| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Collision with vector velocities |
| Difficulty | Moderate -0.5 This is a straightforward application of conservation of momentum in two dimensions with coalescing particles. Students need to apply m₁v₁ + m₂v₂ = (m₁+m₂)v for each component separately, then calculate speed using Pythagoras. It's slightly easier than average because it's a direct formula application with no complications, though the vector notation adds minor complexity over 1D problems. |
| Spec | 1.10c Magnitude and direction: of vectors6.03b Conservation of momentum: 1D two particles6.03d Conservation in 2D: vector momentum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Using conservation of momentum: \(3\begin{pmatrix}6\\-2\end{pmatrix} + 7\begin{pmatrix}-1\\4\end{pmatrix} = 10\mathbf{v}\) | M1 | Attempt at conservation of momentum with correct masses |
| \(\begin{pmatrix}18\\-6\end{pmatrix} + \begin{pmatrix}-7\\28\end{pmatrix} = 10\mathbf{v}\) | M1 | Correct momentum calculation for both particles |
| \(\mathbf{v} = \begin{pmatrix}1.1\\2.2\end{pmatrix}\) m s\(^{-1}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Speed \(= \sqrt{1.1^2 + 2.2^2}\) | M1 | Attempt at magnitude of their velocity vector |
| \(= \sqrt{6.05} = \sqrt{\frac{121}{20}} \approx 2.46\) m s\(^{-1}\) | A1 | cao |
# Question 1:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Using conservation of momentum: $3\begin{pmatrix}6\\-2\end{pmatrix} + 7\begin{pmatrix}-1\\4\end{pmatrix} = 10\mathbf{v}$ | M1 | Attempt at conservation of momentum with correct masses |
| $\begin{pmatrix}18\\-6\end{pmatrix} + \begin{pmatrix}-7\\28\end{pmatrix} = 10\mathbf{v}$ | M1 | Correct momentum calculation for both particles |
| $\mathbf{v} = \begin{pmatrix}1.1\\2.2\end{pmatrix}$ m s$^{-1}$ | A1 | cao |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Speed $= \sqrt{1.1^2 + 2.2^2}$ | M1 | Attempt at magnitude of their velocity vector |
| $= \sqrt{6.05} = \sqrt{\frac{121}{20}} \approx 2.46$ m s$^{-1}$ | A1 | cao |
---1 Two particles, $A$ and $B$, are moving on a smooth horizontal surface when they collide. During the collision, the two particles coalesce to form a single combined particle. Particle $A$ has mass 3 kg and particle $B$ has mass 7 kg .
Before the collision, the velocity of $A$ is $\left[ \begin{array} { r } 6 \\ - 2 \end{array} \right] \mathrm { m } \mathrm { s } ^ { - 1 }$ and the velocity of $B$ is $\left[ \begin{array} { r } - 1 \\ 4 \end{array} \right] \mathrm { m } \mathrm { s } ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the velocity of the combined particle after the collision.
\item Find the speed of the combined particle after the collision.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2009 Q1 [5]}}