How Much Kinetic Energy is Stored in the Spring After a Collision?

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SUMMARY

The discussion focuses on calculating the fraction of kinetic energy stored in a spring after a collision involving a ball and a spring gun. A ball with a mass of 0.25 kg, traveling at 120 m/s, collides with a spring gun of mass 1.8 kg, which is initially at rest. The conservation of momentum is applied using the formula (m1+m2)V_cen. of mass = m1v1 + m2v2, and the kinetic energy of the center of mass is determined to find the energy transferred to the spring. The method outlined ensures that no energy is lost to friction, providing a clear path to the solution.

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A ball of mass 0.25 kg is fired with velocity 120 m/s into the barrel of a spring gun of mass 1.8 kg initially at rest on a frictionless surface. The ball sticks in the barrel at the point of maximum compression of the spring. No energy is lost to friction. What fraction of the ball's initial kinetic energy is stored in the spring?
 
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You'll need inelastic collision conservation of momentum:
(m1+m2)V_cen. of mass = m1v1 +m2v2

And a translation into a center of mass reference frame:
v'1 + v_cen. of mass = v1

If you get the kinetic energy of the center of mass of the system before the collision that will be the kinetic energy that was lost into internal spring potential...

Maybe there is an easier way, but I'm pretty sure this will get you the answer.
 

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