JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2017
- Hard
- MCQ
An object is dropped from a height \(h\) from the ground. Every time it hits the ground it looses \(50\%\) of its kinetic energy. The total distance covered as \(t \to \infty \) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(3\,h\)
Step-by-step Solution
Detailed explanation
\(( K.E.)' =50%\) of \(K.E.\) after hit i.e.,
\(\frac{1}{2}mv{'^2} = \frac{{50}}{{100}} \times \frac{1}{2}m{v^2} \Rightarrow v'\frac{v}{{\sqrt 2 }}\)
Coefficient of restitution \( = \frac{1}{{\sqrt 2 }}\)
Now, total diatance travelled by object is
\(H = h\left( {\frac{{1 + {e^2}}}{{1 - {e^2}}}} \right) = h\left( {\frac{{1 + \frac{1}{2}}}{{1 - \frac{1}{2}}}} \right) = 3h\)
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