Conservation of kinetic energy in two dimensional elastic collisions

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?

• It is stated that it moves along the x-axis – Taylan Dec 10 '18 at 19:04
• yes I mean I just stated it to you, not included in my text – Taylan Dec 10 '18 at 21:36

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.