Free Body Diagram of Mass-Spring System

[chaneth8]

Homework Statement

Draw the Free

Relevant Equations

$F = -kx$

Suppose we are given the 2 following masses 1 and 2, where 1 initially moves at velocity $v_\rm{1}$ and 2 is stationary. Note, however, that 2 is not bolted down to any surface - it is free to move around after collision. What would the free-body force diagram of masses 1 and 2 be, after they collide?

This is what I think it will be - I just want to check because I'm not 100% sure.

Suppose the spring is compressed by length $x$ from its relaxed position after collision. Then the force pushing block 1 to the left is the spring force of magnitude $kx$:

Similarly, from the perspective of block 2, the spring is compressed by by length $x$ too, so it will push block 2 to the right by a force of magnitude $kx$:

The reason I'm not sure if this is correct is because block 2 is allowed to move around - is the only force that is pushing block 2 to the right the spring force, or are there more forces?

I'd appreciate any guidance to this problem.

 

[TSny]

Your diagrams are correct for the horizontal forces acting on the blocks. However, to make the diagrams complete, you should show all of the vertical forces also.

 

[apostolosdt]

I was under the impression that free-body diagrams are only used in static cases to depict all applied forces summing up to equilibrium.

 

[haruspex]

I was under the impression that free-body diagrams are only used in static cases to depict all applied forces summing up to equilibrium.

I'm not.

 

[apostolosdt]

I'm not.

On second thought, you're most probably right; the role of a free-body diagram is to show only the essential parts of the situation. Thank you for correcting me.

 

[kuruman]

I was under the impression that free-body diagrams are only used in static cases to depict all applied forces summing up to equilibrium.

Free body diagrams are used as aids to determine the vector summation on left-hand side of $\vec F_{\text{net}}=m\vec a$ or $\vec{\tau}_{\text{net}}=I\vec{\alpha}.$ In cases where the acceleration (linear or angular) is not known, the vector sum on the left-hand side determines whether the right-hand is or is not zero.