Two identical balls each of mass $M = (1.25 k g)$ and radius R are kept in contact and are connected to two identical unstretched springs each of force constant $k (= 100 N / m) .$ A third ball also of mass M but radius R/2, moving in vertically downward direction strikes with the two balls symmetrically with speed $v_{0} = (1.6 m / s)$ and comes to rest just after the impact. Find the maximum compression in each spring in cm.

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answer is 8.

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Detailed Solution

$I =$ Impulse by each of bigger balls on smaller ball $\sin θ = 2 / 3, \cos θ = \frac{\sqrt{5}}{3}$ By impulse momentum theorem, for smaller ball $2 I \cos θ = M v_{0} \Rightarrow I = \frac{3 M v_{0}}{2 \sqrt{5}}$ Now, $I \sin θ = M v$
(here v=speed acquired by each of the bigger balls after collision)
Applying law of conservation of energy, we get
$\frac{1}{2} M {(\frac{v_{0}}{\sqrt{5}})}^{2} = \frac{1}{2} k x^{2} \Rightarrow x = v_{0} \sqrt{\frac{M}{5 K}} \Rightarrow x = 1.6 \sqrt{\frac{1.25}{5 \times 100}} m \Rightarrow x = \frac{1.6}{20} m \Rightarrow x = 8 c m$

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