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Two small spheres, each of mass m, lie on a smooth table, connected by a taut inelastic string of length 2l. The mid-point of the string is pulled with constant power P, along the table, perpendicular to the imaginary line connecting the balls. What is the velocity of each sphere toward the other, just before the spheres collide?

This question is a part of a series of questions that my colleagues and I have thought of. In each question, the mid-point is pulled with constant something: velocity, acceleration, force, or power. We have found (elegant) answers to the the first three questions, but we have not found the answer to the question involving constant power.

EDIT: The question was closed, possibly because I did not show enough of my own effort to find the answer. So, in an effort to re-open the question, I will explain what I have done so far.

The question with constant velocity can be answered by considering another situation: the mid-point is fixed to the table, and the spheres are projected along the table perpendicular to the string with velocity v, so that the spheres collide. In both situations, the relative motion between the spheres is the same.

The question with constant acceleration can be answered by considering another situation. You're on a planet where acceleration due to gravity is a. No table. With the string horizontal, hold the mid-point and let the spheres fall. In both situations, the relative motion between the spheres is the same.

The question with constant force can be answered by considering another situation. No string. Push each sphere along the table perpendicular to the imaginary line connecting them, with force F/2, for a distance d, which is, in the original situation, the distance traveled by the spheres perpendicular to the imaginary line connecting the spheres from the beginning until their collision. Then push each sphere toward the other sphere with force F/2 for a distance l. In both situations, the total work is the same, and the work done in pushing the spheres parallel to each other is the same, so the work done in pushing the spheres toward each other is the same.

For the question with constant power, I cannot think of a comparable situation which would give me the answer to the question. I can produce a differential equation to model the problem, but the differential equation is quite convoluted and I do not know how to solve it. My intuition tells me that, if there is an elegant answer, it can be found by using a comparison situation, and so the differential equation will not need to be solved.

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closed as off-topic by stafusa, Kyle Kanos, Jon Custer, M. Enns, John Rennie Jan 9 at 16:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – stafusa, Kyle Kanos, Jon Custer, M. Enns, John Rennie
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Please note that this site is not a place to obtain solutions to worked problems. Please see this Meta post on asking homework-like questions and this Meta post for "check my work problems". $\endgroup$ – Kyle Kanos Jan 9 at 11:05
  • $\begingroup$ This is not a homework question. I am a high school math teacher, and my colleagues and I created this question out of pure interest. $\endgroup$ – Dan Jan 9 at 11:09
  • $\begingroup$ Hi Dan, if you take a moment to read the links I provided, you'd see that we don't define "homework" to be the traditional "something assigned by an instructor" but instead it's defined as a question in which you want the method to solve a given problem. Since you're asking us how to solve the given problem, this is therefore a homework question. $\endgroup$ – Kyle Kanos Jan 9 at 11:13
  • $\begingroup$ Hi Kyle, apologies for not carefully reading the links you provided. Now I am more familiar with the expectations of this site. Thank you. $\endgroup$ – Dan Jan 10 at 2:19