Two students on roller skates stand face-to-
face, then push each other away. One student
has a mass of 93 kg and the second student
69 kg.
Find the ratio of the magnitude of the first
student’s velocity to the magnitude of the
second student’s velocity.

Question

Two students on roller skates stand face-to-
face, then push each other away. One student
has a mass of 93 kg and the second student
69 kg.
Find the ratio of the magnitude of the first
student’s velocity to the magnitude of the
second student’s velocity.

anonymous · Answer

Assuming that the wheels are frictionless, zero air drag, and so forth, we can use the Linear momentum equation:

$$m _{1}u _{1}+m _{2}u _{2} = m _{1} v _{1}+m _{2}v _{2}$$

where the "m1" and "m2" are the mass of the two students respectively, and the "u" being their initial velocity and and "v" being the final velocity after they push each other away.

Keep in mind they were at rest in the beginning so the linear momentum is conserved, in which the first part of the equation reverts to "Zero":
$$0 = m_1v_1+m_2v_2$$
From here you can figure out the ratio of the magnitude of the first student's velocity to the second student's magnitude of velocity, as you are given the mass of both.

anonymous · Answer

would they have the same velocity?

anonymous · Answer

Or do I set mv=mv and solve for each... ha idk

anonymous · Answer

No, they would not have the same velocity as it also depends on their mass which is different. 

Yes, set them equal to each other, but remember the original equation was:$$0 = m_1v_1 + m_2v_2$$ so how do you bring the m1v1 or the m2v2 to the other side? You would subtract them right? In which, it will give a negative m1v1 or m2v2 depending on which one you choose to subtract to the other side. The reason why one is negative is because they are moving "away" from each other.

preetha · Answer

Nice work Orion!

anonymous · Answer

Thank you!