How would i calculate m2 in a completely inelastic 2d collision?

Question

Here is the question: Two ice skaters undergo a collision, after which their arms are intertwined, and they have a common velocity of 0.85 m/s [27° S of E]. Before the collision, one skater of mass 71 kg had a velocity of 2.3 m/s [12° N of E], while the other skater had a velocity of 1.9 m/s [52° S of W]. What is the mass of the second skater?

Answer 1

There are three ways to do this that depend on what you have done in class.

1) The motion of the center of mass must be the same before and after the collision

2) The vector sum of the momenta of the incoming masses equals the momentum of joined masses after the collision

These are equivalent and can be solved using cosine law to find sum of momentum vectors or algebra/trig decomposition and adding of the momentum vectors

3) The easiest way is to assume that the problem designer wrote the problem correctly and you can solve this problem doing a momentum balance in just x or just y:

m₁v_1x + m₂v_2x = (m₁ + m₂)v_12x where v₁₂ is the velocity after the collision with 1 and 2 stuck together (The only unknown is m₂ and you can solve for it.

In order to find the x components you need to transform the headings to math angle, θ, so that you can use the equation mv_1x = mv₁cosθ

12: 27°S of E is -27 1. 12° N of E is 12 2. 52º S of W is -128 or 232°

You know everything else.