Physics problems... yeah!?

Question

I need help working out problems but I have the answers so your help is appreciated..

Two brothers, Dustin and Parker, have a combined mass of 168 kg. At an ice skating rink, they stand close together on skates, at rest and facing each other, with a compressed spring between them. the spring is kept from pushing them apart because they are holding each other. When they release their arms, Dustin moves off in one direction at a speed of 0.90 m/s, while Parker moves off in the opposite direction at a speed of 1.2 m/s. Assuming that friction is negligible, find Dustin’s mass.

Answer 1

Dustin's mass is 96 kg

Since there are no external forces (no friction), then the center of mass of the system "Dusting+Parker+spring" is not moving. You can also say that the system momentum is conserved. If we neglect the spring mass, this means that after releasing the spring, Dusting and Parker will move off in opposite directions, and they will have equal products "mass times speed":

m_{D}*v_{D} = m_{P}*v_{P}.

Here, {D} refers to Dustin and {P} refers to Parker. You also know the total mass:

m_{D}+m_{P} = M_{total}.

From these two equations it follows

m_{D} = M_{total}*v_{P}/(v_{D}+v_{P})=
168*1.2/(0.9_1.2)=96.

Answer 2

First the total mass is
m1+m2=168 and
m2= 168-m1
Now lets write an energy conservation equation and equate the potential energy of the spring to kinetic energy of the brothers.
Ps=|Ke1| + |Ke2| when Ps=0

Ke1=Ke2

0.5 m1V1^2=0.5m2V2^2

It follows that
m1V1^2=(168-m1)V2^2
m1(V1^2+V2^2)= 168 V2^2
m1= 168 V2^2/(V1^2+V2^2)
m1= 168 x1.2^2/(0.9^2 +1.2^2)

m1=108 kg
m2= 168- 108=60 kg