What is the Velocity of Gliders A and B in an Inelastic Collision?

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In an inelastic collision between two gliders, the total momentum before the collision is calculated using the formula m1v1 + m2v2, where glider A has a mass of 0.269 kg and a speed of 1.07 m/s, and glider B has a mass of 0.330 kg moving at -0.675 m/s. The combined mass after the collision is (m1 + m2), and since the gliders stick together, they move with a common velocity, v3. The momentum conservation equation is set up as (0.269 * 1.07) + (0.33 * -0.675) = (0.269 + 0.33)v3. Solving this gives the velocity of the combined gliders post-collision, demonstrating the principle of momentum conservation in inelastic collisions.
KU_Mustang
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Given:

Two gliders on an air track collide. Glider A has a mass of 0.269 kg and is moving in the +x direction with a speed of 1.07 m/s. Glider B has a mass of 0.330 kg and is moving in the -x direction with a speed of 0.675 m/s. a) What is the velocity of A and B if the collision is completely inelastic? Calculate the velocities of b) A and c) B if the collision is completely elastic.

Im stuck on solving for the veloctiy of A + B if it in inelastic. I understand how to do it if it is elastic, just not sure what to do when KE is transfered.
 
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If the collision is completely inelastic then you don't need to worry about the energy.

setup the conservation of momentum equation like this:
m1v1 + m2v2 = (m1 + m2) v3
(m1 + m2) because they are stuck together post collision.

you have m1, m2, v1 and v2 so just solve for v3.
 
COMPLETELY inelastic means that the two objects stick together and move as one after the collision so you only have one variable, the common speed after the collision.
Before the collision, A's momentum was 0.269(1.07) kg and B's momentum was (0.33)(-0.675) so the total momentum was (0.269)(1.07)-(0.33)(0.675). Their joint momentum after the collision is (0.260+0.33)v. Set those equal and solve for v.
 
Thank you so much. I really do understand it now.
 
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