Momentum and Collisions with a Spring

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SUMMARY

The discussion focuses on a physics problem involving momentum and collisions with a spring. The spring constant is 200 N*m, and a 5 kg block is compressed 0.5 m before being released. The calculated velocity of the 5 kg block after release is 3.16 m/s, while the 2 kg block achieves a velocity of 7.9 m/s post-collision. The energy lost during the collision is determined to be 37.446 J. The discussion highlights confusion regarding the behavior of the 5 kg block after the collision and the assumptions necessary for solving the problem.

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  • Understanding of Hooke's Law and spring constants
  • Knowledge of conservation of momentum principles
  • Familiarity with kinetic energy calculations
  • Basic concepts of projectile motion
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  • Study the principles of elastic and inelastic collisions
  • Learn about the coefficient of restitution and its application in collision problems
  • Explore advanced projectile motion equations and their derivations
  • Investigate energy conservation in mechanical systems
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to momentum, collisions, and energy conservation.

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Homework Statement


There is a frictionless table 1 meter above the ground. A spring is on one end of the table with a constant of 200N*m that is compressed .5m by a block of 5kg. The block of 5kg is released and then collides with a 2kg block also sitting on the table.
a. Find the velocity of the 5kg block after it is released.
b. Find the velocity of the 2kg block after the colllision.
c. Find the amount of energy lost.
d. Find the distance between the impact points on the ground.

Homework Equations


a. 1/2kx^2 = 1/2mv^2
b. m1v1 + m2v2 = m1v1 + m2v2
c. Change in KE = KEfinal-KEinitial = 1/2mv^2 - 1/2mv^2
d. Projectile Motion? initial velocity off of the table at 7.9m/s for the 2kg and what for the 5kg?

The Attempt at a Solution


a. 3.16m/s
b. 7.9m/s
c. Change in KE = KEfinal-KEinitial = 1/2mv^2 - 1/2mv^2 = 1/2(5)(3.16^2) - 1/2(2)(7.9^2) = 37.446J
d. I was confused at this. I do not see how the 5kg block will hit the ground if it stops after the collision, or does it? If not then it will hit the spring again and be launched with a new velocity after the 2kg block hits?


I need help mostly on part d and think I have a-c although if I'm wrong please tell me. Thanks!
 
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The 5kg block will not stop. The problem is faulty, or has not been stated completely. There is no way you can do parts b, c, and d without making an assumption about energy loss in the first place. Was anything said about the nature of the collision? Is it head on or at an angle (assume head on unless otherwise stated)? Was a coefficient of restitution given?
 

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