Maximum compression of Spring Blocks System

  • Thread starter better361
  • Start date
Block 1 has a mass of 2kg and is moving to the right on a frictionless surface at 10m/s. Block 2 is ahead of Block 1, has a mass of 5 kg, and is moving to the right at 3m/s. Block 2 also has a spring attached to its left end, which has a spring constant of 1120 N/m.

When the blocks collide, the compression of the spring is maximized at the instant both blocks have the same velocity. Find the maximum compression.

The system comprised of the two blocks have no external forces, so the velocity of the center of mass is constant. When the two blocks are moving with the same speed, their speeds from the center of mass's point of view is zero, so the speed at that instant of both blocks must equal the velocity of the center of mass. But,from here, i couldn't find a way to solve this without using conservation of energy. The furthest I got was that the impulse delivered to either block was equal to the integral of 1120 N/m *(x compressed) dt and I don't know how to integrate this.

My question is, is there a way to do it without energy?
 

tiny-tim

Science Advisor
Homework Helper
25,790
242
Well, the problem was from the center of mass and momentum chapter, so I figured the solution must have something to do with their equations. Also I would like to have a better understanding of how to transfer a function of a particular variable to another.

Heres an example:
Consider a 1-dimensional motion of a point mass particle of mass m=1 kg which is moving in a potential with the potential energy V(x)=1/x2 J. Initially the particle has velocity 5 m/s and position x=1 m. Find the position x in m of the particle at t=5 s.
I can find F(x) by taking the negative of the derivative of V(x). But, the differential equation I get from this is M*dv/dt=-1/x^3. And once again, I have an integral that I cannot solve.
 

tiny-tim

Science Advisor
Homework Helper
25,790
242
hi better361! :smile:

(just got up :zzz:)
M*dv/dt=-1/x^3. And once again, I have an integral that I cannot solve.
dv/dt = dv/dx*dx/dt (chain rule) = dv/dx*v

so M*v*dv/dx = -1/x3

so M*v*dv = -dx/x3 :wink:
 
Wow, never thought about that. Thanks
 
Actually, I still have issues with this problem. After I integrate the first time and found the constant, my equation is now v^2=1/x^2+24 (I substituted m=1kg) Then rooted each side and rewrote v as dx/dt. After integrating again, I got 1/24 sqrt(24+1/x^2)*x. Now I don't know if I need to include a constant again, because I already plugged in x and v the first time around. Also, either way my answer is wrong. :(
 

tiny-tim

Science Advisor
Homework Helper
25,790
242
hi better361! :smile:

(try using the X2 button just above the Reply box :wink:)
Consider a 1-dimensional motion of a point mass particle of mass m=1 kg which is moving in a potential with the potential energy V(x)=1/x2 J. Initially the particle has velocity 5 m/s and position x=1 m. Find the position x in m of the particle at t=5 s.
Actually, I still have issues with this problem. After I integrate the first time and found the constant, my equation is now v^2=1/x^2+24 (I substituted m=1kg) Then rooted each side and rewrote v as dx/dt. After integrating again, I got 1/24 sqrt(24+1/x^2)*x. Now I don't know if I need to include a constant again, because I already plugged in x and v the first time around. Also, either way my answer is wrong. :(
(wouldn't it be a lot easier to write it xdx/√(24x2 + 1) = dt ? :wink:)

yes, the second time you integrate, there will be another constant (consistent with x = 1 aat t = 0) :smile:
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top