Forces two objects apply on each other during inelastic collision

[austindubose]

Homework Statement

Values from previous problem:
Pre-collision: v_2kg=6m/s -->, v_3kg=3m/s <--
Post-collision: v_2kg=2m/s <--, v_3kg=5.67m/s -->
Collision time: 0.2 sec

Homework Equations

F=mA=m(Δv/Δt)

The Attempt at a Solution

My instinct to solve this was to use the pre-collision values to find out the forces each puck had on each other, because it seems like post-collision values for velocity would be irrelevant because the pucks are no longer having any force on each other. So, I did the following...

Force of 2kg puck on 3kg puck: F=2kg(6m/s / 0.2 sec) = 60N
Force of 3kg puck on 2kg puck: F=3kg(3m/s / 0.2 sec) = 45N

Is this the correct way to approach this problem? Thanks for your help!

 

[WatermelonPig]

J = Favg*t. Also, Newton's Third Law.

 

[austindubose]

Can you elaborate a bit on that? I'm not completely familiar with that formula. We probably just use different variables because I don't recognize the J.

 

[WatermelonPig]

J = magnitude of change in momentum.

 

[Collaptic]

I think watermelonpig is saying Impulse= Force(average) x Time Interval. I=Ft. I think most physics books call change in momentum, impulse, so maybe that will help. Your also given the time...