AQA M1 2014 June — Question 5 5 marks

Exam BoardAQA
ModuleM1 (Mechanics 1)
Year2014
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeCollision with vector velocities
DifficultyModerate -0.3 This is a straightforward application of conservation of momentum in vector form. Students must equate total momentum before and after collision, then solve a simple linear equation for k. The vector components can be handled separately, making this slightly easier than average but still requiring careful algebraic manipulation.
Spec6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form

5 Two particles, \(A\) and \(B\), have masses of \(m\) and \(k m\) respectively, where \(k\) is a constant. The particles are moving on a smooth horizontal plane when they collide and coalesce to form a single particle. Just before the collision the velocities of \(A\) and \(B\) are \(( 4 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\) and \(( 6 \mathbf { i } - 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\) respectively. Immediately after the collision the combined particle has velocity \(( 5.2 \mathbf { i } - 0.4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). Find \(k\).
[0pt] [5 marks]

Question 5:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Conservation of momentum: \(m(4\mathbf{i}+2\mathbf{j}) + km(6\mathbf{i}-2\mathbf{j}) = (m+km)(5.2\mathbf{i}-0.4\mathbf{j})\)M1 Applying conservation of momentum with correct masses
i component: \(4m + 6km = 5.2m(1+k)\)M1 Forming equation from either component
\(4 + 6k = 5.2 + 5.2k\)A1 Correct equation
\(0.8k = 1.2\)
\(k = 1.5\)A1
Check j component: \(2m - 2km = -0.4m(1+k)\)M1 Verification using second component
\(2 - 3 = -0.4(2.5) = -1\) ✓A1 Confirmed consistent
## Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Conservation of momentum: $m(4\mathbf{i}+2\mathbf{j}) + km(6\mathbf{i}-2\mathbf{j}) = (m+km)(5.2\mathbf{i}-0.4\mathbf{j})$ | M1 | Applying conservation of momentum with correct masses |
| **i component:** $4m + 6km = 5.2m(1+k)$ | M1 | Forming equation from either component |
| $4 + 6k = 5.2 + 5.2k$ | A1 | Correct equation |
| $0.8k = 1.2$ | | |
| $k = 1.5$ | A1 | |
| **Check j component:** $2m - 2km = -0.4m(1+k)$ | M1 | Verification using second component |
| $2 - 3 = -0.4(2.5) = -1$ ✓ | A1 | Confirmed consistent |

---
5 Two particles, $A$ and $B$, have masses of $m$ and $k m$ respectively, where $k$ is a constant. The particles are moving on a smooth horizontal plane when they collide and coalesce to form a single particle. Just before the collision the velocities of $A$ and $B$ are $( 4 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ and $( 6 \mathbf { i } - 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ respectively. Immediately after the collision the combined particle has velocity $( 5.2 \mathbf { i } - 0.4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.

Find $k$.\\[0pt]
[5 marks]

\hfill \mbox{\textit{AQA M1 2014 Q5 [5]}}