| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Direct collision, find final speed |
| Difficulty | Easy -1.2 This is a straightforward momentum conservation problem with particles coming to rest, requiring only direct application of the conservation formula and a simple kinetic energy calculation. Both parts involve routine algebraic manipulation with no conceptual challenges or problem-solving insight needed—significantly easier than average A-level questions. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\pm[3.2v + 2.4 \times (-6)] = 0\) | M1 | Attempt at conservation of momentum; 2 non-zero terms; allow sign errors |
| \(v = 4.5\) | A1 | M1A0 for use of \(mgv\). \(v = -4.5\) is A0 |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(KE = \pm\frac{1}{2}\times3.2\times(\text{their } 4.5)^2\) OR \(\pm\frac{1}{2}\times2.4\times6^2\) | M1 | Attempt at either KE term using their \(v\). Do not allow \(\frac{1}{2}\times3.2\times(\text{their } 4.5\pm6)^2\), or \(\frac{1}{2}\times2.4\times(\text{their } 4.5\pm6)^2\), or \(\frac{1}{2}\times(3.2+2.4)\times(\text{their } 4.5\pm6)^2\), or \(\frac{1}{2}\times3.2\times(\text{their } 4.5-0)^2\), or \(\frac{1}{2}\times2.4\times(6-0)^2\) |
| \(KE_{\text{loss}} = 75.6\text{ J}\) | A1 | Allow \(-75.6\). Note \(\frac{1}{2}\times(3.2+2.4)\times6^2\) or \(\frac{1}{2}\times(3.2+2.4)\times(\text{their } 4.5)^2\) is M1A0 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\pm[3.2v + 2.4 \times (-6)] = 0$ | **M1** | Attempt at conservation of momentum; 2 non-zero terms; allow sign errors |
| $v = 4.5$ | **A1** | M1A0 for use of $mgv$. $v = -4.5$ is A0 |
| | **2** | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $KE = \pm\frac{1}{2}\times3.2\times(\text{their } 4.5)^2$ OR $\pm\frac{1}{2}\times2.4\times6^2$ | M1 | Attempt at either KE term using their $v$. Do not allow $\frac{1}{2}\times3.2\times(\text{their } 4.5\pm6)^2$, or $\frac{1}{2}\times2.4\times(\text{their } 4.5\pm6)^2$, or $\frac{1}{2}\times(3.2+2.4)\times(\text{their } 4.5\pm6)^2$, or $\frac{1}{2}\times3.2\times(\text{their } 4.5-0)^2$, or $\frac{1}{2}\times2.4\times(6-0)^2$ |
| $KE_{\text{loss}} = 75.6\text{ J}$ | A1 | Allow $-75.6$. Note $\frac{1}{2}\times(3.2+2.4)\times6^2$ or $\frac{1}{2}\times(3.2+2.4)\times(\text{their } 4.5)^2$ is M1A0 |
---2 Two particles $A$ and $B$, of masses 3.2 kg and 2.4 kg respectively, lie on a smooth horizontal table. $A$ moves towards $B$ with a speed of $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and collides with $B$, which is moving towards $A$ with a speed of $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. In the collision the two particles come to rest.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $v$.\\
\includegraphics[max width=\textwidth, alt={}, center]{e5ee28f2-5876-4149-9a77-18c5792c1bd8-03_61_1569_495_328}
\item Find the loss of kinetic energy of the system due to the collision.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q2 [4]}}