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.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 8.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

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
$\frac{1}{2} M {(\frac{v_{0}}{\sqrt{5}})}^{2} = \frac{1}{2} k x^{2} \Rightarrow x = v_{0} \sqrt{\frac{M}{5 K}} \Rightarrow x = 1.6 \sqrt{\frac{1.25}{5 \times 100}} m \Rightarrow x = \frac{1.6}{20} m \Rightarrow x = 8 c m$