What is the speed of m2 after the spring is released?

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The discussion focuses on determining the speed of block m2 after a spring is released between two blocks on a frictionless surface. The initial momentum of the system is zero since both blocks are at rest. By applying the conservation of momentum principle, the equation m1v1 = m2v2 is used to find the speed of m2. Given that m1 has a speed of 3.0 m/s, the calculation leads to the conclusion that m2 must have a corresponding speed to conserve momentum. The key takeaway is that the momentum before and after the release remains constant in the system.
bulldog23
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[SOLVED] Conservation of Momentum

Homework Statement


Two blocks with masses m1 = 2.6 kg and m2 = 3.6 kg are at rest on a frictionless surface with a compressed spring between them. The spring is initially compressed by 60.0 cm and has negligible mass. When both blocks are released simultaneously and the spring has dropped to the surface, m1 is found to have a speed of 3.0 m/s. What is the speed of m2?
prob05a.gif



Homework Equations


P=m*v


The Attempt at a Solution


I am trying to figure out the initial momentum of the system, so do I m1v1=m2v2?
 
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bulldog23 said:
Two blocks with masses m1 = 2.6 kg and m2 = 3.6 kg are at rest on a frictionless surface
Why do u have to calculate initial momentum? Isn't it zero!?
Yes, you may apply conservation of linear momentum, after thinking about all the forces acting on the system.
 

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