# Elastic collision between 2 particles in 2D [closed]

A particle with mass $$m_1=m$$ moves along the x-axis at a velocity of $$v_0$$ and collides with another particle $$m_2=4m$$. As a result of the collision $$m_1$$ travels upwards at an angle of $$90 ^\circ$$. (see diagram)

How can I find the velocities after the collision?

My approach with conservation of momentum and kinetic enregy:

Momentum in the x direction: $$m_1v_0=m_1u_1(cos90^\circ)+m_2u_2 => v_0=4u_2cos(\alpha)$$

In the y direction: $$u_1=4u_2sin(\alpha)$$

Kinetic energy... $$v_0^2=u_1^2+4u_2^2$$

I've been going at it for quite a while now haha for some reason I'm always left with too many unknowns.

answers should be: $$u_1=v_0\sqrt{0.6} \\ u_2=v_0\sqrt{0.1}$$

Any help would be appreciated!

Try to square your solutions for $$v_0$$ and $$u_1$$, you'll get a simplification by summing them ; then you'll be able to express $$u_1$$ and $$u_2$$ as a function of $$v_0$$ using this new equation and the one obtained via the conservation of kinetic energy.
• By squaring you have $v_0^2=16u_2^2 cos^2(\alpha)$ and $u_1^2=16u_2^2 sin^2(\alpha)$. To get rid off the angle, you can just compute the sum $v_0^2 + u_1^2=16u_2^2$ Commented Feb 20 at 12:54