Problem: two bodies linked via a spring

[fluidistic]

Homework Statement

Two bodies A and B of mass m_1 and m_2 are connected via a spring of natural longitud l_0 and elastic constant k. Both bodies are free of net force until at an instant t_i something applies a constant force F to the body A, in the direction of B. (see the diagram)
a)Calculate the initial acceleration of the center of mass of the system
b)Calculate the initial acceleration of each of the 2 bodies
c)Calculate the respective accelerations in the instant in which the spring is compressed by a length x.
d)Indicate all the pars of action-reaction forces in the moment in which the spring is compressed by a length x.

Homework Equations

F_{spring}=k \varedelta x
\sum \vec{F}=m\vec{a}.

The Attempt at a Solution

a)Just for fun I calculated the center of mass to be at \frac{l_0}{m_1+m_2} if the origin is situated at body A in instant t_i.
I think I've read somewhere that if an external force is applied, then it will modify the acceleration of the center of mass of the system following Newton's second law.
So \vec{a}_{CM}=\frac{F}{m_1+m_2}i. Am I right?
b)Using Newton's second law, \vec{a}=\frac{F}{m_1+m_2}i for the body A.
And here start my problems. I'm not sure how to find the acceleration of the body B at t_i. I'm tempted to go against my intuition and say that it will be the same as the center of mass of the system, but I don't think it's possible. Please help me going further this.

 

[Doc Al]

a)Just for fun I calculated the center of mass to be at \frac{l_0}{m_1+m_2} if the origin is situated at body A in instant t_i.
I think I've read somewhere that if an external force is applied, then it will modify the acceleration of the center of mass of the system following Newton's second law.
So \vec{a}_{CM}=\frac{F}{m_1+m_2}i. Am I right?

Yes, you are right.

b)Using Newton's second law, \vec{a}=\frac{F}{m_1+m_2}i for the body A.

Not exactly. What's the net force on body A? The mass of body A?

And here start my problems. I'm not sure how to find the acceleration of the body B at t_i. I'm tempted to go against my intuition and say that it will be the same as the center of mass of the system, but I don't think it's possible.

The way to find the acceleration of any system (which could be one mass or both masses) is to apply Newton's 2nd law to that system. Start by identifying the forces acting on the system of interest at the moment of interest.

 

[fluidistic]

Oops! I did it right but get confused when typing here.
b) For the body A, \vec{a}=\frac{F}{m_1}i.
So for the body B, I have to draw a F.B.D. in order to identify the forces acting on it at t_i.
I'm not sure... but at t_i the spring isn't compressed, right? Or it is, by a differential x? If so, then \vec{a}=\frac{k \cdot \Delta x}{m_2} and I think I could even work out \Delta x in function of F. Am I in the right direction?

 

[Doc Al]

b) For the body A, \vec{a}=\frac{F}{m_1}i.

Good.

So for the body B, I have to draw a F.B.D. in order to identify the forces acting on it at t_i.

Right.

I'm not sure... but at t_i the spring isn't compressed, right?

Right. At that moment, the spring has not had a chance to compress.

Or it is, by a differential x? If so, then \vec{a}=\frac{k \cdot \Delta x}{m_2} and I think I could even work out \Delta x in function of F. Am I in the right direction?

Let \Delta x = 0.

(You're doing fine.)

 

[fluidistic]

Thank you very much Doc Al. I've completed the problem now.
I understand better now that it's possible for a body to have an acceleration but with a velocity equal to 0. It happens in springs and pendulums for example.
In this problem, the fact that A has no velocity in t_i is the responsible of the no-compression of the spring and so that B cannot suffer any force thus any acceleration.

 

[Doc Al]

Good.

It's certainly possible for something to have an acceleration but (at least for the moment) zero velocity. Consider a ball tossed straight up. When it reaches the top, its velocity is zero, yet its acceleration is always g downward.

In this problem, the trick is realizing that the force exerted by a spring is simply given by its degree of stretch or compression. No stretch or compression, no force.