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JEE Main 2021 (17 Mar, Shift 1) · Kinetic Energy & Momentum

Kinetic Energy & Momentum question from JEE Main 2021 (17 Mar, Shift 1)

Q13JEE Main 2021 · 17 MAR · SHIFT 1Easy

Kinetic Energy & Momentum

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

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Answer: (c)

$K \propto v^2$ , so $v' = v\sqrt{K'/K}$ .
$v' = 20\sqrt{0.05} = 20 \times 0.2236$
$v' \approx 4.47\text{ m/s}$ , so the answer is (c).

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