AQA M1 2011 January — Question 1 3 marks

Exam BoardAQA
ModuleM1 (Mechanics 1)
Year2011
SessionJanuary
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeCoalescence collision
DifficultyEasy -1.2 This is a straightforward application of conservation of momentum with a single equation to solve. Students substitute given values into momentum before = momentum after, then solve a simple linear equation for m. It requires only recall of a standard formula with no problem-solving insight or multi-step reasoning.
Spec6.03b Conservation of momentum: 1D two particles

1 A trolley, of mass 5 kg , is moving in a straight line on a smooth horizontal surface. It has a velocity of \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when it collides with a stationary trolley, of mass \(m \mathrm {~kg}\). Immediately after the collision, the trolleys move together with velocity \(2.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find \(m\).
(3 marks)

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Conservation of momentum: \(5 \times 6 + m \times 0 = (5 + m) \times 2.4\)M1 Correct momentum equation
\(30 = (5 + m) \times 2.4\)A1 Correct equation
\(m = \frac{30}{2.4} - 5 = 7.5\) kgA1 Correct value of \(m\)
# Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Conservation of momentum: $5 \times 6 + m \times 0 = (5 + m) \times 2.4$ | M1 | Correct momentum equation |
| $30 = (5 + m) \times 2.4$ | A1 | Correct equation |
| $m = \frac{30}{2.4} - 5 = 7.5$ kg | A1 | Correct value of $m$ |

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1 A trolley, of mass 5 kg , is moving in a straight line on a smooth horizontal surface. It has a velocity of $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it collides with a stationary trolley, of mass $m \mathrm {~kg}$. Immediately after the collision, the trolleys move together with velocity $2.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.

Find $m$.\\
(3 marks)

\hfill \mbox{\textit{AQA M1 2011 Q1 [3]}}