AQA M1 2013 June — Question 1 3 marks

Exam BoardAQA
ModuleM1 (Mechanics 1)
Year2013
SessionJune
Marks3
PaperDownload PDF ↗
TopicMomentum and Collisions
TypeCoalescence collision
DifficultyEasy -1.2 This is a straightforward one-step conservation of momentum problem with coalescence. Students simply apply m₁u₁ + m₂u₂ = (m₁+m₂)v with one object at rest, requiring only direct substitution and basic algebra. No problem-solving insight needed, just routine application of a standard formula.
Spec6.03b Conservation of momentum: 1D two particles

1 A toy train of mass 300 grams is moving along a straight horizontal track at a speed of \(2.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). This toy train collides with another toy train, of mass 200 grams, which is at rest on the same track. During the collision, the two trains lock together and then move together. Find the speed of the trains immediately after the collision.

1 A toy train of mass 300 grams is moving along a straight horizontal track at a speed of $2.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. This toy train collides with another toy train, of mass 200 grams, which is at rest on the same track. During the collision, the two trains lock together and then move together.

Find the speed of the trains immediately after the collision.

\hfill \mbox{\textit{AQA M1 2013 Q1 [3]}}