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I have a question where two particles collide with each other in a non central way.

Mass of particle 1 ($m_1$) is $1\text{kg}$ and its velocity $v_1$ is $5\text{m/s}$ (moving along the x-axis) before it collides with particle 2 (which isn't moving). After the collision, the particle 1 will have a momentum of $2\text{kg.m/s}$ in the x axis ($P_{1x} = 2\text{kg.m/s}$) and a momentum of $-3\text{kg.m/s}$ in y axis ($P_{1y} = -3\text{kg.m/s}$).

I am asked to find the kinetic energy of particle 2 after the collision.

I know that kinetic energy is conserved in elastic collisions and thus; $$\frac{1}{2} m_1(v_{1i})^2 + \frac{1}{2} m_2(v_{2i})^2 = \frac{1}{2} m_1(v_{1f})^2 + \frac{1}{2} m_2 (v_{2f})^2$$

But I am confused when it comes to working in two dimensions $(x,y)$. Can you please help me about this by maybe providing with some equations or a helpful link?

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  • $\begingroup$ It is stated that it moves along the x-axis $\endgroup$ – Taylan Dec 10 '18 at 19:04
  • $\begingroup$ yes I mean I just stated it to you, not included in my text $\endgroup$ – Taylan Dec 10 '18 at 21:36
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For energy conservation, the directions of the vectors are not important, as energy is a scalar quantity. For the kinetic energy you can simply plug in everything you have in the text into the equation you stated - as long as the collision is elastic. The directions only matter for the conservation of momenta, this is $$m_1v_{1i}+m_2v_{1i}=m_1v_{1f}+m_2v_{2f}\,,$$ where you need to take care of the directions of the vectors, i.e., the direction of the momenta.

Sometimes it is useful to combine conservation of energy and conservation of momenta to solve for unknown quantities. See for example https://en.wikipedia.org/wiki/Elastic_collision

But be careful - kinetic energy is in general not conserved for inelastic collisions.

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