I got a trouble with solving this problem.

"Two balls with masses $m$ each are connected by spring of length $l$ and stiffness $k$ and initially at rest on the table without friction. One ball acquires initial speed $v$ perpendicular to the line connecting two balls. Determine this speed if the maximum length of spring during motion is $L$."

I thought that it is just about conservation of energy

$$ \frac{m v^2}{2} = \frac{k (L-l)^2}{2}$$ where I assumed that when the stretching is maximum the velocities are zero. But it seems that it is completely wrong and one needs more accurate solution.

I also found some hint that one of the ways to solve this problem is to solve it in the center of mass reference frame. But I don't understand two things:

  1. How to find the speed of center of mass right after the initial moment? It looks like one of the balls has speed $v$ and another is at rest (in the laboratory reference frame), so the speed of center of mass is

$$ v_{com} = \frac{m v}{m + m} = \frac{v}{2} $$

It in its turn means that on the center of mass ref. frame one ball (with speed $v$) moves with speed $v/2$ while another moves with $-v/2$ in the opposite direction.

Am I right?..

  1. Why do the balls continue their motion while the stretching is maximal? In other words, why does not all kinetic energy transform to the potential one?

P.S. And also why is it simpler to introduce COM reference frame? I think we can use both conservation laws (energy and angular momentum) in the laboratory frame, can't we?


closed as off-topic by Kyle Kanos, John Rennie, Jon Custer, Yashas, sammy gerbil Sep 7 '17 at 20:03

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You have a bunch of questions there... Let's take them in turn.

1) speed of COM : your method is correct. This will be the speed of the COM not only right at the start, but afterwards as well (this is the whole point of working with the COM frame... the center of mass is stationary in that frame).

2) why do the balls continue their motion? The configuration has angular momentum! So when the string is fully stretched, the two masses are rotating about the COM, and they will carry some amount of kinetic energy during that rotation. You solve the problem by setting the initial kinetic energy equal to the final kinetic energy plus the potential energy in the spring.

Note - if you did not use the COM, you would have to solve the equations of motion of the two point masses - you don't really know where they will be when the spring reaches maximum stretch, so you don't know the direction of their motion. This makes things a LOT more complicated. In the COM you just have two balls going around a common center, attached by a spring. MUCH simpler.

Can you take it from here?

  • 1
    $\begingroup$ Yes, thank you very much. It was a conceptual question. $\endgroup$ – newt Sep 2 '17 at 17:00

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