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?
Solution
$K.E{._f} = 5\% \,K{E_i}$<br><br>${1 \over 2}m{v^2} = {5 \over {100}} \times {1 \over 2} \times m \times {20^2}$<br><br>${v^2} = {1 \over {20}} \times {20^2} = 20$<br><br>$v = \sqrt {20} = 2\sqrt 5$ m/s<br><br>= 4.47 m/s
About this question
Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force
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