Dumbbell and Spring

A spring dumbbell comprises two balls of mass m m that are connected with a spring of stiffness k k . Two such dumbbells are sliding towards one another on a smooth horizontal surface, the velocity of either is v 0 v_{0} . At some point, the distance between them is L L .

The time in which the distance between them L L again can be expressed as α L v 0 + β m k . \frac{\alpha L}{v_{0}}+\beta \sqrt{\frac{m}{k}} .

What is the sum of constants, α + β \alpha + \beta ?

Assume the collisions are perfectly elastic.


The answer is 3.221.

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1 solution

Talulah Riley
Oct 11, 2020

@Karan Chatrath Here it is

Talulah Riley - 8 months ago

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Thanks for sharing. Nicely done. I made a mistake while solving. Report deleted.

Karan Chatrath - 8 months ago

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