# Common Velocity at Maximum Deformation

1. Jan 19, 2005

### maverick280857

Hello everyone

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

A mass A is traveling with a constant intial velocity $$(v_{A})_{1}$$ and a massless relaxed spring (spring constant = k) is attached to the mass A. A second mass B is traveling with a constant velocity $$(v_{B})_{1}$$ such that $$(v_{A})_{1} < (v_{B})_{1}$$. 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 ($$x_{max}$$) 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:

$$\frac{1}{2}{m_{A}}{(v_{A})_{1}}^2 + \frac{1}{2}{m_{B}}{(v_{B})_{1}}^2 = \frac{1}{2}{m_{A}}{(v_{A})_{2}}^2 + \frac{1}{2}{m_{B}}{(v_{B})_{2}}^2 + \frac{1}{2}kx^2$$

$$m_{A}{(v_{A})_{1}} + m_{B}{(v_{B})_{1}} = m_{A}{(v_{A})_{2}} + m_{B}{(v_{B})_{2}}$$

(setting $$x=x_{max}$$ and $$(v_{A})_{2} = (v_{B})_{2}$$ gives an expression for $$x_{max}$$)

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

2. Jan 19, 2005

### arildno

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 $$\frac{dL}{dt}=0$$ at the time of maximal compression.

4. let the position of the spring tip in contact with A be $$P_{A}(t)$$, whereas the position of the spring tip in contact with B have position $$P_{B}(t)$$
Since $$L(t)=P_{B}(t)-P_{A}(t)$$
$$\frac{dP_{B}}{dt}=V_{B},\frac{dP_{A}}{dt}=V_{A}$$

(Note: this argument assumes that the spring remains straight throughout the motion)

Last edited: Jan 19, 2005
3. Jan 19, 2005

### Gokul43201

Staff Emeritus
An alternative approach :

In the center of mass frame of reference, where $$v({CoM) = \frac{m_Av_A + m_Bv_B}{m_A + m_B}~,$$
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 $~v(CoM)$ .

PS : The term 'deformation', if used without additional qualifiers (by materials scientists, at least) is more commonly used to refer to 'plastic deformation'.

Last edited: Jan 19, 2005
4. Jan 20, 2005

### maverick280857

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