Work Energy Power

JEE Main level, single correct, medium

104 of 183

Question 104 of 183

A boy is rolling a 0.5kg0.5kg ball on the frictionless floor with the speed of 20ms−1.20{ms}^{-1}. The ball gets deflected by an obstacle on the way. After deflection it moves with 55% of its initial kinetic energy. What is the speed of the ball now?

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Answer: (B) 4.4ms−14.4{ms}^{-1}

Given, m=0.5kgm=0.5kg and u=20ms−1u=20{ms}^{-1}
Initial kinetic energy (ki)=12mu2\left({k}_{i}\right)=\dfrac{1}{2}m{u}^{2}
=12×0.5×20×20=100J=\dfrac{1}{2}\times 0.5\times 20\times 20=100J
After deflection it moves with 55% of ki{k}_{i}
∴kf=5100×ki⇒5100×100∴{k}_{f}=\dfrac{5}{100}\times {k}_{i}\Rightarrow \dfrac{5}{100}\times 100
⇒kf=5J\Rightarrow {k}_{f}=5J
Now, let the final speed be ′v′ms−1,'v'{ms}^{-1}, then:
kf=5=12mv2{k}_{f}=5=\dfrac{1}{2}m{v}^{2}
⇒v2=20\Rightarrow {v}^{2}=20
⇒v=20=4.47ms−1\Rightarrow v=\sqrt{20}=4.47{ms}^{-1}

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