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I based my answer around kinetic energy. We can let [itex]v_1[/itex] and [itex]v_2[/itex] be the velocities of the masses relative to the center of mass, and [itex]v[/itex] be the velocity of the center of mass. The total initial energy of the system (before the spring is compressed) as: [itex]E=\frac{m_1}{2}v_1^2+\frac{m_1}{2}v_1^2+\frac{M}{2}v^2[/itex]. (where [itex]M[/itex] is the total mass of the system) My argument is that we want to minimize the amount of energy locked up in the movement of the center of mass so that the energy can be used to give potential energy to the spring. So then:A 2-kg mass and an 8-kg mass collide elastically, compressing a spring bumper on one of them.

part 1) Which was of causing the collision to occur will result in the greater compression of the spring: Giving the 8-kg mass 16J of kinetic energy or giving the 2-kg mass 16J of kinetic energy? (the other mass remains stationary.

part 2) keeping the total kinetic energy to be 16J, how should this energy be divided to obtain the greatest compression of the spring?

[itex]\frac{M}{2}v^2=\frac{m_1+m_2}{2}\cdot \frac{(m_1 v_1+m_2 v_2)^2}{(m_1+m_2)^2}[/itex].

We also have [itex]J=\frac{1}{2}mv^2\rightarrow v=\sqrt{\frac{2 J}{m}}[/itex]. The other velocity is zero, so:

[itex]\frac{(m_1 v_1+m_2 v_2)^2}{2(m_1+m_2)}=\frac{(m_1 v_1)^2}{2(m_1+m_2)}[/itex]

Since the bottom is constant we can just consider the top term

[itex]=m_1^2 v_1^2=m_1^2\frac{2 J}{m_1}=m_1 2 J[/itex]

To minimize this term, we obviously need to use the 2kg mass. The second part can be completed in the same way by setting the center-of-mass's velocity to be zero.

My question is... how can I justify this better? I'm really uncomfortable with energy based arguments