One side of spring of initial, instreached length `l_(0) = 1m`, lying on a frictionless table, is fixed, the other one is fastened to a small puck of mass `m = 0.1 kg`. The puck is given velocity in a direction perpendicular to the spring at an initial speed `v_(0) = 11 m//s`. In the course of the motion, the maximum elogation of the spring is `l = l_(0)//10`. What is the force constant of the spring (in `SI` units) ?
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Answer

`210`

Explanation

From energy conservation
`(1)/(2) mv_(0)^(2) = (1)/(2) Kx^(2) + (1)/(2) mv_(1)^(2)`
`mv_(0)^(2) = Kx^(2) + mv_(1)^(2)`…(1)
form conservation of angular momentum
`mv_(0) l_(0) = m (v_(0)) (1_(0) + x)`
`v_(0) l_(0) = (v_(1)) (l_(0) + (l_(0))/(10))`
`v_(0) l_(0) = ((11 l_(0)/(10)) v_(1)`
`v_(1) = 10 m//s`
Substituting value of `v_(1)` in `(1)`
`(0.1) (11)^(2) = K ((1)/(10))^(2) + (0.1) (10)^(2)`
`0.1 xx(11)^(2) - 0.1 (10)^(2) = K [(1)/(10)]^(2)`
`2.1 = (K)/(100) rArr K = 210`.