1D Elastic Collisions with air gliders

[aeromat]

Homework Statement

Q: Two air track gliders of masses 300g, and 200g move towards each other in opposite directions with speeds of 50cm/s and 100cm/s respectively. Take the direction of the more massive glider as positive.

If the collision is elastic, find the velocity of each glider AFTER collision

[Given answer for the question: [-70cm/s, 80cm/s], but I'm not sure if that is respective to the gliders themselves]


Let A be the 300g plane
Let B be the 200g plane

Homework Equations

[1]---> MAVA + MBVB = MAV'A + MBV'B
[2]---> 1/2MVA^2 + 1/2MVB^2 = 1/2MV'A^2 + 1/2MV'B^2

1/2's are canceled out, so we get:
[2] MVA^2 + MVB^2 = MV'A^2 + MV'B^2

The Attempt at a Solution

Isolated V'B in the momentum conservation equation to get the following:

V'B = \frac{MAVA + MBVB - MAV'A}{MB}
V'B = \frac{(300)(50) + (200)(-100) - 300(V'A)}{200}

Subbed into the kinetic energy conservation equation:

MAVA^2 + MBVB^2 = MAV'A^2 + MB(\frac{MAVA + MBVB - MAV'A}{MB})^2
Let V'A rep x
(300)(50)^2 + (200)(-100)^2 = (300)(x)^2 + 200(\frac{(300)(50) + (200)(-100) - 300(x)}{200})^2

At this step, I am doubting whether or not I should countinue doing it this way. Is there any other algebraic method to get the two velocities after the collision, being it an elastic collision? I carry out this entire question taking up about half a page and then find myself scratching my head over a wrong answer...

 

[clementc]

There's a much neater way of doing this! Try using the difference of 2 squares and dividing the KE conservation equation by the momentum conservation equation