Solve Collision Problem: Two 8kg & 5.9kg Blocks

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SUMMARY

The discussion centers on solving a collision problem involving two blocks with masses of 8 kg and 5.9 kg on a frictionless surface, connected by a spring. After a cord is burned, the 5.9 kg block moves to the right at 20 m/s. Using the conservation of momentum equation, m1v1f + m2v2f = m1v1i + m2v2i, participants calculated the final velocity of the 8 kg block. The correct final velocity for the 8 kg block is determined to be -14.75 m/s, accounting for the direction of motion.

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1. Two blocks of masses 8 kg and 5.9 kg are
placed on a horizontal, frictionless surface. A
light spring is attached to one of them, and
the blocks are pushed together with the spring
between them. A cord holding them together
is burned, after which the block of mass 5.9 kg
moves to the right with a speed of 20 m/s.
What is the velocity of the other mass in
m/s?
Answer in units of m/s.




2. m1v1f+m2v2f=m1v1i+m2v2i



3. the blocks of masses m sub 1 and m sub 2 are connected by a cord with a spring between them. i assumed that they are initially at rest due to the cord. so the initial portion of the above equation would be 0 as their velocities would be 0. solving for v sub 2 final i get m1v1f/m2 =v2f. i get 14.75 m/s. this is incorrect.
 
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Maybe you're failing to take the direction of the velocity into account.
 
yup that was it lol. moving in opposite direction give opposite signs
 

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