Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If the Kinetic energy of a moving body becomes four times its initial Kinetic energy, then the percentage change in its momentum will be :

JEE MainJEE Main 2021Work, Energy and Power

Solution:

$K_{2}=4 K_{1}$
$\frac{1}{2} m v_{2}^{2}=4 \frac{1}{2} m v_{1}^{2}$
$v_{2}=2 v_{1}$
$P=m v$
$P_{2}=m v_{2}=2 m v_{1}$
$P_{1}=m v_{1}$
$\%$ change $=\frac{\Delta P}{P_{1}} \times 100$
$=\frac{2 m v_{1}-m v_{1}}{m v_{1}} \times 100=100 \%$