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2D Collisions

Part of the Mechanics section of Coddy's Physics journey. Lesson 72 of 97.

A collision at an angle: conserve momentum along x, and separately along y.

∑px before=∑px after,∑py before=∑py after\textstyle\sum p_x \text{ before} = \sum p_x \text{ after}, \quad \sum p_y \text{ before} = \sum p_y \text{ after}

Top view. Before: puck A moves along the x axis at 5 metres per second toward puck B at rest. After: A moves at 2.5 metres per second at 60 degrees above the x axis and B at 4.33 metres per second at 30 degrees below it

After the hit, A goes up and to the right, B down and to the right.

No y momentum before: the y momenta after are equal and opposite. Tip to tail, the momenta after add up to the momentum before.

A momentum triangle: the momentum of A before the collision is the horizontal side; the momenta of A and B after it, placed tip to tail, run from the same start to the same end, with a right angle between them

The two momenta after the collision add up to the momentum before.

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Challenge

A 0.2 kg0.2\ \mathrm{kg} puck slides at 6 m/s6\ \mathrm{m/s} along the x axis and hits a 0.3 kg0.3\ \mathrm{kg} puck at rest. Afterwards the 0.2 kg0.2\ \mathrm{kg} puck moves at 3 m/s3\ \mathrm{m/s}, 60∘60^\circ above the x axis. Find the x velocity v2x\relax v_{2x} of the 0.3 kg0.3\ \mathrm{kg} puck.

Try it yourself

Physics diagram: path, ball, vec; arrows v 1 = 6 m/s, v 1f = 3 m/s, v 2x = ?

Solve for v2xv_{2x}

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