How Far Do Colliding Blocks Slide on a Spring?

AI Thread Summary
The discussion centers on a physics problem involving two colliding blocks attached to a spring, where one block is initially at rest and the other is positioned closer to the spring's equilibrium. The user applies conservation of energy to determine the velocity at which the blocks collide and uses Newton's second law to calculate the subsequent acceleration and distance they slide together after the collision. There is some confusion regarding the application of conservation of momentum and the mass used in calculations. Additionally, the user expresses frustration over a lack of responses and considers deleting the thread. Overall, the conversation highlights the complexities of solving inelastic collision problems in the context of spring mechanics.
naianator
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Homework Statement



problems_MIT_boriskor_BKimages_10-mass-spring-two-block-collision.png


A block of mass m is attached to a spring with a force constant k, as in the above diagram. Initially, the spring is compressed a distance x from the equilibrium and the block is held at rest. Another block, of mass 2m, is placed a distance x/2 from the equilibrium as shown. After the spring is released, the blocks collide inelastically and slide together. How far (Δx) would the blocks slide beyond the collision point? Neglect friction between the blocks and the horizontal surface.

Homework Equations


U_spring = 1/2*k*x^2

K = 1/2*m*v^2

F = m*a

v_f^2 = v_0^2+2*a*x

The Attempt at a Solution


E_i = 1/2*k*x^2 = E_f = 1/2*m*v^2 + 1/2*k*x_f^2

at x/2 where m collides with 2m:

1/2*k*x^2 = 1/2*3m*v^2 + 1/8*k*x^2

k*x^2 = 3m*v^2 + 1/4*k*x^2

3/4*k*x^2 = 3*m*v^2

yielding

v = sqrt(k*x^2/(4*m))

by Newtons second law a = -k*x/(3*m)

and using kinematics (v_f^2 = v_0^2+2*a*x)

0 = k*x^2/(4*m) - 2*(k*x/(3*m))*delta(x)

x/4 = 2/3*delta(x)

delta(x) = 3/8*x

Am I way off base with this approach?
 
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Could you possibly explain what you have written , in words ?
It is a bit difficult to understand what you have typed .
 
Qwertywerty said:
Could you possibly explain what you have written , in words ?
It is a bit difficult to understand what you have typed .
I used conservation of energy to find the velocity where the two blocks collide at x/2. I'm not sure about this though, because I don't think that conservation of momentum applies and I used 3m for the mass (assuming that the blocks had collided). Then I used N2L to find the acceleration based on the force (-kx) and mass (3m). I plugged that into the kinematics equation for final velocity and solved for delta x.
 
naianator said:
Yes, sorry, I wasn't having any luck with responses and it was due a half hour ago so I reposted it. Is there a way to delete this thread?
No, but you might be able to edit the title to something that will stop readers wasting their time delving into it.
A moderator could delete it.
 
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