Two identical metal balls connected at the ends of a light spring of force constant k form a dumbbell like structure. The dumbbell rests on a frictionless horizontal floor and third identical ball is placed at distance $l$ from the right ball of the dumbbell. All the three balls are in a line. A fourth identical ball moving with velocity u collides with left ball of the dumbbell
If all the collisions between the balls are elastic and head-on, the minimum distance $l$ so that the rightmost ball acquires speed u is $π u \sqrt{\frac{m}{x k}}$ where $x$ is------

Question

Two identical metal balls connected at the ends of a light spring of force constant k form a dumbbell like structure. The dumbbell rests on a frictionless horizontal floor and third identical ball is placed at distance $l$ from the right ball of the dumbbell. All the three balls are in a line. A fourth identical ball moving with velocity u collides with left ball of the dumbbell
If all the collisions between the balls are elastic and head-on, the minimum distance $l$ so that the rightmost ball acquires speed u is $π u \sqrt{\frac{m}{x k}}$ where $x$ is------

Accepted Answer

In C.M frame of the dumb bell just after collision is as follows

If ball 2 compresses and comes back to mean position and at same time ball 3 travels ‘l’ distance then the given situation can arise. So time taken $= π \sqrt{\frac{μ}{k}} = π \sqrt{\frac{m}{2 k}}$

$s o, l = \frac{u}{2} (t i m e t a k e n) = π u \sqrt{\frac{m}{8 k}}$

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