| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Collision with two possible outcomes |
| Difficulty | Standard +0.3 This is a straightforward momentum conservation problem with two standard scenarios (elastic collision and coalescence). Part (a) requires one application of conservation of momentum with given final velocities. Part (b) requires considering two directions for the combined particle, leading to a simple linear equation solved twice. All steps are routine M1 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| \(2m - 2 \times 3 = m \times (-0.5) + 3 \times 0.5\) | M1, A1 | Equation for conservation of momentum with four terms: \(2m\), 2×3, 0.5m and 3×0.5 regardless of signs. Correct equation with correct signs. |
| \(2.5m = 7.5\) | ||
| \(m = 3\) kg | A1 | Correct mass. Arguments based on the symmetry of the situation that lead to m = 3 can be awarded full marks. Note: Consistent use of \(mg\) instead of \(m\): deduct one mark. Note: Use of all positive signs leads to \(m = -3\), which might be changed to +3 by candidates (M1A0A0). Note: \(m = 3\) can be obtained via \(1.5m = 4.5\), which will usually score M1A0A0. |
| Answer | Marks | Guidance |
|---|---|---|
| \(2m - 2 \times 3 = m \times 0.5 + 3 \times 0.5\) | M1, A1 | Four term equation for conservation of momentum with ±0.5 for both velocities (no marks for 3m×0.5). Correct equation. |
| \(1.5m = 7.5\) | ||
| \(m = 5\) kg or \(2m - 2 \times 3 = m \times (-0.5) + 3 \times (-0.5)\) | A1 | Correct mass for velocity used. |
| \(2.5m = 4.5\) | M1 | Equation for conservation of momentum with opposite sign for the 0.5. |
| \(m = 1.8\) kg | A1 | Correct mass for the velocity used. |
| Total | 5 |
**8(a)**
| $2m - 2 \times 3 = m \times (-0.5) + 3 \times 0.5$ | M1, A1 | Equation for conservation of momentum with four terms: $2m$, 2×3, 0.5m and 3×0.5 regardless of signs. Correct equation with correct signs. |
| $2.5m = 7.5$ | | |
| $m = 3$ kg | A1 | Correct mass. Arguments based on the symmetry of the situation that lead to m = 3 can be awarded full marks. Note: Consistent use of $mg$ instead of $m$: deduct one mark. Note: Use of all positive signs leads to $m = -3$, which might be changed to +3 by candidates (M1A0A0). Note: $m = 3$ can be obtained via $1.5m = 4.5$, which will usually score M1A0A0. |
**8(b)**
| $2m - 2 \times 3 = m \times 0.5 + 3 \times 0.5$ | M1, A1 | Four term equation for conservation of momentum with ±0.5 for both velocities (no marks for 3m×0.5). Correct equation. |
| $1.5m = 7.5$ | | |
| $m = 5$ kg or $2m - 2 \times 3 = m \times (-0.5) + 3 \times (-0.5)$ | A1 | Correct mass for velocity used. |
| $2.5m = 4.5$ | M1 | Equation for conservation of momentum with opposite sign for the 0.5. |
| $m = 1.8$ kg | A1 | Correct mass for the velocity used. |
| Total | 5 | |
---
## Total Marks: 758 Two particles, $A$ and $B$, are travelling towards each other along a straight horizontal line.\\
Particle $A$ has velocity $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and mass $m \mathrm {~kg}$.\\
Particle $B$ has velocity $- 2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and mass 3 kg .\\
\includegraphics[max width=\textwidth, alt={}, center]{a381686b-0b1e-41ba-b88f-be1601e42098-5_220_1157_516_440}
The particles collide.
\begin{enumerate}[label=(\alph*)]
\item If the particles move in opposite directions after the collision, each with speed $0.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, find the value of $m$.
\item If the particles coalesce during the collision, forming a single particle which moves with speed $0.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, find the two possible values of $m$.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2008 Q8 [8]}}