| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Coalescence collision |
| Difficulty | Moderate -0.8 This is a straightforward momentum conservation problem with coalescence. Part (a) requires setting up and solving a single linear equation using conservation of momentum with given speeds. Part (b) is a direct calculation of kinetic energy before and after collision using standard formulas. Both parts are routine applications of standard mechanics formulas with no conceptual challenges or multi-step reasoning required. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt at use of conservation of momentum | M1 | 4 terms implied, i.e. \(m\) and \(km\) included before and after collision. Velocity after collision is the same for \(m\) and \(km\). |
| \(km \times 6 - m \times 2 = (km + m) \times 4\) | A1 | |
| \(k = 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| KE initial \(= \frac{1}{2} \times km \times 6^2 + \frac{1}{2} \times m \times (-2)^2\); KE after \(= \frac{1}{2} \times (km+m) \times 4^2\) | M1 | Attempt at any of the three possible KE terms, unsimplified. \(k\) need not be substituted here. |
| Loss of KE \(= 24m\) J | A1 FT | KE loss \(= 56m - 32m\). FT on their \(k\), KE loss \(= (10k-6)m\), \(k > 0.6\). |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt at use of conservation of momentum | M1 | 4 terms implied, i.e. $m$ and $km$ included before and after collision. Velocity after collision is the same for $m$ and $km$. |
| $km \times 6 - m \times 2 = (km + m) \times 4$ | A1 | |
| $k = 3$ | A1 | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| KE initial $= \frac{1}{2} \times km \times 6^2 + \frac{1}{2} \times m \times (-2)^2$; KE after $= \frac{1}{2} \times (km+m) \times 4^2$ | M1 | Attempt at any of the three possible KE terms, unsimplified. $k$ need not be substituted here. |
| Loss of KE $= 24m$ J | A1 FT | KE loss $= 56m - 32m$. FT on their $k$, KE loss $= (10k-6)m$, $k > 0.6$. |2 Two small smooth spheres $A$ and $B$, of equal radii and of masses km kg and $m \mathrm {~kg}$ respectively, where $k > 1$, are free to move on a smooth horizontal plane. $A$ is moving towards $B$ with speed $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $B$ is moving towards $A$ with speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. After the collision $A$ and $B$ coalesce and move with speed $4 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find $k$.
\item Find, in terms of $m$, the loss of kinetic energy due to the collision.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q2 [5]}}