JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 1)
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?
14.41 ms$$-$$1
19.0 ms$$-$$1
4.47 ms$$-$$1
1.00 ms$$-$$1
Explanation
$$K.E{._f} = 5\% \,K{E_i}$$
$${1 \over 2}m{v^2} = {5 \over {100}} \times {1 \over 2} \times m \times {20^2}$$
$${v^2} = {1 \over {20}} \times {20^2} = 20$$
$$v = \sqrt {20} = 2\sqrt 5 $$ m/s
= 4.47 m/s
$${1 \over 2}m{v^2} = {5 \over {100}} \times {1 \over 2} \times m \times {20^2}$$
$${v^2} = {1 \over {20}} \times {20^2} = 20$$
$$v = \sqrt {20} = 2\sqrt 5 $$ m/s
= 4.47 m/s
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