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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2021
  • Medium
  • Numerical

A ball of mass \(4\, kg\), moving with a velocity of \(10\, ms ^{-1}\), collides with a spring of length \(8\, m\) and force constant \(100\, Nm ^{-1}\). The length of the compressed spring is \(x\, m\). The value of \(x\), to the nearest integer, is ........ .

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Answer & Solution

Correct Answer

(C) \(6\)

Step-by-step Solution

Detailed explanation

Let's say the compression in the spring by: \(y\).

So, by work energy theorem we have

\(\Rightarrow \frac{1}{2} mv ^{2}=\frac{1}{2} ky ^{2}\)

\(\Rightarrow y =\sqrt{\frac{ m }{ k }} \cdot v\)

\(\Rightarrow y=\sqrt{\frac{4}{100}} \times 10\)

\(\Rightarrow y =2\, m\)

\(\Rightarrow\) final length of spring \(=8-2=6\, m\)

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