If the Kinetic energy of a moving body becomes four times its initial Kinetic energy, then the percentage change in its momentum will be :
Solution
K<sub>2</sub> = 4K<sub>1</sub><br><br>${1 \over 2}$mv$_2^2$ = 4${1 \over 2}$mv$_1^2$<br><br>v<sub>2</sub> = 2v<sub>1</sub><br><br>P = mv<br><br>P<sub>2</sub> = mv<sub>2</sub> = 2mv<sub>1</sub><br><br>P<sub>1</sub> = mv<sub>1</sub><br><br>% change = $${{\Delta P} \over {{P_1}}} \times 100 = {{2m{v_1} - m{v_1}} \over {m{v_1}}} \times 100 = 100\% $$
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Linear Momentum and Kinetic Energy
This question is part of PrepWiser's free JEE Main question bank. 158 more solved questions on Rotational Motion are available — start with the harder ones if your accuracy is >70%.