JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
As per the given figure, two blocks each of mass \(250\,g\) are connected to a spring of spring constant \(2\,Nm ^{-1}\). If both are given velocity \(V\) in opposite directions, then maximum elongation of the spring is:
- A \(\frac{ v }{2 \sqrt{2}}\)
- B \(\frac{ V }{2}\)
- C \(\frac{ V }{4}\)
- D \(\frac{ V }{\sqrt{2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{ V }{2}\)
Step-by-step Solution
Detailed explanation
using energy conservation \(\frac{1}{2} mv ^{2} \times 2=\frac{1}{2} kx ^{2}\) \(\frac{1}{4} v ^{2}=\frac{1}{2} \times 2 \times x ^{2}\) \(x =\frac{ v }{2}\)
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