- #1
maverick280857
- 1,789
- 5
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
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