, the magnitude of the impulse received by $A$ in the collision.
\hfill \mbox{\textit{Edexcel M1 2024 Q1}}