Elastic collison of 2 moving object

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In a perfectly elastic collision involving a 0.900 kg object moving east at 2.8 m/s and a 1.30 kg object moving west at 1.4 m/s, both momentum and kinetic energy are conserved. The momentum equation is m1V1i + m2V2i = m1V1f + m2V2f, while the kinetic energy equation is m1V1i^2 + m2V2i^2 = m1V1f^2 + m2V2f^2. By substituting the given values and considering the direction of motion, the velocities after the collision can be determined. A simplification method is suggested to avoid using quadratics by doubling the kinetic energy terms. The discussion emphasizes the importance of setting up simultaneous equations to solve for the final velocities.
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a .900 kg object is moving east at 2.8 m/s.
a 1.30 kg object is moving west at 1.4 m/s.
If the collision is perfectly elastic find the velocities after the collision. (neglect friction)

I know momentum and kinetic energy is conserved but I am really clueless as to what to do.
Please help me start this problem.
 
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Two simultaneous equations here may be useful, one for KE and one for momentum.

Let V1i be object 1's velocity before impact V2i be object 2's veloctiy before impact V1f be object 1's velocity after imapct and V2f be object 2's velocity after impact. m1 is first object's mass, m2 = object 2's mass

Then m1V1i + m2V2i = m1V1f + m2V2f.
m1V1i^2 + m2V2i^2 = m1V1f^2 + m2V2f^2

(Note how I have canceled out the 'half' in K = 1/2 * mv^2 - in effect doubling every term in the 2nd equation, so it's simpler to work with but still doesn't effect the system of equations).

Substitute in the values you are given. Take moving east as moving in the positive direction (and therefore moving west in the negative direction).
 
You can just use Vi - Vf = -(V2i - V2f)

Is a shortcut my professor showed us instead of using quadratics by 1/2 mv^2
 
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