Common Velocity at Maximum Deformation

In summary, the two masses A and B are traveling with different velocities and when they collide, the spring compresses. It is not possible to find the maximum compression of the spring using elementary classical mechanics. However, using more advanced mechanics it is possible to find that the spring is at its maximum compression when the sum of the kinetic energy of the masses is minimum.
  • #1
maverick280857
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Hello everyone

I came across a few (relatively standard) problem in elementary classical mechanics...

A mass A is traveling with a constant intial velocity [tex](v_{A})_{1}[/tex] and a massless relaxed spring (spring constant = k) is attached to the mass A. A second mass B is traveling with a constant velocity [tex](v_{B})_{1}[/tex] such that [tex](v_{A})_{1} < (v_{B})_{1}[/tex]. The mass B collides with the spring compressing it. Find the maximum compression of the spring (one part of the problem). Assume that the surface on which A and B move is frictionless.

This is how I worked it out (and got the right answer): (by the way A is the leading mass and B is the lagging mass initially and the spring is between A and B)

At maximum compression ([tex]x_{max}[/tex]) the system (A+B+spring) will behave (momentarily) as a rigid body and so relative velocity of A and B will be zero. In other words, at maximum compression, A and B will have a common velocity which can be computed by applying the energy momentum equations:

[tex]\frac{1}{2}{m_{A}}{(v_{A})_{1}}^2 + \frac{1}{2}{m_{B}}{(v_{B})_{1}}^2 = \frac{1}{2}{m_{A}}{(v_{A})_{2}}^2 + \frac{1}{2}{m_{B}}{(v_{B})_{2}}^2 + \frac{1}{2}kx^2[/tex]

[tex]m_{A}{(v_{A})_{1}} + m_{B}{(v_{B})_{1}} = m_{A}{(v_{A})_{2}} + m_{B}{(v_{B})_{2}}[/tex]

(setting [tex]x=x_{max}[/tex] and [tex](v_{A})_{2} = (v_{B})_{2}[/tex] gives an expression for [tex]x_{max}[/tex])

Now, it is obvious and logical that the velocities of A and B will be common at maximum deformation but can this proved mathematically? In other words, can I use the energy momentum equations treating momentum conservation as a constraint on the two final velocities and maximize x somehow and arrive at the condition that the velocities are equal at maximum x? I tried doing this but I was not successful.

As I understand a rigorous mathematical proof is not available in the realms of elementary mechanics but is a solution possible using some more advanced mechanics/mathematics?

Thanks and cheers
Vivek
 
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  • #2
Well, I don't find the following argument terribly advanced:

1. At maximal compression, it is certainly true that B's velocity must equal the velocity of the spring tip which is in contact with B.
2. At all times, the velocity of the spring tip attached to A is equal to A's velocity.

3. The length of the spring is a minimum at the time when maximal compression occurs;
that is, if L(t) is the total length of the spring, we must have [tex]\frac{dL}{dt}=0[/tex] at the time of maximal compression.

4. let the position of the spring tip in contact with A be [tex]P_{A}(t)[/tex], whereas the position of the spring tip in contact with B have position [tex]P_{B}(t)[/tex]
Since [tex]L(t)=P_{B}(t)-P_{A}(t)[/tex]
your result follows, since:
[tex]\frac{dP_{B}}{dt}=V_{B},\frac{dP_{A}}{dt}=V_{A}[/tex]

(Note: this argument assumes that the spring remains straight throughout the motion)
 
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  • #3
An alternative approach :

In the center of mass frame of reference, where [tex]v({CoM) = \frac{m_Av_A + m_Bv_B}{m_A + m_B}~, [/tex]
the total momentum is always zero.

Clearly, the spring is at maximum compression when the sum of KEs is minimum. In this frame, this happens when both velocities become zero. So, at this instant, in any other frame, both velocities must be the same, and equal to the velocity of that frame, relative to [itex]~v(CoM)[/itex] .

PS : The term 'deformation', if used without additional qualifiers (by materials scientists, at least) is more commonly used to refer to 'plastic deformation'.
 
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  • #4
Thanks arildno and Gokul. I thought of a similar approach as Gokul's and I also came across such an explanation (though rather contorted and involving "pseudo forces") in a book but I decided to keep myself on the ground frame.

Cheers
vivek
 

What is common velocity at maximum deformation?

Common velocity at maximum deformation refers to the velocity at which an object reaches its maximum deformation or deformation limit. It is the point where the object can no longer withstand the applied force and permanently changes its shape.

Why is common velocity at maximum deformation important?

Common velocity at maximum deformation is important because it determines the maximum amount of force an object can withstand before permanently changing its shape. This information is crucial in engineering and design to ensure the safety and functionality of structures and materials.

How is common velocity at maximum deformation calculated?

Common velocity at maximum deformation is calculated by dividing the maximum force applied to an object by its cross-sectional area. This gives the stress on the object, which can then be compared to its stress-strain curve to determine the common velocity at maximum deformation.

What factors can affect common velocity at maximum deformation?

Common velocity at maximum deformation can be affected by various factors such as the material properties of the object, the shape and size of the object, and the rate at which the force is applied. Temperature and environmental conditions can also play a role in the deformation of an object.

What are some real-life examples of common velocity at maximum deformation?

Some real-life examples of common velocity at maximum deformation include the breaking point of a bone in the human body, the collapse of a building during an earthquake, and the deformation of a car during a high-speed collision. These are all situations where the applied force exceeds the object's maximum deformation limit.

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