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 ........ .
Pick an option to check your answer.
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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