Finding individual masses of ice dancers pushing apart

[Patra]

Homework Statement

Two ice dancers, Boris and Natasha, have a combined mass of 136kg. They embrace and push apart from each other: Boris moving at 0.90m/s and Natasha moving at 1.2m/s in the opposite direction. Find their individual masses.

Homework Equations

p=mv

The Attempt at a Solution

It seems so stunningly simple, yet I am at a loss as to how to attempt it without having a time variable or knowing the mass of one of the skaters.

 

[Sniperman724]

use the equation (m1+m2)vf=mv1+mv2, this is an inelastic collision problem. Maybe that will help a little more than p=mv

 

[Patra]

Thanks, so does that mean it should be something along the lines of (m1+m2)vf=(136kg)(0.90m/s)+(136kg)(1.2m/s)? Should I be using substitution at some point during this problem?

 

[Sniperman724]

no, both ice dancers start together, then push a part from each other
(m1 + m2)v just equals the momentum of the two dancers as a whole because they both initially start out together.
so the only two unknown variables are m1 and m2

 

[NeedsHelp1212]

(136 kg) * 0 m/s = m1 (1.2 m/s) + m2 ( -.9 m/s)

0 = 1.2m + -.9 m Keep in mind they are separate variables

You also have m1 + m2 = 136.

So put those equations on top of each other and i like to use elimination. Here is the work:

1.2m -.9 m = 0
-1.2(m1 + m2 = 136)
Which gives you
1.2 m -.9 m =0
-1.2 m -1.2 m = -163.2
Which simplifies to
-2.1 m = -163.2 (we are solving for m2, i always made m2 second in every equation)
m2 = 77.71 kg
m1= 136- 77.71= 58.29 kg

You can check it and you get there respective momentums which should be equal to be 69.948= -69.939, so its close enough

* And yeah my method is a little different from the other persons. I did this chapter a while ago so i kind of forgot that stuff..lol