| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Multiple sequential collisions |
| Difficulty | Standard +0.3 This is a standard M1 momentum conservation question with three sequential collisions. Parts (i) and (ii) are straightforward applications of conservation of momentum. Part (iii) requires recognizing that maximum speed occurs when B and the combined particle move together after collision (coefficient of restitution = 0), which is a common exam technique. The multi-part structure and need to track multiple collisions adds modest complexity, but all steps follow standard procedures without requiring novel insight. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form |
| Answer | Marks |
|---|---|
| Marks | Guidance |
| B1 | Before momentum, signs different, no \(g\) |
| M1 | Uses momentum conservation, no \(g\) |
| A1 | |
| A1 | B's "after" velocity |
| [4] |
| Answer | Marks |
|---|---|
| Marks | Guidance |
| B1 | |
| M1 | No \(g\) |
| A1 | CD "after" velocity |
| [3] |
| Answer | Marks |
|---|---|
| Marks | Guidance |
| B1 | It cannot be less than the speed of A |
| M1 | Momentum, B and CD particles, essentially 4 non-zero terms with distinct velocities. Letters used at this stage should be checked against values used later. |
| ft cv(v(i)) and v(ii)) | |
| A1√ | |
| A1 | |
| [4] |
## Part i
**Answer/Working:**
$0.8x6 - 0.2x2 (=4.4)$
$0.8x6 - 0.2x2 = 0.8x4 + 0.2v (= 4.4)$
$v = 6$ m s$^{-1}$
| **Marks** | **Guidance** |
|-----------|-------------|
| B1 | Before momentum, signs different, no $g$ |
| M1 | Uses momentum conservation, no $g$ |
| A1 | |
| A1 | B's "after" velocity |
| [4] | |
## Part ii
**Answer/Working:**
After mass = 0.3+0.1
$0.3x5 (+0.1x0) = (0.3+0.1)v$
$v = 3.75$ m s$^{-1}$
| **Marks** | **Guidance** |
|-----------|-------------|
| B1 | |
| M1 | No $g$ |
| A1 | CD "after" velocity |
| [3] | |
## Part iii
**Answer/Working:**
Least final speed $B = 4$
$0.2x6+(0.3+0.1)x3.75 = 0.2x(v \geq 4) + 0.4V$
$0.2x6+(0.3+0.1)x3.75 = 0.2x4+0.4V$
$V = 4.75$ m s$^{-1}$
| **Marks** | **Guidance** |
|-----------|-------------|
| B1 | It cannot be less than the speed of A |
| M1 | Momentum, B and CD particles, essentially 4 non-zero terms with distinct velocities. Letters used at this stage should be checked against values used later. |
| | ft cv(v(i)) and v(ii)) |
| A1√ | |
| A1 | |
| [4] | |
---\includegraphics{figure_4}
Four particles $A$, $B$, $C$ and $D$ are on the same straight line on a smooth horizontal table. $A$ has speed $6\text{ m s}^{-1}$ and is at rest towards $B$. The speed of $B$ is $2\text{ m s}^{-1}$ and $B$ is moving towards $A$. The particle $C$ is moving with speed $5\text{ m s}^{-1}$ away from $B$ and towards $D$, which is stationary (see diagram). The first collision is between $A$ and $B$ which have masses $0.8\text{ kg}$ and $0.2\text{ kg}$ respectively.
\begin{enumerate}[label=(\roman*)]
\item After the particles collide $A$ has speed $4\text{ m s}^{-1}$ in its original direction of motion. Calculate the speed of $B$ after the collision. [4]
\end{enumerate}
The second collision is between $C$ and $D$ which have masses $0.3\text{ kg}$ and $0.1\text{ kg}$ respectively.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item The particles coalesce when they collide. Find the speed of the combined particle after this collision. [3]
\end{enumerate}
The third collision is between $B$ and the combined particle, after which no further collisions occur.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Calculate the greatest possible speed of the combined particle after the third collision. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2016 Q4 [11]}}